Re: [Vo]:circuit diagram
On 03/21/2010 10:59 PM, Stephen A. Lawrence wrote: On 03/21/2010 07:25 PM, Harry Veeder wrote: The capacitor is on the input side. A pick up coil was added later to see if it is possible to close the loop and generate OU. Interesting. Sorry for my confusion; I didn't watch the videos where he had no pickup coil (video 4 already had the extra coil, of course). Is this circuit, which charges the cap, the same as the diagram you showed but with the battery snipped out? If so, it's surprising (to me) that the thing can produce DC on the input side; I don't see an obvious mechanism for rectifying the signal generator output. Ah well actually yes I do. Check out a data sheet for the mosfet; here are links to a couple (it's a standard part, available from a number of manufacturers): http://www.datasheetcatalog.org/datasheet/philips/IRF640_S_1.pdf http://www.datasheetcatalog.org/datasheet/SGSThomsonMicroelectronics/mXtvwts.pdf Note that there's an internal diode between the source and drain. With that, the AC signal feeding the circuit, and the inductor in series with the diode and capacitor, it seems pretty reasonable that the SG could charge the input capacitor. The same argument still applies, in any case -- the power output of the signal generator hasn't been measured. Measure that, compare it with the energy appearing the capacitor and being dissipated in the coil, and *then* see if it still looks like anything funny is going on. Strange coupling is the order of the day when working with AC and coils, but you should always be able to find the source of the energy which comes out of the system. If you can't get the books to balance at least to within the margin of error of the instruments, that's something worth pursuing. But if power in equals power out (within instrument error) then the question becomes, How is the power getting from the input to the output?, rather than, Where is the power coming from?. Instrument error may be pretty substantial when measuring high frequency AC with out of phase volts and amps, by the way. As another aside, you can certainly boost voltage with just a single coil. Put volts across the coil for a while, allowing the current to ramp up, then chop off the drive voltage abruptly. Current continues to flow through the coil, and if the load which the coil sees after the power supply is cut off is high impedence, a large voltage will appear across the load. This is why, for instance, you can get arcing across switch contacts when shutting off power to a large electromagnet, even if you're energizing it with a low voltage source. The fact that the signal generator which is driving this circuit is apparently making square waves, not sine waves, is what makes this relevant. He tried in test 10 but didn't succeed: http://www.youtube.com/watch?v=c7CsBr7ouPE harry - Original Message From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Sun, March 21, 2010 2:51:23 PM Subject: Re: [Vo]:circuit diagram Harry will need to confirm this, but I believe the diagram in question is only for the input side. The output side, which isn't shown, consists of a pickup coil, some related circuitry, and the aforementioned capacitors. If I'm wrong, then I'm confused (no great surprise there). On 03/21/2010 11:56 AM, Michel Jullian wrote: Wait a minute, I see no cap attached to the output on Harry's diagram photo 2discussed here (haven't followed the other discussions), only one capacitor on the input side, in parallel with the battery until the latter is disconnected, which BTW isn't explained on the diagram. Is the diagram not complete? 2010/3/21 Stephen A. Lawrence href=mailto:sa...@pobox.com;sa...@pobox.com: On 03/21/2010 09:55 AM, Michel Jullian wrote: Which voltage? Volts on the caps attached to the output -- right, Harry? But the signal generator is still hooked up, and it's coupled to the output (at least) through the gate capacitance of the FET and the linked inductors of the transformer, and the signal generator's output power hasn't been measured or even estimated. So, there's no reason to believe this rig is doing anything other than transforming and rectifying the output of the SG. As I've already said a boringly large number of times, this is the same general sort of system as Stiffler's circuit, where he had a signal generator capacitively coupled to the system, and it was driving a handful of LEDs. The main innovation here comes from Naudin, and it's the use of a toroidal coil as the primary with a neo magnet on the outside of the coil which twists the core's field to allow the toroidal coil to couple to the pickup coil. 2010/3/20, Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected
Re: [Vo]:circuit diagram
Which voltage? 2010/3/20, Harry Veeder hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected. harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Sat, March 20, 2010 3:59:08 AM Subject: Re: [Vo]:circuit diagram What do you mean, the inductor (10 turns of wire on a core) is connected between the positive end of the supply and one end of the switch (drain of the MOSFET) isn't it? 2010/3/20 Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: The toroid is also wired in differently from the inductor in the wiki diagram, but I suppose that doesn't matter either? harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 1:42:52 PM Subject: Re: [Vo]:circuit diagram The capacitor on your photo 2 is in parallel with the battery so it's part of the converter's input supply. The capacitor in the operating principles diagram of the wikipedia article is the converter's output capacitor, which might as well not be there in steady state is there is no load (once charged it just stays charged at a high voltage, and the Boost's diode never conducts-- so the diode might as well not be there either). So everything to the right of the switch in the boost converter diagram could be removed in no load condition, that's why I say the circuit operates like a Boost converter without a load. Which explains why it steps up the input voltage, that's what Boost converters do. Michel 2010/3/19 Harry Veeder ymailto=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles target=_blank href=http://en.wikipedia.org/wiki/Boost_converter; target=_blank http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian ymailto=mailto: href=mailto:michelj...