# Thyristor not pulling current away I'm trying to use this thyristor to pull current away from another part of the circuit. I expected that triggering the thyristor would cause current to flow away from the base of transistor T1. Instead, after triggering all I get is a voltage drop from the thyristor collector to the gate. Is this because the thyristor resistance is greater than the resistance to the T1 base (i.e. 0) and so all the current continues to go to T1 base? Would putting another resistor before T1 base solve this?

• (1) "... all I get is a voltage drop from the thrysitor collector to the gate." Thyristors have an anode, a cathode and a gate. No collector. Can you edit to fix this? (2) "... Is this because the thyristor resistance is greater than the resistance to the T1 base (i.e. 0) and so all the current continues to go to T1 base?" What's important is the voltage level. We generally don't talk about "resistance" of transistors or thyristors. (3) Add a link to the thyristor datasheet but when you find it check the expected voltage drop across the thyristor when it's turned on. – Transistor May 23 at 20:25
• "2N5054"? You sure about that? – brhans May 23 at 20:25

Conceptually there is nothing wrong with using an SCR to turn off another device such as a transistor. However you must consider the characteristic of both devices to be successful.

First let's consider the transistor circuit you have: simulate this circuit – Schematic created using CircuitLab

Consider in this circuit:

1. You have approximately 24mA of base current flowing into the 2N3904. The BE voltage would be about 0.8V at this base current.
2. You have approximately 11mA of Collector current flowing to light the LED.

This is a very unusual configuration where you have more base current flowing than collector current. Even at very low temperatures (where the Hfe drops) you would not consider an Ib/Ic ratio of less than 10:1. So for the load current you show an Ib of 1mA would seem more appropriate. This would mean raising R1 to about 20k Ohm.

To turn OFF Q1 you need to reduce the Ib current. Your comment that you want 'pull current away' or divert the current is not wrong, but the transistor is NOT simply current driven. You have to consider the BE voltage (it acts much like a forward biased diode). In this sort of circuit you need to hold the BE junction at a voltage that prevents current flowing into the BE junction. Since turning OFF means reducing the Ic to close to zero, you need to reduce the voltage on the BE junction to less than 0.6V. At this voltage, there will be very little current flowing into the BE junction. Typically you would do this using another transistor with a VCE(sat) much less than 0.6V,and this would sink the R1 current while starving Q1 BE of current.

Now let's consider your SCR (I assume you meant a 2N5064 sensitive gate SCR) circuit. One of the characteristics most important in an SCR is the ON state voltage (consider this to be somewhat like the saturation voltage in a transistor).
Dealing with just the SCR circuit: simulate this circuit

From the above you can see that the SCR is not capable of turning off the transistor, since in your case it is unlikely to pull the V(be) less than 0.78V and at this voltage level there will still be considerable current into the transistor base.

Another characteristic of the SCR is that once turned on it remains on until either the power supply is removed or Ik is reduced beneath the holding current. For the 2N5064 this holding current is about 5mA. So that even if your circuit worked there would be no way to turn OFF the SCR unless you remove the 24V supply. I'm not sure if this was your intent (but I'll assume this was you intent).

There are several ways to fix your circuit detailed below: simulate this circuit

I option A, you simply divide the drive voltage to the transistor so it is OFF when the SCR is ON.

In Option B the base voltage required is increased to accommodate the SCR ON state voltage.

In option C since you appear to want only to be able to turn the LED OFF, you could simply use the SCR to lower the voltage available so the LED is OFF. (I'm sure this is not what you actually want to do, but it's there as an option.

• Thanks! I learned a bunch reading this answer. – zeta-band May 23 at 23:23
• @zeta-band same here! Thanks for that great answer Jack! I picked up a lot of good-practice advice from this all. I also appreciate your giving me the benefit of the doubt throughout. Indeed you were right in all of your assumptions. I do want to circuit to be completely off until power is shut of. I'm about to implement A. Thanks a ton! – lbman May 24 at 15:07
• Implemented A. Tested. Works like a charm! Thanks Jack! – lbman May 24 at 22:43

First, you cannot think in terms of "resistance" in the operating ranges of these devices. Thinking of resistance is futile*. The transistor operates because the base-emitter junction is acting like a diode, with a drop of between 0.6V and 0.7V. The SCR acts like a pair of transistors, NPN and PNP, connected base-collector. When it's on, it (very roughly) acts like a saturated PNP in series with a saturated NPN. So it's voltage drop will be (very roughly) the typical 0.2V Vcb drop of a saturated transistor plus the typical 0.7V Vbe drop of a saturated transistor.

Adding a resistor would just make the transistor turn on weakly, but you wouldn't get it to turn off.

Putting in a diode and resistor (D1 and R1 in the schematic fragment below) might work -- the diode will provide a drop as long as R1 pulls some current through it. There should be a combination of values of R2 and R1 that'll turn the transistor on hard when the SCR isn't active, but turn it off when it is.

Test this -- I just dreamed it up; I will not guarantee it. simulate this circuit – Schematic created using CircuitLab

* How could I resist?

• Isn't SCR1 upside-down? – Transistor May 23 at 21:01
• @Transistor Not anymore! Thanks. – TimWescott May 23 at 21:50