How can drive an inductive load very fast?

[tanbakoo]

Hi
I want to drive an inductive load ( 600uH) with -1.5A to +1.5A. Input signal is an unformed signal -10v to +10v (every form and non periodic ).

I do that with an opamp and a pushpull and current feedback of inductor ( a resistor series with inductor so voltage of it equal to current of yoke )

In this circuit when input signal change and current output dont change ( or small change because of inductor ) current feedback generate a larger output voltage ( V=L*di/dt ) for having larger di/dt and current change rapidly.

But it is need to have a very large VCC and VEE such as +200v and -200v and it is not suitable for me.

How can change current of inductive load very fast ( very large di/dt ) with +20v and -20v Power supply ?

Best Regards

 

[srizbf]

Hi
I want to drive an inductive load ( 600uH) with -1.5A to +1.5A. Input signal is an unformed signal -10v to +10v (every form and non periodic ).

I do that with an opamp and a pushpull and current feedback of inductor ( a resistor series with inductor so voltage of it equal to current of yoke )

In this circuit when input signal change and current output dont change ( or small change because of inductor ) current feedback generate a larger output voltage ( V=L*di/dt ) for having larger di/dt and current change rapidly.

But it is need to have a very large VCC and VEE such as +200v and -200v and it is not suitable for me.

How can change current of inductive load very fast ( very large di/dt ) with +20v and -20v Power supply ?

Best Regards

any possibility of driving the load from a current source?

- - - Updated - - -

Hi
I want to drive an inductive load ( 600uH) with -1.5A to +1.5A. Input signal is an unformed signal -10v to +10v (every form and non periodic ).

I do that with an opamp and a pushpull and current feedback of inductor ( a resistor series with inductor so voltage of it equal to current of yoke )

In this circuit when input signal change and current output dont change ( or small change because of inductor ) current feedback generate a larger output voltage ( V=L*di/dt ) for having larger di/dt and current change rapidly.

But it is need to have a very large VCC and VEE such as +200v and -200v and it is not suitable for me.

How can change current of inductive load very fast ( very large di/dt ) with +20v and -20v Power supply ?

Best Regards

any possibility of driving the load from a current source?

- - - Updated - - -

unexpectedly two copies have been posted. please ignore.

 

[chuckey]

I see your problem, your inductor is generating 180V+ due to the Di/Dt, so you need > 180V to actually drive current into it. The only solution I can see is to use a transformer to change the > 180V to something less then 20V. Sounds very complicated.
Frank

 

[FvM]

How can change current of inductive load very fast ( very large di/dt ) with +20v and -20v Power supply ?

You simply can't. You can rather easily make the inductor current drop from 1.5 to 0 A by disconnecting the source and providing a kind of energy absorber, e.g. a Z-diode. But you need to feed energy to the inductor to raise the current to 1.5 A again.

If you don't rely on a specific current waveform, the current reversal can be achieved by a capacitor connected in parallel to the inductor.
Disconnected the 1.5A current source, let the LC circuit perform a half cycle, connect a -1.5A current source. With the help of the capacitor, you don't need a high voltage supply, only a switch that is able disconnect the respective voltage.