Ideas on the cause of sense resistor failure - Page 1
Forum section: electronics / repair

author: juanfermed
author_rank: Regular Contributor
timestamp: October 31, 2016, 05:04:57 pm
content: Hello all,I have several battery charging devices whose output current sense resistor is failing in a very similar manner: all with a small spot exactly in the center of the resistor, always. Please look at the attached images. This device is used to charge 12V lead acid batteries, either from a solar panel or from a wall adapter. But all the resistors failed while using a solar panel. The solar panel is 15W, 18Vmp, 23Voc with a 10m cable.I am attaching only 2 pictures of 2 different resistors, but there are far more devices with the exact same failure. Unluckily it is hard to debug when the failure happens becuause it happens when the device is in the clients hands.ThanksPS: I hope this is the right forum section to ask this question. If not please tell me so I move it.

author: bktemp
author_rank: Super Contributor
timestamp: October 31, 2016, 05:21:10 pm
content: It looks like there was a severe overload, so the resistor exploded. Without knowing the circuit around the resistor it is impossible to tell what caused this overload.If the resistor is connected to the battery it could be a high current flowing from the battery through the resisor into a short somewhere on the pcb.

author: juanfermed
author_rank: Regular Contributor
timestamp: October 31, 2016, 07:28:53 pm
content: Hi, thank you. The implementation is almost the same circuit ashttp://www.ti.com/lit/ds/symlink/bq24650.pdf,page 1. The sense resistor is the one that appears there as Rsr. In "normal" conditions I have not found any inrush current or voltage spikes that may cause it to burn or fail, and the only "shorting" path that I can imagine is if the low side Q2 transitor of the same circuit failed and shorted its drain to source and finally to ground.Do you think a an overvoltage (and not an overcurrent) can cause this failure? If yes, what magnitud of overvoltage should it be?Thanks,

author: cowana
author_rank: Frequent Contributor
timestamp: October 31, 2016, 07:42:20 pm
content: Quote from: juanfermed on October 31, 2016, 07:28:53 pmFor a resistor (which by definition has a fixed resistance), the current is always proportional to the applied voltage. This means it was as much an overvoltage as an overcurrent.If the entire resistor looked burnt, it would point to a long period of high power and over-heating. As it seems fairly localised, I would agree that seems to point to a shorter event - maybe around 5-10 times the rated power for a few seconds?

author: bktemp
author_rank: Super Contributor
timestamp: October 31, 2016, 08:13:17 pm
content: Quote from: juanfermed on October 31, 2016, 07:28:53 pmThere could be an inrush spike if the battery gets connected and C8 and C6 get charged by the battery. If they are small like in the example circuit the current spike is too short to cause any damage, but if there are larger electrolytic capacitors, the current surge could be high enough to burn the resistor.Did you check if Q2 is shorted?

author: JFJ
author_rank: Regular Contributor
timestamp: October 31, 2016, 09:45:02 pm
content: Quote from: juanfermed on October 31, 2016, 07:28:53 pmWhat type of transistor is Q2?In the TI circuit, Q2 appears to be (if my recollection of transistor symbols is correct) a depletion mode N-Channel MOSFET. If that is the type used, then whenever the gate terminal is at zero volts (e.g. at night when there's no output from the solar panel) the transistor's channel would become fully conductive - thereby, shorting the battery to ground via the sense resistor.

author: juanfermed
author_rank: Regular Contributor
timestamp: November 01, 2016, 08:48:10 pm
content: Hey guys,Thanks for your feedback. Ok, so a short-period overload is possibly the cause. I have not finished to test all circuits with this failure, but what I have seen so far is either the transistor Q1 or Q2 has failed, the chip BQ24650 itself, or both. Q1 and Q2 are on the same package (Si7288), so it makes it more difficult to know if it was the high side or low side that failed first. But a test, if either continues to work, will reveal more.I found one board in which the Q2 was shorted and caused every resistor replacement to fail in the same way every time. Unfortunately, the chip BQ24650 was damaged too, so I cannot say which of both (Q2 or BQ24650) caused the failure, but at least the resistor failure in that board was due to its Q2 was shorted.I think it might be that at some point the Q2 fails short circuited, and that produces a huge current from the battery through the sense resistor, through Q2 and finally ground, which last as long as any of the parts in the middle can handle, and that is when the resistor fails. If this was true, the question is why would Q2 fail short circuit?Quote from: JFJ on October 31, 2016, 09:45:02 pmHmm that is interesting, Q2 is an N-Channel MOSFET enhancement mode, the model is Si7288.

