as we know in transistors, current controls current, in JFET voltage controls current, but how about SCR? I have learnt that the gate can be triggered using both voltage and current (called gate triggering voltage or gate triggering current). So can one say SCR is both voltage and current controlled device? In other words can we trigger the gate using both voltage and current sources?

The second question is as follows : as the picture shows a a circuit for gate triggering enter image description here

I can't understand why the resistor is being used? My book says it provides noise suppression and better on off time, but I really couldn't get that.

  • 1
    \$\begingroup\$ Actually, bipolar transistors are controlled by voltage as well - the current is a side effect. The same applies to SCRs \$\endgroup\$ Jan 29, 2021 at 15:18
  • \$\begingroup\$ @Sayan "In other words can we trigger the gate using both voltage and current sources?" Yes, you must use what is called "load line" concept in a diagram Voltage_gate .vs. Current_gate ... max voltage, max current, max power for duty power pulses ... \$\endgroup\$
    – Antonio51
    Oct 5, 2021 at 19:33

1 Answer 1


If I remember right, if you leave the gate floating, currents from the collector can make their way inside the SCR to the gate silicon then proceed to the emitter silicon and cause it to turn on (i.e. spurious turn on due to noise). With the resistor, these collector currents that make it to the base can leave through the resistor without flowing through the emitter. Same thing can happen with BJTs but it is much worse with SCRs since they latch on.

In an SCR it should be current that controls the gate. I don't know why it is not called the base though.

  • 2
    \$\begingroup\$ I upvoted because I think you remembered right! I have six GE C291D SCRs that need quite a lot of gate current to trigger. And, for lower power SCRs, like the homemade SCR at lower left, the gate resistor shunts leakage current to ground. \$\endgroup\$
    – Ed V
    Jan 29, 2021 at 15:09

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.