I have a question regarding a circuit I've been working on for some time. The circuit is based around existing supercapacitor protection modules and uses a TL431 (actually a TLV431 lower power version) precision Zener device. Basically, the circuit works, but I wanted to get some feedback about resistor values, and whether or not my NPN or PNP transistors are doomed to fail.

In summary, the circuit uses the TLV431 and it's reference and sensing resistor divider to set a trip point (somewhere around 2.65 V as I have it at the moment) the TLV431 then starts conducting, pulling current (this is where I get foggy) from the base of the PNP transistor (BC557) turning it on (hopefully in saturation). When it turns on the NPN gets (an unknown amount of?) base current, and turns on sinking the charge current (with a maximum of?) through a resistor and 'around' the capacitor cell.

Eventually I'd like to have this circuit in series with my series capacitor strings so that any stage in the stack never sees more than it's 2.7 V (2.65) so long as I stay under the dissipation capabilities of this circuit on the input. The current modules that I bought from eBay only seem to limit on supplies up to about 500 mA. I'd like to get this up to 2 - 3 or 4 amps. Let me know what you think - the specific question is about R3 value, R4 value, R5 value, and if a resistor is required between the PNP and NPN base. Thanks!

enter image description here

I've seen versions of this circuit that omit the PNP base resistor, but I've added one because without it I think the TLV431 would burn and fail - perhaps someone could also clarify this aspect of the circuit? If anyone else is building protection circuitry around the TLV431/TL431 for 2.7V supercapacitors I'd really appreciate you chiming in!

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    \$\begingroup\$ How can we analyze the circuit if there are no component values? \$\endgroup\$
    – Voltage Spike
    Jul 29, 2019 at 19:43
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    \$\begingroup\$ Hi Ryan. When Q2 starts conducting, it will put full V+ into the base of Q1, probably destroying both transistors. There needs to be a resistor somewhere on the collector or emitter of Q1 to prevent this. \$\endgroup\$
    – rdtsc
    Jul 29, 2019 at 19:51
  • \$\begingroup\$ The component values are generally what's in question! I have R5 as a 10W 1 OHM Resistor. R4 is currently a 470 OHM. To 'power' the TLV431 I've had R3 as a 2.2K and currently a 1K Resistor. I've had various combinations of R1 & R2 - the current trip point is set at 2.65V producing a divided 1.25V at the TLV431 reference/sense pin using a R1=7K and R2=6K \$\endgroup\$ Jul 29, 2019 at 20:08
  • \$\begingroup\$ I would think that Q2 would have the resistor on the collector leg leading into the base of Q1 to limit current... the maximum emitter current for a BC557 is listed at 200mA should I be looking to supply the base of the BD437 with close to this limit to allow it to saturate (will it saturate at say 180mA of base current?) \$\endgroup\$ Jul 29, 2019 at 20:13
  • \$\begingroup\$ Not enough information provided for anyone to help you. What is the charging current for your supercap string? Is the charger current limited? Is the charger voltage limited? (in other words is the charger VC/CC). The BC557 can never saturate in your current configuration, but it can blow up. \$\endgroup\$ Jul 29, 2019 at 21:12

2 Answers 2


First off, for a clamp voltage of 2.65V this circuit really needs the TLV431 because it only drops 1.25V, whereas the TL431 drops 2.5V which doesn't leave enough voltage to turn on Q2.

The BD437 has a saturation voltage of up to 0.8V at 3A with 300mA Base current. The BC557 is only rated for 100mA maximum Collector current, so to get 4A you need a power transistor with higher gain (and preferably lower saturation voltage) at that current.

When Q1 is fully turned on R5 should dissipate most of the power, but at lower current it will be shared between the resistor and transistor. Maximum transistor dissipation occurs when the voltage is split evenly, at a current of ~2A. So you need a transistor that can safely dissipate at least (2.65V/2)*2A = 2.65W.

Unfortunately high gain, high power, low saturation transistors are not very common. The 2SC6082 is one example. It has a typical current gain of over 100 and saturation voltage <0.2V at 4A, so it requires less than 50mA Base drive.

A BC557 can easily handle 50mA at 2.65V. However in your circuit only Q2's Base current is limited by R4, so maximum Collector current could vary widely depending on the individual transistor's gain. To fix this you could put R4 in series with the Emitter and connect the Base directly to the TLV431. The TLV431 drops 1.25V and Q2's Base-Emitter junction drops ~0.6V, leaving 2.65V-(0.6V+1.25V) = ~0.8V across the Emitter resistor. 0.8V / 50mA = 16Ω. This will also protect the TLV431 because the Base current is about 100 times less than the Emitter current.

Assuming Q1's saturation voltage is ~0.2V, R5 should be (~2.65-0.2)/4A = ~0.6Ω. It could dissipate up to ~(2.65V-0.2V)*4A = ~10W so it should be rated for at least 20W - and it will get hot!

  • \$\begingroup\$ Thank you for taking the time to provide this answer ! In fact I am using the lower power tlv431. Your answer is very informative I didn't know the base of the BC557 safe to connect to the TLV431 directly if the current through collector / emitter was limited. I will try and obtain another medium/high power NPN transistor as suggested with a lower saturation. \$\endgroup\$ Jul 30, 2019 at 4:29

Your circuit as you have it has several problems:

  1. The BC557 is unsuitable since it is rated to only 100mA.
  2. Q2 has no collector resistor so is held out of saturation and will get hot (very hot).
  3. Trying to get Q3 into saturation is pointless, you should spread the dissipation between the resistor and the output transistor.
  4. You need to positively turn off the ouptut transitor or leakage may be a problem.
  5. You have very little headroom for the TLV431, though it will work.

I'd suggest the following components may be viable at 2A charging current:


simulate this circuit – Schematic created using CircuitLab

The waveforms would be like this:

enter image description here

The output stage (both resistor and transistor) have to dissipate about 5.3W. power resistors are vey expensive so it's best to keep the BD437 out of saturation and spread the power between them. If you have multiple Supercaps then you are going to need to consider how much power is dissipated when you have multiple units connected together.
Maybe you should consider a fan.

  • \$\begingroup\$ The BD473 was getting quite hot I didn't run my tests for long but will consider the part to be compromised. Thanks for taking the time to respond Jack \$\endgroup\$ Jul 30, 2019 at 4:32

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