@gmail.com;michelj...@gmail.com href=mailto: href=mailto:michelj...@gmail.com;michelj...@gmail.com ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto: href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry __ Looking for the perfect gift? Give the gift of Flickr! href= href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/; target=_blank href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/ __ Looking for the perfect gift? Give the gift of Flickr! href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/ __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
Re: [Vo]:circuit diagram
On 03/21/2010 09:55 AM, Michel Jullian wrote: Which voltage? Volts on the caps attached to the output -- right, Harry? But the signal generator is still hooked up, and it's coupled to the output (at least) through the gate capacitance of the FET and the linked inductors of the transformer, and the signal generator's output power hasn't been measured or even estimated. So, there's no reason to believe this rig is doing anything other than transforming and rectifying the output of the SG. As I've already said a boringly large number of times, this is the same general sort of system as Stiffler's circuit, where he had a signal generator capacitively coupled to the system, and it was driving a handful of LEDs. The main innovation here comes from Naudin, and it's the use of a toroidal coil as the primary with a neo magnet on the outside of the coil which twists the core's field to allow the toroidal coil to couple to the pickup coil. 2010/3/20, Harry Veeder hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected. harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Sat, March 20, 2010 3:59:08 AM Subject: Re: [Vo]:circuit diagram What do you mean, the inductor (10 turns of wire on a core) is connected between the positive end of the supply and one end of the switch (drain of the MOSFET) isn't it? 2010/3/20 Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: The toroid is also wired in differently from the inductor in the wiki diagram, but I suppose that doesn't matter either? harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 1:42:52 PM Subject: Re: [Vo]:circuit diagram The capacitor on your photo 2 is in parallel with the battery so it's part of the converter's input supply. The capacitor in the operating principles diagram of the wikipedia article is the converter's output capacitor, which might as well not be there in steady state is there is no load (once charged it just stays charged at a high voltage, and the Boost's diode never conducts-- so the diode might as well not be there either). So everything to the right of the switch in the boost converter diagram could be removed in no load condition, that's why I say the circuit operates like a Boost converter without a load. Which explains why it steps up the input voltage, that's what Boost converters do. Michel 2010/3/19 Harry Veeder ymailto=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles target=_blank href=http://en.wikipedia.org/wiki/Boost_converter; target=_blank http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian ymailto=mailto: href=mailto:michelj...@gmail.com;michelj...@gmail.com href=mailto: href=mailto:michelj...@gmail.com;michelj...@gmail.com ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto: href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role
Re: [Vo]:circuit diagram
Harry will need to confirm this, but I believe the diagram in question is only for the input side. The output side, which isn't shown, consists of a pickup coil, some related circuitry, and the aforementioned capacitors. If I'm wrong, then I'm confused (no great surprise there). On 03/21/2010 11:56 AM, Michel Jullian wrote: Wait a minute, I see no cap attached to the output on Harry's diagram photo 2discussed here (haven't followed the other discussions), only one capacitor on the input side, in parallel with the battery until the latter is disconnected, which BTW isn't explained on the diagram. Is the diagram not complete? 2010/3/21 Stephen A. Lawrence sa...@pobox.com: On 03/21/2010 09:55 AM, Michel Jullian wrote: Which voltage? Volts on the caps attached to the output -- right, Harry? But the signal generator is still hooked up, and it's coupled to the output (at least) through the gate capacitance of the FET and the linked inductors of the transformer, and the signal generator's output power hasn't been measured or even estimated. So, there's no reason to believe this rig is doing anything other than transforming and rectifying the output of the SG. As I've already said a boringly large number of times, this is the same general sort of system as Stiffler's circuit, where he had a signal generator capacitively coupled to the system, and it was driving a handful of LEDs. The main innovation here comes from Naudin, and it's the use of a toroidal coil as the primary with a neo magnet on the outside of the coil which twists the core's field to allow the toroidal coil to couple to the pickup coil. 2010/3/20, Harry Veeder hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected.
