Trying to understand this design choice - Page 1
Forum section: electronics / repair

author: Jonny
author_rank: Regular Contributor
timestamp: May 25, 2020, 09:37:14 am
content: Q2 exploded. Trying to understand why they've configured it this way to decide on best part to replace with. I'm sure the answer is blindingly obvious but I'm stumped as to the role Q2 plays. When Q1 is turned on then Q2 will turn on and turn off Q1? What if Q2 was PNP?Appreciate your thoughts

author: wraper
author_rank: Unknown
timestamp: May 25, 2020, 09:44:32 am
content: IMHO it's overload protection. Some general purpose BJT like BC547 should do the job.

author: PKTKS
author_rank: Super Contributor
timestamp: May 25, 2020, 10:46:48 am
content: Looks like OCP (overcurrent protection)It is not very common using 2 drivers like thisbut without the rest of the schema hard to say.Usually a single BJT drains the output driverOverlooking the BJT at bottom "drains" the baseof the BJT on the upper side by limiting theamount of current it outputs causing a fastcurrent drop.IMHO looks like crap. it may oscillate andcause some weird things.Paul

author: xavier60
author_rank: Super Contributor
timestamp: May 25, 2020, 12:12:23 pm
content: The 2 transistors' purpose could be as a constant current sink. Q1 should have been damaged also.

author: JohnPi
author_rank: Regular Contributor
timestamp: May 25, 2020, 05:26:20 pm
content: Its hard to be 100 % confident without further information about the schematic and connections, but it seems to be a current sink (drawn strangely):R2 & R?150k may be an R divider to feed Q1 collector voltage back to the CPU.R?4k7 is probably the 'turn-on' signal -- it also provides base drive to Q1.See Q1's emitter goes to GND via a 10 Ω R1. This voltage is monitored by Q2 (basically compared to the 0.7 VBE needed t turn it on). If the voltage tends to exceed 0.7 V, Q2 will turn on and shunt base current away from Q1, thus tending to turn it off.In the end, it all acts as a current sink at a level of about 0.7/10 = 70 mA. Depending on beta of Q1 and the voltage applied by the CPU to the 4k7, there may not actually be sufficient base current available. Note the base of Q1 is at about 2*0.7 = 1.4 V. So, if CPU is at 3.3 V, you get a base current of (3.3-1.4)/4.7k = 400 uA; you'd need a guaranteed beta of 70mA/0.4mA = 175 to make this work. This is ignoring the driver resistance from the CPU (maybe 1k ?).I would suggest changing the 4.7k to ~ 1k to ensure it actually regulates as intended.It's hard to see how Q2 could have been damaged; its collector current is limited to << 1 mA by 4k7; it is possible the circuit oscillates, or Q1 became damaged and passed the input voltage through, but that is probably unlikely.

author: Jonny
author_rank: Regular Contributor
timestamp: May 26, 2020, 07:09:11 am
content: Thanks for all your input. Kinda makes sense that it could be current limiter so Q2 turns off Q1 at a given point. I should have mentioned this is part of a circuit that drives the window motor (now not moving) in a car door, but the H-Bridge is a separate section. Q1 is not damaged so still unsure how Q2 popped. R1 has also gone high as a result of whatever happened.I think I need to get hold of the car and find out where Q1 collector actually goes to.Thanks guys

author: 2N3055
author_rank: Super Contributor
timestamp: May 26, 2020, 07:32:01 am
content: It exploded because base current of Q2 is not limited. Once you have more than 0,5-0,6 V drop on 10 Ohm resistor, B-E diode of Q2 will shunt the current to ground.. boom..That will happen if you charge /discharge large capacitors for instance or have momentary hard short on power buses with large capacitors. In time needed for circuit to start limiting, current is not limited at all...

author: wraper
author_rank: Unknown
timestamp: May 26, 2020, 08:37:35 am
content: Quote from: 2N3055 on May 26, 2020, 07:32:01 amMost likely transistor failed because R1 failed first as said in the post above yours.

author: 2N3055
author_rank: Super Contributor
timestamp: May 26, 2020, 02:00:49 pm
content: Quote from: wraper on May 26, 2020, 08:37:35 amI didn't see that but what I said still holds..

author: JohnPi
author_rank: Regular Contributor
timestamp: May 28, 2020, 02:13:26 am
content: Probably not. If R1 failed open, then the circuit would limit Q1's current to a very small level -- because any emitter current in Q1 would flow into Q2's base, pulling Q1's base down and cutting it off.If R1 failed short, you would not have a current limit. Is the function of the circuit to provide a constant current, or that just a current limit against some _other_ fault ?