How does this PNP circuit work? (Overvoltage protection?) - Page 1
Forum section: electronics / beginners

author: MaximilianMM
author_rank: Contributor
timestamp: January 04, 2022, 10:36:45 am
content: Hi,I came across this circuit, while reverse-engineering a device. It looks like a overvoltage protection circuit but has a different arrangement.This circuit drives an optocoupler for serial communication, the input ranges from 15V-24V (idle), depending on the used actuator.What is the purpose of this circuit and how does it work, I am pretty sure it is supposed to be a overvoltage protection for the Optocoupler, right? If so, why this weird arrangement?Thanks in advance.Edit: I don't know the exact specifications for the used Zener diode (The Mfr. No isn’t correct).

author: xavier60
author_rank: Super Contributor
timestamp: January 04, 2022, 11:00:06 am
content: When the voltage is high enough, about 7V, current flows out of Q5's Base through R10 and D4A to ground.Q5 partly turns on, passing current through the opto's LED.As the current approaches 0.5mA the voltage across R7 approaches 0.6V, partly turning on Q4 which prevents further increase in Q5's Base current. It becomes a constant current source for the LED.The value or R10 is unusually high.

author: MaximilianMM
author_rank: Contributor
timestamp: January 04, 2022, 11:55:03 am
content: Thanks for the good explanation, you helped my allot.So, the Zener voltage determines at which voltage (I+) the LED turns on? This circuit than also works like a comparator, right?How does this circuit cope with high voltage? Could such a circuit be designed to withstand 230V?I would like to design a circuit, which could withstand such voltages in the worst case.

author: xavier60
author_rank: Super Contributor
timestamp: January 04, 2022, 12:13:56 pm
content: Quote from: MaximilianMM on January 04, 2022, 11:55:03 amYes, the Zener voltage rating determines the threshold voltage. For predictable operation, low Zener current should not be allowed to drive Q5.To withstand 230V, Q5 needs to have high voltage spec like 400V or more. The circuit should be redone to use NPN transistors so that there is a large selection or transistors to choose from.What's the intended detection voltage? If it's much lower than the 230V expected maximum, the design become difficult trying to avoid excessive Zener current.

author: MaximilianMM
author_rank: Contributor
timestamp: January 04, 2022, 12:39:52 pm
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 04, 2022, 01:02:15 pm
content: Actually, the circuit is already designed to cope with 230V. Q5 is rated for 500V and the reason for R10 being 220K now makes sense. It will dissipate a bit less tan 250mW at 230V.

author: Ian.M
author_rank: Super Contributor
timestamp: January 04, 2022, 01:23:43 pm
content: Here's what a Mouser search for smaller parts with enough VCBOturned up as possible:https://mou.sr/3qK3WqZI limited it to abs(Vcbo)>460V and abs(Ic)<=500mA.Oddly enough, there appear to be more PNPs than NPNs. You'd still have to trawl through that lot to find parts with enough gain.You are almost certainly going to run into problems with Q5's SOA at high voltages, (hot-spotting and secondary breakdown) so it is unlikely that a (near) constant LED current can be safely maintained up to the circuit's peak withstand voltage. The fix there is to add another Zener and series resistor, with a Zener voltage well above the normal operating range of the circuit, to the base of Q4 and a diode between Q4 base and Q5 emitter, so Q4 turns on during overvoltage, holding Q5 hard off.You still have to deal with reverse polarity of course - wrap it in a bridge so it responds to AC or add a high voltage low leakage diode in series with the whole circuit to simply stand-off reverse voltage.

author: xavier60
author_rank: Super Contributor
timestamp: January 04, 2022, 09:57:09 pm
content: Is the 230V AC voltage?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 05, 2022, 07:43:21 am
content: A similar function can be implemented with depletion mode MOSFETs, which are also available in high voltages; importantly, they do not require base bias current, so can be placed anywhere in a loop, including back-to-back, so that complements (P-ch) are not required.Tim

author: MaximilianMM
author_rank: Contributor
timestamp: January 05, 2022, 04:13:11 pm
content: Quote from: xavier60 on January 04, 2022, 09:57:09 pmYes, but the circuit also has an input protection with reverse voltage protection. So it's only a half wave.Quote from: T3sl4co1l on January 05, 2022, 07:43:21 amThat also sounds like a good solution.In the next days I will have a look on the circuit of the device I would like to control, as it supports both 24V and 230V. Than I know how they have done it.

