enter image description here


I understand that one of the method to turn-off an scr is by connecting a transistor parallel to it in order to short it. I would like to turn on the transistor(BC547B)by touching the wires with my finger(see photo) and make the scr turn-off. Through the base the current may be about 8 microamper,so the max hfe will be 1.6mA(X200)and that current will be,only if i will connect the collector of the transistor before the R2(second position).

My question are: Does it enough current(1.6mA)to make the SCR turn-off?

What is the minimum current that i need to pass through the transistor in order to turn-off the scr?

Should i connect the collector wire of the transistor as the 1 position or the second position?(before R2 or after the R2,LED?). Thanks.

  • \$\begingroup\$ For a fun hobby project you would want to use a mosfet not a bjt, to use skin conduction \$\endgroup\$ – sstobbe Oct 10 '17 at 18:30

The transistor must conduct essentially all of the load current in order to reduce the SCR current below its specified holding value. In this case, the LED is the load, so the transistor must handle whatever current you're putting through it.

And position 1 is the correct connection. You wouldn't want the transistor to put a short circuit directly across your power supply.

| improve this answer | |
  • \$\begingroup\$ If connection will be as position 1,why that the current will prefer to flow through the transistor rather than the SCR?besides that,i don't understand the idea of bypass the scr,since,unless there will be a serious voltage drop of the source by shorting the scr,the transistor is in parallel to the scr,so the scr will get,anyway,the current he needs.The only change in the circuit will be that the current consumption will increase.The hFE of the transistor is X200,so only 1.6mA max will pass through it,so it doesn't matter if it is directly across the power supply,does it? \$\endgroup\$ – xchcui Oct 10 '17 at 13:28
  • 1
    \$\begingroup\$ The current will flow through the transistor because the Vce of a saturated transistor is significantly less than the Vf of the SCR. \$\endgroup\$ – Dave Tweed Oct 10 '17 at 13:40
  • \$\begingroup\$ In my case,the transistor does not in a saturate state,since the base current is only 0.008mA,does it? \$\endgroup\$ – xchcui Oct 10 '17 at 14:41
  • \$\begingroup\$ Exactly my point -- you need to drive the transistor into saturation in order to accomplish your goal. A single BJT is not likely to have enough gain to do that based on a finger touch. \$\endgroup\$ – Dave Tweed Oct 10 '17 at 16:32
  • 1
    \$\begingroup\$ It seems like the same 10mA will flow through the R2,led,scr plus 1.6mA through the transistor,while the scr,obviously,won't turn-off. \$\endgroup\$ – xchcui Oct 11 '17 at 13:40

With a typical small signal gain of about 330 (per datasheet) and a measured base current of 0.008mA for the BC547B I get a collector current of 2.6mA.

The C106 (per datasheet) has a holding current of typical 0.2mA and max 3mA.

This might not be enough to divert all of the LED current (I assume at least 10mA) to turn off the SCR.

To overcome this limitation you should amplify the turn off signal further, e.g using a darlington configuration.


simulate this circuit – Schematic created using CircuitLab

| improve this answer | |
  • \$\begingroup\$ So,what should i need?a base current of at least 0.7mA,so the transistor will be saturated?and let a current flow of 14mA through it?(assuming that the R2 is 480Ω) \$\endgroup\$ – xchcui Oct 10 '17 at 14:48
  • \$\begingroup\$ Calculate what the base current should be to drive the transistor into saturation and then give it at least three times that so that saturation is guaranteed with temperature, voltage and load variation, etc. \$\endgroup\$ – Transistor Oct 10 '17 at 20:50
  • \$\begingroup\$ Note that the saturation voltage of a Darlington pair is significantly higher than that of a single BJT, often >1V. Have you ever actually used this circuit to do commutation of an SCR? \$\endgroup\$ – Dave Tweed Oct 11 '17 at 12:48
  • \$\begingroup\$ No,i haven't actually try this circuit and i understand that the voltage drop of a darlington pair is higher than a single bjt,so what does it mean?,that it won't turn-off the scr?BTW,the voltage drop of the scr,refer to the datasheet,is max 2.2V. \$\endgroup\$ – xchcui Oct 11 '17 at 13:40
  • \$\begingroup\$ Actually, the "you" in my previous comment was referring to @try-catch-finally. But regarding what it means, have you been following my comments at all? Especially the one regarding the relationship between Vce(sat) and the Vf of the SCR? \$\endgroup\$ – Dave Tweed Oct 11 '17 at 13:47

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.