Lab 23 RS

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Inverting OP Amp

1. Construct the inverting amplifier according to the wiring diagram below.

Here is the data sheet for the 741 Op Amp

File:LM741CN OpAmp.pdf

Use [math]R_1 = 1k\Omega[/math] and [math]R_2 = 10 k\Omega[/math] as starting values.

2. Insert a 0.1 [math]\mu[/math]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 [math]R_2[/math] = 10 k[math]\Omega[/math], 100 k[math]\Omega[/math], 1M[math]\Omega[/math]. Keep [math]R_1[/math] at [math]1k\Omega[/math].

I have used the following values of [math]R_1[/math] and [math]R_2[/math] (as was suggested by Dr Forrest at the lecture)

[math]R_1 = (10.02 \pm 0.02)\ k\Omega[/math]
a) [math]R_2 = (99.0 \pm 0.2)\ k\Omega[/math]
b) [math]R_2 = (198.5 \pm 0.2)\ k\Omega[/math]
c) [math]R_2 = (800.0 \pm 2.0)\ k\Omega[/math]

So my theoretical gain of OP Amp would be:

a) Gain1[math]= \frac{R_2}{R_1} = \frac{99.0 \pm 0.2}{10.02 \pm 0.02} = (9.88 \pm 0.03)[/math]

b) Gain2[math]= \frac{R_2}{R_1} = \frac{198.5 \pm 0.2}{10.02 \pm 0.02} = (19.81 \pm 0.04)[/math]

c) Gain3[math]= \frac{R_2}{R_1} = \frac{800.0 \pm 2.0}{10.02 \pm 0.02} = (79.84 \pm 0.26)[/math]

Below is my measurements and gain calculation for the case a) [math]R_1 = (10.02 \pm 0.02)\ k\Omega[/math] and [math]R_2 = (99.0 \pm 0.2)\ k\Omega[/math]

Below is my measurements and gain calculation for the case b) [math]R_1 = (10.02 \pm 0.02)\ k\Omega[/math] and [math]R_2 = (198.5 \pm 0.2)\ k\Omega[/math]

Below is my measurements and gain calculation for the case c) [math]R_1 = (10.02 \pm 0.02)\ k\Omega[/math] and [math]R_2 = (800.0 \pm 2.0)\ k\Omega[/math]

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

[math]G_{dB} \left(\frac{V_{out}}{V_{in}}\right) = 20\cdot \log_{10} {\frac{V_{out}}{V_{in}}}[/math]

Here the error calculation as usual and for this specific case is:

[math]dG_{dB}(x) = \frac{\partial G_{dB}(x)}{\partial x}\cdot dx = \frac{20}{x\ \ln 10}\cdot dx[/math]

where [math]x[/math] and [math]dx[/math] are corresponding gain and error of gain from the tables above

Impedance

Input Impedance

1. Measure [math]R_{in}[/math] 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:

[math]R_{inp} = \frac{V_{inp}}{I_{inp}}[/math]

and from my real circuit inside the shaded region:

[math]I_{inp} = \frac{V_{inp}-V_1}{R_1}[/math]

so finally my input impedance becomes:

[math]R_{inp} = \frac{V_{inp}}{V_{inp}-V_1}\ R_1[/math]

Below is the table with my measurements and input impedance calculations for four asked different cases. Here I have used the meter instead of oscilloscope to reduce the error in my measurements of input voltage.

As we can see the input impedance equals the resistor value [math]R_1[/math] for low frequency [math]f=100\ Hz[/math] and increase up to [math]33\ k\Omega[/math] for high frequency [math]f=10\ kHz[/math].