Re: [Vo]:circuit diagram
The capacitor is on the input side. A pick up coil was added later to see if it is possible to close the loop and generate OU. He tried in test 10 but didn't succeed: http://www.youtube.com/watch?v=c7CsBr7ouPE harry - Original Message From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Sun, March 21, 2010 2:51:23 PM Subject: Re: [Vo]:circuit diagram Harry will need to confirm this, but I believe the diagram in question is only for the input side. The output side, which isn't shown, consists of a pickup coil, some related circuitry, and the aforementioned capacitors. If I'm wrong, then I'm confused (no great surprise there). On 03/21/2010 11:56 AM, Michel Jullian wrote: Wait a minute, I see no cap attached to the output on Harry's diagram photo 2discussed here (haven't followed the other discussions), only one capacitor on the input side, in parallel with the battery until the latter is disconnected, which BTW isn't explained on the diagram. Is the diagram not complete? 2010/3/21 Stephen A. Lawrence href=mailto:sa...@pobox.com;sa...@pobox.com: On 03/21/2010 09:55 AM, Michel Jullian wrote: Which voltage? Volts on the caps attached to the output -- right, Harry? But the signal generator is still hooked up, and it's coupled to the output (at least) through the gate capacitance of the FET and the linked inductors of the transformer, and the signal generator's output power hasn't been measured or even estimated. So, there's no reason to believe this rig is doing anything other than transforming and rectifying the output of the SG. As I've already said a boringly large number of times, this is the same general sort of system as Stiffler's circuit, where he had a signal generator capacitively coupled to the system, and it was driving a handful of LEDs. The main innovation here comes from Naudin, and it's the use of a toroidal coil as the primary with a neo magnet on the outside of the coil which twists the core's field to allow the toroidal coil to couple to the pickup coil. 2010/3/20, Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected. __ Connect with friends from any web browser - no download required. Try the new Yahoo! Canada Messenger for the Web BETA at http://ca.messenger.yahoo.com/webmessengerpromo.php
Re: [Vo]:circuit diagram
03/21/2010 11:56 AM, Michel Jullian wrote: Wait a minute, I see no cap attached to the output on Harry's diagram photo 2discussed here (haven't followed the other discussions), only one capacitor on the input side, in parallel with the battery until the latter is disconnected, which BTW isn't explained on the diagram. Is the diagram not complete? I assumed you were following the video updates and would have known. harry __ Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com
Re: [Vo]:circuit diagram
So the voltage which rises after disconnection of the battery is that of the single capacitor shown on the diagram, which was initially in parallel with the battery? 2010/3/22 Harry Veeder hlvee...@yahoo.com: The capacitor is on the input side. A pick up coil was added later to see if it is possible to close the loop and generate OU. He tried in test 10 but didn't succeed: http://www.youtube.com/watch?v=c7CsBr7ouPE harry - Original Message From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Sun, March 21, 2010 2:51:23 PM Subject: Re: [Vo]:circuit diagram Harry will need to confirm this, but I believe the diagram in question is only for the input side. The output side, which isn't shown, consists of a pickup coil, some related circuitry, and the aforementioned capacitors. If I'm wrong, then I'm confused (no great surprise there). On 03/21/2010 11:56 AM, Michel Jullian wrote: Wait a minute, I see no cap attached to the output on Harry's diagram photo 2discussed here (haven't followed the other discussions), only one capacitor on the input side, in parallel with the battery until the latter is disconnected, which BTW isn't explained on the diagram. Is the diagram not complete? 2010/3/21 Stephen A. Lawrence href=mailto:sa...@pobox.com;sa...@pobox.com: On 03/21/2010 09:55 AM, Michel Jullian wrote: Which voltage? Volts on the caps attached to the output -- right, Harry? But the signal generator is still hooked up, and it's coupled to the output (at least) through the gate capacitance of the FET and the linked inductors of the transformer, and the signal generator's output power hasn't been measured or even estimated. So, there's no reason to believe this rig is doing anything other than transforming and rectifying the output of the SG. As I've already said a boringly large number of times, this is the same general sort of system as Stiffler's circuit, where he had a signal generator capacitively coupled to the system, and it was driving a handful of LEDs. The main innovation here comes from Naudin, and it's the use of a toroidal coil as the primary with a neo magnet on the outside of the coil which twists the core's field to allow the toroidal coil to couple to the pickup coil. 2010/3/20, Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected. __ Connect with friends from any web browser - no download required. Try the new Yahoo! Canada Messenger for the Web BETA at http://ca.messenger.yahoo.com/webmessengerpromo.php
Re: [Vo]:circuit diagram
