How does this PNP circuit work? (Overvoltage protection?)

[ledtester]

The threshold voltage is between 5-10V. (The exact value isn't really important)

The circuit doesn’t need to work for 230V, it would just be nice to design it in a way, so it doesn’t get destroyed if such high voltages occur.

Another device you might want to look at is a Bourns TBU ("transient blocking unit").

As I understand it they can withstand 450 Vrms and will trigger within 1 microsecond on an over-current condition. When triggered they only let a small current through and then they'll reset when the voltage across them drops to a low enough (like 15V).

Here's a Bourns training/marketing video explaining their operation and use:

https://youtu.be/E6iEILeVr7A

[langwadt]

Sorry the +21 is supposed to be +21V, I forgot the V.

This voltage is generated by a DC-DC converter (24V-21V, galvanic isolated) and so the voltage should never be over ~21V.
In a fault condition there could be 230V AC on the Input (MTH6 & MTH7), but the Diode D10 should prevent an overvoltage condition on the left circuit. So thats why Q17 still doesn’t make sense to me.

if MHT6 is pulled (enough) below ground (MHT7) Q17 turns of Q18

[Terry Bites]

This is a current mirror / source https://electronics.stackexchange.com/questions/539580/pnp-transistor-constant-current-source

[langwadt]

This is a current mirror / source https://electronics.stackexchange.com/questions/539580/pnp-transistor-constant-current-source

that's  a current source not a current mirror, and only half the circuit in question

[MaximilianMM]

Thanks for all your replies.
I will most probably stick with the same design, as it is used by all devices which I looked into, so it should work just fine.

But the function of Q17 still doesn’t make sense to me.

[xavier60]

D10 blocks high positive voltage being applied to V+ but high negative voltage still gets through and is applied to  Q18 and although the polarity is correct for Q18, the dissipation would likely be too high for it.
Q17 and associated divider senses the negative voltage and turns off Q18. No current, no dissipation.

[MaximilianMM]

Now I understand it. Many thanks to you all, you helped me allot.

[MaximilianMM]

I think I need some more help on the other part of the circuit.

The initial circuit I posted was for receiving commands, this circuit is for sending commands. It’s very similar. I don’t understand how it should work. It’s obviously again a constant current source but I don’t understand the function of R19. It must have something to do with overvoltage protection, as this circuit can also see up to 325V on the input. At this voltage Q12 wouldn’t last very long, so it needs to be turned off somehow. Can somebody try to explain this circuit to me.

[xavier60]

R19 is a load resistor so that over 400uA needs to flow through the optocoupler's transistor before the constant current circuitry is activated.
It prevents small leakage currents prematurely turning things on.

[MaximilianMM]

So, to turn on Q12 you need a voltage drop, over R19, of roughly 1.2V (U_R18 + U_Q12BE). Which would mean a current of ~0.44 mA through R19. Did I understand that correctly?

But how does this circuit limit the current through the optos transistor? Q10 essentially shorts out R19 to control the current through R18. So, there must be a huge current through the optos transistor, right? In normal operation that wouldn’t be problem because the input has a current limit of ~20 mA. (Circuit I posted further up). But if 325V from mains are connected and the opto turns on, the circuit will certainly be damaged/destroyed, right?

[xavier60]

So, to turn on Q12 you need a voltage drop, over R19, of roughly 1.2V (U_R18 + U_Q12BE). Which would mean a current of ~0.44 mA through R19. Did I understand that correctly?

But how does this circuit limit the current through the optos transistor? Q10 essentially shorts out R19 to control the current through R18. So, there must be a huge current through the optos transistor, right? In normal operation that wouldn’t be problem because the input has a current limit of ~20 mA. (Circuit I posted further up). But if 325V from mains are connected and the opto turns on, the circuit will certainly be damaged/destroyed, right?

Q12 actually starts turning on with 0.6V across R19 but the current is very low at this point. Maximum current is 30mA when the voltage across R18 limits at 0.6V, 0.44mA through R19.
So there is a limit of 30mA available to the optocoupler's transistor. The optocoupler's transistor will likely sink  only few milliamps anyway.
The optocoupler's transistor current can be approximately calculated from its CTR specification and the its LED current.
Q12 will have the most stress, passing 30mA at 325V but it will not be continuous, only at times when the optocoupler is being driven.

[MaximilianMM]

So, when there is a maximum voltage drop across R19 of 0.44mA*2700 Ω= ~1.2V the remaining voltage (still ~324V) needs to be dissipated across Q14 and Q15 or am I missing something? The 30mA current limit just applies for the collector current of Q12.

[xavier60]

So, when there is a maximum voltage drop across R19 of 0.44mA*2700 Ω= ~1.2V the remaining voltage (still ~324V) needs to be dissipated across Q14 and Q15 or am I missing something? The 30mA current limit just applies for the collector current of Q12.

Almost all of the 324V will be across Q12 and Q14. The 2 series connected diodes in Q6 will clamp the voltage at about 1.4V causing something less than 1V to be across the optocoupler's transistor.

[MaximilianMM]

Thank you for the explanation!
As long as the optocoupler isn’t turned on too many times the circuit should hold up just fine, but if I send telegrams constantly with 325V at the input, the power dissipation will definitely destroy this circuit. Maybe I could add some circuitry to disable the optocoupler if high voltage is present.