Output Impedance

1. Measure [math]R_{out}[/math] for the 10 fold and 100 fold amplifier at ~100 Hz and 10 kHz frequency. Be sure to keep the output ([math]V_{out}[/math]) undistorted

Again the equivalent circuit I am going to use is:

And my output impedance is:

[math]V_{out} = V - I_{out}\cdot R_{out}[/math]

But now I am going to use the load resistor [math]R_L[/math] to measure the output circuit:

[math]I_{out} = \frac{V_{out}}{R_L}[/math]

By graphing the current on the x-axis and the measured voltage [math]V_{out}[/math] on the y-axis for several values of the load resistance [math]R_L[/math] we can find the output internal impedance of our amplifier as the slope of the line [math]V_{out}=V_A - I_{out}\cdot R_{out}[/math]

Below are my measurements and current calculation for different cases. Here I have used the meter instead of oscilloscope to reduce the error in my measurements of output voltage.

Below are the plots of the [math]V_{out}[/math] as function of [math]I_{out}[/math]

As it follows from the plots above the output impedance (the slope of the line) are:

1) 10 fold amplifier at ~100 Hz frequency: [math]R_{out} = (8.68\cdot 10^{-11} \pm 112.3)\ \Omega[/math]
2) 10 fold amplifier at ~10 kHz frequency: [math]R_{out} = (6.70\cdot 10^{-11} \pm 120.4)\ \Omega[/math]
3) 80 fold amplifier at ~100 Hz frequency: [math]R_{out} = (-1.65\cdot 10^{-11} \pm 136.7)\ \Omega[/math]
3) 80 fold amplifier at ~10 kHz frequency: [math]R_{out} = (8.35\cdot 10^{-11} \pm 218.8)\ \Omega[/math]

As we can see the output impedance is essentially the zero. Unfortunately the calculated error is up to [math]220.\ \Omega[/math]. To get the better results I need dramatically improve my measurements, and particularly, improve the measured error in [math]V_{out}[/math]

[math]V_{io}[/math] and [math]I_{B}[/math]

[math]V_{out}= -\frac{R_2}{R_1} V_{in} + \left ( 1 + \frac{R_2}{R_1}\right)V_{io} + R_2 I_B[/math]

Use the above equation and two measurements of [math]V_{out}[/math], [math]R_1[/math], and [math]R_2[/math] to extract [math]V_{io}[/math] and [math]I_B[/math]. I will use two different values of [math]R_2[/math] to make two different measurements. [math] V_{in}[/math]=0 (grounded).

Below are my measurements. I have used the meter instead of oscilloscope to reduce my measured error in measured small output voltage.

1) [math]f=1\ kHz[/math] [math]R_1 = (10.06 \pm 0.01)\ k\Omega[/math], [math]R_2 = (99.5 \pm 0.1)\ k\Omega[/math]:  [math]V_{in}=(0.05 \pm 0.05)\ mV[/math] [math]V_{out}=(6.9 \pm 0.05)\ mV[/math]
2) [math]f=1\ kHz[/math] [math]R_1 = (10.06 \pm 0.01)\ k\Omega[/math], [math]R_2 = (199.6 \pm 0.1)\ k\Omega[/math]:  [math]V_{in}=(0.05 \pm 0.05)\ mV[/math] [math]V_{out}=(14.1 \pm 0.05)\ mV[/math]

Now I can construct 2 equations with 2 unknowns [math]V_{io}[/math] and [math]I_B[/math].

[math] (1 + (99.5 \pm 0.1)/(10.06 \pm 0.01))\cdot V_{io} + (99.5 \pm 0.1)\ k\Omega \cdot I_B = (6.9 \pm 0.05)\ mV[/math]

[math] (1 + (199.6 \pm 0.1)/(10.06 \pm 0.01))\cdot V_{io} + (199.6 \pm 0.1)\ k\Omega \cdot I_B = (14.1 \pm 0.05)\ mV[/math]

or

[math] (10.891 \pm 0.014)\cdot V_{io} + (99.5 \pm 0.1)\cdot 10^3\ I_B = (6.9 \pm 0.05)\cdot 10^{-3}[/math]

[math] (20.841 \pm 0.022)\cdot V_{io} + (199.6 \pm 0.1)\cdot 10^3\ I_B =(14.1 \pm 0.05)\cdot 10^{-3}[/math]

with [math]V_{io}[/math] given in Volts and [math]I_B[/math] given in Amperes.