On 03/21/2010 07:25 PM, Harry Veeder wrote: The capacitor is on the input side. A pick up coil was added later to see if it is possible to close the loop and generate OU. Interesting. Sorry for my confusion; I didn't watch the videos where he had no pickup coil (video 4 already had the extra coil, of course). Is this circuit, which charges the cap, the same as the diagram you showed but with the battery snipped out? If so, it's surprising (to me) that the thing can produce DC on the input side; I don't see an obvious mechanism for rectifying the signal generator output. The same argument still applies, in any case -- the power output of the signal generator hasn't been measured. Measure that, compare it with the energy appearing the capacitor and being dissipated in the coil, and *then* see if it still looks like anything funny is going on. Strange coupling is the order of the day when working with AC and coils, but you should always be able to find the source of the energy which comes out of the system. If you can't get the books to balance at least to within the margin of error of the instruments, that's something worth pursuing. But if power in equals power out (within instrument error) then the question becomes, How is the power getting from the input to the output?, rather than, Where is the power coming from?. Instrument error may be pretty substantial when measuring high frequency AC with out of phase volts and amps, by the way. As another aside, you can certainly boost voltage with just a single coil. Put volts across the coil for a while, allowing the current to ramp up, then chop off the drive voltage abruptly. Current continues to flow through the coil, and if the load which the coil sees after the power supply is cut off is high impedence, a large voltage will appear across the load. This is why, for instance, you can get arcing across switch contacts when shutting off power to a large electromagnet, even if you're energizing it with a low voltage source. The fact that the signal generator which is driving this circuit is apparently making square waves, not sine waves, is what makes this relevant. He tried in test 10 but didn't succeed: http://www.youtube.com/watch?v=c7CsBr7ouPE harry - Original Message From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Sun, March 21, 2010 2:51:23 PM Subject: Re: [Vo]:circuit diagram Harry will need to confirm this, but I believe the diagram in question is only for the input side. The output side, which isn't shown, consists of a pickup coil, some related circuitry, and the aforementioned capacitors. If I'm wrong, then I'm confused (no great surprise there). On 03/21/2010 11:56 AM, Michel Jullian wrote: Wait a minute, I see no cap attached to the output on Harry's diagram photo 2discussed here (haven't followed the other discussions), only one capacitor on the input side, in parallel with the battery until the latter is disconnected, which BTW isn't explained on the diagram. Is the diagram not complete? 2010/3/21 Stephen A. Lawrence href=mailto:sa...@pobox.com;sa...@pobox.com: On 03/21/2010 09:55 AM, Michel Jullian wrote: Which voltage? Volts on the caps attached to the output -- right, Harry? But the signal generator is still hooked up, and it's coupled to the output (at least) through the gate capacitance of the FET and the linked inductors of the transformer, and the signal generator's output power hasn't been measured or even estimated. So, there's no reason to believe this rig is doing anything other than transforming and rectifying the output of the SG. As I've already said a boringly large number of times, this is the same general sort of system as Stiffler's circuit, where he had a signal generator capacitively coupled to the system, and it was driving a handful of LEDs. The main innovation here comes from Naudin, and it's the use of a toroidal coil as the primary with a neo magnet on the outside of the coil which twists the core's field to allow the toroidal coil to couple to the pickup coil. 2010/3/20, Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: yes. You are aware that the the voltage keeps rises even after the battery is disconnected. __ Connect with friends from any web browser - no download required. Try the new Yahoo! Canada Messenger for the Web BETA at http://ca.messenger.yahoo.com/webmessengerpromo.php
Re: [Vo]:circuit diagram
What do you mean, the inductor (10 turns of wire on a core) is connected between the positive end of the supply and one end of the switch (drain of the MOSFET) isn't it? 2010/3/20 Harry Veeder hlvee...@yahoo.com: The toroid is also wired in differently from the inductor in the wiki diagram, but I suppose that doesn't matter either? harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Fri, March 19, 2010 1:42:52 PM Subject: Re: [Vo]:circuit diagram The capacitor on your photo 2 is in parallel with the battery so it's part of the converter's input supply. The capacitor in the operating principles diagram of the wikipedia article is the converter's output capacitor, which might as well not be there in steady state is there is no load (once charged it just stays charged at a high voltage, and the Boost's diode never conducts-- so the diode might as well not be there either). So everything to the right of the switch in the boost converter diagram could be removed in no load condition, that's why I say the circuit operates like a Boost converter without a load. Which explains why it steps up the input voltage, that's what Boost converters do. Michel 2010/3/19 Harry Veeder ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles target=_blank http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry __ Looking for the perfect gift? Give the gift of Flickr! href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/ __ Looking for the perfect gift? Give the gift of Flickr! http://www.flickr.com/gift/
Re: [Vo]:circuit diagram
