I have been looking for more information on how to connect two thyristors in parallel in order to balance the load. From what I have found there are two options: either using balancing resistors, which is inefficient, or using reactors (inductors in series with the thyristor).

So far, everything is clear to me. However, I can't find any information on how to calculate the inductance required for each inductor. Since both thyristors are identical, the unbalanced current shouldn't be that high, therefore the inductance shouldn't be that high either (compared to a parallel connection of two different thyristors). I also assume that both inductors would have the same spec, since both thyristors are identical.

Also, can two separated inductors be used as reactors? Or do those two inductors have to be on the same common core?

Let's say I have two thyristors of 220 V and 1 A and the desired output is 1.5 A. How do I calculate the values for these inductors?

Here's a schematic of what I am trying to achieve:

Thyristors in parallel

  • 2
    \$\begingroup\$ Do you really need to parallel SCRs? Very large SCRs are available, I used 1800 amp SCRs 40 years ago. \$\endgroup\$
    – Mattman944
    Oct 27, 2022 at 20:05
  • \$\begingroup\$ @Mattman944 preferably, yes. There are two reasons behind it. Heat dissipation and stock availability. \$\endgroup\$ Oct 27, 2022 at 21:12
  • \$\begingroup\$ @Mattman944 For the context: The SCRs will be used in an AC motor controller. Therefore the SCRs will be either SMT or THT in order to get a small design. \$\endgroup\$ Oct 27, 2022 at 21:43
  • \$\begingroup\$ How little? Having to parallel SMT thyristors sounds like a thermal management problem more than one of total current ratings. \$\endgroup\$ Oct 27, 2022 at 23:50
  • \$\begingroup\$ Surely you would use a single CT inductor .To make the question get better answers why dont you state the turn on time of your SCRs \$\endgroup\$
    – Autistic
    Oct 28, 2022 at 1:02

3 Answers 3


Also, can two separated inductors be used as reactors? Or do those two inductors have to be on the same common core?

Balancing was done (in my youngest time ...) with ... transformer.
So coupling your two inductors would be "enough".
Inductors used alone would limit di/dt.

Here is a simulation showing the need of "balancing" currents.
Note that this can be done only with TWO SCRs.
If you use 4 SCRs paralleled, this must be done with 3 transformers.

First simulation, when no balancing is used ... Just an anodic resistance of 1 Ohm difference.
The currents are "very" different.

enter image description here

Second simulation, adding a "transformer" in the "anodic" circuit.
Are you convinced? The currents are now quasi equal.

enter image description here

NB: the big SCR uses a "bunch" of little SCR paralleled ...
No balancing is needed ... because there are on the same "die",
and/or specifications are selected very "nearest".
And wiring of these PCBs is a very special kind of "wiring" ...

Evaluation of "inductors/transformer" for "reasonable" result.

Link to file made with microcap v12.

Example. See the inductor's value and "sharing" fraction.

enter image description here

enter image description here

One could make LL bigger ...

enter image description here

  • \$\begingroup\$ Thanks for the answer however, main question was how do I calculate the minimum inductance required for those two inductors? \$\endgroup\$ Oct 29, 2022 at 16:42
  • \$\begingroup\$ Note that L1 & L2 is ... a transformer, and some mH would be enough. Just passing a large current. Note that sense of coupling is very important. \$\endgroup\$
    – Antonio51
    Feb 4, 2023 at 8:50
  • \$\begingroup\$ is there a way to calculate the inductance for that? What's the optimal mH required? \$\endgroup\$ Feb 5, 2023 at 15:44
  • \$\begingroup\$ I don't have notes for now, but this could help ... youtube.com/watch?v=-hwnGHOe9f8 \$\endgroup\$
    – Antonio51
    Feb 5, 2023 at 21:37
  • \$\begingroup\$ I add also the file (interactive) in answer for simulation "evaluating" ... \$\endgroup\$
    – Antonio51
    Feb 5, 2023 at 22:23

Since both thyristors are identical, the unbalanced current shouldn't be that high

Your assumption -- but they aren't matched in any meaningful way, as far as I know. (Maybe they can be special ordered that way?)

Normally, diode junctions (SCRs are about two in series, when on) vary by some 10s of mV between individuals of given type, at same current and temperature. 10s of mV additionally for differences in temperature. And that gives, well, you'll want to refer to the VTM curve or dynamic resistance figure to see how much voltage mismatch translates to what current mismatch. Needless to say, a, say, 10% voltage mismatch, will result in much more than a 10% current mismatch.

The inductors serve two purposes:

  1. Provides some voltage drop to ballast (i.e., dominate over) the difference in VTM
  2. Somewhat isolates the transient event as they turn on (i.e., one can potentially turn on, say, some 100s ns before the other, and both still see similar turn-on conditions.)

(Hm, I don't actually know offhand whether this matters. Is SCR turn-on effectively independent of VA? I would expect it's at least slower near VTM.)

Somewhat as a corollary, inductance also limits dI/dt, which may be needed with some sources/loads. (Likewise a snubber may be needed to absorb the energy stored in the inductor during reverse recovery.)

