Reducing the Voltage to a square root of itself

[Zook104]

Hello

I am pretty new to electronics so I haven't heard of most devices or circuits.

I was wondering if there was a method of reducing a voltage to a square root of itself. The value of the input voltage would not be a fixed value.

I am also looking for a method of squaring the value of the voltage.

Basically I have a set up with two input voltages Va and Vb. I want the output, Vc, to be equal to the square root of Va minus the square root of Vb, all squared.

Any and all help with my problem would be greatly appreciated!

Many Thanks!

 

[GreenGiant]

What you are trying to do is going to be nearly impossible without a microcontroller, using an arduino you could do this... fairly easily, though its not going to be a cake walk.

Do you need to have any current behind these voltages or is it all just going to be signals?

 

[Zook104]

I thought an an arduino might pop up at some point.

No I am not using just signals and no current at this point.

 

[Arouse1973]

What you are trying to do is going to be nearly impossible without a microcontroller, using an arduino you could do this... fairly easily, though its not going to be a cake walk.

Do you need to have any current behind these voltages or is it all just going to be signals?

Sorry it is possible. Square root circuits have been around longer than micros. All you need is an op-amp an n-channel FET put in the feed back loop of the op-amp. Set up the amp as an inverting amp and connect the gate to 0V the source connects to the output and the drain to the inverting input. You will also need to connect the non inverting terminal to 0V and you will need a resistor between the input and the inverting terminal. This works because the current to voltage conversion of a Mosfet is a square root function.
Thanks
Adam

 

[Arouse1973]

Oh and if you want a logarithmic response the replace the FET with a NPN BJT. Handy for amplifying small signal and attenuating large signal.
Thanks
Adam

 

[BobK]

You could use a log amp, divide by 2 then an exponential amp.

Bob

 

[(*steve*)]

Of course, a problem you might get is that the output is nothing like what you want, even if the circuit operates strictly correctly.

For example, here are some potential input and output voltages of two circuits where the output is the square root of the input:

Circuit 1

10mV --> 31mv
100mV --> 100mV
250mV --> 158mV
500mV --> 224mV

Circuit 2

10mV --> 50mv
100mV --> 158mV
250mV --> 250mV
500mV --> 354mV

Voltages are not absolute numbers. OK, 0V might represent 0 (but it equally might not). More importantly, what voltage represents unity? In the above examples one circuit assumed 100mV, the other 250mV. There is no particular reason why one is more right than the other.

In order to do this in a practical sense (and this applies to any non-linear transformation), you need to provide more parameters.

If you don't, your question is as ambiguous as "what is the answer to this 10 - 7 = ?".

(the answer is -- quite naturally -- 9, or 3, or 1, or and of an infinite number of answers...). You need to understand the problem before you can give an answer.

 

[Arouse1973]

Hi Steve what is Circuit two doing, you don't really explain. And how the hell can 10-7 equal 1.

 

[(*steve*)]

Hi Steve what is Circuit two doing, you don't really explain. And how the hell can 10-7 equal 1.

Both circuit 1 and circuit 2 produce an output which is the square root of the input.

10 - 7 is 1 in octal. (so you need to know what 10 represents)

The two circuits differ only in what is represented by 1.

 

[Arouse1973]

Both circuit 1 and circuit 2 produce an output which is the square root of the input.

10 - 7 is 1 in octal. (so you need to know what 10 represents)

The two circuits differ only in what is represented by 1.

f

Thanks Steve.
Adam