yes. You are aware that the the voltage keeps rises even after the battery is disconnected. harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Sat, March 20, 2010 3:59:08 AM Subject: Re: [Vo]:circuit diagram What do you mean, the inductor (10 turns of wire on a core) is connected between the positive end of the supply and one end of the switch (drain of the MOSFET) isn't it? 2010/3/20 Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: The toroid is also wired in differently from the inductor in the wiki diagram, but I suppose that doesn't matter either? harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 1:42:52 PM Subject: Re: [Vo]:circuit diagram The capacitor on your photo 2 is in parallel with the battery so it's part of the converter's input supply. The capacitor in the operating principles diagram of the wikipedia article is the converter's output capacitor, which might as well not be there in steady state is there is no load (once charged it just stays charged at a high voltage, and the Boost's diode never conducts-- so the diode might as well not be there either). So everything to the right of the switch in the boost converter diagram could be removed in no load condition, that's why I say the circuit operates like a Boost converter without a load. Which explains why it steps up the input voltage, that's what Boost converters do. Michel 2010/3/19 Harry Veeder ymailto=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles target=_blank href=http://en.wikipedia.org/wiki/Boost_converter; target=_blank http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian ymailto=mailto: href=mailto:michelj...@gmail.com;michelj...@gmail.com href=mailto: href=mailto:michelj...@gmail.com;michelj...@gmail.com ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto: href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry __ Looking for the perfect gift? Give the gift of Flickr! href= href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/; target=_blank href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/ __ Looking for the perfect gift? Give the gift of Flickr! href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/ __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
Re: [Vo]:circuit diagram
2010/3/19 Harry Veeder hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry - Original Message From: Harry Veeder hlvee...@yahoo.com To: vortex-l@eskimo.com Sent: Thu, March 18, 2010 10:46:19 PM Subject: Re: [Vo]:circuit diagram Ok, I gave him the wiki reference. Harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Thu, March 18, 2010 7:34:49 PM Subject: Re: [Vo]:circuit diagram Nothing mysterious about this circuit, it's a silly boost converter without a load. See: target=_blank href=http://en.wikipedia.org/wiki/Boost_converter; target=_blank http://en.wikipedia.org/wiki/Boost_converter 2010/3/18 Harry Veeder href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: - Original Message From: Jed Rothwell ymailto=mailto: href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com href=mailto: href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com ymailto=mailto:jedrothw...@gmail.com; href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com To: href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com; ymailto=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Thu, March 18, 2010 5:22:20 PM Subject: Re: [Vo]:circuit diagram Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed I am ignorant about electronics but I don't see what the fuss is about since it is all DC current. If you know the resistance and the voltage can't you safely infer that as the voltage rises and falls so does the current? No, V=R*I works only on a pure resistor. An inductor or a capacitor obey different laws. I still think that in certain simple circuits voltage measurements can serve as a pretty good indicator of current and power. Not here. Michel __ Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
Re: [Vo]:circuit diagram
I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry __ Looking for the perfect gift? Give the gift of Flickr! http://www.flickr.com/gift/
Re: [Vo]:circuit diagram
The capacitor on your photo 2 is in parallel with the battery so it's part of the converter's input supply. The capacitor in the operating principles diagram of the wikipedia article is the converter's output capacitor, which might as well not be there in steady state is there is no load (once charged it just stays charged at a high voltage, and the Boost's diode never conducts-- so the diode might as well not be there either). So everything to the right of the switch in the boost converter diagram could be removed in no load condition, that's why I say the circuit operates like a Boost converter without a load. Which explains why it steps up the input voltage, that's what Boost converters do. Michel 2010/3/19 Harry Veeder hlvee...@yahoo.com: I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry __ Looking for the perfect gift? Give the gift of Flickr! http://www.flickr.com/gift/
Re: [Vo]:circuit diagram
The toroid is also wired in differently from the inductor in the wiki diagram, but I suppose that doesn't matter either? harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Fri, March 19, 2010 1:42:52 PM Subject: Re: [Vo]:circuit diagram The capacitor on your photo 2 is in parallel with the battery so it's part of the converter's input supply. The capacitor in the operating principles diagram of the wikipedia article is the converter's output capacitor, which might as well not be there in steady state is there is no load (once charged it just stays charged at a high voltage, and the Boost's diode never conducts-- so the diode might as well not be there either). So everything to the right of the switch in the boost converter diagram could be removed in no load condition, that's why I say the circuit operates like a Boost converter without a load. Which explains why it steps up the input voltage, that's what Boost converters do. Michel 2010/3/19 Harry Veeder ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: