Mosfet gate breakdown survivable? - Page 1
Forum section: electronics / beginners

author: justinjja
author_rank: Regular Contributor
timestamp: September 09, 2018, 01:17:25 am
content: This circuit was running fine at 30v, when cranked up to 35v the pnp transistor and p channel mosfet both started over heating,realized I'm missing a resistor under the first 2N5551.Do you think I damaged the mosfet by breaking down the gate?I shut it down before it got too hot. (magic smoke still in side...)

author: amyk
author_rank: Super Contributor
timestamp: September 09, 2018, 02:29:02 am
content: That part you're using only has a 20V Vgs. If it survived it would probably die soon.In any case a blown gate is definitely not "survivable", and you can check by measuring with a multimeter.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 09, 2018, 02:40:09 am
content: You only need 10V. Put a e.g. 12V zener (1N5242 say) in parallel with the pull-up resistor, and don't omit the emitter resistor of the first transistor.Also, move the series resistor to between the emitter follower and MOSFET gate. Use a smaller value like 47 ohms. This dampens possible oscillations.Tim

author: justinjja
author_rank: Regular Contributor
timestamp: September 09, 2018, 07:53:20 pm
content: Mosfet is fine currently, Just wondering if is going to fail soon.Actually a resistor after the emitter doesn't work, Needed to use 2 series resistors before the collector as a voltage divider.And the 500R series resistor is gone now.Thanks,Justin

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 10, 2018, 03:10:09 am
content: You probably used the wrong values then. This is the preferred method:Note the input voltage is logic level (0 and 5V) only, and the resistor ratios, and zener diodes used.Note also this is a rather inefficient method, especially if you're going to be cranking it up to 80V. Just use a high side gate driver chip like IR2101, and don't waste time with poor-performing P-channel FETs.Tim

author: Zero999
author_rank: Super Contributor
timestamp: September 10, 2018, 07:52:24 am
content: The the voltage driving the BJT stable and regulated, such as from an MCU?If so, why not add an emitter resistor to the input transistor, which should also speed it up?

author: justinjja
author_rank: Regular Contributor
timestamp: September 15, 2018, 02:19:37 am
content: Been a while since college,but every way I tried to solve the circuit with an emitter resistor instead of a voltage divider didn't work out.Am I solving that wrong?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 15, 2018, 07:29:18 am
content: A also see a problem, though the voltages and problem might be a little different:The case with low signal at the input should be OK. The first NPN is blocking and everything is fine.With a high input signal the first NPN would be turned on hard, delivering around 45 mA collector current.This would result in quite some base current for the PNP at the upper gate buffer, that is than clamped by the zener.The 2N4403 is than working in inverted mode, with base current (around 30 mA) flowing from base to collector and that trough the zener.Those 45 mA would likely blow the initial NPN (due to some 60 V drop) - though no transistor specified (an BD140 could survive).So having that extra series resistor at the collector is a good idea, in addition to the Zener.Another point would be that the 2 zener stabilized parts could be wired in series. This could save some current and 1 resistor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 15, 2018, 08:17:00 am
content: Check your models. Something is seriously screwed up, pinned out wrong, etc.Or are you solving that by eye? You seem to have forgotten the C-B junction of the upper PNP, which clamps the gate drive voltage through the zener. It works out alright.I forget what the input was intended to be, might've been TTL not CMOS. Supply was originally 20-40V or thereabouts, so yes the transistor (2N3904 BTW) would run pretty hot at 80V, and would have to be upgraded besides (MJE340 maybe?). In any case, the 100 ohms isn't a requirement. Adjust values until currents and voltages are right.Likewise if slower gate drive is acceptable and more efficiency is desirable, the resistor values can be increased, or the pull-up resistors changed to current sources, etc.Tim

author: justinjja
author_rank: Regular Contributor
timestamp: September 15, 2018, 01:31:30 pm
content: ya I was solving that in Brain-spice lolmissed the zenner in your circuit,I was originally going to try something with a zenner, but I didn't have any on hand while building this,so had to come up with a way to do it without them.In other news while debugging my mcu code and plugging/unplugging the 80v supply into my breadboard...I missed once and connected 80v to ground killing off the 3 p channel mosfets in question.