author: MaximilianMM
author_rank: Contributor
timestamp: January 07, 2022, 05:10:58 pm
content: Today I reverse engineered the device, which can definitely handle both 230V AC and 24V DC on the data line.The circuit is very similar, in fact the used the exact same components as the circuit I posted before, but the layout is a bit different.[ Attachment Invalid Or Does Not Exist ]This circuit doesn’t make sense to me, shouldn’t be Q11 always be turned off as the base will always be at a higher voltage than the emitter (due to the voltage divider of R20 and R21)?The normal operation voltage on I+ is around 24V. But the circuit should also handle 325V DC on I+, at least for short durations.How is this circuit supposed to work and what’s the difference to the circuit I posted before? Which one is better?Edit: I messed up. The circuit is the exact same, as posted before.

author: xavier60
author_rank: Super Contributor
timestamp: January 07, 2022, 09:43:55 pm
content: That wouldn't work very well at all. R20 appears to be in the wrong place.Can you post photos?

author: MaximilianMM
author_rank: Contributor
timestamp: January 08, 2022, 10:20:57 am
content: I messed up. The circuit is the exact same, as posted before.Maybe I should give you a little bit of context:The two devices I reverse engineered is a smart blind motor and the corresponding controller for it. The motors can either be controlled via said controller (24V telegrams) or via a normal wall switch (push button). (230V on I+ => motor moves up; 230V on I- => motor moves down)These two operation modes may never be operated at the same time.For a university project I build my own hardware controller and wrote my own software to control these smart motors. Now I would like to improve my hardware design, so it could handle the 325V DC on the bus line, that’s why I reverse engineered these commercial products, to get an idea how they have done it.What do you think about the circuit? Would you implement it in the same way, or would you choose a different approach? (e.g: depletion mode MOSFETs as Tim suggested)I also don’t know how this circuit copes with a continuous voltage of 325V DC on I+, as the motor may only be operated with a push button, in “AC mode”. As long as I choose components with a high enough power rating it shouldn’t be a problem, right?

author: xavier60
author_rank: Super Contributor
timestamp: January 08, 2022, 11:05:16 am
content: Quote from: MaximilianMM on January 08, 2022, 10:20:57 amThe mention of the 325V DC troubles me greatly. I'm assuming it's being derived from mains power which is able to supply dangerously high energy levels. Worst case I can think of is a circuit breakdown resulting in the opto's LED being zapped with the 325V DC which could cause catastrophic damage to the opto, possibly resulting in HV being applied to the isolated side.The only advice I am comfortable giving at the moment is to have a simple HV rated resistor in series with the LED for 325V DC operation.

author: xavier60
author_rank: Super Contributor
timestamp: January 08, 2022, 11:16:32 am
content: I would also feel comfortable with anEnhancementDepletion Mode MOSFET configured as a CC source plus an extra HV rated resistor in series to limit the fault current in case of a breakdown.https://www.radiolocman.com/shem/schematics.html?di=599185https://www.microchip.com/en-us/product/LND250#document-table

author: MaximilianMM
author_rank: Contributor
timestamp: January 08, 2022, 12:07:43 pm
content: To be clear the 325V DC on I+ would be a fault condition. My circuit doesn’t need to work for these high voltages, it just shouldn’t get destroyed (at least immediately) if it is applied to it.The opto just needs to work up to~24V DC on I+.Another thing I must consider is the speed of the circuit, as it is used for communication. I don’t know if a MOSFET would therefore be a good option(due to the higer capacitance).An additional current limiting resistor on Q5s Collector is also good idea.I am pretty confident in this circuit, since it is used by the motor, which is explicitly designed to handle this voltage. But maybe some additional protection wouldn’t be a bad idea, since I really don’t want to destroy the circuit on the low-voltage side, let alone the devices which are connected to it.

author: xavier60
author_rank: Super Contributor
timestamp: January 08, 2022, 12:23:54 pm
content: Quote from: MaximilianMM on January 08, 2022, 12:07:43 pmA 10K HV resistor could be placed in series with Q5's Collector to limit the fault current in case of a breakdown.With serial, try to keep the low/high threshold voltage close to half way, 12V in this case, which the existing circuit looks like it does anyway.

