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In our experiments, we use coils to generate magnetic fields. In order to have a stable magnetic field, we actively regulate the current in the coil (typically between 10 to 60 A) by using a PID and power MOSFETs (IXYS IXFN210N30P3, 8 of them in parallel). The current is measured via a LEM current sensor.

The PID outputs a voltage that is directly fed into the gate of the MOSFETs to control the current in the coil. The PID uses a THS3091 op-amp that can output a max. current of 250 mA to drive these MOSFETs. The coil has an inductance of 300 μH and resistance of 75 mΩ. The voltage across the coil is given by a 12 V power supply that can deliver up to 100 A current. We also have a varistor to protect the circuit from the surge.

I would like to know if there is any better way to drive these MOSFETs to achieve a fast feedback and how to optimize the gate resistor value? Also, any other suggestions to improve the circuit are welcome. Here is the circuit:

enter image description here

sudipta
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    What do you mean by "fast feedback"? How quickly are you wanting the current to change? – Andy aka Dec 31 '22 at 16:14
  • Currently, it takes about half a millisecond to reach from 10A to 15A. We would like to reduce it by a factor of 10. – sudipta Dec 31 '22 at 16:18
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    Just showing an op-amp without its feedback is almost entirely meaningless. Could you please _edit your question_ to show the relevant circuitry around the op-amp? This would include the current sense for the coils, how that is fed back to the op-amp input, and your current command input. Looking at your given values, I'm not sure how you expect to get there from here: it takes 1.5ms to get a 300\$\mu\mathrm H\$ coil from 0A to 60A with 12V applied, irrespective of any control system. – TimWescott Dec 31 '22 at 16:24
  • Yes, you are right, adding some resistance would help to lower the time but unfortunately that's difficult for us. Going from 0A to 60A would take 1.5ms, but we are mostly interested in small changes in currents, for example 10A to 15A. This should be faster. – sudipta Dec 31 '22 at 16:46

1 Answers1

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adding some resistance would help to lower the time

Forget about adding resistance; it'll just slow things down even more.

Currently, it takes about half a millisecond to reach from 10A to 15A. We would like to reduce it by a factor of 10

So you want a 5 amp change in 50 μs. That requires this voltage step change: -

$$V = L\cdot \dfrac{di}{dt} = 300\space \mu\cdot \dfrac{5}{50\space\mu} = 30 \text{ volts}$$

The voltage across the coil is given by a 12 V power supply

That's going to be a big problem; you need to change the voltage by 30 volts to get a change of current in the inductor of 5 amps.

I thought decreasing time constant would lead to faster rise and fall time.

Look at this graph: -

enter image description here

A larger value resistor does lead to a quicker resolution of the steady-state situation but it hardly helps because, the final steady-state current is severely restricted in the process.

Andy aka
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  • I mean adding a resistor to the coil, that should decrease the time constant L/R. With higher voltage of 30V, the rise time improves slightly but the circuit becomes unstable and it starts ringing. – sudipta Jan 01 '23 at 12:25
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    Adding a resistor to the coil will not improve things. A parallel resistor will bypass current from the coil (unwanted) and a series resistor will prevent current rising to the values you need. No, adding a resistor is a bad idea all-round. Decreasing the time constant is NOT the same as changing the current more quickly. – Andy aka Jan 01 '23 at 12:31
  • Could you please explain why is that? I thought decreasing time constant would lead to faster rise and fall time. – sudipta Jan 01 '23 at 12:41
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    The time constant isn't something directly related to time (as in if we decrease it things happen more quickly). The time constant is directly related to how quickly the current acquires a steady state value. I'll add a graph in my answer... – Andy aka Jan 01 '23 at 12:51