Plate Resistance

<< < (2/3) > >>

k7yoo:
I think we have pretty well put the subject to rest but....
Modulated impedance is strictly the DC load--Plate V/ Plate I
RF Load impedance is based on the duty cycle of the class of service, that is why the multiplier varies between class C, B, AB1, etc. You will note that the more linear you get, the lower the multiplier. The old 50's & 60's handbooks never mentioned this and I remember wondering why the Pi-Net values for my homebrew amps (Linear) never came out right. It wasn't until I actually READ the books and the theory that the light came on in my head.
The Tonne Pi-Net calculator is really great and it has helped me with several projects. I keep it on my laptop for network calculations--beats the slide rule.

AB2EZ:
Fred et. al.

The reason that the modulation-related plate impedance (i.e., V/I) is different from the r.f.-related plate impedance (i.e., k V/I, where k is a number between 0.5 and 1) is because the r.f. current waveform is not a sine wave... unless the output tube is running in Class A.

[Also, to avoid some confusion: The modulation-related plate impedance is the impedance that the modulation transformer secondary looks into (i.e., in that case, the r.f. output tube is playing the role of the load, and the modulator is playing the role of the source). The highest power transfer from the modulator to the r.f. output tube occurs when this load matches the effective output impedance of the modulator. The r.f.-related plate impedance is the optimal r.f. load to place at the output of the r.f. output tube (i.e., in that case, the r.f. output tube is playing the role of the source), to get maximum power output from the r.f. output tube]

To be more precise, the r.f. - related plate impedance (the value of the RF load impedance that produces maximum output) can be calculated as follows

1. Assume that the r.f. load (e.g. the pi network) has a much higher impedance at the fundamental operating frequency than it has at any harmonics of the fundamental operating frequency.

2. If the above is the case, then the peak voltage produced across the r.f. load (e.g., the input to the pi network) is 

V(peak)= Z(load) x I (fundamental);

where Z(load) is the impedance of the load at the fundamental frequency, and I (fundamental) is the amplitude of the fundamental frequency component of the plate current (which, in general has many harmonics, because the plate current is not a sine wave).

We want to ensure that V(peak), the peak voltage across the r.f. load, does not exceed the DC plate voltage, V(DC)... so that we don't bottom out the net voltage on the plate of the tube. This also leads to maximum output power at the fundamental frequency.

Thus we should set the r.f. load to a value that makes  V(peak) = V(DC) = Z(load) x I(fundamental).

Solving for Z(load), we find that the optimal r.f. load impedance is V(DC) / I(fundamental)

3. We can easily measure the average value of plate current, I(DC),  but we need to know I(fundamental)... the fundamental frequency component of the plate current.

With some calculus, which I won't attempt to produce here, we can compute I(fundamental) if we know two things: a) the average plate current (which we can easily measure), and b) the shape of the plate current waveform: Class A, Class B, Class C with duty cycle xx (radians), etc.

For a Class A (sine wave superimposed on a DC value) plate current waveform, where the sine wave amplitude = the DC level, then I(fundamental) = I(DC).

For a Class B waveform, I(fundamental) = (pi/2) x I(DC) ~ 1.6 x I(DC)

For a Class C waveform, depending upon the details of the shape (particularly the width) of the current pulse that flows each cycle, I fundamental ~ 2 x I(DC)

Thus, you obtain the formulas for the optimal load impedance that were referred to by Skip

For Class A operation: Z(load) = V(DC) / I(DC)

For Class B operation: Z(load) = V(DC) / [1.6 x I(DC)]

For Class C operation: Z(load) ~ V(DC) / [2 x I(DC)]

etc.

Best regards
Stu

flintstone mop:
ahhh Yes,
The early days before computers and software that you can plug numbers into. I'm a technician that only knows shortcuts and being able to contact people who are in the know. I don't know why I didn't try use ohms law and get into the ballpark. The Internet has made it pretty easy to figger this stuff out.
Stu, I couldn't get past Algebra I. Me and math STILL do not get along!!
My hat's off to you and the others here.

Fred

K3ZS:
http://www.qsl.net/n9bor/MultiMods.htm

This website has useful mods to the AF67 pi-net for matching to 50 ohms.

flintstone mop:
Thanks Robert
I made those mods and that's what started troubleshooting woes of the AF67. With the help of this board and the Mighty Fine Junk 259 I found that the coil taps called out and the values for loading caps were a little out in "left field". They weren't plug n play steps and it magically puts out 50 watts. I now have close to 40 watts on all bands with the extra time mentioned above. I have a B+ supply that's sagging when keyed and sags more when the modulator pubes are in the circuit. NOW the RF output drops to 35 watts under this additional load. BEFORE the mods, I was lucky to get 25 watts RF out with mod tubes in place. The output dropped even lower when I had 6550's in for modulators. Beautiful audio, but I couldn't bear the lowered RF outpoot.
 
I built a power supply from an ER article using back-to-back 240 vac transformers. I do not have the refernce here, but it was a quick way to build a 500vdc supply.
I think I will go to e-Pay and get a Heath HP23B P.S. and modify for B+ needed by the Elmac. The Heathkit P.S. won't sag under load. And I'll get the full 50 watts.
But common cents tells me that I'm driving a linear and with reduced power of 12-15 watts, there is plenty of modulation and the ER built P.S. will suffice, for now.

I still love electronics and this great hobby of Ham Radio and your guys and gals
Merry Christmas
PHRED and Family

Navigation

[0] Message Index

[#] Next page

[*] Previous page