Here I have a switching power supply diagram that I found on the internet. I have a slightly different one for which I would like to adopt this same scheme for starting process so that I can keep the control circuit completely isolated from the input, but I can't understand how this works.

The auxiliary source, which comes out of the secondary of the toroidal transformer, is responsible for powering the control circuit. At the moment the source is turned on, there is no energy in the secondary, so when turning on the source, the control circuit theoretically should not work and the PSU shoudn't start. I imagine that when turning on the source, as the primary is energized, the circuit does something to induce a sufficient current in the secondary to start the IC.

What I want to know is how this happens.

enter image description here


1 Answer 1


It looks like R8 and R14 provide some bias to Q1 and Q4 so they conduct slightly just with application of power. They start self-oscillating by virtue of positive feedback provided by the winding of transformer TR1 that connects to TR2.

The oscillations couple through TR2 to the diodes D11 and D12 to bring up the auxiliary supply that powers the controller IC1. From then on normal operation commences.

The bias from R8 and R14 is overdriven by the signals from the controller through TR2 so they don't affect normal operation.

For your second question relating to biasing the IGBT devices, R9 and R14 would cause any small bias current to pass through the winding of the transformer without developing any significant gate bias to Q1 and Q2. The drive network from the transformer to the transistors would have to be changed (note that the first diagram using capacitance coupling from the transformer to the transistors). It is non-trivial to do that without compromising the drive during normal operation.

  • 1
    \$\begingroup\$ The first clue to this was the comment at the bottom of the schematic: "Vcc is about 22.5 V when operating normally - less if running in self-excited/startup mode" Then I noticed the bias resistors R8 and R14. \$\endgroup\$
    – Dave Tweed
    Commented Feb 11 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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