Feb 26: Op Amp (Part II)

Op Amp Current Source

Looking at the current through the variable resistor (R2):

Following the Golden Rule, V- = V+ = Vin.

Using Ohm's Law: I1 = V-/R1 which must equal I1 using Golden Rule (#2) since current doesn't go through inputs of the op amp. 

So: I2 = V-/R1. 

Here, the current flowing through R2 is independent of value of R2 over a wide range of resistances, which means this is a good (relatively constant) current source as the resistance of the load changes (vs. voltage source e.g. batteries which are good at maintaining a constant voltage across a load resistor independent of the load's resistance).

Op Amp Current Source, Experimentally (8-5)

Vin  = (1k/(15k + 1k))*12 V = (1/16)*12 V = 0.75 V = V+ = V- (Golden Rule)

V- = IR 
0.75 V = I (180)  --> I = 0.004 amps or 4 mA

Observations:
-When R-load = 0 Ohms, configuration like a follower. 
-The current was calculated to be constant at around 4 mA.
-We observed 4.2 mA using the multimeter, but as we adjusted (increased) the resistance using the potentiometer, (or when R-load) became too large (where Vout > 12 V), we did not observe this constant current behavior and instead measured lower current values. 

Note: 
-This current source has a disadvantage of requiring a "floating" load (neither side connected to ground, like our 10 k load). 
-Also has significant speed limitations, which causes problems in situations in which either output current or Z-load varies quickly. 
-But precise and stable vs. our simple transistor current source


Photo-diodes
-when light absorbed at the p-n junction of a diode, photo-current generated (reverse process to LEDs)
-Size of photocurrent approximately linearly proportional to intensity of illuminating light so a diode can serve as a light detector
-To detect, we "converted" photocurrent to voltage
-But to use a resistor (for this I to V converter), 2 disadvantages: photodiode not a compliant current source as it can only sustain about 0.5 to 2 V. Also, as voltage is allowed to develop across the photodiode, response can vary (so can't really use that linear proportionality).

Solution? 
Phototransistor
-"photodiode with gain"
-when wired, each photon received causes about 100 electrons worth of current


Op Amp Current-to-Voltage Converter

-improved I-to-V converter
-small feedback resistor used to reduce sensitivity of device
-regulates/maintains current (controlled by light)
-also have a faster response time (weakness of photodiodes) by using a different type of junction ("PIN junction" vs. p-n junction)


Current-to -Voltage Converter, Experimentally

-Used a 10 k Ohm resistor for feedback loop to lower Vout (versus a 100 k to avoid saturation)

Average Input Photocurrent (middle/halfway of peak-to-peak amplitude)

Vout = -3.44 V = -Ipd (10000 Ohms)

Ipd = 0.34 mA

Percentage Modulation

-Measured from peak to peak over middle (of peaks) to the zero line

(-3.28 V / -3.44) * 100% = 96 %

-At the summing junction, Vx = 0. (Golden Rule #1)


Summing Amplifier

-Currents through input resistor add at the summing junction and flow through the feedback resistor (also proven by Golden Rule #2, in which there is no current going into the op amp inputs).

So: I-fb = I1 + I2

-Summing junction is a virtual ground (Golden Rule #1) so:

Vout = 0V + -IfbRfb = -(I1+I2)Rfb (Ohm's Law)

Vout = -(V1/R1 + V2/R2) Rfb using Ohm's Law again

And if R1 = R2 (Say they = R), then:

Vout = -(Rfb/R)*(V1 + V2) 

Here, we see that V1 and V2 and added with a gain factor, making this a summing amplifier that also inverts the signal. 

Question: Can you design a circuit that subtracts two voltages?

Using Golden Rules, Ohm's Law, and Superposition 

I1 = (V1 - V-)/R1

I2 = (V2 - V+)/R2

If = (V- - Vout)/R3

V+ = V- (Golden Rule)

Voltage Divider

V+ = (R4/(R2+R4))*V2

Now applying superposition (adding up Vout when V1 is short and when V2 is short to get Vout of the circuit shown above):

Case 1: V2 short (i.e. V2 = 0)

Vout(a) = -V1 (R3/R1) like feedback because I1 = If where I1 = V1/R1 and If = -V_out(a)/R3 (negative value because we assumed V2 = 0 so V+ = 0 so V- = 0)

Case 2: V1 short (i.e. V1 = 0)

Here, V+ = (R4/(R2+R4))*V2           

Also, V+ = V- = (R1/(R1+R3))*Vout_b

So: Vout(b) = ((R4/(R2+R4))*V2*((R1+R3)/R1)

Using superposition, Vout = Vout(a) + Vout(b) 

so:

