Operational amplifiers are DC-coupled high-gain electronic voltage amplifier with a differential input (two inputs) and a single output. It produces an output voltage that is typically a hundred thousand times larger than the voltage difference between the input terminals, i.e. it is a differential amplifier.Op amps had their origins in analog computers, where they were used to do mathematical operations in many linear, non-linear and frequency-dependent circuits.
They solve the problem of connecting a component with high Rout, such as a microphone, to a component with low Rin, such as a loudspeaker, and still produce a high desirable Vout.
An ideal op-amp has the following important characteristics:
- Infinite differential mode gain
- Zero common mode gain
- Infinite input impedance
- Zero output impedance
These ideals can be summarized by the two "golden rules":
I. With negative feedback in place, the output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
II. The inputs draw no current.
Where Gdiff is the Gain of the amplifier, usually of the order of a few hundred thousands.
Op-amp 411 (HH Lab 8-1)
This exercise was mainly to learn to plug-in an op-amp in the board, and to understand its basic function.
We used +/- 12V as power supply for the Op-amp.
It was almost impossible to get the output to come to 0, since amplification of the signal by the op-amp is much greater than my ability to manipulate the brush of the potentiometer to go to the 0-point.
It was hard to verify the claim "G=200V/mV", however the gain was very huge.
From the scope, even though the signal going into V+ seems to be sitting at 0 (blue line), the Vout from the op-amp jumps between +11.8V and -10.2V with the slightest movement of the brush. This shows the huge gain in sensitivity gained from an op-amp circuit, and why it is useful for controlling some signal at a particular point (e.g. temperature control :).
The maximum output voltage will be ~1V less than the +ve supply voltage and the minimum will be ~1V greater than the -ve supply voltage. I noticed that the +ve and -ve power supplies were >12V and >-12V respectively. This is the reason why the Vout on the +ve side is >11V, and on -ve side is <-10V.
Inverting Amplifier (HH Lab 8-2)
Negative feedback acts as a "correcting" influence, stabilizing the system at the expense of lowering the gain of the system.
An inverting amplifier amplifies the differential signal, and puts a 180° phase change on it (hence the name).
Gain of the output is given by:
G = - R2/R1 = -10k/1k = -10
This can be verified from the sine wave on the scope, the Vout (yellow) is 10 times Vin (green) and is 180° out of phase. The maximum output swing is 20Vpp.
The gain remains -10 until the frequency is raised to ~100kHz, when the gain starts to drop-off (probably due to some internal component of the op-amp - transistor not fast enough). At 10 MHz, the sine wave is completely distorted. At frequency > 100 kHz, triangle wave starts to lose its linearity.
On adding a 1kΩ in series with the input, the input, as well as the output signal, becomes half. Showing that the input impedance is 1kΩ, which is not very high.
I tried putting 100Ω, 10Ω, and 1Ω in parallel with the output signal, but no visible change is noticed in the output signal. Showing that the output impedance is much less than 1Ω.
Non-inverting Amplifier (HH Lab 8-3)
A non-inverting amplifier has a bigger amplification and there is no phase difference between Vout and Vin.
G = 1 + R2/R1 =1+10k/1k = 11
From the scope, it can be seen that the gain is ~11, and the signals are in phase.
On connecting a 1MΩ resistor in series with the input signal, the input and output signal becomes half, showing that the input impedance is ~1MΩ.
Like the inverting amplifier, the output impedance in this case should also be very small, since they are essentially the same circuit.
I forgot to find the f3DB. But had I found it, the C can be calculated by
C = 1/2piRf3DB = 1/2pi x 0.5M x f3DB
C = 1/2piRf3DB = 1/2pi x 0.5M x f3DB
Follower (HH Lab 8-4)
This is like the non-inverting amplifier, but with R2 0 and R1 infinite.
G = 1 + 0 = 1
So, the Vout is the same as the Vin and in phase.
This can be seen on the scope, where Vout is in yellow and Vin is in green.
Op Amp Current Source (HH Lab 8-5)
This can be shown experimentally by monitoring the current across R2, and by changing its value.
In the HH Lab 8-5, we are asked to monitor the current in the branch by connecting a potentiometer in place of R2, with an ammeter in series. By varying the resistance, we should observe that the current through the ammeter remains fairly constant. However, we blew up the fuse in our ammeter, and Prof. T Bauer suggested we measure the potential drop across our resistor, and calculate the current.
V- = V+ = Vin = 1/16 x 15V = 0.9375VIR1 = V-/R1 = Vin/R1 = 0.9375V/180Ω = 5.21 mA
IR2 = IR1 = 5.21 mA
I replaced the potentiometer with a 1kΩ resistor, and connected a voltmeter across it. The reading was 4.25V. If I connect a second 1kΩ resistor across the first, I'll expect the voltage to become halved, if the current in the branch is constant. And thats exactly what we observed, the potential difference is 2.12V. I tested further with a few other resistors, and got the expected voltage drops, proving that the current in the branch is constant.