author: mzzj
author_rank: Super Contributor
timestamp: September 15, 2018, 03:19:06 pm
content: Quote from: amyk on September 09, 2018, 02:29:02 amI have tested several different types of mosfets and all of them survived at least 80 volts to gate. Some non-logic level ones over 100 volts.(Anyways its something to avoid as it's different case with hot die, fast pulse rates and long term.)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 15, 2018, 09:34:01 pm
content: I've heard actual failures from 30 to 80V, with time being a factor as well (near actual breakdown, there is some charge injection into the gate oxide, causing a change in Vgs(th)). Note that hot-plugging transients can easily double that to 160V for a few microseconds (more if ceramic capacitors are present), which will toast anything.Tim

author: bson
author_rank: Unknown
timestamp: September 15, 2018, 10:42:29 pm
content: Quote from: justinjja on September 15, 2018, 02:19:37 amHmm. Well, time to dust off the college cobwebs.Looking at the top half of the bridge, the resistance into the base of the 2N3904 is hFE*RE≈ 100*100 ≈ 10kΩ. Assuming a 0.65V VBEdrop, 4.35V/10kΩ = 0.43mA. IREdepends on the VBEdrop, which is narrow range, and 1/hFEwhich is small since hFEis big, so the tolerance of ICis easily within a few percent when using a high-hFEtransistor like the 2N3904. In reality it might have an hFEaround 250-300, so the tolerance of REwill dominate, but 100 is a safe minimum.The current across the collector resistor (680Ω) above depends on ICin the 2N3904, which matches IREwithin 1/hFE(≈ 1%) as outlined above. 45mA*680Ω = 30.5V, so VCcan dip that much below +Batt. This VCbecomes the input of the class B driver pair (2N4401, 2N4403). The gate voltage will be within one VCE(sat)each of +Batt and +Batt-12V (the common produced by the ≈ 12V 1N5242B zener). The 1kΩ resistor to the far left sets the zener current.The bottom side of course just mirrors the top, for the other half of the bridge.A problem is the circuit is dependent on a clean logic input signal, if it's not clean the MOSFETs can end up sitting in the ohmic region, which in turn could kill them if they can't dissipate enough heat. Maybe stick an inverter in front of it if there's also a connector, wiring, etc and it's not clear what the origin of the logic input is at all times.A 30V swing seems a bit excessive to me also... that's like 1.3W over the 680Ω resistor. I can't see any reason for such a large swing; it just generates heat.I'd either reduce the current (change the 100Ω RE on the 2N3904, to say 330Ω), or reduce the swing to ≈ 12V by changing 680Ω to 270Ω. That makes the dissipation 0.54W. Drop the dissipation to under half a watt, say 0.3W, but reducing the current to 0.3W/12V = 25mA. This puts REon the 2N3904 at 4.35V/25mA = 174Ω. Say 180Ω to err on the low side for IC.Oh, wait, it's not a bridge - it's a basic push-pull. So two of these circuits will make a bridge, driven at opposing phase...

author: Zero999
author_rank: Super Contributor
timestamp: September 15, 2018, 10:57:41 pm
content: Quote from: justinjja on September 15, 2018, 02:19:37 amThe circuit I posted should work. It's a simple common emitter amplifier with the gain and therefore the output voltage set by the ratio of the collector to the emitter resistor, followed by a push-pull emitter follower, to lower the output impedance. There should be no need for a zener, other than to protect against something going wrong, which which case opt for a higher voltage, then this circuit will usually produce, such as 16V. Try simulating it with a single MOSFET first. If it's not working, then there's a problem with your models. Breadboard it with a much lower voltage, say 24V or 48V and a lower current load.Whoops, I forgot that Vgate will be a VBEdrop lower, due to the emitter follower loss, but I hope you get the general idea.