Square Root Circuit

Thread Starter

belxiors

Joined Jan 27, 2017
7
Hello!

So i'm looking for a circuit that can make a square-root of a voltage. I know it's pretty complex, and i'm aware of log solutions, TI note 31 etc ... But none of them are accurate and they all need a very specific specification.

So does anyone know about some good solutions, or has any new ideias?

Thanks!
 

joeyd999

Joined Jun 6, 2011
4,287
Hello!

So i'm looking for a circuit that can make a square-root of a voltage. I know it's pretty complex, and i'm aware of log solutions, TI note 31 etc ... But none of them are accurate and they all need a very specific specification.

So does anyone know about some good solutions, or has any new ideias?

Thanks!
What's the application?
 

ian field

Joined Oct 27, 2012
6,539
Hello!

So i'm looking for a circuit that can make a square-root of a voltage. I know it's pretty complex, and i'm aware of log solutions, TI note 31 etc ... But none of them are accurate and they all need a very specific specification.

So does anyone know about some good solutions, or has any new ideias?

Thanks!
There's probably something in the NS application notes floating about somewhere online.

The appnotes are pretty old - so you'll have to research modern replacements for some of the listed semiconductors.

The principles have been around since they thought analogue computers could compete with digital.
 

Alec_t

Joined Sep 17, 2013
10,584
Somebody has to say it, so it might as well be me. Have you considered using a microcontroller rather than an analogue circuit?
 

MrChips

Joined Oct 2, 2009
19,727
Simple.

Take a voltage generator y.
Feed it into both inputs of a multiplier circuit to give y^2.
Compare the input voltage x with y^2.
Feed the error voltage back into the y generator.
y = √x
 

Papabravo

Joined Feb 24, 2006
12,684
Simple.

Take a voltage generator y.
Feed it into both inputs of a multiplier circuit to give y^2.
Compare the input voltage x with y^2.
Feed the error voltage back into the y generator.
y = √x
You want to be really careful about the gain required in the vicinity of the origin.
 
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