Lab 23 RS
Inverting OP Amp
1. Construct the inverting amplifier according to the wiring diagram below.
Here is the data sheet for the 741 Op Amp
Use and as starting values.
2. Insert a 0.1
F capacitor between ground and both Op Amp power supply input pins. The Power supply connections for the Op amp are not shown in the above circuit diagram, check the data sheet.Gain measurements
1.) Measure the gain as a function of frequency between 100 Hz and 2 MHz for three values of
= 10 k , 100 k , 1M . Keep at .
I have used the following values of and (as was suggested by Dr Forrest at the lecture)
a) b) c)
So my theoretical gain of OP Amp would be:
a) Gain1
b) Gain2
c) Gain3
Below is my measurements and gain calculation for the case a) and
Below is my measurements and gain calculation for the case b)
and
Below is my measurements and gain calculation for the case c) and
2.) Graph the above measurements with the Gain in units of decibels (dB) and with a logarithmic scale for the frequency axis.
Below my plot of gain as function of frequency. Here
Here the error calculation as usual and for this specific case is:
where
and are corresponding gain and error of gain from the tables aboveImpedance
Input Impedance
- Measure for the 10 fold and 100 fold amplifier at ~100 Hz and 10 kHz frequency.
I am going to measure the input and output impedance of my amplifier using the following equivalent circuit:
where the shaded region is my actual amplifier constructed before:
From equivalent circuit the input impedance is:
and from my real circuit inside the shaded region:
so finally my input impedance becomes:
Below is the table with my measurements and input impedance calculations for four asked different cases
As we can see the input impedance equals the resistor value for low frequency and increase up to for high frequency .
Output Impedance
- Measure for the 10 fold and 100 fold amplifier at ~100 Hz and 10 kHz frequency. Be sure to keep the output ( ) undistorted
Again the equivalent circuit I am going to use is:
And my output impedance is:
But now I am going to use the load resistor
to measure the output circuit:
By graphing the current on the x-axis and the measured voltage
on the y-axis for several values of the load resistance we can find the output internal impedance of our amplifier as the slope of the line
Below is my measurements and current calculation for the case f = 100 Hz, 10 kHz and gain = 10 (here nothing change with frequency)
Below is my measurements and current calculation for the case f = 100 Hz, 10 kHz and gain = 80 (here nothing change with frequency)
Below is the plots of the
as function of
As it follows from the plots above the output impedance (the slope of the line) are:
1) 10 fold at ~100 Hz and 10 kHz frequency:2) 100 fold amplifier at ~100 Hz and 10 kHz frequency:
As we can see the output impedance is essentially the zero. Unfortunately the calculated error is up to
. To get the better results I need dramatically improve my measurements, and particularly, improve the measured error in
I did some numerical calculations how the final errors depends on the measured error. Foe example if I will be able to improve my error measurements up to me results for the case of 80 gain are:
and my plot is:
So my output impedance becomes:
that is big improvement in my final answer.
On the other side the improvement in resistor error measurements doesn't change my final answer a lot.
and
- (need to be checked)
- (I have derived)
Use the above equation and two measurements of
, , and to extract and . I will use two different values of to make two different measurements. =0 (grounded).
Below are my measurements:
1), : 2) , :
Now I can construct 2 equations with 2 unknowns and .
or
with
given in Volts and given in Ampere.
To solve these equations I can use the matrix method. Let's do it in general to be able to handle the error propagation.
We have:
and the two solutions are:
- :
Substituting the corresponding coefficients in general solutions
and by doing math and handling the error propagation we find:
From results above we see that we have the reasonable small values for and but my error is too large to believe in results. The main source of error is when we are doing the difference operation in the formulas above (the results are small quantities which is less then total error). We need some special techniques to be able to measure this small quantities or we need some spacial instrumentations to be able to decrease the experimental error dramatically.
Now we will put in a pull up resistor
as shown below.Instead of the current
we have the current- (to be checked)
- (I have derived)
Use the same technique and resistors from the previous section to construct 2 equations and 2 unknowns and extract , keep =0.
Below are my measurements:
1), : 2) , :
and to equations becomes:
with
given in Volts and given in Amperes.
The two solutions are:
and by doing math and handling the error propagation we find:
The offset Null Circuit
- Construct the offset null circuit above.
- Adjust the potentiometer to minimize with .
- Use a scope to measure the output noise.
Capacitors
- Revert back to the pull up resistor
Capacitor in parallel with
- Select a capacitor such that when = 10 kHz.
- Add the capacitor in parallel to so you have the circuit shown above.
- Use a pulse generator to input a sinusoidal voltage
- Measure the Gain as a function of the frequency and plot it.
Capacitor in series with R_1
- Select a capacitor such that when = 1 kHz.
- Add the capacitor in series to so you have the circuit shown above.
- Use a pulse generator to input a sinusoidal voltage
- Measure the Gain as a function of the frequency and plot it.
Slew rate
Measure the slew and compare it to the factory spec.
Power Supply Rejection Ratio
- Set = 0.
- Measure while changing
Output voltage RMS noise
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