I'll pass that along. But the capacitor looks like it is in the wrong place to be a booster converter with or without a load. compare photo 2: http://tinyurl.com/ycw4xm4 with operating principles target=_blank http://en.wikipedia.org/wiki/Boost_converter Harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Fri, March 19, 2010 4:54:02 AM Subject: Re: [Vo]:circuit diagram 2010/3/19 Harry Veeder href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter I said it was a boost converter _without a load_. as none of them will recharge the input source(cap) while being operated. Ive tried. This is because he hasn't tried removing the load. If you do, in the course of one oscillation cycle, the input source first sources current, and then sinks current. Note there is a hidden component in the circuit which is important to understand where the inductor's current flows to and from in this no load operation, that's the MOSFET's output capacitance. The IRF640's antiparallel diode is another hidden component which plays an important role, it prevents the drain voltage from going below zero. Michel And you wont find any dc/dc converters with magnets on the coil core. ;] Harry __ Looking for the perfect gift? Give the gift of Flickr! href=http://www.flickr.com/gift/; target=_blank http://www.flickr.com/gift/ __ Looking for the perfect gift? Give the gift of Flickr! http://www.flickr.com/gift/
Re: [Vo]:circuit diagram
BTW I think you may need to be signed into facebook to see this image. (At any rate, I couldn't see it until I signed in.) On 03/18/2010 12:15 PM, Harry Veeder wrote: http://www.facebook.com/photo.php?pid=3513566l=e812a3f42eid=676517267 http://tinyurl.com/yewd4sf Harry __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at http://downloads.yahoo.com/ca/internetexplorer/
Re: [Vo]:circuit diagram
So what we've got here is a toroidal coil with a neo magnet outside it. The neo magnet, by saturating /part/ of the ferrite core, essentially cuts the toroid, so it's no longer going to be a closed system; when we put a current through the coil, we're going to see a chunk of its field leaking out of the torus. The battery juice is chopped by the signal generator, so we've got something close to a square wave going in. There's also going to be some capacitive coupling of the signal generator's output across the FET's gate. Predicting exactly how much power is going into this coil is not going to be easy; the only way to know for sure is to measure it. (The *power*, not the voltage!) Finally, since the coil has been cut by the neo magnet, a pickup coil placed next to it will act as the secondary to a transformer which consists of the two coils, the core(s), and the neo magnet. What *voltage should we expect to see induced in the pickup coil? == I have no idea! First, this configuration is hard to understand to start with, and the coupling between the two coils is confusing, at best. Second, he's feeding it with square waves, not nice smooth easy to understand sine waves; when the FET cuts off, there's going to be an induced voltage spike which may be a lot larger than the original input voltage. That spike may very well be what's charging his caps to his unexpectedly high voltage. So, any claim that the induced voltage isn't what we would expect is just silly. Unless somebody can come up with a serious analysis of this configuration and make some plausible predictions, it's just anybody's guess what kind of volts should be seen in the pickup coil. And, once again, all he needs to do to get a rough handle on the energy budget is measure power in and power out. He's studiously avoided doing that. Ergo, either he's a lot more ignorant than the level of sophistication exhibited here would suggest, or he doesn't want to know what the energy budget is. (Or he's actually done it and just isn't admitting it.) On 03/18/2010 12:15 PM, Harry Veeder wrote: http://www.facebook.com/photo.php?pid=3513566l=e812a3f42eid=676517267 http://tinyurl.com/yewd4sf Harry __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at http://downloads.yahoo.com/ca/internetexplorer/
Re: [Vo]:circuit diagram
Sorry, he has now corrected the depiction of the toroid windings. click the second photo: http://www.facebook.com/album.php?aid=156403id=676517267l=2a69ff7aae http://tinyurl.com/ycw4xm4 You should be able to access the photos without having to sign into facebook, so I don't know what the problem is. - Original Message From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Thu, March 18, 2010 1:03:38 PM Subject: Re: [Vo]:circuit diagram BTW I think you may need to be signed into facebook to see this image. (At any rate, I couldn't see it until I signed in.) On 03/18/2010 12:15 PM, Harry Veeder wrote: href=http://www.facebook.com/photo.php?pid=3513566l=e812a3f42eid=676517267; target=_blank http://www.facebook.com/photo.php?pid=3513566l=e812a3f42eid=676517267 http://tinyurl.com/yewd4sf Harry __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at href=http://downloads.yahoo.com/ca/internetexplorer/; target=_blank http://downloads.yahoo.com/ca/internetexplorer/ __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at http://downloads.yahoo.com/ca/internetexplorer/
Re: [Vo]:circuit diagram
