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I'm building an H-bridge circuit designed to send a pulse through an electromagnet, and then rapidly collapse the field of the electromagnet coil as soon as the transistors switch off. Minimizing the field collapse time is a critical parameter to the application.

I am using the following components:

N-channel MOSFET - IRF3205
On-state resistance Rds(on): 8 mΩ
Voltage Vgs highest: ±20 V
Voltage Vds Typical: 55 V
Current Id continuous: 110 A
Current Idm pulse: 390 A

RFUH20TJ6S Schottky diode
https://www.mouser.com/datasheet/2/348/rfuh20tj6s-1920871.pdf
Reverse Voltage: 600 V
Non-repetitive forward surge current: 120 A
Average Current: 20 A

enter image description here

The idea is that we have a single half-bridge driver to drive the gates of the two MOSFET with 12 V. When the Gates have 12 V applied, the MOSFETs conduct power through the electromagnet coil.

When the MOSFETs are switched off, the back-EMF from the inductive load is conducted through the diodes, through the braking resistor, which rapidly collapses the field. I assumed that the Diodes themselves do not need to be rated to handle 60 A, because they only need to handle the current of the back-EMF, which should be very small due to the braking resistor. The larger the value of the Braking Resistor, the smaller the field collapse time. However, I have been having trouble with the circuit. The MOSFETs are failing (drain to source conducting) and the diodes seem to be failing as well, sometimes without even applying any gate voltage.

Is this a valid design for the intended function? Where are there issues/things missing?
Any and all feedback is welcome.

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  • \$\begingroup\$ 150kΩ, is that a typo? How many volts is that peak (voltage at turn-off is current at turn-off times resistance)? Is 450mH really ~0.1Ω as well? Not impossible, but fairly impressive for such a large value. \$\endgroup\$ Mar 27, 2023 at 5:13
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    \$\begingroup\$ Whenever I hear "field collapse", something is very wrong. U=L*di/dt and IL=NAB. To bake your "field collapse" to zero, you need to push a lot of current in opposite direction. Please simulate your circuit. \$\endgroup\$
    – winny
    Mar 27, 2023 at 11:16
  • \$\begingroup\$ @winny For clarity, instead of using the braking resistor, I could push an equivalent reverse current to make the inductor collapse? So I could redesign this to use a full-bridge circuit and effectively cancel the current. If I had a pulse with duration of 10ms in forward direction, applying a 10 ms pulse in reverse direction would effectively discharge the inductor. Is that correct? \$\endgroup\$
    – Flywheel
    Mar 27, 2023 at 16:09
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    \$\begingroup\$ Something like that. Ideally, you'd want to recycle the energy stored in the coil if it happens repeatedly, but if it's rare, feel free to have a voltage rail opposite of what you charge it with, move the energy to there, making the field go to zero (the inductor will not "go to zero", it will remain in place unless you melt it). Please simulate it. \$\endgroup\$
    – winny
    Mar 27, 2023 at 16:23
  • \$\begingroup\$ @winny Got rid of the braking resistor, and just used the reverse current approach you described. Works awesome!! And you can control the amount of damping by varying the length of the reverse pulse. Pretty neat. \$\endgroup\$
    – Flywheel
    Mar 28, 2023 at 2:14

2 Answers 2

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I assumed that the Diodes themselves do not need to be rated to handle 60A, because they only need to handle the current of the back-emf, which should be very small due to the Braking Resistor.

That's not how inductors work. Look at the inductance equation: \$v(t) = L \frac {d}{dt} i(t)\$. That says (possibly after some math), that the inductor current does not change across the instant that you turn off the FETs. The only thing that the higher braking resistance does is increase the voltage, and speed up how fast the current decays.

What a high braking resistance does buy you is that fast current decay -- the total amount of energy dissipated in the diodes will go down as the resistance goes up. But at the instant that the switches open, the current will be the full inductor current.

