I'm very stuck and I've run out of solutions. I designed a buck converter in ltspice which I then physically made. It worked decently at high voltages, but at low voltages, the gate signal just becomes warped and alot of heat is dissipated. That's not even the main issue, the issue is that the circuit only works if I'm using an oscilloscope and scope probe to monitor the signal on my microcontroller which provides the pulse signal (in this case a teensy 3.5).

I've attached an updated schematic of my simulation:

Current theoretical simulation

The DC-DC converter that I would use physically is much simpler than the LTM8057. I'm simulating with the LTM8057 to see if the problems I was having with my buck converter was DC-DC converter related or otherwise.

I've narrowed down the issue to the gate driver (TC4432). The thing has no reference voltage and then uses my Teensy/scope probe has a high impedence to ground. This causes my Teensy's to blow up (I've fried 3 so far). Reading some online sources suggests that the gate driver should have a reference to the source side of the circuit (on the right of the MOSFET). My idea was to just link the Vout- of the LTM8057 to the right side (source) of the MOSFET. But no matter what I do, the circuit doesn't run. It only runs in the above configuration where even at maximum duty cycle, I only get 9 V on the load:

Voltage across 10 ohm load

I have no idea what to do now. I don't want to waste time soldering another circuit. The last circuit I built confirmed that there was something badly wrong:

Efficiency of most recent physical buck converter

At lower duty cycles, it becomes horribly inefficient. It only worked at higher duty cycles because I was scope probing the Teensy to monitor the output PWM signal. I thought it was a common ground issue, but rearranging my grounds didn't work. So I've fallen back to simulations again.

So finally my question is: In the above simulation, how do I get full range of voltage across my 10 ohm load?. Then I can try to recreate it physically.


EDIT: If the gate needs to be driven with a higher side voltage more than or equal to the source (24 V), then why does this following schematic work? The isolated DC-DC converter has been swapped out with an isolated pulsed supply with its negative output to the source.  This works even though the voltage is 12 V.

  • 1
    \$\begingroup\$ Shouldn't Vout- of the dcdc converter and the GND pin of your driver along with decoupling caps be connected to the source pin of your MOSFET? If not the gate voltage depends on the output voltage. Also, the PWM signal would have to be floating as well. \$\endgroup\$
    – ocspro
    Mar 9, 2020 at 15:14
  • \$\begingroup\$ Link to the FET driver datasheet, but I think ocspro has it. Your 12V isolated supply isn't isolated because you grounded it. As a result your Vout is limited to 12V - Vgs. \$\endgroup\$
    – user16324
    Mar 9, 2020 at 15:28
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    \$\begingroup\$ TC4432 is normally an 8-pin package part (at least the one from Microchip). You have a N-channel MOSFET in the high-side of your 24V buck circuit with a V_GS threshold of 10V, meaning you need 34V on the gate to turn it ON. You do not have a 34V supply in your circuit and as such the MOSFET is 'pinching off' the input DC. You need to rethink your circuit somewhat. \$\endgroup\$ Mar 9, 2020 at 15:35
  • \$\begingroup\$ @ocspro I tried that (connecting both vout- and pin 4 of my gate driver to the source side of my MOSFET), I either get an error or the source side is short circuited to ground. Could you explain abit more what you mean? And the decoupling capacitors are on the Vdd side (pin 3) of my gate driver. Why are they involved in this? \$\endgroup\$ Mar 9, 2020 at 16:01
  • \$\begingroup\$ @AdamLawrence The LTSpice model of the TC4432 is a 5 pin, Input (1), Output (2), Positive Supply (3), Negative Supply (4) and UV_Lockout (5). I'm new to the power electronics scene. Can rethinking my circuit be as simple as using a P-channel MOSFET instead? \$\endgroup\$ Mar 9, 2020 at 16:03

1 Answer 1


To turn off that FET, the gate driver Vout needs to be up at 24 volts.

This circuit does not provide that.

  • \$\begingroup\$ But the MOSFET I chose can be driven at 12 V, that's why the signal of the gate driver is around 12 V. \$\endgroup\$ Mar 9, 2020 at 16:07
  • \$\begingroup\$ He's referenceing the driver to the source, so it's fine. (I'm talking about the last spice schematic shown) \$\endgroup\$ Mar 9, 2020 at 18:55

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