I came across an old circuit from work that discharges a capacitor (C1) into a step-up transformer to a low-impedance load. This event is triggered by a pulse into an SCR, whom has a parallel RC filter across its gate-cathode terminals. Both the Pulse amplitude and the capacitor's voltage may be in the 12V to 30V range, but they may not be equal.

I can fathom the filter capacitor (C2) -- this may prevent high-frequency noise from prematurely triggering the SCR -- but I cannot imagine the use of the resistor (R2).


  1. Perhaps R2 simply provides a bleed path for C2? That way, after Pulse has gone back to 0V C2 doesn't hold the gate open longer. But from what I've seen, C1 discharges in the microsecond (or less) timescale whereas Pulse is high for a few milliseconds -- long after the SCR's holding current has diminished.

That's the only thought I have so far. It just seems like R2 would bleed some current from the SCR gate's and load down the Pulse supply more than necessary. Can anyone think of a reason that it's in there?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Try simulating it. \$\endgroup\$
    – Andy aka
    Commented May 19, 2018 at 16:51
  • \$\begingroup\$ Without seeing the pulse generator it is hard to say. Maybe R2 is needed to protect the SCR gate from over-voltage or over-current. Maybe it is simply used to lower the impedance of the SCR input so that weak signals do not fire the SCR. A 500pF cap is only effective against high-frequencies. But R2 lowers the impedance from DC on up. \$\endgroup\$
    – user57037
    Commented May 19, 2018 at 17:44

1 Answer 1


R2 is a "gate-cathode resistor". Visually, the schematic makes it look like R2 is about C2, but it's really about the SCR's performance. A gate-cathode resistor allows some of the SCR anode current to bypass the cathode, making for stabler operation and better high-speed performance.

See, for example, page 1-9 of this app note. For more informal description, see this web page.

The action of the resistor can be seen with respect to the two transistor analogy of the SCR. It shows that a low external resistance between the gate and cathode bypasses some current around the gate junction. Accordingly a higher anode current is required to initiate and maintain conduction. It is particularly found that low current high sensitivity SCRs are triggered at very low current levels and therefore an external gate-cathode resistance is required to prevent triggering by thermally generated leakage current in the gate region. However the gate cathode resistance bypasses some of the internal anode current caused by the rapid rate of change of the anode voltage (dv/dt). It also raises the forward break-over voltage by reducing the efficiency of the NPN transistor region thus requiring a somewhat higher avalanche multiplication effect to initiate the triggering. The current that bypasses the gate junction also affects the latching and holding currents.

It can therefore be seen that the effects of using the gate cathode bypass resistor include:

◦ Increase the dv/dt capability.

◦ Retain gate damping to assure the maximum repetitive peak off-state voltage VDRM capability.

◦ Lower the turn-off time, tq.

◦ Raise latching and holding current levels

  • 1
    \$\begingroup\$ "Low-current, high-sensitivity SCRs are triggered by such a low current through the gate junction that a specified external gate-cathode resistance is required in order to prevent triggering by thermally-generated leakage current." I think that's spot-on what I was looking for, especially since the SCR is a low-current, "sensitive-gate" type. \$\endgroup\$ Commented May 19, 2018 at 18:55

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.