[T3sl4co1l]

Would like to see foldback added, so current drops at high voltage, eventually going to ~0 so that it stays completely off (or, as off as the cascode opto configuration allows -- this could be arranged differently I think).

Tim

[MaximilianMM]

Would like to see foldback added, so current drops at high voltage, eventually going to ~0 so that it stays completely off (or, as off as the cascode opto configuration allows -- this could be arranged differently I think).

Tim

I also think that this circuit could be arranged differently. The constant current part of this circuit doesn’t make sense in my opinion, as the voltage source is current limited either way. A good overvoltage protection would be more helpful.

[MaximilianMM]

I rearranged the circuit. What do you think about that? It should disable Q31 if the voltage is above ~34V. The only problem is R41 as it has to dissipate a lot of power. Maybe I could use a PTC for R41.

[xavier60]

I rearranged the circuit. What do you think about that? It should disable Q31 if the voltage is above ~34V. The only problem is R41 as it has to dissipate a lot of power. Maybe I could use a PTC for R41.

The main problem is that it doesn't provide any current gain for the opto.

[xavier60]

A simple idea which will cause progressive foldback is to put a resistor in series with the Base of Q10 in the original design. Move the top of R17 to the Base of Q10. If the current is reduced by too much at normal operating voltages, the value of R18 could be reduced to compensate.

[MaximilianMM]

I rearranged the circuit. What do you think about that? It should disable Q31 if the voltage is above ~34V. The only problem is R41 as it has to dissipate a lot of power. Maybe I could use a PTC for R41.

The main problem is that it doesn't provide any current gain for the opto.

What do you mean by that?
Just to be clear, this circuit just needs to work for max.~24V. So, it would be ok to turn the circuit off completely, if that value is exceeded.

A simple idea which will cause progressive foldback is to put a resistor in series with the Base of Q10 in the original design. Move the top of R17 to the Base of Q10. If the current is reduced by too much at normal operating voltages, the value of R18 could be reduced to compensate.

Like so? I don’t think that would work, right?

[xavier60]

The way I understand it, the transmitter needs to be capable of sinking more current than the receiver can supply for it all to all work properly. Although we don't yet know how much current the opto's transistor is capable of sinking, I'm certain that it's not enough. The 0.44mA needs to enable the 30mA sink.
Yes, you have those resistors positioned correctly. Start with 1.5K for R44.

[xavier60]

If that last idea cant be made to work properly, that is it being able to sink the correct current while be able to also protect itself, the same idea as the circuit in Reply #29 could be used.

[MaximilianMM]

The way I understand it, the transmitter needs to be capable of sinking more current than the receiver can supply for it all to all work properly.

Yes, that’s absolutely right.

If that last idea cant be made to work properly, that is it being able to sink the correct current while be able to also protect itself, the same idea as the circuit in Reply #29 could be used.

That’s what I tried to do, when posting my circuit (reply 41). It’s the exact same circuit, as the one which provides the voltage for the bus. I only changed the current limiting resistor to get a higher current, so that the bus-voltage collapses for a low-signal.
The datasheet of the optocoupler states a maximum collector current of 50mA, so it should be fine.
But there must be reason, why the manufacturer has chosen this design.

[MaximilianMM]

Little Update:

I tested the two circuits (reply 32 and reply 41) with 230V AC. (With input protection posted in reply 29)

Circuit from reply 32:
The circuit seems to handle 230V AC just fine if the Optocoupler Q15 is turned off.
If the Q15 is turned on permanently, the circuit only last for maybe half a second, before the magic smoke escapes.
If I pulse the optocoupler (simulating a transmission) the circuit holds up a little longer (about 7 sec.).

Circuit from reply 41:
I only tested it with the optocoupler tuned on permanently. The circuit only lasts for about half a second (same as the other circuit).

What do you think is the problem? Is Q30 too slow, for turning off Q31?  I thought the circuit would work, as Q30 should turn off Q31 if the voltage exceeds ~30V.

[xavier60]

Little Update:

I tested the two circuits (reply 32 and reply 41) with 230V AC. (With input protection posted in reply 29)

Circuit from reply 32:
The circuit seems to handle 230V AC just fine if the Optocoupler Q15 is turned off.
If the Q15 is turned on permanently, the circuit only last for maybe half a second, before the magic smoke escapes.
If I pulse the optocoupler (simulating a transmission) the circuit holds up a little longer (about 7 sec.).

Circuit from reply 41:
I only tested it with the optocoupler tuned on permanently. The circuit only lasts for about half a second (same as the other circuit).

What do you think is the problem? Is Q30 too slow, for turning off Q31?  I thought the circuit would work, as Q30 should turn off Q31 if the voltage exceeds ~30V.

Just the circuit in  reply 32. Confirm that only half wave of the 230VAC is applied with the correct polarity?
And if all of that is correct, the dissipation Q12 will be in the order of watts at 30mA.
The dissipation in Q14 is unknow because we dont know Q15's On current. I dont understand how it survives in the original application.
Are you certain that high voltage is expected to be applied to it?