I designed a Half-Bridge power supply using Gate Driver Transformers to drive my two MOSFETs. It's working great, I winded the two transformers around very small toroids with some turns, I didn't calculated anything, they just worked!

Now, I'm trying to understand how to select a proper gate driver transformer. My switching frequency is 100kHz so I'm aiming for a rising/falling times of about 300-400ns or even lower. I went to DigiKey and found this 1:1 ratio transformer. It claims a primary inductance of 10mH.

On this page I found this picture:

enter image description here

According to this my inductance should be between 0.5 and 2mH so 10mH is too much. I tried to do some calculations using di = v dt / L. I'm powering my driver transformer using a 6A peak MOSFET driver (MCP1407) with a 12V output, at 100kHz and with 50% duty cycle:

di = 12V * (5*10^-6 / 10*10^-3) = 6mA!!!

Only 6mA? I'm pretty sure I'm really missing something big, this drivers have high peak currents to rapidly charge/discharge the gate capacitance of the MOSFET.

So my questions are:

  1. Isn't 10mH a really big inductance for a fast transition?
  2. According to the picture, the inductance for 100kHz should be between 500uH and 2mH, why can't it be lower (or higher)?
  3. How do you exactly select a gate driver transformer and how can you calculate how fast the current/voltage will rise on the MOSFET's gate to ensure proper rise/fall times?
  • \$\begingroup\$ Could you please separate your observations from your questions. Put you questions at the end of your observations and list them with a question mark at the end (?). Right now your questions and observations are jumbled together. It will make it much easier for us to answer your questions in a sequence, by the order you list them. \$\endgroup\$
    – user105652
    Commented May 12, 2016 at 3:01
  • \$\begingroup\$ @Sparky256 how about now? Thank you for the suggestion! :) \$\endgroup\$
    – Andres
    Commented May 12, 2016 at 3:07
  • \$\begingroup\$ That 6mA figure is the magnetizing current. Any current you pass through to the secondary is extra on top of that. So the high primary inductance saves power. (It may also imply a high leakage inductance, which would limit the power you can deliver to the secondary) \$\endgroup\$
    – user16324
    Commented May 12, 2016 at 10:19
  • \$\begingroup\$ @BrianDrummond. Thanks for the info. I am implying those hints in my answer. \$\endgroup\$
    – user105652
    Commented May 12, 2016 at 19:00

1 Answer 1


I do not know the details of your mosfets, so these answers are general guidelines. You seem to already have a good grip on the important issues:

1) You are correct in assuming 10mH is way too high of an inductance for any switching power supply. 10mH is so high you could use it in audio circuits. Your chart (Table 3) for guidelines is good. Stay below 4mH if possible, but it is a good value if you want minimum power consumption on these drive transformers @100KHZ.

2) For a given frequency, small ferrite toroid cores will have a limited range of inductance in which they are efficient at resonating with the frequency and transferring power to the secondary windings. At 500uH it will work but the idle current will be on the high side.

At 2mH the idle current will be much lower, but if the toroid is very tiny and you go above 2mH you may saturate the core, causing the primary to act like a short circuit.

Current limiting and / or fusing of the primary driver IC or mosfet is recommended. I would select a toroid at least 1cm diameter with a cross section of 2 x 5 mm. Avoid way over sized toroids as they have a lower resonate frequency, about 30KHZ to 50KHZ.

3) For 12 volts and minimum inductance that is safe for the mosfet driver, 6 turns on the primary and 2 (or 4) x 6 turns on the secondary will do just fine if you wind it yourself. Each set of 6 turns should wrap around the entire toroid to avoid eddy currents that create 'dead spots' in the core.

Be sure to put a 10 or 22 ohm 1/4 watt resistor right at the gate of each mosfet. This reduces ringing at the drain pin and parasitic oscillation. The basic rise/fall time for a given mosfet is specified by the manufacturer.

To calculate the combined gate rise/fall time use: 2 x pi x R x L x C x 12(volts), where L is the inductance of your secondary winding, C is the mosfet gate capacitance, and R is the series resistor I mentioned. Don't forget your dealing with square waves with sharp rise and fall times (a few 100nS at most).

There is no 'perfect' rise and fall time, just the fastest times possible with inductive values between min and max limits as per Table 3, and the resistor to prevent distortion at the mosfet drain pins.

4) For a combination of efficiency and enough drive current for the mosfets, I recommend starting out with a toroid or 'pot' core with a primary and secondary inductance of about 2mH. I would ask you to buy a LCR meter if you wind your own, but they are expensive.

  • \$\begingroup\$ Thank you for such a great a answer! I just a have a question, let's suppose I have a 2mH primary inductance, how can I know how fast the rising/falling edge will be? Does the inductance matter or only the gate capacitance? \$\endgroup\$
    – Andres
    Commented May 13, 2016 at 1:24
  • \$\begingroup\$ Because you are using standard power mosfets with a somewhat predictable and consistent gate capacitance (300 to 1000pf), the inductance rules because it set a very low drive impedence, capable of a wide range of capacitive loads. Wire gauge should be 20 to 22awg, to keep the wire impedence low. Once you buy a mosfet you can do the math based on its specs, but engineers back it up by using an oscilloscope to check the signal at the drain pin. You should not need a snubber diode if the mosfet is rated at least 100volts D-S. \$\endgroup\$
    – user105652
    Commented May 13, 2016 at 1:36

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