Is dead time needed for a MOSFET?

Question

Is it necessary to implement "dead time" (anti-shoot through) for MOSFETs? F=50kHz

Answer 1

In the bottom circuit, dead time is absolutely essential. Without it, you will have small pulses of extremely high current being short circuited through the FETs, and it may well result in rapid destruction of your device.

In the top one, it's a bit harder to say. Technically, if you turn both on, they produce a counteracting field in the transformer and there is therfore no inductance, only resistance opposing the flow of current. In reality, the coils will not be perfectly matched, and there is leakage inductance so the current will build up more slowly than the second case. It is a much less bad case of shoot through, but still not nice.

The frequency is not really important to the question, you require dead time regardless. It will typically be the case that you need about 3x the RC time constant of the MOSFET gate capacitance and the total drive resistance.

Answer 2

Now, for basic purposes, almost definitely yes. Don't worry about it. Leave a few hundred ns dead time and it's probably fine. This can be generated by many MCUs' timers, if you're going for digital control, or say from a common logic drive signal by delaying the rising edge with a timer network (series resistor, shunt capacitor and diode), then sharpening it with a schmitt trigger (usually also a gate driver; a 74HC family gate will be sufficient drive for transistors of this size, at 5V supply).

^{simulate this circuit – Schematic created using CircuitLab}

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Devil's advocate: no. But maybe. It depends.

There are edge cases where you might not.

The most common case, I think, is where delaying turn-on would cause unwanted forward-bias of the body diode or whatever. This occurs when the load is inductive (as in most switching converters), and when synchronous rectification is being used (hence the MOSFET at all). Which appears to be applicable in the second circuit (half-bridge feeding an LC filter). This is unwanted because the diode has reverse recovery, drawing excess current very briefly (about t_rr), increasing power dissipation and ringing (because the Q1-Q2-Cin loop necessarily has stray inductance, proportional to the length of that loop; or in the push-pull (transformer coupled, first circuit) case, strays include leakage, which is to say, winding length of the primary halves in the transformer).

This situation can be avoided by setting dead time very close to zero, or even slightly negative. The downside is, shoot-through can draw a lot of current too. But we know precisely what that current is, because it rises at \$\frac{dI}{dt} = \frac{L_s}{V_\textrm{IN}}\$, so we can set the stray inductance \$L_s\$ (including artificially increasing this inductance -- there can indeed be reason to do this!) such that, over the expected range of timing variation, current stays within adequate limits (say, a certain fraction of nominal maximum load current?). The energy stored in that inductance then gets dissipated in a snubber, which can be a simple RC, or a rate (dV/dt) or peak-clamp RCD type.

Basically, the insight is that, a diode acts as a short circuit until it turns off, so what you intended as dead-time can actually result in shoot-through anyway because of the diodes. In that case, it is better to shoot through the MOSFETs (channel current) with timing controlled by the circuit -- if and when the timing can be controlled more precisely than the recovery time.

This applies not just to DC converters, but AC as well, as a class D amplifier (say for mains power, audio output, etc.) also likely has to deal with reactive loads, i.e. sometimes drawing negative current through the top or bottom transistor, activating either body diode from time to time. The shorter or nonexistent dead time also

Anyway, to be clear, this is also not something to trifle with. Note the constraint: if stray inductance is nonexistent (in practice, this means a few to tens of ~nH, requiring timing precision of maybe a few ns), then any overlap will quickly draw massive amounts of current through the transistors, probably destroying them. Hence the first half of this answer!

Answer 3

Yes, you definitely need to ensure that both MOSFETs are not on at the same time.
It seems fairly easy to do and has already been covered in these answers.
How can I calculate the dead time to prevent shoot-through?
How is dead time in a half bridge implemented

Answer 4

Is it necessary to implement "dead time" (anti-shoot through) for MOSFETs? F=50 kHz

Both circuits are types of push-pull drivers for switch mode power supplies and, both circuits require a small amount of dead-time (anti-shoot-through) to avoid MOSFET conduction overlap.

The 2nd circuit is probably more obvious than the first of course.