Current reversal in push/pull?

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.

Reply to
Fred Bloggs
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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 . . . .

Reply to
Fred Bloggs

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 . . . .

Reply to
Fred Bloggs

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

Reply to
Pooh Bear

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 . . . . . .

Reply to
Fred Bloggs

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

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

Reply to
Fred Bloggs

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.

Reply to
Fred Bloggs

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

Reply to
Pooh Bear

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

martin

Reply to
martin griffith

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

Reply to
Pooh Bear

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

Reply to
Pooh Bear

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

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Michael A. Terrell
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Reply to
Michael A. Terrell

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

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Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
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Reply to
John Woodgate

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

Reply to
Jeffery Dean

"Jeffery Dean"

** HUH ???

Scotty needs to hear about this.

.......... Phil

Reply to
Phil Allison

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

Reply to
martin griffith

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

Ditto for non-public figures ;)

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Thanks, Frank.
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Reply to
Frank Bemelman

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.

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Tony Williams.
Reply to
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?

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 Thanks,
    - Win
Reply to
Winfield Hill

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

Reply to
Harry Dellamano

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