Feb 20: Operational Amplifiers

Operational Amplifiers

-Used for amplifying signals 

ex.) microphone and speaker
Microphone (high output impedance) vs. speaker (low input impedance) so the already small voltage by microphone further attenuated

Output Impedance 
-a measure of device's ability to provide current to load
-so lower the output impedance of source, better its ability to supply current to a load
-Summary: when load connected to source, output impedance of source must be smaller than input impedance of load, i.e. Zout << Zin

Output voltage
Vout = Gain * Vin, G = gain of voltage 


Differential Amplifiers
-to measure difference between two input signals
-minimizes unwanted "pickup" of signals:

So, here: Vout = G(V+ - V-)

Characteristics of an Ideal Differential Amplifier:

-large gain

-zero common mode gain (ideally, V+ = V- (golden rule which will be mentioned later). In other words, if two input nonzero identical voltages --> expect 0 V at output)

-large (infinite) input impedance and small (zero) output impedance. Again, Zout << Zin.

Operational Amplifiers

-class of amplifiers that almost follow the aforementioned characteristics and behaviors

-require both positive and negative power supply

-We used 12 V (-12 V and 12 V) supply, but what we amplify is the difference between the two input voltages (non-inverting (+) and inverting (-)). The difference is ideally very small, but the gain factor is so large that the signal is able to be amplified to a significant (usable) amount.

Open Loop Gain of an Op Amp, Experimentally (Lab 8-1)

-Decoupling: decoupled power supplies (12 V) with small ceramic capacitors (0.1 uF) placed as close to op amp as possible to keep power line constant locally (and avoid "parasitic oscillations"). If placed further away, other high frequency along the way may interfere with stability of power supply. 

Our circuit

-Used a 10 k potentiometer to vary the resistance

-When applied 0 V, got 12 V as output (because we can't get exactly 0 V, and even a small difference in input voltage gets amplified because of the large gain).

-The 411 specification claimed "Gain = 200 V/ 1mV", meaning 1 mV of input would result in 200 V of output (or proportional to that), but we couldn't get 200 V as output because we "hit" the rail (+12V/-12V power lines).

Feedback

-"feeding" some portion of output back into input of system

-positive vs. negative feedback

• positive feedback: output fed back to reinforce input 
• negative feedback: output fed back to partially "cancel" some of input; correcting influence to stabilize system (though this decreases the gain).

Op Amp Golden Rules

-Assume negative feedback.

1) With negative feedback in place, output of op amp will do what is necessary to keep voltage difference between inputs equal 0

i.e. V+ approx = V-

ex) Suppose V- begins to drift slightly before ground (so V- - V+ < 0). The output (Vout = -(R2/R1)*Vin) will generate positive voltage, which some is fed back to input and effectively raises V- back closer to ground.

2) Due to their very high input impedance, inputs of an op amp will neither "source" or "sink" appreciable currents. 

i.e. op amp does not draw any current

Inverting Amplifiers

-inverts and amplifies signal

How?

By Golden Rule #1: V+ = V- = 0 

(V+ = 0 because non-inverting input at ground, and by the rule, V+ - V- = 0).

Then, using Ohm's Law, Vin/R1 = I1

Golden Rule #2: I1 = I2 (no current going through inputs of op amp)

By Ohm's Law again:

Vout = 0 - I2*R2 = -(Vin/R1)*R2 

or Vout = -(R2/R1)* Vin

where Vout is inverted and is amplified by a gain factor of R2/R1.


-Even though Vout may theoretically be very large, the output swing is limited by the supply voltages. Max output voltage about 1 V less than the positive supply voltage and minimum output voltage about 1 V greater than the negative supply voltage. 

-Input impedance of inverting op amp configuration:
Zin = Vin/Iin = R1

Why?
Note the virtual ground on the circuit above. The feedback in this circuit keeps that node at 0 V (mentioned before when discussing negative feedback. Tries to keep Golden Rule #1). So the input voltage would be expressed as R1 * I1, in which R1 is the input resistance/impedance.
*Note that the op amp itself has a very high input impedance (Rule #2) so this configuration may not be the best if one needs a high input impedance.


Inverting Amplifier, Experimentally

Theoretical Input: 1 V

Theoretical Output: -10 V (because R2 = 10k and R1 =1k)

Gain =  10 V /1 V = 10

Experimental Input: 1.20 V

Experimental Output: -9.6 V

Experimental Gain: 9.6 V / 1.20 V = 8 

Linearity: maintains linearity; if input a straight line, output a straight line (so no differentiating, etc.) 
(shown by triangle wave)

Frequency Effects?
-At high frequencies (we tried 3 MHz), there was no amplification, and we also observed a phase shift.

