As I was designing the power supply for my tube amp, I was trying to reduce the negative transients of the choke input supply, I tried to add a capacitor in front of it (C1 in the schematic).


It reduced the transients, but also reduced the peak current of the rectifier. I tried experimenting with different values (in the Duncan's PSU designer) and it looks like 1uF is better than other common values.

Rectifier current with the cap 1nF (effectively no cap): 1nF

Rectifier current with 1uF: 1uF

Rectifier current with 2uF: 2uF

I guess the capacitor is compensating for the choke current (power factor?). Anyone knows the proper theory why does this happen? Also, does anyone know a formula that can be used to calculate the optimal capacitance?

I tried this with Multisim too and saw similar results.

  • 1
    \$\begingroup\$ What do you meanly "negative transient"? I can't see any. Also although the capacitor is reducing the peak at the middle of each cycle it increases the positive leading edge transient. It would be easier to see if you just showed 1 or 2 cycles rather than ~10. A choke input filter is often recommended because it is more gentle on the rectifier tube. \$\endgroup\$ – Kevin White Jun 15 '15 at 1:57
  • \$\begingroup\$ @KevinWhite I meant the PIV across the tube when this power supply is with lower load (which would happen in case of a failure of one of the output tubes or if the rectifier warmed up faster). The cap increases the positive leading edge transient but reduces the peak (no cap - 0.55A, 1uf - 0.45A). \$\endgroup\$ – Pentium100 Jun 15 '15 at 2:56
  • 1
    \$\begingroup\$ Those graphs tell you nothing. To determine that something has reduced a full peak-to-peak view is needed. \$\endgroup\$ – Andy aka Jun 15 '15 at 7:42
  • \$\begingroup\$ @Andyaka, well this is a current trough one rectifier, so it is either zero or what is shown in the graph. \$\endgroup\$ – Pentium100 Jun 15 '15 at 10:08
  • \$\begingroup\$ I would like to know what RMS current you get in these situations, I think it would be mostly the same for all cases. \$\endgroup\$ – akaltar Jun 26 '15 at 15:56

You are harmonicly tuning up L1 and C 1 to reject 2F PSU ripple . Most of the ripple occurs at 2F on a full wave single phase rectifier. The lower amplitude higher order terms can easily be dealt with by C2 and the rest of the filter. What you are doing is valid .Your actual optimum value of cap wont be your calculated value because the inductor is not ideal especially with significant DC going through it .You should be able to get 2 symmetrical rounded peaks on your current waveform .This gives lower line current harmonics and better power factor .When I did my work I placed C1 across L1 .It should not really matter because C2 is large compared to C1 .C1 sees significant AC volts so should be a quality metal film type . What hum your Audio project has will sound less buzzey .I experimented with lower inductance and more capacitance figuring that it would be cheaper and lighter .This technique is called a resonant choke rectifier .

  • \$\begingroup\$ Half a Henry! That's a lot of inductance. Good 'ol tube circuitry. :) \$\endgroup\$ – rdtsc Dec 4 '15 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.