To solve these equations will use the matrix method. Let's do it in general to be able to handle the error propagation. We have:

[math]\left( \begin{array}{cc} a_1 & b_1\\ a_2 & b_2 \end{array} \right)\left( \begin{array}{c} V_{io} \\ I_B \end{array} \right) = \left( \begin{array}{c} c_1 \\ c_2 \end{array} \right)[/math]

and the two solutions are:

[math]V_{io} = \frac{\left| \begin{array}{cc} c_1 & b_1\\c_2 & b_2 \end{array} \right| }{\left| \begin{array}{cc} a_1 & b_1\\a_2 & b_2 \end{array} \right| } = \frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1}\ \ [/math]: [math]I_B = \frac{\left| \begin{array}{cc} a_1 & c_1\\a_2 & c_2 \end{array} \right| }{\left| \begin{array}{cc} a_1 & b_1\\a_2 & b_2 \end{array} \right| } = \frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}[/math]

Substituting the corresponding coefficients:

[math]V_{io} = \frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1} = \frac{(6.9 \pm 0.05)\cdot 10^{-3}\cdot (199.6 \pm 0.1)\cdot 10^3 - (14.1 \pm 0.05)\cdot 10^{-3}\cdot (99.5 \pm 0.1)\cdot 10^3}{(10.891 \pm 0.014)\cdot (199.6 \pm 0.1)\cdot 10^3 - (20.841 \pm 0.022)\cdot (99.5 \pm 0.1)\cdot 10^3}[/math]

[math]I_B = \frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1} = \frac{(10.891 \pm 0.014)\cdot (14.1 \pm 0.05)\cdot 10^{-3} - (20.841 \pm 0.022)\cdot (6.9 \pm 0.05)\cdot 10^{-3}}{(10.891 \pm 0.014)\cdot (199.6 \pm 0.1)\cdot 10^3 - (20.841 \pm 0.022)\cdot (99.5 \pm 0.1)\cdot 10^3}[/math]

and by doing math and handling the error propagation we find:

[math]V_{io} = (-0.257 \pm 0.114)\ mV [/math]
[math]I_B = (97.5 \pm 12.0)\ nA [/math]

[math]I_{io}[/math]

Now we will put in a pull up resistor [math]R_3= \frac{R_1 R_2}{R_1+R_2}[/math](measured) as shown below.

I have constructed [math]R_3[/math] from two resistor in parallel identical to my [math]R_1[/math] and [math]R_2[/math] used in my initial circuit.

a) [math]R_3= (10\ k\Omega)||(100\ k\Omega) = (9.20 \pm 0.01)\ k\Omega[/math](measured)
b) [math]R_3= (10\ k\Omega)||(200\ k\Omega) = (9.66 \pm 0.01)\ k\Omega[/math](measured)

Instead of the current [math]I_B[/math] we have the current [math]I_{io}[/math]

[math]V_{out}= -\frac{R_2}{R_1} V_{in} + \left ( 1 + \frac{R_2}{R_1}\right)V_{io} + R_2 I_{io}[/math]

Use the same technique and resistors from the previous section to construct 2 equations and 2 unknowns and extract [math]I_{io}[/math], keep [math]V_{in}[/math]=0.

Below are my measurements. I have used the meter instead of oscilloscope to reduce my measured error in measured small output voltage. [math]f=1\ kHz[/math].

