You should, at a minimum get the following: A small 'solderless breadboard':
http://www.radioshack.com/product/index.jsp?productId=2734155 http://www.mpja.com/prodinfo.asp?number=4443+TE http://www.allelectronics.com/make-a-store/item/PB-400/SOLDERLESS-BREADBOARD-400-CONTACTS//1.html http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_20601_-1 I put a link to Radioshack in there, just because they're everywhere, which is really their only saving grace. That way you won't have to 'mail order' (dating myself) anything, if you don't want to. Don't try to build an entire system on one of these. Just use it to check out a circuit block, like that anode driver stage. Get a roll of 22AWG solid wire and a 9V battery clip, so you can use a 9V battery to power a test circuit. MPSA42s and MPSA92s are pretty cheap, but for trying out transistor concepts with any transistor. The PN2222 and 2N3904 are commonly found NPNs, and can be substituted for the A42, as long as the supply is under 40V. Also a 2N3906 is a common PNP, and sub for the A92. Note, this is only to try out concepts. The other transistors won't handle nixie voltages. Back to the basic transistor lessons. In any bipolar transistor. One with a Base, Emitter, and Collector. Applies to both NPN & PNP. The current out of the emitter is the sum of the base current and collector current. Lets place 5V at the base of that MPSA42 (NPN). The total drop from base to ground is that 5V. 0.7V will drop across the BE junction, therefore 4.3V has to be across that 10K resistor. By ohms law, the current is 4.3V/10K = 430uA. Now suppose your idea was true, and the collector current was 50 times (assuming an hfe ~50), that or 21.5mA. Then the emitter current is the sum of the base and collector, and then ~21.5mA. If that was true, the the drop across the 10K resistor would be 215V. If the transistor survived that, that means the emitter is 210V above the base; a reverse bias of 210V. The transistor would turn OFF. Not because it died (which it would), but because the forward current flow is no longer there. In that configuration, if the 10K resistor was the intended load, it would be called a 'emitter follower'. That means the emitter will follow the base. The voltage gain is roughly 1, or no voltage gain at all. Its used for current gain. But there's a real reason why the emitter follows the base, and its not just because the BE junction limits the voltage drop. At least not as a simple diode. As you raise the voltage at the base, from just under the the 'knee' of 0.7V (ideally) current starts to flow thru the BE junction. This will, in turn, cause a flow in current from C to E, by the factor of hfe, due to the physics of a bipolar transistor (don't ask me, I only use 'em, I don't design 'em). That's current 'Ib' coming into the base, current 'Ic' (=hfe*Ib) coming into the collector, and the summed current 'Ie' (=Ib+Ic=[hfe+1]*Ib) coming out of the emitter. This current, Ie, will cause a voltage drop across that 10K resistor. This is actually a negative feedback source, since as the resistor voltage rises, it lowers the drop across the BE junction. That will in turn reduce both Ib and Ic, hence lower Ie, and the resistor voltage drops. Like all negative feedback, this tends to balance things out, which magically is right at the 'knee' (0.7V across BE). The practical result is that the 10K resistor looks like a lot higher resistance at the base, by the factor of hfe+1, or 510K (assuming and hfe of 50). So get some parts, breadboard them up, and play a little. Since you only have a DVM, just use a wire to flip that base from 0V to +5V, or use a small trimpot, and watch the voltage track it. Start with just the NPN side, tying the collector to the positive supply. And just for fun and yucks adjust the supply voltage from 5V, to 9V, to 18V, to 27V, while keeping the base at 5V, and see what happens. Then rig the whole anode drive circuit. Instead of driving a nixie or neon, use a LED, and keep the voltage low (under 24V). Measure how all the differnt parts work. > On Jan 24, 2:50 pm, will <ossumguyw...@gmail.com> wrote: > OK, I think I understood some of that. You're saying that the amount > of current the MPSA42 will let through is defined by some relationship > with (Vin-Vdrop)/Rground, or (5-.7)/10000=.00043A. This is the amount > of current across base to emitter, yes? But, if the Hfe is 50, doesn't > that mean that I get up to 50*.00043A or .0215A across the collector > to emitter? How did you get that 1/50th of that goes across the BE and > 49/50 of that goes across the CE? > > Also, I really can't afford to just "try stuff out". I have a limited > supply of parts, which is why I carefully plan everything beforehand. > -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To post to this group, send an email to neonixie-l@googlegroups.com. 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