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I have built a vacuum tube amplifier. Its power supply consists of a transformer, a vacuum tube rectifier and choke input filter.

Everything works well, except that sometimes when I turn it off (mechanical DPDT switch), it generates some sort of transient on the mains input that makes some switching power supplies to turn off temporarily. Also sometimes, but quite rarely, the fuse blows when I turn the amp off.

This is probably due to the magnetic field collapsing in the core of the transformer and generating a high voltage spike (maybe the choke adds a bit too).

How do I prevent it? I probably should connect a capacitor in parallel with the primary of the transformer or across the switch (I have seen this done in other equipment), but how do I calculate the value?

EDIT:

Circuit diagram of the power supply:

schematic

simulate this circuit – Schematic created using CircuitLab

Rectifier is a dual diode vacuum tube. The transformer obviously has 5V and 6.3V secondaries for heaters.

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  • \$\begingroup\$ An RC filter? Do you have a scope capture of the waveform? \$\endgroup\$
    – winny
    Aug 9, 2021 at 20:25
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    \$\begingroup\$ Most likely the issue is on the secondary, the primary just reflects it. You should unveil some schematics. \$\endgroup\$ Aug 9, 2021 at 20:53
  • \$\begingroup\$ In addition to an LC filter, you might connect MOV varistors across the primary, and to ground, to both reduce transients and to protect the transformer from surges. Use varistors rated just above the peak line voltage (i.e., 180 V peak or 120 VAC, 360 V peak or 240 VAC), and at least a few hundred joule energy rating. \$\endgroup\$ Aug 9, 2021 at 22:06
  • \$\begingroup\$ These old days tube rectifiers had a bleeder resistor connected in parallel to the DC link output. \$\endgroup\$ Aug 10, 2021 at 11:18

1 Answer 1

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The recently inserted schematic shows that you have a big inductor with DC current. That DC will not stop suddenly if you try to break it, but generates as much voltage as needed to decay gradually. It's seen as voltage peak in the primary. The peak is as high as needed to let the current continue as an arc in the switch and die gradually as the inductor energy is dissipated.

A fix: Insert a diode between the left end of the inductor and the GND. It doesn't short the DC voltage if the cathode is up. I would try it at first.

The inductance doesn't suddenly sink more current, the arc only lets the current decay gradually. The arc can be so big that it shorts also the mains AC - that's your current peak. The suggested diode allows the inductor current to decay with no generated extra voltage => no arc.

All parts which are in parallel with your transformer to the right from SW1 enjoy in your current system a voltage peak. You have been lucky-the insulation in the transformer is still OK.

There may be certain non-technical, but decisive reasons why no semiconductor parts can be included. That excludes the diode (except if you can insert a tube). Another idea is to insert a RC series circuit in parallel with the transformer primary.

The capacitor must be so big that it can contain the energy of the inductor (= 0.5 * L * I^2) with low enough voltage to prevent arcing.

The capacitor has energy 0.5 * C * U^2. As reversed U=sqrt(2 * energy/C).

The resistance must be so big that the current peak isn't too high for the switch nor fuses when the amp is turned ON. It must be small enough or it can neutralize the effect of the capacitor. The max primary current * R must be lower than the arcing limit voltage in the switch. The resistor dissipates energy continuously from the mains AC. That must also be taken into the account.

It's well possible that a good compromise doesn't exist. More possibilities:

  • new design
  • mask a diode to look some 100 years old component. It must still be documented!
  • have an old style light bulb in parallel with the transformer primary. One made for an electric oven can do the job (a guess, not calculated). Obviously the heaters of the tubes connected to other windings are far too light load to absorb the pulse.
  • have SW1 which has the halves in separated spaces and far away from the protective earth (the arc occurs but cannot make a short). One pole switching makes it easier to achieve.
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  • \$\begingroup\$ Placing a capacitor in front of the inductor (making it a CLC filter) would raise the voltage of the power supply and increase noise. This type of filter (cathode of the rectifier straight to the inductor) is the method describes in the datasheed to the rectifier tube and allows the power supply to produce more current without overloading the tube (capacitor input filter produces large current peaks). \$\endgroup\$
    – Pentium100
    Aug 10, 2021 at 7:12
  • \$\begingroup\$ How much current is expected to be there? There's not a lot of space there for a second big rectifier tube, so I'll have to find a small one that's still good enough. (I know I could use a silicon diode, but I do not want to, I made this amp with the idea of "how they would have made it in the past") Can't I just put a RC circuit somewhere (in parallel to the primary or wherever) to absorb the transient? \$\endgroup\$
    – Pentium100
    Aug 10, 2021 at 7:32
  • \$\begingroup\$ OK I removed the capacitor. To be effective it may have to be so big that it really can cause voltage increase. In old tube rectifiers that was taken in to the account, but if your transformer outputs high enough voltage for your current filter the capacitor at the left end of the inductor cannot be recommended. \$\endgroup\$
    – user136077
    Aug 10, 2021 at 7:32
  • \$\begingroup\$ @Pentium100 I suggested RC series circuit in parallel with the primary in my already removed comment. I insert some facts to the answer. \$\endgroup\$
    – user136077
    Aug 10, 2021 at 7:34
  • \$\begingroup\$ How do I calculate the required capacitance and resistance? What I found on the internet assumes I am switching the load very fast (PWM and such). \$\endgroup\$
    – Pentium100
    Aug 10, 2021 at 7:46

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