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As far as I know, a typical MOSFET is driven by a circuit like below.

a typical mosfet drive circuit

I read several MOSFET driver IC datasheets. It looks like they have a more complicated internal structure.

typical circuit by zcgd3003

For example, the ZXGD3003 IC has two different inputs and source & sink outputs.

Can you please give me internal structure of an IC like this.
Why do they take two inputs and how do they give separate source and sink outputs?

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2 Answers

This is kind of a broad question, but fortunately FET (and IGBT/ESBT) switching is not that hard to understand, so I'll cover all the bases.

MOSFETs require quite some charge to be pumped in or out of the gate in order to switch. This means that any MOSFET driver simply tries to provide a path with as low impedance as possible to some voltage above or below the source potential. This is the primary function of a MOSFET driver.

However, not all circuits are created equally. When you use a MOSFET to switch an inductive or especially a capacitive load too quickly, it will cause very high dI/dt or dV/dt spikes in your circuit, which can be a source of interference. This doesn't always have to be a discrete circuit element - parasitic properties of the MOSFET and surrounding components can cause problems with voltage and current spikes.

MOSFETs themselves are also not necessarily symmetric devices. If you just short the gate to the source to turn it off, and short the gate to some positive voltage to turn it on, the transition period between truly on and off may differ on the rising and falling edge.

Lastly, your MOSFET source may not be referenced to a fixed voltage (e.g. in an H-bridge with only N-channel MOSFETs). The source of the FET is floating, so your driver 'ground' needs to be floating as well and the driver should have its own bootstrapped supply.

All these things cause the complexity in the drivers that you're interested in. Those separate sourcing and sinking outputs are there to be able to drive the MOSFET differently on rising and falling edge - potentially to match rising and falling times, or to compensate for parasitic elements throwing a wrench in your circuit. Some drivers have bootstrapped high-side drivers that can be referenced to very high voltages. Some very high performance drivers need to have additional internal drivers as well (two 'layers' of FET drivers) which can cause very high propagation delays, which may again be compensated for with external components. You may find these kinds of drivers to be even more complex still.

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The specific IC you mention seems to have exactly two BJTs in one package, much like in your first picture. The reason why I think so is the picture for the "Switching time test cicuits" in the data sheet.

Actually, the longer I look at it, I believe that the IC you mention offers access to all the terminals of the two internal transistors:

PNP, C: Pin GND
PNP, B: Pin IN1
PNP, E: Pin SOURCE
NPN, C: Pin VCC
NPN, B: Pin IN2
NPN, E: Pin SINK

You could actually verify this using a diode tester or, better, a transistor tester (beta meter).

Other gate driver ICs will hoften have an input driven with a logic signal (e. g. 0 V, 5 V) and an output capable of driving something between 10  and 15 V. Those would of course use a level shifter before the output driver.

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