Current reversal in push/pull?

Sep 18, 2005 39 Replies

The bridge drive alternates the polarity of voltage across the solenoid- so the current will attempt to reverse. If you put the solenoid inside a full-wave rectifier bridge with (+) and (-) terminals connected to solenoid and AC terminals connected to opamp outputs, then the voltage doubles and is rectified- but you lose two diode drops.

You can try this- assuming any other problems- there is no diode drop: View in a fixed-width font such as Courier.

. . 2V . --- . | | . | | . | | . V | | . --- -- -- . | | + - . | | >----|+\\ . -- -- | >-|>|---+-KKKKKK--+----+ . +-|-/ | | | . | | | | . +------------+---| -+ . +-|+/ . | . --- . gnd . . . .

Maybe we're taking about two different things- the differential voltage across the solenoid is 2x the voltage drive of the op-amp pair: one output goes to Vin and the other to -Vin, this makes a magnitude of

2xVin across the solenoid. A safe way to do this is: View in a fixed-width font such as Courier.

. . 2V . --- . | | . | | . | | . V | | . --- -- -- . | | + - . | | >-+----|+\\ . -- -- | | >-|>|---+-KKKKKK--+----+ . | +-|-/ | | | . | | | | | . | +------------+---|--+ . +-|+/ . | . --- . gnd . . . .

Actually, your push-pull ( actually a bridge configuration ) is used quite specifically in order to be able to reverse current flow. It doesn't actually double the voltage available, merely offers a bi-directional output.

Single output gives a swing of 0 to Vsupply. Dual output gives 0 to +Vsupply or -Vsupply.

Tie one connection to the solenoid to ground or supply ( whichever makes most sense ). Drive it from one op-amp only.

I hope you have catch diodes on the op-amps output to deal with the inductive 'flyback' pulse when the solenoid is de-activated btw.

Graham

I suppose if he is just railing them out- this will work well too: View in a fixed-width font such as Courier.

. . 2V . --- . | | . | | . | | . V | | . --- -- -- . | | + - . | | >-+----|+\\ . -- -- | | >-|>|---+-KKKKKK--+----+ . | +-|-/ | | | . | | | | | . | | +---|--+ . - >----+---------------------|+/ . 2 . . . . . .

Make that : View in a fixed-width font such as Courier.

. . 2Vdd- some things . --- . | | . | | . | | . V Vdd | | . --- | -- -- . | | | + - . | | >-+----|+\\ . -- -- | | >-|>|---+-KKKKKK--+----+ . | +-|-/ | | | . | | | | | | . | | | +---|--+ . - >----+---------------------|+/ . 2 . . . . . .

I just noticed he said a *single* positive supply-"I have an app where I am driving a solenoid using a dual op amp in bridge configuration to double the effective voltage from a single positive supply." So he's sunk. I can't even understand them half the time- the basis for their question is so off-the-wall they are unintelligible.

The voltage *does not* double !

A single driver will product a drive voltage = Vsupply.

A bridge will produce a drive voltage of Vsupply or -Vsupply.

A bridge driver arrangeemnt with rectifier bridge will produce a drive voltage of Vsupply - 2*Vf. I.e. *less* than the single driver.

Graham

I canna repeal the laws of ffisiks..........

martin

Your circuit understood. I 'assumed' that the OP was driving the op-amps to rail ( which would make sense ).

In any event the gain can be simply increased - if that's what's needed - by adjusting the feedback but a newbie might indeed not realise that.

Usual problem - not enough detailed info from OP.

Regards, Graham

Exactly.

If we knew the relay voltage and the supply voltage it would make more sesne.

As you say, with a single supply there's a limit to what's do-able here.

Graham

Not as funny, now that James Doohan is dead. :(

? Michael A. Terrell Central Florida

I don't agree. For a public figure to be remembered with affection can't be bad.

Regards, John Woodgate, OOO - Own Opinions Only. If everything has been designed, a god designed evolution by natural selection. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I have an app where I am driving a solenoid using a dual op amp in bridge configuration to double the effective voltage from a single positive supply.



In terms of current flow, can someone please clarifiy how this is different from using a single amp?



I need the emiited field to be unidirectional and not reverse.



Is this possible with a push/pull driver?



Thank you,



Jeffery


"Jeffery Dean"

** HUH ???

Scotty needs to hear about this.

.......... Phil

I disagree, Scotty, did one hell of a lot for getting youngsters interested in engineering. I hope that he would be happy for people to use his quotes.

martin

"John Woodgate" schreef in bericht news: snipped-for-privacy@jmwa.demon.co.uk...

Ditto for non-public figures ;)

Thanks, Frank. (remove \'q\' and \'.invalid\' when replying by email)

We've been here before Fred, a year or so ago. A transient capacitive voltage doubler, generating 2xVs to pull the solenoid in, dropping back to 1xVs for holding.

Tony Williams.

Nice. I handed my technician a design for a box with ten small microwave relays -- except the crazy relays want tight connections I suppose, and each one takes 5W of coil power! That's 50W for the set, and it's got me worried about heat. I can't add a fan, because it's going on a quiet optics bench. My tech's waiting for parts, so I've got a few days to improve the coil-power scene.

Tony, did you come up with a slick circuit?

Thanks, - Win

Yea Tony, let's see that X2 relay driver again. I can do it with a FET bridge, diode and 2*V storage cap but yours was much slicker. Regards, Harry

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