Ballasting behaves same as paralleling any other diodes, so, a fraction of a volt should be sufficient. Even a couple volts might be acceptable, as an inductor is non-dissipative at AC (assuming these SCRs will be used in anti-parallel pairs as a TRIAC, or that you're actually using TRIACs as such).

Strictly speaking, the inductors can be coupled, by using a center-tapped winding with the load connected to the tap and the two SCRs at either end (in RF, this is known as a 0° power combiner). But this will not limit dI/dt once both turn on, and the coupling may cause hazardous voltage in the instant before both have turned on. Best to use independent inductors, then.

At 50Hz and 1A, 1V drop is afforded from 1Ω reactance, or 3.2mH.

Given the reference to SMT components, the cure [large, expensive inductors] sounds much worse than the disease. If your problem is thermal, consider a MOS type SSR: the resistive characteristic can have much lower voltage drop. It will be more expensive than SCRs (but probably still cheaper than the inductors, hah).

I have no idea about availability issues; without location and budget given, I'm assuming what is typical for me and most readers, i.e., western markets.

  • \$\begingroup\$ the SCRs will be used in parallel and in the same direction, so it won't be like a triac configuration. The load is a 160V DC uniploar stepper motor. Hence why I attached the image to the question. That's exactly what I want to accomplish. I split across the two SCRs. \$\endgroup\$ Oct 28, 2022 at 19:26
  • \$\begingroup\$ About MOS type SSRs, would they be able to switch on and off in 1mS, and also switch off at the 0 AC crossing? \$\endgroup\$ Oct 28, 2022 at 19:30
  • \$\begingroup\$ @SumutiuMarius Do you know they'll turn off at the zero crossing? Given motor inductance and all? \$\endgroup\$ Oct 28, 2022 at 20:15
  • \$\begingroup\$ the SCRs would be turned on by a PWM signal from the microcontroller and let run till they turn off automatically when the AC goes into negative. \$\endgroup\$ Oct 28, 2022 at 20:37
  • \$\begingroup\$ @SumutiuMarius But you said DC... would there not be a rectifier? Wait, are you just doing this half-wave?! \$\endgroup\$ Oct 28, 2022 at 21:01

So, after doing some digging through my physics books, I think I found the answer.

We start with the general formula for voltage at the terminals of an ideal inductor:


From this, we can say that L is:


We know that VRMS is:

$$V_{RMS}=\frac{1}{\sqrt{2}}\cdot V$$

Since we know that the SCRs won't conduct electricity at the same time for a brief period of time, the intensity through a single SCR is the total intensity of the parallel pair (in my case, 10A). We also know that the SCR will only conduct electricity on the positive side of the sine wave. Therefore VRMS is:


So, our VRMS is 125 V. Our frequency is 50 Hz, and that is one cycle in 0.02 s. Knowing this, we can calculate the total inductance for the parallel pair of SCRs:


From this, we calculate that L is 0.25 H. Also, we use two inductors for our pair of two SCRs, therefore we divide the inductance by the number of inductors coupled, resulting that both inductors need to be L = 0.125 H and 10 A.

Please correct me if I am wrong. Maybe I have misunderstood something.

  • \$\begingroup\$ You've mixed up formulas; in the first, V and I are time-domain signals \$v(t)\$ and \$i(t)\$. Capitals are usually used to denote either frequency-domain values, or peak or average or other reference values -- constants, not functions. But often they are still used (for convenience or shorthand) to refer to functions (I'm guilty of this myself..). It is clear from the derivative that these are functions; the expression isn't meaningful otherwise. But then there is no meaning of the Vrms expression, if the same V is meant. This is an AC sinusoidal steady-state relationship. \$\endgroup\$ Nov 1, 2022 at 6:16
  • \$\begingroup\$ Finally, your calculated L invokes cycle time... arbitrarily? And is apparently calculated for handling the full voltage drop, which doesn't sound like it's intended by the application. For reference, the AC S-S relation is \$X_L = 2 \pi F L\$, for frequency in Hz, L in H, and \$X_L\$ in Ω. \$\endgroup\$ Nov 1, 2022 at 6:18
  • \$\begingroup\$ One cycle for AC is done in 0.02s (50 cycles per second -> 50Hz -> 1 cycle/0.02s), and that's our dT. \$\endgroup\$ Nov 1, 2022 at 19:53
  • \$\begingroup\$ I'm not sure the expression has been understood correctly: "dt" isn't a finite difference in time. It's an infinitesimal, and the "d[..]/dt" expression is a calculus operator, meaning, take the slope of the function. The slope of a sine wave is non-constant -- a single number cannot be assigned to it -- it's another function. \$\endgroup\$ Nov 2, 2022 at 5:36
  • \$\begingroup\$ In your formula: XL=2πFL, we have two values we don't know. We don't know the inductance we need. We also don't know what XL this inductor will have since we don't know what inductance we need. \$\endgroup\$ Nov 3, 2022 at 20:01

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