Is this TL431-based overvoltage limiter circuit alright?

I have a supply rail that is 15 V nominally +/- 5%. So the highest nominal voltage is 15.75 V. Due to external fault currents of up to 200 mA, the voltage on this rail can rise anomalously. The fault currents are cut out by additional circuitry after several 100 µs, but during this time, clamping diodes dump the fault current into the supply rail.

The devices on this rail have a lowest absolute maximum rating of 18 V. I have figured that the window between 15.75 V and 18 V is too tight for a simple limiter based on Zeners/TVS diodes. So I thought about using a cheap generic shunt regulator such as the TL431 and came up with the following circuit (which limits approximately to 16.5 V as set by R1 and R2):

I would like to ask if this circuit would work as expected, as I have never used such shunt regulators before and don't know their pitfalls.

Specifically, since the reference pin voltage is below 2.5 V during normal operation, the TL431 is not in regulation. When an overvoltage strikes, would the regulator turn on quickly enough as in the simulation to limit the rail voltage? Also the datasheet mentions regions of instability due to cathode-to-anode capacitance (e.g. page 17 and 18). As the TL431 is not in regulation, could that region change and cause instability also in my circuit?

• @verbal kint is an expert on all such things. || It looks ok to me. However , the 10uF would be expected to ramp at 0.5us per volt or about 0.75 us to swing the 1.5 V shown . you are seeing about 100 us so it is probably not dominated by the 10 uF - but i suggest removing the capacitor in simulation to see how the TL431 responds by itself. Dec 6, 2021 at 9:27
• @ATCSVOL without any capacitance there, the voltage rise is beyond 21 V and for a fault current rise time of 1 µs. However, this is unrealistic, as there is capacitance on the rail that will limit the rail voltage rise rate. For 10uF, I calculate a rise rate of 0.15A/10uF=0.015V/µs=1.5V per 100 µs which is exactly the value in the sim. Dec 6, 2021 at 9:32
• TL431 requires a max. Bias current of 1mA. R3 should have a value that allows this 1mA to flow without turning Q1 on. Dec 6, 2021 at 9:39
• Instead of pulling the rail low with all the force of the universe (or, let's say, all the force your BJT can or cannot handle), you might consider a transistor that disconnects the external load. Far less power might be needed and you might be able to use a small-ish transistor. Dec 6, 2021 at 10:08
• @zebonaut Please see my edit of the first paragraph. Such mechanism is already present, but the used switches need some time to turn off. Dec 6, 2021 at 10:27

TL431 will work fine in this application, with a few caveats:

When VREF<2.5V it is off but it will still draw a small cathode current to power the internal circuitry. When VREF=2.5V, in order to regulate, it needs to be able to draw a cathode current at least equal to "minimum cathode current for regulation" which is 1mA.

So R3 should be around 470-560 ohms to make sure these conditions are met.

The other potential problem is instability. TL431 is a feedback system, and adding a follower transistor increases open loop gain. Stability of the loop depends on what kind of capacitance and ESR you have on your power rail. It you only have a 10µF MLCC it's probably going to be unstable. A large aluminium cap like >200µF will be fine. So you need a TL431 spice model with accurate open loop behavior. To make it stable, you may need to add a small cap like 100pF between VREF and Cathode, and a 330R base resistor on your PNP.

In addition, BC857C won't handle the current and dissipation: 15V*200mA = 3W so you need a TO220/TO129 device with a small heat sink, or a DPAK which uses the ground plane as heat sink.

A simpler circuit that is stable would be this one, but the zener will be less accurate, so you'll need a bit of extra margin:

• Thanks bob, I will play around a bit with R3 and capacitors as you suggest. The fault current will be cut out after a few 100 µs, and 200 mA is the absolute worst case, so the BC857 would be good enough. I have also considered Zener circuits (even such with an additional opamp), but in the end, I would need very specific voltage values or tight zener tolerance or several tight resistors to make it work so I thought it's a good opportunity to get into the TL431 ;) Dec 6, 2021 at 9:36
• Why does the voltage on the power rail rise? Is it due to protection diodes dumping current into the rail? Dec 6, 2021 at 9:56
• It looks like a) capacitors around ref-cathode generally make stability worse in the sim. b) If C1 is in the "stable" range, it will be fine. Reducing C1 to 1uF puts it too close to the unstable region and I see oscillation. So C1 must be sufficient. With respect to R3: Reducing it has minimal consequences. I suppose, the TL will draw the necessary supply out of Q1 but still not saturate it. So the TL will hover near a "just-shy-of-on" state. However smaller R3 has no detrimental effect, so I'll go there anyway. PS: yes diodes dump fault current into the rail. Dec 6, 2021 at 10:03
• Yeah the compensation cap can make it worse. Don't forget the ESR of the cap, 10µF MLCC have just a few mOhms so you can neglect it, but 10µF caps have a few ohms ESR so you can't. Dec 6, 2021 at 10:05

I would like to ask if this circuit would work as expected, as I have never used such shunt regulators before and don't know their pitfalls.

The TL431 is not particularly fast device so, the saviour here is the 10 uF capacitor in your circuit (C1). C1 really does need to be here to slow things down enough so that the TL431 can handle the slower surge.

But, given that surges are short lived (especially ESD) then, you could probably get-away with just using C1. However, it all depends on what "threat" you are trying to cope with. If it's ESD (EN61000-4-2) then you should model a charged 150 pF (8 kV) in series with a 330 Ω resistor. If it's an indirect lightning (EN61000-4-5) then it might be a waveform like this: -

Where VNpk can be 1 kV, 2 kV etc. depending on the particular part of the standard to be met. Image from: -

And, the series current limiting resistance might only be 2 Ω (depending on application). If all you are interested in is the 200 mA surge mentioned in your question then ignore my extra ramblings.

• This is not really for EMI, but to make the users inputs more sturdy against accidental overvoltage. The device cuts out the ports when it detects and out-of-range-voltage but this takes several 100 µs to complete. The faster EMI events you mention are mostly absorbed by series resistance, clamp-to-rail-diodes and rail capacitance. Dec 6, 2021 at 9:45
• Sure @tobalt - I hear you but, I like to throw a few extra thoughts in just in case you might not have taken a bigger picture. It seems you haven't of course! Dec 6, 2021 at 9:48