Square root circuit

With the resistors connected to the inverting input, the op amp looks like an open circuit for output voltages below the threshold (where the diodes don't conduct).

Thus its resistor is effectively removed from the circuit. Once V_out crosses V_th, the op amp holds its end of the resistor at V_th, just like throwing a switch. That results in a gain change in the voltage divider, but ideally no voltage step.

It's a similar idea to the two-op-amp absolute value circuit, e.g. Fig

4.46 of AoE II. (iirc you don't have AoE III.)

Cheers

Phil Hobbs

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Reply to
Phil Hobbs
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I don't have AoE. I'm not able to visualize the correct circuit. Are you referring to the output resistors? So the output is taken from the inverting input to the opamp which means through the diodes?

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Rick
Reply to
rickman

Here's the simplest example: gain of 1 for Vout < Vth (op amp railed) and then gain of 0.5 from there on up, when the diode conducts and closes the feedback loop.

You can make the slope change the other way by reversing the diode.

At the threshold, the current through the resistor is zero even when the loop is closed, so there's no voltage step in the output.

Vin 0-----RRRR---------------*------------0 Vout | 10k | R 10k R R R | | *------>|-----------* | | | | \ | *------| -\ | | \ | | >-------* | / Vth 0------| +/ | /

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

Look at it this way, If you figured out what the meaning was, the rest of us were way ahead of you, we didn't need you commenting. We know your comment was just an effort to make us think you are smart. Keep trying. Mikek

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Reply to
amdx

One more thing.

A 324 is actually an excellent choice for a circuit like this in a low-speed application like thermal control. As JT reminds us periodically, its input transistors are lateral PNPs, which have some huge base-emitter breakdown voltage, and so doesn't need diodes across its inputs.

Some op amps have those diodes, and so won't work in this circuit.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Ok, that is clear now. In the original circuit I missed where Vin was being applied.

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Rick
Reply to
rickman

What is the correct diagram?

Reply to
Simon S Aysdie

Fear not George, there are only 11 components and the squaring/square rooting is inherent in the topology. All the passives can be 5% or even

10% tolerance parts.

piglet (but prefers hunny to haycorns)

Reply to
piglet

Like what I posted for Rick, except with a few more op amp/diode/resistor sets hung off the output.

It's what Spehro posted, except with the resistors moved to the other sides of the diodes. (If I recall the posted circuit correctly.)

(NB this isn't my circuit, I'm just fooling around with his. I think it's pretty, though.)

Cheers

Phil Hobbs

Reply to
Phil Hobbs

Take the end of the resistor that is connected to the output of the opamp, and connect it to the inverting input. It acts like a battery (Vbr) and switch in series, diode polarity determines on which side the switch is on. I must admit it's a fun circuit. A wonder I hadn't seen it before.

George H.

Reply to
George Herold

OK after my experience with the analog multiplier I'm leery* of the behavior near zero.

George H. (leary is not right...)

Reply to
George Herold

On Fri, 10 Jul 2015 21:12:47 +0800, Techman Gave us:

You are an idiot and yet another punk putz. FOAD.

Reply to
DecadentLinuxUserNumeroUno

Behavior near zero is fine but as presented there is some ripple near full scale. Increasing C3 may fix that but further reduce bandwidth.

piglet

Reply to
piglet

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