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Lets say I have a CC source and set it to 10 mA and connect it to a coil. If I induce a reverse voltage in the coil with a magnet, the CC source should regulate the current to 10 mA by increasing the voltage. Lets say the magnet induces a voltage high enough so that the current increases over 10 mA, will the CC regulated source be able to invert the voltage? If there is such a current source, can you tell me which kind of converter can do that? Is there something ready I can buy? Do I have to run two converters in parallel with reversed polarity and protection diodes?

I am talking about low voltages. 0-20 V.

Edit: I think I found a paper which discusses exactly the kinds of solutions I was searching for:

Active vibration damping using self-sensing, electrodynamic actuators

C Paulitsch, P Gardonio and S J Elliott

oi:10.1088/0964-1726/15/2/033

winny
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x3oo
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  • We need more details, such as what voltage you are going to induce. Also, recommendations for specific products are off-topic. – Null Feb 09 '22 at 22:40
  • i mean, are there module i can buy. like boost or buck converters in general, which are able to do this. – x3oo Feb 09 '22 at 22:45
  • If I understand you correctly, what you're asking is basically, "Is there a constant current source which is capable of acting as a load instead of a power source?" – Tanner Swett Feb 10 '22 at 00:01
  • A two quadrant CC source? – winny Feb 10 '22 at 07:23

2 Answers2

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You can certainly create a bipolar CC circuit with a bipolar supply.

But it is not certain that you can drive a reactive load with current to match your input voltage slew rate.

Proof

Here I used an ideal Op Amp with no current limit but using +-50V supplies. I supplied a slow square wave bipolar voltage reference to an arbitrary inductive load with some DCR. The Op Amp still clipped the output voltage try to reach V=LdI/dt with dt=0 but current limited by the + feedback R and supply voltage in this Howland CC design.

This simple linear analog cn be used to understand what needs to happen with a SMPS CC circuit driving a coil. The dV/dt of the input signal but slow enough for the design to follow with a bipolar current to match scaled by R=V/I.

There are also stability oscillation risks to driving a reactive load with fast changing reference signal. Conjugate impedance matching is involved in compensation and/or reducing the open loop gain. It is not a coincidence that speaker power Amps have low open loop gain. I had to reduce this ideal Op Amp gain to 1k to avoid oscillation.

I hope that helps get you started.

Tony Stewart EE75
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    Wow thats very nice of you. I meant that the current in the coil always flows in the same direction even if the induce voltage switches. My initial idea was if i could dampen a vibrating speaker by using a CC power supply. If the speaker membrane goes one way, the CC-Source would just increase the voltage to counteract the induced voltage, but i guess your common CC-Source cant reverse polarity to counteract the too much current.

    Basically i would like to actively dampen a speaker without using an external sensor.

     – x3oo Feb 09 '22 at 23:39
  • That sounds like a different problem. Damping Factor is due to the power Amp being as close to a true voltage source as possible by lowering output impedance. Thus Damping factor = impedance ratio of speaker/driver. 100:1 is nominal, 1000:1 is more desirable and 30:1 is poor.. – Tony Stewart EE75 Feb 10 '22 at 01:13
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I think you are misunderstanding what a current source does because it will not fight against any induced voltage in your example; it will continue to produce a constant current if it is ideal or, if the source is not ideal, it will continue to feed a constant current if the induced voltage is not too extreme.

A current source is by definition a source that has infinite output impedance therefore, no matter what terminal voltage is induced by the external magnet the current source will neither create an additive nor a subtractive current. The current source will continue to produce 10 mA and, the induced coil terminal voltage will be the same as if the coil were unconnected to the current source or was open circuit.

A practical current source will however only be able to cope with so-much induced voltage under extreme circumstances and, when that happens, it's likely that the induced voltage will be clipped at its peaks.

Andy aka
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