author: xavier60
author_rank: Super Contributor
timestamp: January 09, 2022, 12:19:39 am
content: Better still, a high voltage PTC.https://au.element14.com/vishay/ptccl05h100sbe/ptc-thermistor-1-6k-th-600v/dp/3020841?gclid=Cj0KCQiAieWOBhCYARIsANcOw0waR77P_vKsIxpzstwM_XyX9pAVC-4qBlrBBMkNEXQieZ_xQrjhATEaAkFBEALw_wcB&mckv=sH8LfqJzK_dc|pcrid|430179663787|pkw|ptccl05h100sbe|pmt|p|slid||product||pgrid|100894312984|ptaid|kwd-336991970617|&CMP=KNC-GAU-GEN-SKU-MDC-NEW

author: MaximilianMM
author_rank: Contributor
timestamp: January 09, 2022, 12:47:35 pm
content: Quote from: xavier60 on January 09, 2022, 12:19:39 amWouldn’t be a PTC to slow acting, to protect the opto?I also reverse engineered another part of the circuit and I’m really struggling to make sense of it.As it is connected to the output of a DC-DC converter I assume it is for current limiting, but the circuit doesn’t make sense to me.Edit: I fixed a mistake in the circuit. Thanks xavier60 for the hint!

author: xavier60
author_rank: Super Contributor
timestamp: January 09, 2022, 01:20:59 pm
content: The PTC might not prevent damage to the opto but will prevent it from receiving enough energy to rupture it. A series HV resistor could be added for full protection.Maybe the Base of Q16 connects to the Emitter of Q18?

author: MaximilianMM
author_rank: Contributor
timestamp: January 09, 2022, 01:59:33 pm
content: Quote from: xavier60 on January 09, 2022, 01:20:59 pmYes, you’re right. I updated the schematic. Now the circuit makes more sense to me, but I still don’t understand the function of Q17.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 09, 2022, 02:30:54 pm
content: Foldback.Tim

author: MaximilianMM
author_rank: Contributor
timestamp: January 09, 2022, 03:41:32 pm
content: Quote from: T3sl4co1l on January 09, 2022, 02:30:54 pmThanks for the answer Tim.I don’t understand in which scenario Q17 will be enabled. To enable Q17, there must be a current of 0.15mA through R23 (0.6V/3900Ohm), but this current will never occur, as it is limited through the 220K resistor.Am I missing something?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 09, 2022, 04:03:50 pm
content: Evidently when "V+" is pulled below about 34V below "+21". Mind, I have no idea what these supplies are in relation to, or what's connected to them; it seems like the intent is to protect Q18 when MTH6 is pulled low, and judging by the MOV on it, this could happen to some hundreds of volts in either direction?Tim

author: MaximilianMM
author_rank: Contributor
timestamp: January 09, 2022, 05:19:31 pm
content: 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.

author: ledtester
author_rank: Super Contributor
timestamp: January 09, 2022, 05:47:59 pm
content: Quote from: MaximilianMM on January 04, 2022, 12:39:52 pmAnother 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

author: langwadt
author_rank: Super Contributor
timestamp: January 09, 2022, 05:51:13 pm
content: Quote from: MaximilianMM on January 09, 2022, 05:19:31 pmif MHT6 is pulled (enough) below ground (MHT7) Q17 turns of Q18

author: Terry Bites
author_rank: Super Contributor
timestamp: January 09, 2022, 06:01:58 pm
content: This is a current mirror / sourcehttps://electronics.stackexchange.com/questions/539580/pnp-transistor-constant-current-source

author: langwadt
author_rank: Super Contributor
timestamp: January 09, 2022, 06:24:12 pm
content: Quote from: Terry Bites on January 09, 2022, 06:01:58 pmthat's a current source not a current mirror, and only half the circuit in question

author: MaximilianMM
author_rank: Contributor
timestamp: January 10, 2022, 09:53:29 am
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 10, 2022, 10:36:03 am
content: 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 10, 2022, 04:24:29 pm
content: Now I understand it. Many thanks to you all, you helped me allot.

author: MaximilianMM
author_rank: Contributor
timestamp: January 18, 2022, 11:37:23 am
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 18, 2022, 11:58:46 am
content: 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 18, 2022, 01:10:38 pm
content: 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?

