# Determine Relative Error from 741 Output

#### Ian787

Joined Aug 1, 2013
7
Hello all

I wonder if someone would be able to help me

I have the following circuit This circuit is a thermocouple which is outputs the voltage V8 at 600°C induces a voltage of 45.093mV which is to be amplified to output a voltage of 6Volts. the circuit is tuned to the correct output voltage by potentiometers R4 and R2.

V4 is the reference junction of the thermocouple which is held at 25°C which induces a voltage of 1.495mV.

The tolerance of all resistors in your circuit is ±1% of their normal value. The digital display has a quantitation error of ±10mV. The tolerance of the EMF of the thermal couple is ±0.75% of its table value. Assume all errors are random and uncorrelated.

My question is how do I determine the % error at the output of the amplifier taking into account the above?

I think I need to take the errors of the components, display and thermocouple and add them together like this

Δε/ε = √(ΔVthermocouple/Vthermocouple)+(ΔR/R)+(ΔVDisplay/VDisplay)

But I am not completely sure. If there is anyone who is knowledgeable on error it would be greatly appreciated

Many thanks

#### Ian787

Joined Aug 1, 2013
7
Hi bertus

Apparently, according to my exercise. The offset of the op-amp can be neglected on this occasion.

Do you happen to know how to calculate the relative error?

#### MrAl

Joined Jun 17, 2014
10,871
Hello there,

I believe what you are after can be found using the total differential.

The error for a three variable problem for example, where we want to calculate the max error in the diagonal of a box given the length, width, and height, and the max error in each of these is 5 percent, would go like this...

In the following Dx, Dy, and Dz, are the non infinitesimal changes and are usually written like this:
Δx, Δy, Δz

or in words:
"delta x", "delta y", "delta z".

First calculate the diagonal based on the given sizes, x0,y0, and z0:
Fo=sqrt(x0^2+y0^2+z0^2)

Now take the three partial derivatives, Fx, Fy, and Fz:
Fx=x0/Fo
Fy=y0/Fo
Fz=z0/Fo

Now compute the total differential:
dF=Fx*dx+Fy*dy+Fz*dz

which can be written:
dF=dx*x0/Fo+dy*y0/Fo+dz*z0/Fo

and now divided by Fo:
dF/Fo=(1/Fo^2)*(x0*dx+y0*dy+z0*dz)

and the percent error in each size is:
Dx/x0, Dy/y0, Dz/z0

and these are roughly equal to:
dx, dy, dz

respectively, so we can rewrite dF/Fo as:
dF/Fo=(1/Fo^2)*(x0^2*Dx/x0+y0^2*Dy/y0+z0^2*Dz/z0)

and the max error is the abs() of that, so we have:
|dF/Fo|<=(1/Fo^2)*(x0^2*abs(Dx/x0)+y0^2*abs(Dy/y0)+z0*abs(Dz/z0))

so for the percentages given we have:
|dF/Fo|<=(1/Fo^2)*(x0^2*0.05+y0^2*0.05+z0*0.05)

and so:
|dF/Fo|=0.05

which is the maximum percentage error in the calculation of the diagonal.

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• Ian787

#### Ian787

Joined Aug 1, 2013
7
Hi Mr Al

Thank you so much for your reply, this has been a big help!

In the part where you said:

Fo=sqrt(x0^2+y0^2+z0^2)

Does this simply add together all the error? In my research I have found other sources that obtain the error by using the that equation

#### WBahn

Joined Mar 31, 2012
29,469
What you are looking for, and what MrAl has largely provided for you, is called "propagation of errors". Look for information on that and study it.

Basically you want to determine how the output changes with a small change from each component and then, because they being treated as random and uncorrelated, you add them orthogonally using the Pythagorean sum.

• Ian787

#### MrAl

Joined Jun 17, 2014
10,871
Hi Mr Al

Thank you so much for your reply, this has been a big help!

In the part where you said:

Fo=sqrt(x0^2+y0^2+z0^2)

Does this simply add together all the error? In my research I have found other sources that obtain the error by using the that equation
Hello,

The square root of sum of squared errors is a different method. This method is the approximation of the max error by the total differential.

Fo is the function. The function is what you are investigating for total error.
For the box, we could define:
x0=width
y0=length
z0=height
and then we want to know the error in the 3d diagonal so we calculate that diagonal:
Fo=sqrt(x0^2+y0^2+z0^2)

and this Fo is the diagonal and we want to know what the error is in that when we have errors in the three dimensions x0, y0, and z0.

Let us do another example though which will not have that same function.

We have a cone, height h0 and radius r0. We want to know the total error in the volume when the measured height can be off by at most 1 percent, and the measured radius can be off by at most 2 percent.

First we need to come up with a formula for what it is we are concerned with. In this case, it is the volume of the cone which is:
V=(1/3)*pi*r^2*h

We will be working with this function now which as you see is different than before with the box which was sqrt(x^2+y^2+z^2).

The initial values of r and h are called r0 and h0, and sometimes these values are given but here they are not.
So now we have:
Vo=(1/3)*pi*r0^2*h0

Ok, so we need first partials with r0 and h0, so first we do:
pr0=dVo/dr0=(2*pi*h0*r0)/3
ph0=dVo/dh0=(pi*r0^2)/3

Next we form part of the result as:
pr0*dr+ph0*dh

Now Dr=dr/r0, and Dh=dh/h0, and we want to substitute Dr and Dh for dr and dh, and to maintain equality we have to multiply Dr*r0 first and multiply Dh*h0 so we end up with:
pr0*Dr*r0+ph0*Dh*h0

We then just divide by Vo:
(pr0*Dr*r0+ph0*Dh*h0)/Vo

Recalling Vo, pr0, and ph0 and substituting those into this equation and simplifying, we get:
2*Dr+Dh

Before we go any further, we can already see that the error in the radius affects the total error twice as much as the error in the height does.

Finally since we said that the radius can vary by at most 2 percent and the height 1 percent, we have:

2*Dr+Dh=2*0.02+0.01=0.05

so the error in volume is at most 5 percent.

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• Ian787