problem with IR2110 : oscillation of the high side gate drive

Hi,

I'm trying to drive a H-bridge with 2 IR2110 drivers, but I have problems with the high side MOSFET gate drive. I'm using a 20kHz PWM drive. The circuit I'm testing is in attachment ("IR2110.txt"). On the file "Gate_voltage.png", there is the voltage on M1 gate (green), and the voltage on M3 gate (blue). The drive of the low side works well (M3) but as you can see on the green curve, the drive of the high side doesn't work.

I think I have an idea of where does the problem come from, but I would like to be sure. The control of the MOSFET that I want is :
- When I want to increase the current in the motor, VH1 and VL2 are high, so M1 and M3 are ON, and M2 and M4 are off. In my simulation, VH1 and VL2 are on for 10µs.
- Then there is the second part of the chopping, where the current decay through a recirculating path. With my control, all the MOSFET are off during this time. So the current is supposed to decay through the diodes D2 and D3, correct me if I'm wrong.

The problem with this control is that the bootstrap capacitor C2 of the IR2110 can never charge, because the control never bring the ground to the source of M1. So I'm assuming that's why my high side drive does not work properly. What do you think about that ? Is that the reason my circuit does not work ?

And now I would like to know : is there a way to get the IR2110 working with my kind of control ? I need to keep this control, because this is a requirement of my client...

Thank you,
Maxime.

First comment, you don't show the actual Vgs but the gate voltage against ground. Vgs would be a voltage difference.

Secondly, you are right that the applied control scheme doesn't gurantee reliable bootstrap operation. It's also bad in terms of circuit losses, beacuse the MOSFET body diodes are used as rectifiers. You better use synchronous control of all transistors.

Thank you for your answer FvM. I am aware that this control scheme will dissipate a lot of power in the diodes, and that's why I put external diodes with good power dissipation capability.

What do you call synchronous control ? You mean when the low side transistors are driven with an opposite signal than the high side ? In this case I don't see why it is better in terme of circuit losses. Let's imagine I have a positive current flow M1 to M3. Then, M1 and M3 both turn off, and M2 and M4 turn on. The current flow is still is the same direction because of the energy stored in the inductor. But, M2 and M4 are n-channel MOSFET, so the current cannot flow through then because it is in the bad direction. So, for me, the only way for the current is trough the diodes D2 and D3, and the power dissipation will be the same. I'm aware that there must be something I don't understand, but I don't see what.

As my client want to keep this control scheme, I looked for other MOSFET driver, and I found out that only isolated driver could work. The reference ISO5000 from Texas Instrument seems to be OK, but there are some points I don't understand about it. The power supply of the power side of the isolated driven is shown as a perfect voltage source, with its negative terminal connected to the source of the transistor. How can I do that for the high side MOSFET ? Does it mean I have to generate a new isolated power supply, with its negative terminal referenced to the source of the high side MOSFET ?

Thank you.

The error in reasoning is to think that MOSFETs don't conduct in quadrant III (negative Vds), but they do.

I finally convinced my client to use a synchronous control of all transitors, so I'm back with the IR2110.

I did some tests with the component, and there is something I don't understand about it : when I disable the chip by pulling the Shutdown (SD) pin high, LO is at 0V, which is OK, but HO is at about 15V. And I have the same voltage on the VS pin. Is this normal ? I use a 15V supply voltage for the IR2110 (Vcc). I guess the 15V I have on HO comes from this supply, but I don't see why. Why don't I have 0V on HO ? I understand that my transistor is OFF because its Vgs is 15 -15 = 0V, so it seems not to be a big deal, but I would like to be sure and to understand it.

Thank you.

I think to measure the voltage on the HO pin you need to measure between VS and HO. Which if both are 15V then as you figured you get 0V difference. If I understand right, the voltage on VS comes from the bridge itself so if you have 15V on VS(to GND) then when HO(to GND) is off it will also be 15V because HO is a floating high-side output.

Add a 470ohm resistor between gate and source of each mos and a 1nF capacitor serie with a 100ohm resistor across load.

What is the purpose of the 470ohms gate to source resistor and of the 1nF - 100ohms RC ? Is it what we call a snubber circuit ?