author: xavier60
author_rank: Super Contributor
timestamp: January 18, 2022, 01:40:04 pm
content: Quote from: MaximilianMM on January 18, 2022, 01:10:38 pmQ12 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 18, 2022, 02:01:05 pm
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 18, 2022, 02:15:56 pm
content: Quote from: MaximilianMM on January 18, 2022, 02:01:05 pmAlmost 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 18, 2022, 02:53:25 pm
content: 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 18, 2022, 06:24:57 pm
content: 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

author: MaximilianMM
author_rank: Contributor
timestamp: January 19, 2022, 04:08:48 pm
content: Quote from: T3sl4co1l on January 18, 2022, 06:24:57 pmI 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 20, 2022, 03:51:40 pm
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 20, 2022, 09:49:30 pm
content: Quote from: MaximilianMM on January 20, 2022, 03:51:40 pmThe main problem is that it doesn't provide any current gain for the opto.

author: xavier60
author_rank: Super Contributor
timestamp: January 20, 2022, 10:04:30 pm
content: 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 21, 2022, 11:19:48 am
content: Quote from: xavier60 on January 20, 2022, 09:49:30 pmWhat 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.Quote from: xavier60 on January 20, 2022, 10:04:30 pmLike so? I don’t think that would work, right?

author: xavier60
author_rank: Super Contributor
timestamp: January 21, 2022, 11:30:54 am
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 21, 2022, 11:53:05 am
content: 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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 21, 2022, 12:15:32 pm
content: Quote from: xavier60 on January 21, 2022, 11:30:54 amYes, that’s absolutely right.Quote from: xavier60 on January 21, 2022, 11:53:05 amThat’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.

author: MaximilianMM
author_rank: Contributor
timestamp: January 30, 2022, 11:21:32 am
content: 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.

author: xavier60
author_rank: Super Contributor
timestamp: January 30, 2022, 11:51:11 am
content: Quote from: MaximilianMM on January 30, 2022, 11:21:32 amJust 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?

author: xavier60
author_rank: Super Contributor
timestamp: January 30, 2022, 11:59:17 am
content: Did you mean the circuit in Reply 44 you tried next?Q33 should be protected but the opto BJT current is still unknown.Edit: I meant to say Reply 44.

author: MaximilianMM
author_rank: Contributor
timestamp: January 30, 2022, 01:19:58 pm
content: Quote from: xavier60 on January 30, 2022, 11:59:17 amNo, I mean the circuit in reply 41. I can test this circuit relative easily, as this circuit is also used in a similar configuration in another part of the circuit. (posted in reply 18).Quote from: xavier60 on January 30, 2022, 11:51:11 amI can confirm this. I will also measure it, but my differential probe has to arrive first.Quote from: xavier60 on January 30, 2022, 11:51:11 amIn the original device (motor) the optocoupler is always turned off if the motor is used with a normal wall switch (230V AC). So, the circuit can withstand that voltage without any problem, as I also confirmed it with my tests.The motor will never turn on the optocoupler when AC-power is connected, as the motor only responds to telegrams and never initiates them. So, the solution works for them.However, I am designing the corresponding controller for the motor. As the controller initiates the communication, I cannot be sure that there is no overvoltage situation on the bus. And again, my circuit doesn’t need to work with AC-power, I just want to make it resilient to AC-power. And my current circuit does that, as long as the optocoupler isn't turned on.But I just noticed a mistake with my test with the circuit from reply 41. I didn’t use the protection circuit in that test and therefore also the negative half wave was applied to the circuit. I will test the circuit 41 again.

author: MaximilianMM
author_rank: Contributor
timestamp: January 30, 2022, 02:22:17 pm
content: I just tested the circuit from reply 41 again (with reverse voltage protection) and now I seems to work just fine. I think this could be the final solution for my problem.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 12:09:31 pm
content: The circuit works in principle, but the voltage doesn’t get pulled down enough. I think I have to go with the initial design.I also tried the circuit (reply 44)as suggested by xavier60, but that circuit doesn’t work either.I need to somehow add a foldback to this circuit. The best way would probably be, to open the emitter of Q14, if the voltage is above ~30V. How could I achieve this?