Vout = -V1(R3/R1) + ((R4/(R2+R4))*V2*((R1+R3)/R1)

And if R1=R2=R3=R4,

Vout = -V1 + V2 or V2-V1

Summing Amplifier as a Digital to Analog Converter

-can use an op amp to build a digital to analog converter

-let 6 V represent binary 1 and 0 V represent binary 0

-Following circuit produced analog output that is proportional to 4-bit input number

Summing Amplifier, Experimentally

-circuit above sums a DC level with the input signal

I = I1 + I2 = V1/R1 + V2/R2 = 5V/10k + 12V/22k = 1 mA

Output signal offset range: +/- 10 V

Resistance of circuit:

((1/10k)+(1/(10k+22k)))^-1 + 10k = 6875 Ohms

Vout = -(10k/6875) (5V + 12V) = -24.7 V but rails so cutoff at around +/- 10.6 V

Op Amp Limitations

-Real op amps not like idealized versions

-mild violations

so modified golden rules:

1) With negative feedback, output of op amp will try to do whatever is necessary to keep the voltage difference between inputs equal to a very small difference (called offset voltage or Vos usually about a few mV).

2) Due to their very high input impedance, inputs of op amp will sink a very small current called input bias current (I-bias usually about 3 pA)

Another limitation: small amount of output current that they can supply (for 411, at most 25 mA to a load)

Also: open loop gain "rolls off" or becomes distorted at high frequencies... so gain curve:

-This roll off affects/limits op amp's ability to respond to high f signals resulting in a limited slew rate or limited "response time"

-Slew rate defined as the maximum rate at which the op amp's output can change

Note:

-roll off in gain intentional

-output of op amp fed thru a low pass filter with a f-3db of 100 Hz

-to keep op amp stable (or else subjected to spontaneous high f oscillations which could result in stray capacitance and resulting phase shifts... after feedback unintentionally becomes positive at high f, which the modified golden rule #1 doesn't support).

Op Amp Limitations, Experimentally

Slew Rate

1. Square wave input at 1 kHz... rapid rise and fall

Measured slew rate by looking at locally linear region then took slope (delta V/delta t)

slewing up:

delta V/delta t = 1.8 V/ 139.0 ns = 1.29E7 V/s

slewing down:

delta V/delta t = 2.0 V/ 139.0 ns = 1.5E7 V/s

which agreed with 411 claim of 15 V/us 

2. Sine input (not as rapid rise and fall)

Frequency at which output waveform began to distort: 100 kHz (qualitative; harder to see than it was with square waves)

Minimum slew rate = 2*pi*A*f = 2*pi*10V*100kHz = 6.3E6 V/s 

slewing up:

4V/5.56us = 7.2E5 V/s

slewing down: 

3.36V/5.56us = 6E5 V/s

Comparable to 741's claim of a typical slew rate of about 0.5 V/us (5E5 V/s).

Slew rate for square wave input greater (makes sense because of the more instantaneous or rapid change in voltage)

Op Amp Integrator

-Recall with an RC integrator circuit,

Vout << Vin for the circuit to be a decent integrator

-Op amp integrator removes that restriction.

Vout = -(1/RC)*integrate(Vin(t)dt)

holds true for a wide range of frequencies

Problem: Op amp works so well so if input signal has even a small dc offset, the integrating circuit will integrate over time to give a very large result, leading to op amp drifting to one of the supply rails

Solution: Add a large by-pass resistor into feedback loop. The DC current will flow through the resistor and not accumulate on the capacitor, avoiding the accumulation of the small dc offset mentioned above.

Op Amp Integrator, Experimentally

-unforgiving op amp integrator esp with bias currents of picoamps or offset voltages below a mV...

-integrator above sensitive to small dc offsets (gain at DC ~100)

1 kHz square wave

Vpp = 10V

Vout = 2.56 V (peak to peak) = area of input

f = 1 kHz

Z = 1/omegaC = 1.6E4 ohms

R= C ||10Mohms = 1.6E4 Ohms

Vout = (-1/(RC))*(0.5*delta t* height in V of one square in square wave), where delta t = period .... because impedance of capacitor is frequency dependent so Vout also freq or time dependent.

Using these equations, we were not able to get close to the experimental value, but we will work on this!

UPDATE/Corrected:

Our 1kHz square wave experiment worked as expected because the output voltage (peak to peak) should equal the area of one of the "squares" (since area of one "segment"/half of the triangle, which makes another (right) triangle corresponds to one of the squares of V-input.)

So: 

Vout = -1/(RC) * Area of square

Vout,calculated = (-1/RC) * 0.5 * (1/1kHz) * 5 V = -2.5 V.

(https://myclassbook.files.wordpress.com/2014/01/op-amp-integrator-waveforms.jpg)