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Old 10th March 2013, 10:07 PM   #1
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Default PSUD Questions

I have 2 questions about PSUD:

1. "constant current" is the load choice for output transformer ?
If so, how do I factor in the DCR of the primary ?

2. "V(load 1)" is the plate voltage of the output tube ?

Last edited by HP8903B; 10th March 2013 at 10:12 PM.
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Old 10th March 2013, 10:37 PM   #2
DF96 is offline DF96  England
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PSUD2 uses the power transformer effective secondary resistance. This is where the primary resistance comes in. You can calculate it, or PSUD2 will do it for you.

V(load 1) will be the PSU output voltage.
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Old 10th March 2013, 10:38 PM   #3
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I meant the DCR of the OPT primary.
I am designing the PS for single end amp with output transformer.
Which should I use for the "load" in PSUD ?
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Old 10th March 2013, 10:44 PM   #4
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Also, I am build stereo amp. The second LC filter is dedicated to each channel.
How can I combine the two separate PS after the second LC filter ?
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Old 10th March 2013, 10:54 PM   #5
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Quote:
Originally Posted by manp111 View Post
I meant the DCR of the OPT primary.
I am designing the PS for single end amp with output transformer.
Which should I use for the "load" in PSUD ?
Figure out how much steady-state current the amp will draw, divide that into the supply voltage, and you have a first approximation of the load resistance as seen by the supply. E = IR also means E/I = R. Remember to convert the current from milliamps to amperes!

For example, consider a 300B single ended stage running at 70 mA no-load current for B+ = 350V. The supply current is 0.07 Amps, so dividing that into 350 V means the supply sees a resistance of 5000 ohms. Plug that into PSUD and simulate.
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Old 10th March 2013, 10:55 PM   #6
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I see.
Just to be clear, I would choose "resistive load" right ?
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Old 10th March 2013, 11:00 PM   #7
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Quote:
Originally Posted by manp111 View Post
Also, I am build stereo amp. The second LC filter is dedicated to each channel.
How can I combine the two separate PS after the second LC filter ?
Simulate the two channels as one channel and make the appropriate corrections when breaking out the second LC pair. Once you're happy with the values, then multiply L by 2 and divide C by 2.

For example, the above amplifier would be 2500 ohms load for a pair, then simulate with that load. You might get something like L = 5H, C = 40 uF (this is off the top of my head, so don't hold me to those values) for a 2500 ohm load, so for each 5000 ohm load you'd use a 10H inductor and a 20 uF capacitor.

Last edited by DSP_Geek; 10th March 2013 at 11:02 PM.
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Old 10th March 2013, 11:00 PM   #8
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Quote:
Originally Posted by manp111 View Post
I see.
Just to be clear, I would choose "resistive load" right ?
Correct.
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Old 10th March 2013, 11:08 PM   #9
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So a pair of 2A3 would draw 120mA at 250V.
Then the "resistive load" value is 2083 ohm.

The power transformer is 300V-0-300V. I entered 300V (31ohm)
Choke input:
1st L = 10H 95ohm
1st C = 120uF
2nd L = 10H 95 ohm
2nd C = 120uF
Load = 2083 ohm

Rectifier is a pair of 6CJ3

The V(R1) = 235V RMS.

That seems too low.
Is that accurate ?

Last edited by HP8903B; 10th March 2013 at 11:10 PM.
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Old 10th March 2013, 11:55 PM   #10
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Accurate? Not sure, I've only used PSUD for sand-state sims so I'm not experienced with tube circuits.

The voltage doesn't look too far off from what one would expect intuitively, though. You have 31 + 95 + 95 = 220 ohms in series with the load, so you already lose 10% of 300 V by voltage division, and the rectifier is probably good for the remaining 35 volts lost. The result is reasonable.

The checkboxes to the left of the simulation curves let you examine the voltage at various points, so a look at voltages before and after various parts of the supply might prove instructive.
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