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 12:27:41 pm
content: The circuit in reply 44 has reduced current. R42 needs to be reduced, maybe to 15Ω.It has foldback behavior. The current drops with increasing voltage. I haven't calculated the dissipation of it.Or you can add protection to the original design that you have just posted like Q17 does in reply 29.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 01:14:39 pm
content: Quote from: xavier60 on January 31, 2022, 12:27:41 pmI don’t really understand how that should work. Q35 also needs to be disabled, to protect it against the high power dissipation.Quote from: xavier60 on January 31, 2022, 12:27:41 pmI tried this in the circuit below, but still Q41 will be enabled by the 220K resistor and therefore the power dissipation in Q41 and Q42 would be too high.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 01:36:25 pm
content: A simple way to limit the current through Q41 is by putting a resistor in series with its Emitter. Start with 1KΩ.This could have the side effect of increasing the turn off delay of the opto's BJT. It might not be a problem.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 01:44:52 pm
content: Yes, that could be a solution, but this would also effect Q40s base current. Isn’t there a simple way to disconnect Q41s emitter if the voltage is above 30V?

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 02:14:37 pm
content: Quote from: MaximilianMM on January 31, 2022, 01:44:52 pmQ41 needs to pass 440uA for it to work. The resistor value might need some adjustment.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 02:28:11 pm
content: There is a way to turn off most of the current.Instead of Q38, use an NPN BJT to pull to ground the Base of Q41.In case the opto's BJT is already on when the high voltage appears, put a resistor driving the Base of Q41 so that the protecting BJT doesnt need to sink the full 30mA to pull down the Base of Q41. Try a value that you already use like 5.1K.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 02:41:43 pm
content: Quote from: xavier60 on January 31, 2022, 02:28:11 pmHow would I drive the base of the new NPN-Transistor? And where should I put that 5K1 resistor?

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 02:50:06 pm
content: Quote from: MaximilianMM on January 31, 2022, 02:41:43 pmWith the same 220K/5.1K divider, upside down.Put the 5.1k resistor in series with the Base of Q41. Connect the Collector of the protection BJT directly to the Base of Q41.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 03:27:19 pm
content: Surely you don’t mean it the way I understood it.In this configuration Q41 will never turn on.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 03:36:50 pm
content: Quote from: MaximilianMM on January 31, 2022, 03:27:19 pmMostly like that but drive the Base of Q38 with a 220K/5.1K divider.We want Q38 to sense the higher than normal operating voltage.Actually I think the bottom divider resistor should be something lower than 5.1K.The protection needs to operate at something a bit higher than 30V.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 03:48:21 pm
content: By "divider" I mean another pair of resistors other than R50 and R47.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 03:54:55 pm
content: Ah, now I understand it.I tested it with 5K1 and the cutoff voltage is around 27V. That seems to work. Which specifications does the NPN transistor Q38 need to have?The Base will always be at a much higher voltage than the collector, isn’t that a problem?

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 04:01:49 pm
content: Yes, that's it. Keep in mind that increasing the temperature of BJT reduces the Base to Emitter threshold voltage.You dont want the protection to operate too close to 24V when the circuitry is hot.

author: MaximilianMM
author_rank: Contributor
timestamp: January 31, 2022, 04:07:28 pm
content: Yes, that’s a good point. I will test this configuration.Thank you for your help, Xavier.

author: xavier60
author_rank: Super Contributor
timestamp: January 31, 2022, 04:08:31 pm
content: Whatever is the SMD version for the BC548 will do.The Base voltage being higher than the Collector is a normal operating condition.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 31, 2022, 06:34:14 pm
content: Why not do it like this?Apologies for using fewer than Seven transistors this time.The thermistor (3500 is the 'B' parameter) and divider stuff (R7-R9) can be removed if you don't mind the tempco. See the difference without:(Series 0, 37.5, 85°C)With:No effort has been made to adjust the current/voltage settings/breakpoints between these, of course, or to review this thread and see what specs are required for the exact application. These are mainly set by R1, R3 and R4.I found that the extra diode (D2) was necessary to get consistent saturation with the opto model I used; YMMV. This, as well as the foldback, further increases the saturation voltage of the circuit (for I_L < peak current), but it wasn't great to start with so I'm just going to assume that isn't a big priority. The diode can be removed if R5 is decreased, and perhaps more LED drive current (lower R6 / higher V2). More bias current may also be needed, i.e., decrease R4 and R3 in proportion, but mind that more bias current also dissipates more power at high voltages. The values shown here draw about 0.2mA at 100V so the dissipation is pretty negligible.Tim