AM-912 Plate and Screen Supply Question

[EmpireAir1]

I am trying to engineer a power supply for the AM-912 (amplifier section for the T-302) after converting it from Class C to Class AB as described here:

http://www.g1ogy.com/www.n1bug.net/tech/am912.html

The 4X150A requires a plate voltage of 2000 V up to 150 MHz and 1250 V from 150 to 500 MHz, all at 250 ma.  The screen voltage is +300 V at 7 ma.  I found a 400 VA toroidal transformer with dual 450V secondaries here:

http://www.antekinc.com/AN-4T450.pdf

If I series connect the secondaries for 900V, can I center tap the secondary through a diode (1N4008) to a voltage divider and derive the screen voltage?

Appreciate any help...

Regards,
Wallace, KI4DVV

[w4bfs]

I believe you can series the 400V sec windings if they are in phase ... using a bridge rectumflyer should give you 1150V or so with 675V availiabe at the windings join point (the defacto center tap) ...the extra diode is unnecessary .... if the Antek hi-pot info is correct (3kV winding to winding) there should be no other concerns ... by the way, I think I have a couple of these rf cavities (or something close plus some new A suffix goof ball tubes) let me know if you are interested ... 73 ... John

oops...I meant 575V ... oh heck!

[AB2EZ]

Wallace

"If I series connect the secondaries for 900V, can I center tap the secondary through a diode (1N4008) to a voltage divider and derive the screen voltage?"

I will answer this specific question in my 3rd post (below)... but first let's look at the configuration of the plate supply... which has an impact on the answer to your specific question.

I suggest that you use the bridge rectifier design that John recommends if 1000 volts of B+ is sufficient. [A voltage doubler design is discussed in my next post ]



See the schematic attached below:

In the schematic, each 100 ohm resistor (R1 and R2) represents an estimate of the resistance of the associated secondary winding.

In the schematic, R3 (4000 ohms) represents the plate load (1000V / 250ma = 4000 ohms); and R7 (40000 ohms) represents the screen load (300 volts / 40,000 ohms = 7.5ma). [If the peak screen current is going to be larger, you may want to put an electrolytic capacitor in parallel with the 300 volt Zener diode.]

1. Make sure that each leg [D1, D2, D3, and D4] of the bridge uses enough 1000 PIV 3A diodes in series to handle 2x the output voltage of the supply. In this case, the output voltage of the supply is ~1000 volts... so you will need at least two (2) diodes in series for each of the four (4) legs of the bridge.

2. Use a pair of capacitors in series for the filter capacitor. In this case, each capacitor (C1 and C2) should be rated at a minimum of 500 volts working voltage (preferably more). If you use commonly available 450 volt-rated capacitors, then I suggest that each 50uF capacitor shown in the schematic (C1 and C2) should consist of two (2) 100uF 450 volt capactors in series. Furthermore, the two (2) bleeder resistors (R4 and R5), shown as 100,000 ohm resistors, should be implemented with a 50,000 ohm balancing / bleeder resistor across each of the total of four (4) 100uF capacitors.

3. Connect the CT of the secondary (the point where the two windings are connected together) to the midpoint of the two capacitors. [This will keep the voltage across the lower capacitor fixed at 1/2 of the total B+, even though you are drawing current from that capacitor to feed the screen]

4. I suggest that you use a Zener diode (two 150 volt 5 watt Zener diodes in series) and a 15000 ohm resistor to produce the 300 volt screen supply. (500 volts - 300 volts) / 15,000 ohms = 13.3 mA of available current, while maintaining 300 volts of regulated output.

A simulation of the output of this supply is shown in attached slide number 2. [The simulation program is SwCAD III.].

The B+ for the plate is 1.05 kV with about 60 volts of peak-to-peak 120Hz residual ripple (6% of the average B+). If you want less ripple, you will need to increase C1 and C2. I usually try to keep the peak-to-peak ripple to 2% or less... but you probably won't notice any hum on your signal.

The voltage at the point where C1 and C2 come together is (as expected) half of the B+ (i.e., roughly 524 volts)

The output voltage for the screen is (as expected) 300 volts (with no significant 120Hz residual ripple).

In slide number 3, I show the simulated voltage at the point where R1, D3, and D4 come together, and also at the point where R2, D2, and D1 come together. You can see that the peak voltage across any of the diodes is about 1.05 kV in this simulation.


Stu

[AB2EZ]

Wallace

Here is the voltage doubler design (attached .jpg slides)

1. Slide 4 shows the schematic

2. Slide 5 shows the voltages at the following places (starting at the top): B+ (1873 volts, with about 48 volts of peak-to-peak residual 120Hz ripple); junction of C1 and C2; junction of C2 and C3, output of screen supply (300 volts) with a 10uF smoothing capacitor (C4).

3. Slide 6 shows the voltage at the junction of D1 and D2.

The diodes now have 1900 volts of PIV across them

Capacitor C1 now has about 975 volts (peak) across it, Capacitor C2 has about 550 volts (peak) across it, and Capacitor C3 has about 420 volts (peak) across it.

[If I run the simulation a little longer (i.e. 5 seconds instead of 2 seconds), the peak voltage across C2 increases to 570 volts, and the peak voltage across C3 decreases to 401 volts]

[AB2EZ]

Finally, in response to your specific question

"If I series connect the secondaries for 900V, can I center tap the secondary through a diode (1N4008) to a voltage divider and derive the screen voltage?"

You can connect the center tap to the input of a separate rectifier + voltage divider + filter capacitor... if that is what you wish/need to do to obtain your screen supply... but to design it properly, and to understand the power that will be lost in the voltage divider... you need to know the voltage on the center tap of the transformer, with respect to ground in whatever configuration the transformer is used for powering the plate supply.

For the case of the bridge rectifier (at least the one I suggested in my earlier post): the voltage between the CT of the transformer and ground is half the B+ (already rectified), including half of the residual ripple. The circuit I showed for extracting a 300V screen supply is, in fact, a voltage divider connected to the center tap of the transformer... but it does not require an additional diode for rectification.

For the case of the voltage doubler:

Attached below are:

a. Another copy of the schematic of the voltage doubler

b. A simulation of the waveforms on the top of the transformer (blue), the bottom of the transformer (the waveform that has the largest swing), and the CT of the transformer (whose voltage is halfway between the other two). [In the case of the top and the bottom, the voltage shown is the voltage on the right side of the 100 ohm resistor that represents the winding resistance of the transformer]

As you can see, the waveform at each of these points (with respect to ground) has a DC (average) value of approximately 1kV, and has an AC component that would need to be filtered out. Since each waveform has a DC component, there is no need for an additional diode.

Having said that, I believe that the screen supply that already exists in the schematic represents a more efficient (less wasted power) approach, than using a voltage divider with a 1000 volt (average value) source.

Best regards
Stu

[EmpireAir1]

Thank you gentlemen for your responses.  Stu, I appreciate you taking the time to develop and run the simulations.  This was far more input than I was expecting.

Regards,
Wallace, KI4DVV