# How does a capacitor and resistor affect an SCR's firing angle?

Greetings! High schooler here studying SCRs. I've been looking at sites and books on the concept of increasing an SCR's firing angle, and what I've noticed with these references is that they show you how to do it, but they never really explain why it was able to do that in the first place!

The circuit that is confusing me right now is this one: My guess on how the firing angle increases: I actually calculated this circuit's firing angle and got 95 degrees!

So what I assume is that because a capacitor lags the current, it takes longer for the SCR (which needs a gate-triggering current) to activate. However! With an angle more than 90 degrees, doesn't that mean all the values of the source voltage have already been tried and not enough for the circuit to produce a gate-triggering current? What more with angles after 90 degrees, which have less voltage values since it passed its maximum?

I think there might be a concept of capacitors I am unfamiliar with? How can an SCR still exhibit a voltage waveform like this:

(Values of circuit if you're curious) R1 = R3 = Rload = 1kΩ, R2 = 500Ω, C = 10uF, Igt = 10mA, Vsource = 160 Vrms

## 4 Answers

You are assuming that the voltage across the capacitor will be "in phase" with the sinewave voltage applied across the circuit (AC source). That isn't true.

The true waveform across the capacitor is another sinewave of lower amplitude but, importantly, it will lag the AC source voltage by anything up to 90 degrees: - Trace A (blue) I'm using to represent an applied voltage (AC source) and the green trace (Trace B) will be the resulting voltage across the capacitor for specific values of R and C. As you can see the peak capacitor voltage occurs way after the peak of the applied voltage and if the SCR trigger threshold were set about 0.7, the SCR would trigger here: - As you can see the trigger point occurs after the voltage peak.

This is not really a very practical half-wave phase control, however it should work with a very sensitive gate SCR.

Normally you'd have a diode in series with the RC network (to prevent reverse biasing the gate) and a trigger device in series with the gate.

As it sits, the capacitor will be close to zero volts after the negative half-cycle since the gate will break down at negative 7-9V.

The SCR will turn on when the gate is positively biased and perhaps 50-100 uA gate current is flowing for a sensitive-gate SCR. The gate voltage at that point will be something like 1V, which is typically very low in comparison to the AC source voltage during (almost) the entire positive half cycle. The voltage on the other side of R3 might be a couple tens of volts, which is still less than the mains voltage more than perhaps 100usec before the end of the half cycle.

So the capacitor will continue to charge, albeit at a lower rate, after the peak right until close to the end of the positive half-cycle or until the SCR triggers..

What decides whether the SCR turns on or not is (usually) not the amount of voltage between anode and cathode, as you seem to suggest, with your "SCR voltage" bottom plot, but the amount of current into the gate terminal (and out of the cathode). The {R1+R2,C} network create, across C, a delayed version of the voltage across the input terminals. That delayed sinusoidal voltage across C creates (thanks to R3) an equally delayed sinusoidal current into the gate of the SCR, and that current is what may turn on the SCR, if and when it exceeds a certain threshold. Since the delay of the sinusoidal current into the gate is adjustable (with R2), so is the turn on time (if it happens) of the SCR. And, since an RC network can delay a voltage by up to 90º, that is the delay range you also have for the turn-on time.

Once the SCR is turned on, a gate current is no longer needed, and the SCR stays on until the current from anode to cathode goes below a certain threshold, very small.

An I said "usually" because you can also turn an SCR on by increasing the voltage between anode and cathode, but that does not seem to be what you want to do.

• Oh, thank you! Your explanation was definitely the best, because you were able to give out the flow of the circuit. Though, I got to ask: does the circuit actually follow an order where it goes from R1 and R2, and then capacitor, and then R3? I had always thought there was already voltage in R3 before the capacitor charged up. Or does current tend to go to the capacitor first before going into R3? – iheartchococake Dec 2 '20 at 15:28
• @iheartchococake resistors don't have voltage "in" them, so the question doesn't make sense. Current goes in all the ways it can go - same as for any circuit - and it can't go into the thyristor until the voltage is high enough to turn the thyristor on. – user253751 Dec 2 '20 at 15:29
• @user253751 Oh yeah, apologies. I was just wondering since current can go anywhere in the circuit, how did we come to realize that the capacitor would be charged up first, rather than having the current in the circuit pass through R3 first – iheartchococake Dec 2 '20 at 15:50
• @iheartchococake because the voltage isn't high enough to turn the thyristor on – user253751 Dec 2 '20 at 16:19

The key with SCRs is always charge in the TRIAC "gate". R1, R2, and C can be thought of a normal RC circuit for charging. R3 is the only thing delivering charge to the TRIAC.

So the instantaneous voltage of the waveform isn't exactly what matters, here, but rather thee amount of charge it lets you push into the TRIAC. It could well be that you haven't reached that critical charge density to turn the TRIAC on until later in the waveform.

Lowering the resistance values will allow for higher current at lower voltages and help move that firing time forward in the half-wave. That's what I would try.

Best of luck--keep us posted.