By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. On 03/18/2010 04:30 PM, Harry Veeder wrote: Sorry, he has now corrected the depiction of the toroid windings. click the second photo: http://www.facebook.com/album.php?aid=156403id=676517267l=2a69ff7aae http://tinyurl.com/ycw4xm4 You should be able to access the photos without having to sign into facebook, so I don't know what the problem is. It's still not letting me see the pictures until I log in, and in fact it doesn't reliably let me see them at all if I use Firefox under Linux. MSIE under Windows works a lot better for this. Truth be told I don't see that the difference makes any difference -- they both look correct to me. The current goes down the inside, up the outside, and after winding halfway around the coil, the wire comes out to jump 180 degrees around, then goes down the inside, up the outside, around the other half of the coil. The tangential direction of the current is the same all the way around, and reversing the two halves cancels the lateral current which otherwise would result in a B field along the axis of the torus equal to that which one gets with a single turn around the torus. Looks like the two diagrams are more or less mirrored, but either one seems like it should be fine. In fact I had pictured something a little different, which was winding a single layer of wire which worked its way all the way around the torus, followed by a second layer which works its way all the way back. But it really doesn't make much difference; the residual field due to an imperfect winding geometry isn't really at issue here. It's surely swamped by the distortion imposed on the field by the external neo magnets. - Original Message From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Thu, March 18, 2010 1:03:38 PM Subject: Re: [Vo]:circuit diagram BTW I think you may need to be signed into facebook to see this image. (At any rate, I couldn't see it until I signed in.) On 03/18/2010 12:15 PM, Harry Veeder wrote: href=http://www.facebook.com/photo.php?pid=3513566l=e812a3f42eid=676517267; target=_blank http://www.facebook.com/photo.php?pid=3513566l=e812a3f42eid=676517267 http://tinyurl.com/yewd4sf Harry __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at href=http://downloads.yahoo.com/ca/internetexplorer/; target=_blank http://downloads.yahoo.com/ca/internetexplorer/ __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at http://downloads.yahoo.com/ca/internetexplorer/
Re: [Vo]:circuit diagram
On Mar 18, 2010, at 12:56 PM, Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. I agree with all the above. Thanks to Harry Veeder. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:circuit diagram
Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed
Re: [Vo]:circuit diagram
- Original Message From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com; vortex-l@eskimo.com Sent: Thu, March 18, 2010 5:22:20 PM Subject: Re: [Vo]:circuit diagram Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed I am ignorant about electronics but I don't see what the fuss is about since it is all DC current. If you know the resistance and the voltage can't you safely infer that as the voltage rises and falls so does the current? I still think that in certain simple circuits voltage measurements can serve as a pretty good indicator of current and power. Harry __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
Re: [Vo]:circuit diagram
Nothing mysterious about this circuit, it's a silly boost converter without a load. See: http://en.wikipedia.org/wiki/Boost_converter 2010/3/18 Harry Veeder hlvee...@yahoo.com: - Original Message From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com; vortex-l@eskimo.com Sent: Thu, March 18, 2010 5:22:20 PM Subject: Re: [Vo]:circuit diagram Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed I am ignorant about electronics but I don't see what the fuss is about since it is all DC current. If you know the resistance and the voltage can't you safely infer that as the voltage rises and falls so does the current? No, V=R*I works only on a pure resistor. An inductor or a capacitor obey different laws. I still think that in certain simple circuits voltage measurements can serve as a pretty good indicator of current and power. Not here. Michel
Re: [Vo]:circuit diagram
On 03/18/2010 05:58 PM, Harry Veeder wrote: - Original Message From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com; vortex-l@eskimo.com Sent: Thu, March 18, 2010 5:22:20 PM Subject: Re: [Vo]:circuit diagram Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed I am ignorant about electronics but I don't see what the fuss is about since it is all DC current. If you know the resistance and the voltage can't you safely infer that as the voltage rises and falls so does the current? This isn't DC (and the load is mostly inductive, not resistive). Check out the scope shots, and look at that circuit diagram again. He's feeding the AC output of the signal generator to the gate of the FET which in turn chops the DC from the battery to produce AC. It's asymmetric AC (goes up from zero then back down to zero) but it's AC none the less; you can view this sort of chopped signal as a symmetric AC signal *plus* a DC signal, where the DC signal is the average, or offset, voltage. It's the AC component of the signal which is going to go through the transformer and out the other side. The power consumed will be the integral of the product of the input voltage and the input current. Power is coming from the battery, where the voltage is (nearly) fixed but the current is varying wildly (in a rough square wave with, I suspect, some pretty substantial peaks), and power is coming from the signal generator, which is producing a neat square wave of *voltage* but, depending on the capacitance of the FET gate and the frequency, may be providing current in a rather squirrely waveform. The actual voltage wave forms are square waves, rather than sine waves. I said (in some earlier post) that those are harder to understand. They can be viewed as being a sum of an infinite progression of ever higher frequency sine waves, and it's the extremely high frequency components which are buried in the square wave which make its behavior