Analog square rooter

MrAl said:

Hello again drpower,

Sorry, i didnt notice the link but now i see it and yes it leads to some interesting circuit stuff. I'd be happy to take a better look at this. They say it works with a wide input range, so that's a very good thing. I still dont know what your input range has to be however, so you may want to mention that.

Just for starters, a squaring circuit can be used to make a square rooter circuit by placing the squaring circuit in the feedback loop of an op amp. It's an old trick used with various functions to get the inverse function. The theory is something like this...

We have a function f(t) and we want F(t) ie the inverse function. By placing f(t) in the feedback loop the output has to be equal to the inverse function in order for the inputs (one f(t) and the other the users input) to become equal. That's because when the output is equal to the inverse function it goes through the feedback loop and then becomes equal to the input again. So the only way the circuit can work (if it is stable) is to output the inverse function. It works with other types of functions too, in short the circuit does:
input1=users input
input2=f(F(t))=input1 (because f(t) is in the feedback path)
So, output=F(t)


I can provide more on this a little later, and i'll look over the schematics you've posted and be back a little later with some more comments.

You should know that there has been more done with squaring (and thus square rooter) circuits done in the not too distant past that you may also want to look into, as well as ready made IC chips that do this function quite well (like 0.3 percent accuracy or around that). I still want to look over the circuits you've posted though.

Back a bit later today...

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MrAl said:

Hello again,

The other circuit with U7 in it is a more straight forward approach to log, 0.5x, antilog, as you noted. It requires a DC bias because the diode doesnt really do a natural log, it does a natural log times a constant (using an approximated diode model), and another scaling constant comes in so we dont have to operate the diode at ridiculous currents like 4 amps and 10 amps, and a constant related to the particular diode, so the bias makes up for three constants. The center section op amp also divides the signal by 2, which provides the division required for the square root. You'll have to double check that part of the circuit to make sure it doesnt oscillate with a gain that low.

Because that lower section of U7 only computes a constant from another constant, it can be replaced by a 0.6v DC voltage reference, which you may want to make adjustable from say 0.5v to 0.7v or so. We can come up with an adjustment strategy if required.

If you'd like to look into the math behind this we can do more of that too. It starts from the diode model equations.

Click to expand...

MrAl said:

Hi again drpower,

One thing i almost forgot to mention is that with that lower U7 op amp section, if that is changed to a constant voltage you may loose some temperature compensation. That means the circuit may drift more with temperature. That's if they used that section to help temperature compensate that is.

Now on to the feedback system with the squaring circuit in the feedback...

See the attached diagram. First we'll look at this intuitively, then mathematically. Im saving the log anti log circuit theory for next time because this got quite large.



The upper circuit is a general feedback system. The input is applied at 'a' and the output
appears at 'x'. 'A' is the feedforward gain, 'f(x)' is the feedback function. f(x) takes
'x' as input and the output of f(x) appears at 'b'. The round circle is a subtractor,
which computes the difference a-b and the difference appears at 'e' so:
e=a-b

This gets amplified by the amplification factor 'A' and we get 'x':
x=A*e

Since x feeds f(x):
b=f(x)

That's the general system. When f(x) is a squaring circuit, then f(x) becomes:
f(x)=x^2

and we get the system shown on the bottom.

To view how this works intuitively, we start with all variables equal to zero.
This means a=0, b=0, e=0 and x=0. We'll start with a gain for A of 1000.

Now we apply an input at 'a' as some positive voltage, say 4 volts. This means
e jumps up to 4v because b is still zero. That 4v gets amplified to 4000v
(remember this is theory, in practice this would saturate) but x doesnt appear
as 4000v right away because there is also a small delay in A, so say we see
initially 1v at x (it ramps up as time goes on depending on the speed of the
op amp). That 1v now appears at the input of x^2 and so b becomes 1^2=1v.
Now we still have a equal to 4v but now b is equal to 1v, so e changes to
4-1=3v. Note that e lowered a little already. By this time, x becomes larger.
Say it went up to 1.1v in that time. Now the input of x^2 is 1.1v and so
b becomes 1.1^2=1.21v. Note now b is even larger than before.
Now the output at e is 4-1.21=2.79v and so it went down a little more.
x is still ramping up, so say now it became 1.2v. That means b becomes 1.44v,
and this process continues until the circuit reaches equilibrium, which is
when e*A=1.9995000625 approximately. That's not exactly the square root of 4,
but it is close. The reason it is not exact is because A is not infinite in
a real life op amp.
If we let A get larger, say 10000, we get a better result at the output:
1.999950000625v which is closer, and if we make A ten times larger (100000)
we get 1.99999500000625 which is even closer yet.
So we see that the larger we make the op amp gain A, the better result we get.
That's the basic way negative feedback systems work.


Now we'll calculate this system output using some math...

Following the signal flow shown by the arrows from left to right, we have:
e=a-b
x=A*e=A*(a-b)

and for the lower feedback section we have:
b=x^2

so inserting this last equation into the previous one for x we get:
x=A*(a-x^2)

We now expand this and get:
x=A*a-A*x^2

and collect terms on the left we get:
x-A*a+A*x^2=0

and this is just a quadratic in x so we solve that and get two possible solutions:
x1=-(sqrt(4*a*A^2+1)+1)/(2*A)
x2=(sqrt(4*a*A^2+1)-1)/(2*A)]

In terms of the op amp, A is the open loop gain typically very large, so the first
solution will be negative so the second solution is the correct solution here:
x=(sqrt(4*a*A^2+1)-1)/(2*A)

Since A is very large typically 100000 for most op amps, in theory we make this infinite,
so we take the limit of this last equation as A approaches +infinity and we get:
x=sqrt(a)

which is simply the square root of the input. Thus, we've managed to use a squaring
circuit to obtain the square root.


Next we can look at the diode equations and the theory behind the log antilog circuits.

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