Driving circuit again with 1 kHz
-Measured input impedance of this amplifier circuit by adding a 1 kOhm resistor in series with the input.
-Created a divider (without output signal inverted):

 where R1 = 1 kOhm

So: Vin = 1 V and Vout = -4.9. (The gain was cut in half: 5 vs. 10.)
This implied: R = 1.3 kOhms

Trying to Measure Output Impedance
-No blocking capacitor needed because resistor has very low resistance
-Zout very low
-We used a 1 Ohm resistor in series with op amp output--> Vout = 10 V
Without the resistor --> Vout = 10 V
-This shows that even with a low resistance resistor (1 Ohm), most of the voltage is dropped across the load, meaning the output impedance of this circuit is VERY low.

Non-Inverting Amplifier

Using the Golden Rule #1:
Vin = V- = V+
(V- = Vin and by rule, V+ - V- = 0)

By Ohm's Law: I1 = V-/R1 (= Vin/R1 because V- = Vin)

By Golden Rule #2:
I1 = I2 (no current thru inputs of op amp)

so I2 = Vin/R1

Then, Vout = V- + (I2)*R2  = Vin + (Vin/R1)*R2 = Vin(1+(R2/R1))



Note:
-Output signal in phase with input signal (non-inverted).
-Again, configuration has very low output impedance (for low signals).
-Because input signal connected to input of op amp, input impedance of configuration very high (Golden Rule #2) but not as stable as inverting amplifier at high gain.

Non-Inverting Amplifier, Experimentally




Rf = 10k
Rg = 1k
So expected gain = (1 + (10/1)) = 11

Experimental Data:
Input V = 2.16 V
Output V = 22.2 V 
Gain = 22.2 V/2.16 V = 10.3

Measuring Input Impedance
-at 1 kHz by putting a 1 MOhm resistor in series with the input

-We placed the probe at the output NOT the input because compared to the large input impedance, the resistance of the probe is lower, which means that what we would measure here would then be the 10 MOhms, the resistance of the probe not of the circuit. Therefore, we put the probe at the output, where the output impedance is lower than the resistance of the probe, meaning the output impedance would dominate. When we had input voltage of 1V, we got 11 V as the output. To find the f-3dB point, we looked for the frequency where the output voltage became 0.707 times its maximum, which was 7.7 V. We put the probe at the output because we can't technically measure the 3dB point at the input because of the relatively low resistance of the probe (vs. the input impedance of the circuit). So, the f-3dB was 28.4 kHz, which we used to calculate Cin (omega = 1/RC so C = 1/(R*omega)), which was 5.6 pF. Then, at 1 kHz, the output impedance was calculated to be (Z = 1/omega*C, where omega = 1 kHz) 2.84E7 Ohms or 28 MOhms. 


Op Amp Follower
-like considering non-inverting amplifier when R2 --> 0 and R1 --> inf (with gain of 1)
-called a follower because output signal a replica of input signal ("follows" whatever input is doing)
-this configuration has a very high input impedance and low output impedance. Thus, it acts as a buffer between the source and load, which makes it easier for the source to drive the load without attenuation.
In other words, if we add a load to the output of the follower, the source can drive the load with minimal attenuation because the follower is like the non-inverting amplifier without the resistors.
-nearly ideal follower but not able to supply large currents


Op Amp Follower, Experimentally

Its Zin and Zout are identical to those of the op amp non-inverting amplifier without amplification (because gain = 1).

Output Impedance
-really small
-Why is it so small? due to feedback
-Tried adding a 1 kOhm resistor in series with the output of the follower

Trials:

1. Feedback #1 without load --> no change in signal

2. Feedback #1 with load --> essentially 2 kOhms outside the input --> R-load = 2kOhms --> signal cut in half

3. Feedback #2 without load --> no change in signal because all of voltage drop to load --> Rth very small

4. Feedback #2 with load --> no change in signal --> all of voltage drop to load --> Rth very small

These trials show that the output impedance is very low due to the feedback. Even after adding the 1 kOhm resistor (not load), if it's "inside" the configuration (with feedback loop around it), there would be no change in signal, implying almost all of the voltage drop is across the load.

                                     Trials 1, 2, 3                                    Trial 4