Difference between revisions of "Lab 23 RS"

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As we can see the gain starts to go up at about 1 kHz, reaches maximum at about 7 kHz and then goes down. At 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.
+
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=
 
=Slew rate=

Revision as of 04:03, 22 April 2011

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Inverting OP Amp

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

TF EIM Lab23.png

Here is the data sheet for the 741 Op Amp

File:LM741CN OpAmp.pdf


Use R1=1kΩ and R2=10kΩ as starting values.

2. Insert a 0.1 μF 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 R2 = 10 kΩ, 100 kΩ, 1MΩ. Keep R1 at 1kΩ.


I have used the following values of R1 and R2 (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) R1=(10.02±0.02) kΩ and R2=(99.0±0.2) kΩ

Gain t01.png

Below is my measurements and gain calculation for the case b) R1=(10.02±0.02) kΩ and R2=(198.5±0.2) kΩ

Gain t02.png


Below is my measurements and gain calculation for the case c) R1=(10.02±0.02) kΩ and R2=(800.0±2.0) kΩ

Gain t03.png


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 x and dx are corresponding gain and error of gain from the tables above


Gain p03.png

Impedance

Input Impedance

  1. Measure Rin 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:

Draw01.png

where the shaded region is my actual amplifier constructed before:

Draw02.png

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.

T rinp01.png


As we can see the input impedance equals the resistor value R1 for low frequency f=100 Hz and increase up to 33 kΩ for high frequency f=10 kHz.

Output Impedance

  1. Measure Rout for the 10 fold and 100 fold amplifier at ~100 Hz and 10 kHz frequency. Be sure to keep the output (Vout) undistorted


Again the equivalent circuit I am going to use is:

Draw011.png

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 RL 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 Vout on the y-axis for several values of the load resistance RL we can find the output internal impedance of our amplifier as the slope of the line Vout=VAIoutRout


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.

T rout01.png


Below are the plots of the Vout as function of Iout


P rout 00.png


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 220. Ω. To get the better results I need dramatically improve my measurements, and particularly, improve the measured error in Vout

Vio and IB

Vout=R2R1Vin+(1+R2R1)Vio+R2IB

Use the above equation and two measurements of Vout, R1, and R2 to extract Vio and IB. I will use two different values of R2 to make two different measurements. Vin=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 Vio and IB.

(1+(99.5±0.1)/(10.06±0.01))Vio+(99.5±0.1) kΩIB=(6.9±0.05) mV
(1+(199.6±0.1)/(10.06±0.01))Vio+(199.6±0.1) kΩIB=(14.1±0.05) mV

or

(10.891±0.014)Vio+(99.5±0.1)103 IB=(6.9±0.05)103
(20.841±0.022)Vio+(199.6±0.1)103 IB=(14.1±0.05)103

with Vio given in Volts and IB 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:

(a1b1a2b2)(VioIB)=(c1c2)

and the two solutions are:

Vio=|c1b1c2b2||a1b1a2b2|=c1b2c2b1a1b2a2b1  : IB=|a1c1a2c2||a1b1a2b2|=a1c2a2c1a1b2a2b1


Substituting the corresponding coefficients:

Vio=c1b2c2b1a1b2a2b1=(6.9±0.05)103(199.6±0.1)103(14.1±0.05)103(99.5±0.1)103(10.891±0.014)(199.6±0.1)103(20.841±0.022)(99.5±0.1)103
IB=a1c2a2c1a1b2a2b1=(10.891±0.014)(14.1±0.05)103(20.841±0.022)(6.9±0.05)103(10.891±0.014)(199.6±0.1)103(20.841±0.022)(99.5±0.1)103

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]

Iio

Now we will put in a pull up resistor R3=R1R2R1+R2(measured) as shown below.

I have constructed R3 from two resistor in parallel identical to my R1 and R2 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)

TF EIM Lab23a.png

Instead of the current IB we have the current Iio

Vout=R2R1Vin+(1+R2R1)Vio+R2Iio


Use the same technique and resistors from the previous section to construct 2 equations and 2 unknowns and extract Iio, keep Vin=0.


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

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:

(10.891±0.014)Vio+(99.5±0.1)103 Iio=(1.5±0.05)103
(20.841±0.022)Vio+(199.6±0.1)103 Iio=(2.8±0.05)103

with Vio given in Volts and IB given in Amperes.


Substituting the corresponding coefficients into general solutions derived above:

Vio=c1b2c2b1a1b2a2b1=(1.5±0.05)103(199.6±0.1)103(2.8±0.05)103(99.5±0.1)103(10.891±0.014)(199.6±0.1)103(20.841±0.022)(99.5±0.1)103
Iio=a1c2a2c1a1b2a2b1=(10.891±0.014)(2.8±0.05)103(20.841±0.022)(1.5±0.05)103(10.891±0.014)(199.6±0.1)103(20.841±0.022)(99.5±0.1)103

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 R3 I was able to reduce my input offset current Iio from about 100 nA (case without R3) down to about 8 nA (case with R3). By doing that I did my amplifier more perfect (input current for perfect OP AMP is zero). Unfortunately my error for Iio 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 Iio, 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

TF EIM Lab23 b.png

1) Construct the offset null circuit above.

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

R1=(10.06±0.01) kΩ, R2=(199.6±0.1) kΩ, R3=(9.20±0.01) kΩ

2) Adjust the potentiometer to minimize Vout with Vin=0.

3) Use a scope to measure the output noise.

My initial values of Vout was Vout=(2.8±0.05) mV. Using potentiometer I was able to reduce my Vout up to zero values. Basically doing that I have reduced my input offset voltage Vio 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 Vout was small. But if I will start with big initial Vout (for example by adjusting potentiometer) we can observe that noise becomes smaller as we reduce the absolute value of Vout

Capacitors

Revert back to the pull up resistor

Capacitor in parallel with R2

TF EIM Lab23 c.png

1) Select a capacitor such that 1ωC2R2 when ω= 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 R2 so you have the circuit shown above. 3) Use a pulse generator to input a sinusoidal voltage Vin 4) Measure the Gain as a function of the Vin frequency and plot it.

Below my measured voltages and gain calculation.

Table gain c2 01.png


Below my plot of gain as function of frequency. Also I overplay here my previous measurements of gain without capacitor C2.

Plot gain c2.png

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 R2 value. So at hight frequency the gain = R2R1 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 R1

TF EIM Lab23 d.png

1) Select a capacitor such that1ωC1R1 when ω= 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 R1 so you have the circuit shown above. 3) Use a pulse generator to input a sinusoidal voltage Vin 4) Measure the Gain as a function of the Vin frequency and plot it.

Below my measurements and gain calculation:

Table gain c1.png


Below my plot of gain as function of frequency. Also I overplay here my previous measurements of gain without capacitor C2.

Plot gain c1.png


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 starts to conduct the AC signal very well. So we are effectively reduce the R1 resistor values and as follow increase the gain = R2R1. 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 Vin = 0.
  2. Measure Vout while changing Vcc

Output voltage RMS noise ΔVRMSout

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