Plate Resistance

[flintstone mop]

Hello
I am trying to rework the final tank circuit on my AF-67. Single 6146 RF P.A. @ 500vdc. I am using my MFJ 259 to look into the tank circuit to get it to play in 50 ohms. I need the value of the plate resistance to simulate the tube in circuit. I am also discovering a very flaky multi-section switch that brings in different taps from the plate tank coils.
This was a valuable tip from someone on this forum, maybe K2DK. So we don't stress the final with overloads while experimenting. Just adjust the values of the coil or loading to make the MFJ display 50 ohms and almost 0 reactance was the trick.

Thanks
Fred

[WD8BIL]

Hi Phred......... Merry Christmas !!

your 500VDC/ (final loaded plate current) will be close enuf !!

ie 500vdc/135ma= 3700 ohms

[AB2EZ]

Fred

To simulate the effective RF plate resistance of the tube (which is, as you know, not the same as the audio plate modulation load resistance presented by the tube)... a good approximation for a plate-modulated class C output stage is:

Effective RF plate resistance ~ .6 [V/I],

Where V is the plate voltage at carrier, and I is the plate current at carrier.

For example, if V=600 volts, and I = 0.125 amps, the the Effective RF plate resistance ~ 2880 ohms.

Don't forget to remove the resistor when you are done. 

I used the method you described to test the tank circuit of my homebrew linear amplifier. In the case of a linear amplifier (as opposed to a plate modulated class C amplifier), one needs to use the plate current, I, that flows at peak output power. So, for my linear amplifier, I used R=  0.6 x 3000 volts / 0.8 amperes = 2250 ohms; where .8 amperes is the plate current at peak output power. I forgot to remove the 1/4 watt resistor... so it disappeared in a small puff of smoke when I applied power to the amplifier at a later time.

Best regards
Stu

[k7yoo]

Here is a software link to get a great pi net values calculator               http://tonnesoftware.com/
Use the PI-EL calculator and enter 50 ohms for the image impedance-normally this is 200. This will provide Pi-net values (as opposed to Pi-l)

The old handbooks all list the formula for class C plate load impedance calculations as plate V divided by 2X plate current. Based on this your rig is running at 2000 ohms. Later books have more details and provide more exact values. I personally use 2X for class C and 1.5 or 1.35 multiplier for linear class AB1 operation. It seems to be close enough.
The Elmac does not have the correct tank values for a 50 ohm load on 160 & 80-way too much L. I usually replace the coil with a toroid and add padding caps to bring the values in line with 50 ohm values. The problem is that the variable cap values are such that it is tricky to get a good tuning range on 160 with padders. It has been 6 or 7 years since my last conversion and I will have to open a rig up to see what I did. I do know that when properly done 160 & 80 will have the same power output as 40, with a 50 ohm load.
Skip

[flintstone mop]

Thanks Skip, Stu, Buddly

Your values were very different from my ballpark guess of 5000. I saw the specs for the secondary of the mod transformer and it stated 5K, so I stuck a 4700 ohm resistor and it got me pretty close. NOW I can determine what is going on with 80/160M. Yous guys got me to an exact value and the party begins. This is a nice brain massager project using test equipment and electronic tricks from this forum.
Thanks

Fred

[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

[K3ZS]

I usually run 500V at 100mA, I think that gives me about 30W out on my AF67.   I made a fixture so that I can use a series light bulb (no CFL's here) to knock down the power.  That way you can tune up to full power on the AF67 and make the linear happy.   I have heard that it isn't good to lightly load it at reduced power when using the AM modulator, maybe someone would know if this is true.   I had to experiment with bulbs, have one for 80M and 40M, depending on the linear.   Mine is a GOnset GSB-101 and has an untuned input but uses a series coil to the cathodes.  For the light bulb I just used a regular electrical octagon box and lightbulb fixture.    I put on two coax receptacles and a switch to short out the bulb when not using the linear.   Its cheap and adds to the glow of the tubes while the bulb flashes as you talk.

[flintstone mop]

hey Bob,
There's something neater than light bulbs!!
Someone here mentioned using an NPN power transistor and a 1meg pot and control the screen of the 6146. I can turn power up and down to the 12-15 watts that I need when driving a leanyear without worrying about loading.
fred

[Steve - WB3HUZ]

This should probably do ya Fred.

http://www.amwindow.org/tech/htm/drivepot.htm

[K3ZS]

Fred and Steve, the pot and transistor looks like a good idea.     With incandescents being outlawed, it looks like a good solution.   There is an empty space across from the modulation level that some use for drive pots.    One question I have, do you have to lower your audio gain on the modulator when decreasing the screen voltage to prevent overmodulation?    I am using the zener diode method of hard limiting, mainly to protect the mod transformer.    I was wondering if this would still be effective when lowering the screen voltage.    I would think if the plate voltage stays the same it would still be OK.    I started my career in electronics after the tube era, still rusty on the fine points of tube based electronics.

[flintstone mop]

YUP the screen is the best way to control that particular tube. And definitely yes, reducing the modulation. The modulator will be in heaven and the plate voltages all stay the same.
Thanks Steve, I have used that ciruit many times for DX100's, to control drive instead of a wirewound pot wearing out.
Fred