Designing 80V CMOS Circuit

Question

I wanted to make a DAC (0-80 V) using a PWM input (6 kHz).

You might ask why I did not make it with a single N-MOS. The reason is, I tried a single as well. But power dissipation could be a very big problem, if I need more than 100 DAC channels in a circuit, so I decided to use C-MOS to get away from this problem. Since I need high voltage, I chose PHC2300 and built up the circuit.

The problem starts here. I tried at the beginning with 24 volts. It seems to be working. But afterwards, if I increase the voltage to 80 V, the complimentary MOSFET does not work anymore. I tried several times; it happens again and again. Do you have any idea what the reason could be?

Answer 1

I think that the complication of making this circuit work might be better served by using an analogue op-amp gain stage such as the LTC6090. It will work up to a combined power rail of 140 volts. The only down side is that it will need a small negative voltage rail to accommodate the input common mode range needed. This can be served by a small DC-DC convertor from the logic supply of the PWM source.

So, you filter the PWM at logic level to produce a DC signal then you use the op-amp to amplify the dc signal up to the voltage level you need.

I'm suggesting this because the complication of the drive circuit for the P channel MOSFET is probably more than using an op-amp (and dc-dc converter). Plus the likely performance is going to be better. If you do want to use a push-pull mosfet driver circuit consider this question OR use an LTC4444 - it can drive two N channel FETs rather than an N and P channel FET. You might find it difficult to obtain a high-side P channel FET driver rated at 100 volts. I'm not saying they don't exist, just hard to find.

Answer 2

You've missed an important point. The data sheet specifies the gate-source voltage as a maximum of 20 volts. Look at your circuit. Pretend, just for the moment, that the 20k resistor were zero ohms. Then, with a 0 volt input, what would the gate-source voltage on the p-type be? Yup - 80 volts.

With the n-type off, the full power supply voltage is applied to the p-type gate, since there is no current flowing in the 20k. At 24 volts, this means that the p-type was always on, but the voltage was still something the p-type could handle (violating a max rating does not guarantee failure - it just means the device is no longer guaranteed to work. There's a difference.). The presence of the 20k meant that when the n-type turned on you didn't get shoot-through and destroy one or both transistors, so the output sort of worked.

When you went to 80 volts, though, it was more than the p-types could handle. Their gate junctions were destroyed and they failed open.

You say you've tried several different MOSFETs. I think that, if you go back and try them at 24 volts, you've killed them all.

You're also lucky you didn't destroy your PWM source.

Answer 3

Another important point: this circuit can't work as shown. Note that the source of the upper device is connected to the supply rail. That device will always be turned ON as soon as the supply voltage exceeds the highest level of the drive signal PLUS the gate threshold voltage of the FET.

This circuit topology is simply not suited for the supply voltage that you want to operate at.

Note that you can make this work by using an off-the-shelf half-bridge driver chip. These chips take all the pain out of level-shifting your drive signals for the upper device(s).