One thing to note, measured IR2 = 4.25V/1k = 4.25 mA, much smaller than IR1. This shows that 411 is not ideal, and some current seeps into the input of the op amp.
Current to Voltage Converter (HH Lab 8-6)
When light is absorbed at the p-n junction of a photodiode, a photo-current is generated, which is approximately linearly proportional to the intensity of the illuminating light. Hence a photodiode can be used as a light detector. The biggest output voltage of a photodiode is 0.5V. Note, that one end of the photodiode is grounded while the other end is at virtual ground. I believe this helps it to get over the problem of non-linear current as voltage develops across the photodiode.
The photodiode signal was getting saturated by the overhead lights, so I held my hands to cast shadow on the diode.
Average DC output level = 8V
Vpp of the signal = 6V
Percentage modulation = 6V/8V x 100% = 75%
Input photocurrent = 6V/1MΩ = 6 μA
One interesting point is, the photocurrent goes in the opposite direction in this case (which is something I don't understand).
At the summing junction, I saw what looked like the inverse of the function seen at the Vout (????). Would have made more sense for it to be virtual ground.
Phototransistor
A phototransistor is a photodiode with a gain. Instead of releasing one electron per photon, this arrangement releases about 100 photons per electron.Average DC output level = 1.2V
Vpp of the signal = 2.4V
Percentage modulation = 2.4V/1.2V x 100% = 200%
Input photocurrent = 2.4V/100kΩ = 24 μA
This current gain makes a phototransistor a much more sensitive light detector.
At the summing junction, the voltage was 0.
In this setup, the current flows in the conventional direction.
I'm not convinced whether I have calculated the right percentage modulation in either of the cases, since they seem to be awfully big.
Summing Amplifier
Here,
Ifb = I1 = I2
Vout = 0V - IfbRfb = -(I1+I2)Rfb
Vout = -(V1/R1 + V2/R2)RfbFor R1=R2=R,
Vout = -Rfb/R (V1+V2)
This was put to practice in the following exercise, where a DC offset was added to an AC signal. The Vout is 180 out of phase as expected.
Limitations Of Op amps (HH Lab 9-1)
Modified Golden rules:
1. There is a very small voltage difference between the inputs, called the offset voltage VOS.
2. The inputs take in a very small current, called the input bias current Ibias.
Also, the open loop gain of the op amp "rolls off" at high frequencies, which limits its ability to respond to high frequency signals. This results in limited slew rate for the op amp, which is defined as the maximum rate which the output of the op amp's output can change. This effect can be observed from an op amp follower responding to a square wave input. For the 411 op amp,
with a 2 Vpp 1 kHz signal,
slew up rate = ΔV/Δt = 3V/250ns = 12 V/μs
slew down rate = 3V/500ns = 7 V/μs
The sine wave's amplitude begins to drop at 1.2 MHz.
For the 741 op amp,
slew up rate = 3V/4μs = 0.75 V/μs
slew down rate = 2.5V/4μs = 0.63 V/μs
The sine wave's amplitude begins to drop at 75 kHz.
Integrator (HH Lab 9-2)
Here,
V+ = V- = 0I(t) = Vin(t)/R
Vout(t) = -q(t)/C
I(t)=dq(t)/dt
This result holds true for a range of frequencies. The circuit on the right is an integrator with a DC gain of 100. On the scope, the input square wave is in green and the output triangle wave is in yellow. Notice the slight dc offset in the -ve direction.
A small dc offset, over time will integrate to give a very large result, drifting to one of the supply rails. By putting a large by-pass resistor in the feedback loop, the 10 MΩ in this case, this problem can be avoided, as the dc current will through the resistor and not accumulate on the capacitor. If the 10 MΩ is taken off, a larger dc offset can be seen, on the scope below.
On playing with the "Offset" of the function generator, I noticed something interesting. As the +ve offset is increased, the output moves to the negative direction, and the bottom part of the triangle gets flatter and flatter, until the whole signal turns into a dc line, as can be seen from the video. This is because the output of an op amp cannot be bigger than the power supply, i.e. +/- 12V. So any signal bigger than that gets cut off!Audio Amplifier
The gain for the signal is:
G = 1 + (100k/1k+Rvar)
So out Gain could be varied between ~ 2 to 20.
Also, the output voltage can go from GND to 3V. This is visible in the video below. When I snap, the signal is driven to the rail, so that it gets cut off in the bottom mostly.
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