Electronics > Beginners
Mosfet gate breakdown survivable?
mzzj:
--- Quote from: amyk on September 09, 2018, 02:29:02 am ---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.
--- End quote ---
I 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.)
T3sl4co1l:
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
bson:
--- Quote from: justinjja on September 15, 2018, 02:19:37 am ---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?
--- End quote ---
Hmm. 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 VBE drop, 4.35V/10kΩ = 0.43mA. IRE depends on the VBE drop, which is narrow range, and 1/hFE which is small since hFE is big, so the tolerance of IC is easily within a few percent when using a high-hFE transistor like the 2N3904. In reality it might have an hFE around 250-300, so the tolerance of RE will dominate, but 100 is a safe minimum.
The current across the collector resistor (680Ω) above depends on IC in the 2N3904, which matches IRE within 1/hFE (≈ 1%) as outlined above. 45mA*680Ω = 30.5V, so VC can dip that much below +Batt. This VC becomes 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 RE on 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...
Zero999:
--- Quote from: justinjja on September 15, 2018, 02:19:37 am ---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.
--- End quote ---
The 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 VBE drop lower, due to the emitter follower loss, but I hope you get the general idea.
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