What this means is that while you do need diodes that can handle the full current for short periods of time, you don't need diodes that can handle the full current continuously. In a perfect world, you'd look for diodes that have pulsed current ratings, and select ones that will work in your circuit -- but it's still up to you to understand the nature of the current pulse that you're feeding to your diodes.

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I'm building an H-Bridge circuit designed to send a pulse through an electromagnet, and then rapidly collapse the field of the electromagnet coil as soon as the transistors switch off.

Your words tell me that you don't understand how an inductor behaves. You appear to believe that you can control the current flowing from an inductor when the circuit MOSFETs deactivate; you can't. At the moment the MOSFETs deactivate, the current that was previously flowing will continue to flow but decay towards zero amps.

So, it 1 amp was flowing when the MOSFETs deactivate the 1 amp will flow through the braking resistor. Given that the braking resistor is 150 kΩ an impossible-to-survive peak instantaneous voltage of 150,000 volts will occur and, we know that this won't happen because the MOSFETs will break-down and destruct. Here's the back-emf current flow: -

enter image description here

So, look at it another way: -

N-channel Mosfet - IRF3205 Voltage Vds Typical: 55V

To prevent the 55 volts from being exceeded requires a 55 Ω braking resistor and, that's when the inductor current has risen to only 1 amp at the time the MOSFETs deactivate. This is a million miles away from your expectations I suspect.

I get the feeling that your inductor current can rise to dozens of amps and, of course this means a much smaller resistor to prevent the 55 volts becoming exceeded.

I assumed that the Diodes themselves do not need to be rated to handle 60A, because they only need to handle the current of the back-emf, which should be very small due to the Braking Resistor.

No, this is a flawed perception of what will actually happen.

The Mosfets are failing (Drain to Source conducting)

This will certainly happen with a brake resistor of 150 kΩ.

Is this a valid design for the intended function? Where are there issues/things missing?

I think your expectations need to be realigned; the peak current that flows when the inductor MOSFETs are deactivated IS the same current that is flowing prior to deactivation. If this is 1 amp then your braking resistor cannot be greater than 55 Ω. If it's 10 amps then the resistor cannot be greater that 5.5 Ω.

In other words, to rapidly collapse the field of an electromagnet coil necessitates a very large back emf <-- that's the only way the energy can be burnt off quickly. If you want to collapse 10 amps to zero amps in (say) 10 milliseconds, the rate of change of current is 1000 amps per second. This necessitates a back emf of this value: -

$$V = L\dfrac{di}{dt} = 0.45\times 1000 = 450 \text{ volts}$$

And, that means a braking resistor of 450/1000 = 0.45 ohm: (ohm's law). Even then, you should simulate to see how long the current takes to reach (say) it's 1 % value.

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  • \$\begingroup\$ Shouldn't the diodes in the circuit prevent the 55V from being exceeded, because the back-emf will flow through the diodes rather than the MOSFETs? \$\endgroup\$
    – Flywheel
    Mar 27, 2023 at 13:36
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    \$\begingroup\$ @Flywheel The diodes don’t disconnect the mosfets. They still are subject to the voltage on the terminals of the inductor. And for quick field collapse, this voltage has to be large. The diodes are just set up to be self-commutating switches, as an alternative to mosfets. \$\endgroup\$ Mar 27, 2023 at 13:48
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    \$\begingroup\$ @Flywheel when the MOSFETs deactivate, the inductor terminals produce a back-emf. That's a voltage of the opposite polarity to when the inductor was charged via the MOSFETs. That back emf rises to a level to maintain the same current current flow seen at the point in time just before the MOSFETs deactivated. The current flow is through both diodes and the braking resistor. Given that the braking resistor is 150 kohm, you can see that the inductor MUST have to generate a really high back-emf voltage to force that current. That will kill your MOSFETs. \$\endgroup\$
    – Andy aka
    Mar 27, 2023 at 14:20

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