You are confusing the Z that the Final sees with the Z that the modulator sees. The RF final is looking into a load reflected by the tank circuit
Typically: the Z that a class C final sees for its RF load is (Ep/Ip)*2 This "Z" is used in determining the proper reactance for the tuned circuit so they will have the proper "Q" OTOH, the modulator is looking at a load that is determined directly by The RF stage ratio of Ep/IP. For some reason most people look at modulator power as a determination of modulation percentage and this is some what misleading. It is really easier to just say that the peak audio voltage delivered by the modulator must equal or exceed the DC voltage on the final RF amp for 100% modulation. If the DC voltage is 600V on the RF class C final then the modulator must be capable of at least 600V peak or 1200 Volts peak to peak, because for 100 percent modulation the DC on the final must vary down to 0 Volts and up to twice the DC value. If a XMTR is unloaded to draw less plate current (Ip) the plate voltage will remain at the same voltage. If there is less current being pulled by the final but the voltage is the same then the Z seen by the modulator is higher than before. The higher resistance that the audio is looking into makes it easier to produce the needed voltage. It is easier for the one modulator (the same mod XFMR ratio) to modulate a 60 watt rig with 600V than a 100 watt rig with 600 Volts. Matching the impedance is not necessary. It is often confused with the statement that says "when the load Z equals the source Z there will be maximum power transfer". This is a true statement. But power transfer is not the issue here. Getting the voltage on the final to swing from 0 to 2*DC with out distortion, that is the issue. Of course we won't to do that at a power level determined by our needs. The concern was how to reduce the power from a particular rig. Is unloading the final tank all that is necessary? Perhaps it is. But there are concerns about this. Because it will make the rig easier to modulate and over modulation can occur if not monitored. This occurs because the modulator is not working into as heavy a load as before and the audio voltage at the output will rise as the load resistance goes up. Here is why. The modulator can be looked upon as a source of voltage with a certain amount of internal resistance. Voltage Source (modulator fixed input with no load) = 900Volts peak Internal resistance is a virtual value not equal to any one thing but a result of the circuit design and tubes. Internal resistance = 2000 OHMS Load resistance (EP/IP of the RF final) = 4000 ohms Total resitance is 6000 ohms The internal resistance will drop the voltage at the load to (4000/6000 * 900)= 600V peak With out changing the audio input level let's raise the load resistance to 8000 ohms (equivalent to reducing plate current to 50 percent by unloading the rig). Total resistance = 10000 ohms Voltage at he load (8000/10000 * 900) = 720 volts peak This increase in modulator audio output voltage is a result of unloading the rig. The audio level at the input will need to be reduced to keep from over modulating. BTW if the load is reduce to 2000 ohms (equal to the internal resistance) Total resistance = 4000 ohms Load voltage = 450 V peak 1/2 the capability of the max output voltage Maximum power is transferred at a lower voltage output but a greater current would be drawn by the modulators and probably a lot of distortion. SEE http://www.qsl.net/wa5bxo/power.html Hope this clears some things up. John -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jim Wilhite Sent: Tuesday, February 12, 2008 9:00 PM To: Discussion of AM Radio in the Amateur Service Subject: Re: [AMRadio] RE: Using a Ranger as a driver Ok John, I have but a few observations and questions that might steer everyone through this. I am having trouble understanding how the Z the modulator see is greater if the plate load impedance goes down. If the Plate Z is = to the Ep divide by 2 times the Ip and you change one value how does the Z remain the same. Even if you maintain the same ratio doesn't the plate Z change as a result? In your example you change both at the same ration which maintains the same impedance. Ep 600 Ip 200 ma. Zp= 600 divided by 2 times .2 = 600/.4 = 1500 Ep 300 Ip 100 ma. Zp = 300 divided by 2 times .1 = 300/.2 = 1500 ______________________________________________________________ Our Main Website: http://www.amfone.net AMRadio mailing list Searchable Archives: http://www.mail-archive.com/[email protected]/ List Rules (must read!): http://w5ami.net/amradiofaq.html List Home: http://mailman.qth.net/mailman/listinfo/amradio Help: http://mailman.qth.net/mmfaq.html Post: mailto:[email protected] To unsubscribe, send an email to [EMAIL PROTECTED] with the word unsubscribe in the message body.