1) [math]R_1 = (10.06\pm0.01)\ k\Omega[/math],[math]R_2 = (99.5\pm0.1)\ k\Omega[/math],[math]R_3 = (9.20\pm0.01)\ k\Omega[/math]:  [math]V_{in}=(0.05 \pm 0.05)\ mV[/math],[math]V_{out}=(-1.5 \pm 0.05)\ mV[/math]
2) [math]R_1 = (10.06\pm0.01)\ k\Omega[/math],[math]R_2 = (199.6\pm0.1)\ k\Omega[/math],[math]R_3 = (9.20\pm0.01)\ k\Omega[/math]:  [math]V_{in}=(0.05 \pm 0.05)\ mV[/math],[math]V_{out}=(-2.8 \pm 0.05)\ mV[/math]

and to equations becomes:

[math] (10.891 \pm 0.014)\cdot V_{io} + (99.5 \pm 0.1)\cdot 10^3\ I_{io} = (-1.5 \pm 0.05)\cdot 10^{-3}[/math]

[math] (20.841 \pm 0.022)\cdot V_{io} + (199.6 \pm 0.1)\cdot 10^3\ I_{io} =(-2.8 \pm 0.05)\cdot 10^{-3}[/math]

with [math]V_{io}[/math] given in Volts and [math]I_B[/math] given in Amperes.

Substituting the corresponding coefficients into general solutions derived above:

[math]V_{io} = \frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1} = \frac{(-1.5 \pm 0.05)\cdot 10^{-3}\cdot (199.6 \pm 0.1)\cdot 10^3 - (-2.8 \pm 0.05)\cdot 10^{-3}\cdot (99.5 \pm 0.1)\cdot 10^3}{(10.891 \pm 0.014)\cdot (199.6 \pm 0.1)\cdot 10^3 - (20.841 \pm 0.022)\cdot (99.5 \pm 0.1)\cdot 10^3}[/math]

[math]I_{io} = \frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1} = \frac{(10.891 \pm 0.014)\cdot (-2.8 \pm 0.05)\cdot 10^{-3} - (20.841 \pm 0.022)\cdot (-1.5 \pm 0.05)\cdot 10^{-3}}{(10.891 \pm 0.014)\cdot (199.6 \pm 0.1)\cdot 10^3 - (20.841 \pm 0.022)\cdot (99.5 \pm 0.1)\cdot 10^3}[/math]

and by doing math and handling the error propagation we find:

[math]V_{io} = (-0.208 \pm 0.112)\ mV [/math]
[math]I_{io} = (7.7 \pm 11.2)\ nA [/math]

So by properly choosing the resistor values [math]R_3[/math] I was able to reduce my input offset current [math]I_{io}[/math] from about [math]100\ nA[/math] (case without [math]R_3[/math]) down to about [math]8\ nA[/math] (case with [math]R_3[/math]). By doing that I did my amplifier more perfect (input current for perfect OP AMP is zero). Unfortunately my error for [math]I_{io}[/math] becomes compared with the calculated values of the quantity by itself. But that is because the used method was (the error really comes from the solution of equations). Probably using different method to calculate [math]I_{io}[/math], for example using scope to measure input voltage and current, we can reduce the measured error. But even with error calculated above we can clearly see the relative reductions of the input offset circuit.

The offset Null Circuit

1) Construct the offset null circuit above.

I have used the gain 20 case with resistors values from previous section:

[math]R_1 = (10.06\pm0.01)\ k\Omega[/math], [math]R_2 = (199.6\pm0.1)\ k\Omega[/math], [math]R_3 = (9.20\pm0.01)\ k\Omega[/math]

2) Adjust the potentiometer to minimize [math]V_{out}[/math] with [math]V_{in}=0[/math].

3) Use a scope to measure the output noise.

My initial values of [math]V_{out}[/math] was [math]V_{out}=(-2.8 \pm 0.05)\ mV[/math]. Using potentiometer I was able to reduce my [math]V_{out}[/math] up to zero values. Basically doing that I have reduced my input offset voltage [math]V_{io}[/math] so my amplifier becomes more perfect.