author: bktemp
author_rank: Super Contributor
timestamp: November 02, 2016, 07:16:04 am
content: I don't like synchronous switching regulators for battery chargers, because if something goes wrong, it will short out the battery.You probably have to do some measurements on a working circuit:Do the mosfets get hot when running? They shouln't at 15W, therefore it is unlikely that they failed because of overcurrent.What is the worst case input voltage? 23V should be ok fo the 28V maximum input voltage, the mosfets are rated for 40V.The only failure mode I can imagine is a voltage/current spike when the battery gets connected to the circuit.

author: Neilm
author_rank: Super Contributor
timestamp: November 02, 2016, 07:35:58 pm
content: Do all the failures have both damaged chip and MOSFET? If so consider the following:-1 What frequency is the switcher running at?2 What is the capacitance of the MOSFETs used?Could it be that the chip is running so fast and having to work so hard it is not fully switching off the MOSTFETs? This would mean that either the chip would fail and turn on both FETs at the same time or the FET would degrade until they fail. This would short the battery to 0V and over current the sense resistor.

author: bktemp
author_rank: Super Contributor
timestamp: November 02, 2016, 08:49:36 pm
content: Quote from: Neilm on November 02, 2016, 07:35:58 pmPWM frequency: 600kHzGate charge: 10nCThe driver has a 30ns deadtime and 10nC is quite low, so it shouln't be a problem unless something else is wrong.

author: SaabFAN
author_rank: Frequent Contributor
timestamp: November 03, 2016, 09:33:34 pm
content: I had my fair share of dying BQ24650 chips.Most of the time they stopped working after some voltage-spike on the battery-side. Either by suddenly disconnecting the battery (I wanted to simulate a loose connector next to a running engine and wiggled the clamp at the pole, causing some sparks), or when I connected a Battery-Pulser.Most of the time one of the internal regulators of the chip died, causing parts or the entire chip to stop working.In the end I switched to a different chip from Linear Technology that was also easier to solder.What I would check in your circuit:- Is there enough dead-time between the ON-Signal of the MOSFETs / Is there any overlap of HIGH-Level on the gates?- Drive-Signals with nice sharp rising and falling flanks?- Is there oscillation with just the battery connected? - Explanation: Current from the battery to the input capacitors stops rather suddenly when the caps are charged and could lead to a small voltage-spike due to the changing current in the inductor, which in turn makes the chip think there's a Solar-Cell connected, which then makes it start the battery-detection cycle that in turn shorts the battery, makes the voltage at the input drop, turns the switch off, turns the current-sink of the inductor off and voila: You basically have a parasitic Step-Up Converter with the sense-resistor as the current-limiterThis could be stopped by adding some capacitance to the MPPSet-Pin, as well as setting the MPPSet-Voltage high enough. - This could explain the way the resistor burned out: Short current-spikes that heat up the spot with the highest resistance.I have to admit, the last point is a bit far fetched but I wouldn't rule it out

author: bktemp
author_rank: Super Contributor
timestamp: November 09, 2016, 10:04:52 am
content: Quote from: SaabFAN on November 03, 2016, 09:33:34 pmMy guess what happens and causes the ic or the mosfets to fail:If you connect the battery, there is a large inrush current flowing from the output back into the circuit charging every capacitor. Because of the inductor the voltage overshoots both at output side (at the current sense terminals because of inductance of the wiring between the circuit an the battery) and at the input (PH pin) because of the stepdown inductor).This voltage spike can easily reach 3x the dc voltage.To avoid this problem, you need to connect capacitors with a larger ESR in parallel to the ceramic capacitors. 100uF electrolytic caps at input + output should prevent any voltage spike when the battery gets connected.

author: SaabFAN
author_rank: Frequent Contributor
timestamp: November 10, 2016, 02:09:11 am
content: Unfortunately, that didn't help in my case. I had 330µF + 2x 10µF ceramic at the output and 2x220µF + 2x10µF at the input of the charger and the battery-pulser (Brand name Megapulser) still managed to kill it. Only long wires from the charger to the battery were able to prevent that.The newer designs therefore all carried a lowpass-filter that eliminated the really fast pulses - I measured risetimes of less than 5ns for a 12V amplitude pulse with 150MHz scope-probes and a Philips PM3320A! With a battery connected, the pulse has a trail of sinusoid waves that differs in frequency depending on the battery and how good the battery is.