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I'm trying to drive a capacitive load with P and N mosfets in a half H bridge. The problem is that my control voltage is 1.8V, and the H bridge is switching 20V. My original plan was to use npn's to switch the drive to the mosfets:

schematic

simulate this circuit – Schematic created using CircuitLab

The problem is that in this configuration, the Vgs will be 20V at times, far exceeding the 12V max listed for many mosfets. Is there a solution for this?

Edit:

Now that I think about it, will this work?

schematic

simulate this circuit

Are there any major problems, except slower turn on/off?

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    \$\begingroup\$ There are many ways to do it but here is a very easy one; I don't suppose you ever want to have both M1 and M2 on at the same time right? so SW1 and SW2 open and close together right? then you can just swap around M1 and M2 so that the bottom one is a p-channel and the top one is an n-channel and then connect their gates together and drive them from one npn-based transistor amplifier like the ones you've got two of now. \$\endgroup\$ – Vinzent Jan 3 '18 at 23:39
  • \$\begingroup\$ But doesn't that allow shoot-through, and also how does that solve the Vgs problem? \$\endgroup\$ – BeB00 Jan 3 '18 at 23:42
  • \$\begingroup\$ No your current circuit allows shoot-through. Vincent is suggesting using them in follower mode, though you will not be able to drive to the rails. Your 2nd circuit is fine assuming you do not want to PWM modulate it at kHz. YOu have to make sure your dead time is long enough to allot the turn offs to happen before you turn on the other side. \$\endgroup\$ – Trevor_G Jan 3 '18 at 23:44
  • \$\begingroup\$ No that actually prevents shoot-through, what you've got now however is very prone to shoot-through. it solves the problem because the voltage on the source is going to follow the voltage on the gate, it is also called a source-follower. \$\endgroup\$ – Vinzent Jan 3 '18 at 23:45
  • \$\begingroup\$ Yeah, the PWM will be driven with dead time. Im still not sure I understand vincents circuit, can you draw it? \$\endgroup\$ – BeB00 Jan 3 '18 at 23:45
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The circuit I proposed in the comments is this one

schematic

simulate this circuit – Schematic created using CircuitLab

Using the fets as source followers instead of as common drain

My original comment:

"There are many ways to do it but here is a very easy one; I don't suppose you ever want to have both M1 and M2 on at the same time right? so SW1 and SW2 open and close together right? then you can just swap around M1 and M2 so that the bottom one is a p-channel and the top one is an n-channel and then connect their gates together and drive them from one npn-based transistor amplifier like the ones you've got two of now"

But the second one you posted will also work (using voltage-dividers)

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  • \$\begingroup\$ Oh right, yeah the output needs to swing to the rails. \$\endgroup\$ – BeB00 Jan 4 '18 at 0:00
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There are lots of ways to do it, none of them particularly pretty though. It really depends on the voltages and the speeds you want to switch at.

Here is a "simplified" discrete version that uses emitter follower buffers to beef up the gate charge currents.

schematic

simulate this circuit – Schematic created using CircuitLab

But as you can see it gets complicated fast and takes real-estate. As such, switching to an integrated MOSFET driver IC quickly becomes a more attractive solution.

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I tried my best to do this with minimal components. The 4v7 zeners regulate the voltage drop across the FET gates.

I added diodes on the Source and Emitter nodes of the transistors to ensure the OFF state, always good practice especially with FETs.

The series resistors on the Collector of Q2 and Q3 set the correct Zener current needed. Without them the Zeners will get hit with the full 20v and burn up! If the value is too high the Zeners will not reach the breakdown voltage (FETs won't turn on fully).

Whats also good about this circuit is that you can increase the supply voltage without changing any of the components.

enter image description here

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  • \$\begingroup\$ Decrease R1 and R2 (5k) to speed this thing up... I got it robustly simulating at 40k. If you decrease these resistors further and add more diodes on the Source it will help prevent shoot through by making the OFF transition time faster than the ON transition time. \$\endgroup\$ – Tony Aug 20 '19 at 2:11

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