This is a follow on from a previous question here.

I asked if replacing a MOSFET with a lower RDSon would reduce the heat generated by the MOSFET and the answer I got was that FETs with lower RDSon have higher gate capacitance.

This would increase the time it takes for the FET to fully switch on and could potentially increase the power dissipation.

My follow on question is this:

Would replacing the circuitry before the FET with a dedicated high-side driver provide any benefit?

I would be looking to decrease the switching time if I were to use a FET with lower RDSon (and in turn higher gate capacitance?)

In order to control the MOSFET from a microcontroller pin, there is already a level shifter from 3.3V to 12V which in turn then goes to a push-pull transistor configuration. This seems a fairly solid design to me and I'm not sure if there is any benefit in changing this for a driver. A cost increase would not be an issue in this design. Any direction would be appreciated.

Full circuit details are in the previous question, but here is the schematic as requested:

enter image description here

  • \$\begingroup\$ Show your current schematic. \$\endgroup\$
    – Andy aka
    Mar 10, 2021 at 11:13
  • \$\begingroup\$ I purposefully avoided repeating the information from my previous question. All the details including schematic are in the linked question. Should I copy and paste the details from there? \$\endgroup\$
    – ChrisD91
    Mar 10, 2021 at 11:14
  • \$\begingroup\$ First analyse the relative proportion of power dissipation from static power (Ron) and switching losses. Then improve the larger source of losses. \$\endgroup\$
    – user16324
    Mar 10, 2021 at 13:57

1 Answer 1


PWM switching purposes (VGS @ -12V, PWM is 20% @ 200us). The FET is switching in to a 200uH, 10ohm load

There is no freewheel diode, so when the driver brings the gate to +12V to turn off the FET, current in the inductive load keeps flowing and brings down the FET drain to whatever voltage is negative enough to exceed its Vds rating and make it avalanche. This means most of the energy stored in the inductor ends up dissipated in the FET.

Solution: add a freewheel diode, like a schottky or a fast rectifier. Either between pins 5 and 6 of the connector (lower losses, slower current drop at turnoff), or between drain and ground (higher losses in the 10R resistors, faster current drop).

At 10kHz there is no need for a special FET driver. If it doesn't switch fast enough, decrease value of R25 to give more base current to the BJTs. But don't make it switch too fast, you're not making a 300kHz switching converter, so there's no need to make more high frequency noise than necessary.


Since this is to generate a magnetic field for a metal detector, I wonder if the avalanche bit is actually part of the design, to get the current and magnetic field to fall very quickly. Here's a simulation:

enter image description here

In this case, a freewheel diode would completely screw things up by removing the sharp edge.

EDIT: How to get rid of the heatsink.

enter image description here

I've used a NMOS because better high voltage devices are available. You can use a PMOS, it will have a bit higher losses, just flip the schematic upside down. In fact you can use your existing circuit since the only change is a cap and a ferrite bead.

Here it goes:

The FET turns ON.

Current rises slowly in load inductor L2, until it reaches the desired value.

FET turns OFF. L2 is now in series with C1, and we have a LC series resonant circuit. It does exactly one half-period at its resonance frequency, and quickly flips polarity of current through L2, which makes the desired quick change in magnetic field.

At the end of the resonance half-period, voltage on C1 is back to zero and current in L2 has reversed direction. The FET's body diode turns on, and the energy stored in the inductor returns back to the power supply.

It returns about 90% of the energy stored in the inductor to the power supply, not that bad.

L3/R12 model a ferrite bead that could be necessary to tame a burst of ringing when the FET body diode turns off, because FET body diodes tend not to have the best fast/soft recovery.

The resonant frequency of L2/C1 is not important, but the peak voltage is, and peak voltage depends on the value of C1. So, C1 value should be just right to have the peak voltage as high as possible to get a sharp change in current, but not high enough to push the FET into avalanche, since the point was to avoid dissipating power in the FET. A proper capacitor type should be used (probably high voltage film or C0G).

Note increasing the power supply voltage will decrease total power used, since after implementing this energy recovery trick most of the power goes in heating the internal resistance of the 200µH coil. A higher supply voltage means it will take less time to reach desired current, so while instantaneous power dissipation will remain the same, it will be shorter, so lower total amount of energy wasted.

Here is a simpler version, with a switch instead of the FET. There is no body diode, so the LC oscillation goes on, but if there was a body diode, it would stop when the FET drain voltage goes to zero, indicated by the arrow.

enter image description here

  • 1
    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Mar 10, 2021 at 16:54
  • 1
    \$\begingroup\$ Hi, after reading through your explanation and the circuit, I think that what is essentially happening is that you still get the big burst of EMF from the collapse of the magnetic field in L2 but the current isnt dissipated solely across the FET. What happens is that in the resonant half period, the cap blocks L2 current going straight to ground and so the current is reversed and goes to L3 which suitably converts the current to a magnetic field. When the voltage across the cap eventually becomes zero, then the current in L3 is reversed and the FET diode conducts. \$\endgroup\$
    – ChrisD91
    Mar 10, 2021 at 19:36
  • 1
    \$\begingroup\$ When you refer to L1 do you mean L3? I intend to implement this circuit as soon as i can get the caps and the ferrite beads. Thanks \$\endgroup\$
    – ChrisD91
    Mar 10, 2021 at 19:37
  • 2
    \$\begingroup\$ I've added a simplified version in the answer. \$\endgroup\$
    – bobflux
    Mar 10, 2021 at 19:57
  • 2
    \$\begingroup\$ Thanks! It's microcap, I really like this software, unfortunately they went out of business, but they were kind enough to make the full-featured version totally free. \$\endgroup\$
    – bobflux
    Mar 10, 2021 at 20:05

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