peculiar. In particular, those high frequencies are likely to leak through the FET gate, which can be viewed as a tiny capacitor. The current leaking through the FET will be visible as a current drain from the signal generator -- but that hasn't been measured (or, at any rate, that measurement hasn't been shown). Without knowing the current drain from the batteries and the signal generator, we know exactly *nothing* about how much power is being provided to the circuit. I still think that in certain simple circuits voltage measurements can serve as a pretty good indicator of current and power. Yes, but this circuit is anything but simple. With a partly inductive load and AC voltage, you don't necessarily know, a priori, whether the current and voltage are even going to be in phase with each other. An aside: When people talk about RMS voltage, they're talking about taking the square root of the mean squared voltage. That's a useful measurement in exactly one case, which is the one you're thinking of: A purely resistive load. As long as the load is (or acts like it is) purely resistive, the current will be linear in the voltage, and the instantaneous power will consequently be proportional to the square of the voltage. The average power will, then, be proportional to the average of the squared voltage. And the square root of the average power will be the equivalent DC voltage which would put the same power as your AC waveform into a given resistive load. But that equivalent DC voltage -- the RMS voltage -- might behave very differently from the original AC wave form if the load is not purely resistive. This circuit is an example where knowing the RMS voltage applied to the circuit doesn't tell you much of anything at all. Harry __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
Re: [Vo]:circuit diagram
Ok, I gave him the wiki reference. Harry - Original Message From: Michel Jullian michelj...@gmail.com To: vortex-l@eskimo.com Sent: Thu, March 18, 2010 7:34:49 PM Subject: Re: [Vo]:circuit diagram Nothing mysterious about this circuit, it's a silly boost converter without a load. See: target=_blank http://en.wikipedia.org/wiki/Boost_converter 2010/3/18 Harry Veeder href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: - Original Message From: Jed Rothwell ymailto=mailto:jedrothw...@gmail.com; href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com To: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com; ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Thu, March 18, 2010 5:22:20 PM Subject: Re: [Vo]:circuit diagram Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed I am ignorant about electronics but I don't see what the fuss is about since it is all DC current. If you know the resistance and the voltage can't you safely infer that as the voltage rises and falls so does the current? No, V=R*I works only on a pure resistor. An inductor or a capacitor obey different laws. I still think that in certain simple circuits voltage measurements can serve as a pretty good indicator of current and power. Not here. Michel __ Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com
Re: [Vo]:circuit diagram
Here is a reply from Magluvin who is also a member of overunity.com: This is not a boost converter as none of them will recharge the input source(cap) while being operated. Ive tried. And you wont find any dc/dc converters with magnets on the coil core. ;] Harry - Original Message From: Harry Veeder hlvee...@yahoo.com To: vortex-l@eskimo.com Sent: Thu, March 18, 2010 10:46:19 PM Subject: Re: [Vo]:circuit diagram Ok, I gave him the wiki reference. Harry - Original Message From: Michel Jullian ymailto=mailto:michelj...@gmail.com; href=mailto:michelj...@gmail.com;michelj...@gmail.com To: ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Thu, March 18, 2010 7:34:49 PM Subject: Re: [Vo]:circuit diagram Nothing mysterious about this circuit, it's a silly boost converter without a load. See: target=_blank href=http://en.wikipedia.org/wiki/Boost_converter; target=_blank http://en.wikipedia.org/wiki/Boost_converter 2010/3/18 Harry Veeder href=mailto: href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com ymailto=mailto:hlvee...@yahoo.com; href=mailto:hlvee...@yahoo.com;hlvee...@yahoo.com: - Original Message From: Jed Rothwell ymailto=mailto: href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com href=mailto: href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com ymailto=mailto:jedrothw...@gmail.com; href=mailto:jedrothw...@gmail.com;jedrothw...@gmail.com To: href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com; ymailto=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com href=mailto: href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com ymailto=mailto:vortex-l@eskimo.com; href=mailto:vortex-l@eskimo.com;vortex-l@eskimo.com Sent: Thu, March 18, 2010 5:22:20 PM Subject: Re: [Vo]:circuit diagram Stephen A. Lawrence wrote: By the way, I should say Thanks! for taking the time to post all these here. It's interesting, even if I don't believe for a minute that it's OU. Someone should communicate the gist of the comments here to the author of the video. Tell him to invest in an ammeter, for crying out loud. - Jed I am ignorant about electronics but I don't see what the fuss is about since it is all DC current. If you know the resistance and the voltage can't you safely infer that as the voltage rises and falls so does the current? No, V=R*I works only on a pure resistor. An inductor or a capacitor obey different laws. I still think that in certain simple circuits voltage measurements can serve as a pretty good indicator of current and power. Not here. Michel __ Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now http://ca.toolbar.yahoo.com.
