HCPL-3120 gate driver

Question

I am trying to use a gate driver HCPL-3210 for an IGBT but one of the big disadvantages that I can notice is that each chip needs its own power supply of 15 V. So as a total I would need 4 power supplies for 4 gate drivers (to implement a single phase inverter).

As an alternative way to implement it. I tried to connect all the chips in a certain way such that one power supply will be used. But after testing this way I got a very strange output signal.

Below pictures show the circuit diagram that I used & the result at the output.

I am looking forward for your kind reply & help.

Answer 1

There are many things wrong with the circuit and I can't see how it's working at all with the schematic shown as there doesn't seem to be any clock driving the bridge.

• U1 and U4's.LEDs are short circuited to ground (1). They can never turn on. None of the opto-LEDs have current limiting resistors. They will burn out.

• With V1 connected to ground the most you can possibly get out of any of the isolators is 18 V via D1 (3) so point (2) can be 18 V max.

• If (4) is at 0 V and U1 turns on then Q1 will turn on pulling (4) towards 600 V. This will immediately cut off the supply to (2) as D1 will be reverse biased. The 0.1 μF capacitor will power U1 for a small fraction of a second and then it will shut off switching Q1 off, allowing (4) to go back to ground. This will power up U1 again and the whole cycle will start again.

• You need close to 600 V to drive the transistor bases. You have only 18 V available.

You have no clock signal. How is it working at all?

So as a total I would need 4 power supplies for 4 gate drivers ...

You only need three isolated supplies as the bottom two share the same ground reference.