My measured output noise at this point becomes:

[math](\Delta V_{out})_{RMS} = \frac{1.0\ mV}{2\sqrt{2}} = 0.35\ mV[/math]

It's difficult to see the noise reduction because my relative compensation of [math]V_{out}[/math] was small. But if I will start with big initial [math]V_{out}[/math] (for example by adjusting potentiometer) we can observe that noise becomes smaller as we reduce the absolute value of [math]V_{out}[/math]

Capacitors

Revert back to the pull up resistor

Capacitor in parallel with [math]R_2[/math]

1) Select a capacitor such that [math]\frac{1}{\omega C_2} \approx R_2[/math] when [math]\omega[/math]= 10 kHz.

[math]C_2 \geq \frac{1}{\omega R_2} = \frac{1}{2\pi\times 10\ kHz\times 199.6\ k\Omega} \approx 80 \ pF [/math]

I will pick up:

[math]C_2 = 100\ pF[/math]

so my break point becomes:

[math]f = \frac{1}{2\pi R_2 C_2} = \frac{1}{2\pi\times 199.6\ k\Omega\times 100\ pF} = 8.0\ kHz[/math]

2) Add the capacitor in parallel to [math]R_2[/math] so you have the circuit shown above. 3) Use a pulse generator to input a sinusoidal voltage [math]V_{in}[/math] 4) Measure the Gain as a function of the [math]V_{in}[/math] frequency and plot it.

Below my measured voltages and gain calculation.

Below my plot of gain as function of frequency. Also I overplay here my previous measurements of gain without capacitor [math]C_2[/math].

As we can see the gain are dropping down at about calculated above frequency. It's can be easily understood. At this point the capacitor starts to conduct AC signal reducing the effective resistors [math]R_2[/math] value. So at hight frequency the gain = [math]\frac{R_2}{R_1}[/math] starts to drop below the usual value without capacitor. It's sometimes very useful because by decreasing the gain at high frequencies we can prevent unwanted possible oscillations at this point ,

Capacitor in series with [math]R_1[/math]

1) Select a capacitor such that[math] \frac{1}{\omega C_1} \approx R_1[/math] when [math]\omega[/math]= 1 kHz.

[math]C_1 \approx \frac{1}{\omega R_1} = \frac{1}{2\pi\times 1\ kHz\times 10.0\ k\Omega} \approx 16 \ nF [/math]

I will pick up:

[math]C_1 = 15\ nF[/math] (can compose like 10nF || (10nF + 10nF))

and my break point becomes:

[math]f = \frac{1}{2\pi R_2 C_2} = \frac{1}{2\pi\times 10.0\ k\Omega\times 15\ nF} = 1.06\ kHz[/math]

2) Add the capacitor in series to [math]R_1[/math] so you have the circuit shown above. 3) Use a pulse generator to input a sinusoidal voltage [math]V_{in}[/math] 4) Measure the Gain as a function of the [math]V_{in}[/math] frequency and plot it.

Below my measurements and gain calculation:

Below my plot of gain as function of frequency. Also I overplay here my previous measurements of gain without capacitor [math]C_2[/math].

As we can see the gain starts to go up at about 1 kHz, reaches maximum at about 7 kHz and then goes down. It can be understood as follow. At break point frequency calculated above the capacitor [math][/math] starts to conduct the AC signal very well. So we are effectively reduce the [math]R_1[/math] resistor values and as follow increase the gain = [math]\frac{R_2}{R_1}[/math]. As frequency goes up the competing process of decreasing gain starts to work. And at some point the total gain starts to go down.

Slew rate

Measure the slew and compare it to the factory spec.

Power Supply Rejection Ratio

1. Set [math]V_{in}[/math] = 0.
2. Measure [math]V_{out}[/math] while changing [math]V_{cc}[/math]

Output voltage RMS noise [math]\Delta V_{out}^{RMS}[/math]

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