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The inductance of the electromagnet is about 0.05H and the resistance is under 50 Ohm and I need about 2 to 5 Amperes to drive it.

I need to be able to change the amplitude and also the frequency at least up to 1kHz.

I have a function generator (SM5075) and a couple of power amplifiers ( https://www.bksv.com/-/media/literature/Product-Data/bp1928.ashx & https://www.bksv.com/-/media/literature/Product-Data/bp1948.ashx)

I need to control the current as the force is proportional to the current in the electromagnet. The impedance will be proportional to the frequency of the current source.

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  • \$\begingroup\$ You have some nice B&K kit so what's the problem really? \$\endgroup\$
    – Andy aka
    Commented Jun 26, 2019 at 14:01
  • \$\begingroup\$ I am not sure how to maintain the constant current. The power amplifier is providing different currents at different frequencies. \$\endgroup\$
    – Diyanko Bhowmik
    Commented Jun 26, 2019 at 14:07

2 Answers 2

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Look at examples of an AGC circuit Automatic Gain Control (AGC) circuits Theory and design paper

or

Understanding Automatic Gain Control

Basically you need to sense the AC current with a sense resistor, amplify it and convert it to a DC value, the use a variable gain op-amp with a FET driven by the filtered DC value of the AC current to keep the AC current at a constant value. The filtering of the DC conversion of the AC current will need to have a time factor in order to be responsive enough to changes but not too responsive as to cause oscillations.

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In this case a simple open loop control should suffice. If you know all the impedances of your system (electromagnet, wiring etc.) you should be able to find a analytic expression for your needed voltage at a given current and frequency.

So just dial in that voltage and it should work, at least as long as the electromagnet doesn't saturate too much so your system becomes nonlinear.

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  • \$\begingroup\$ But I need to make sure that the current supplied remains constant even when I change the frequency which in turn will change the impedance. That will not happen this way, right ? \$\endgroup\$
    – Diyanko Bhowmik
    Commented Jun 26, 2019 at 14:05
  • \$\begingroup\$ You need to change the voltage too if you change the frequency. But that would be by far the easiest solution given that you can easily calculate what voltage you need. \$\endgroup\$
    – raduur
    Commented Jun 26, 2019 at 14:20
  • \$\begingroup\$ But I need some way where it will happen automatically. An open loop is not really the way for me. Thanks, though. \$\endgroup\$
    – Diyanko Bhowmik
    Commented Jun 26, 2019 at 14:21

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