We can make OpAmp followers
The Vout is perfectly the same as Vin! This is a case of the noninverting amplifier.
Zin is still large and Zout is still small.
We can also make current sources
Where DVM is the ammeter.
We expected a current of a few mA and got 3.8 mA.
Once the max voltage is reached, the current can no longer be regulated.
The extra 10 kOhm resistor is needed to increase the max voltage. Without it, the max voltage is a mere 3/4 V, leaving the current unregulated.
We can make a current to voltage converter
The LED is used instead of an LPT100, but it behaves the same way, converting light into current.
Pictures by Hannah
Note that we see a signal oscillating at 60 Hz, as expected!
The average DC output is 125 mV.
Percent modulation is AC Vpp / DC = 40 mV/125 mV = 30%
This output level corresponds to an input photocurrent of (using DC only)
V=IR
125 mV = I * 10 MOhm
I = 1.25E-9 A
Recall that we are using an LED. With a real photodiode, we'd get VDC closer to 10 V
This circuit is preferable to a circuit without the OpAmp, since this circuit also amplifies the voltage.
The summing junction is a virtual ground, and we do see 0V at the summing junction experimentally!
We can also use a phototransistor
Second picture by Hannah
VDC = 4V
I = VDC/R = 4 V / 100 kOhm = 4E-5 A
AC Vpp = 4V
Percent modulation is 100%
Summing junction is again virtual ground.
We can make a summing amplifier
Photo and video by Hannah
This sums a DC level with the input, basically adding a DC offset
In the video, Vin is yellow and Vout is blue. As the pot is turned, the DC offset changes and Vout moves up and down. The offset is between +/- 8 V as expected. The Vout is inverted, but the amplitude of the wave is the same.
As shown in class, the following circuit is subtract two voltages
However, there are some limitations of the OpAmps!
The first being the slew rate.
Using a 1 kHz square wave, varying the Vpp, the slope of Vout at the transitions in the square wave will determine the slew rate. Note that the slew up and down might be different!
Vpp (V) | Voltage change (V) | Time up (ns) | Slew up (V/s) | Time down (ns) | Slew down (V/s) | Picture |
---|---|---|---|---|---|---|
1 | 1 | 200 | 5E6 | 200 | 5E6 | |
3 | 3.2 | 320 | 1.1E7 | 300 | 1.3E7 | |
5 | 5.4 | 450 | 1.2E7 | 400 | 1.4E7 | |
7 | 7.2 | 650 | 1.1E7 | 530 | 1.4E7 |
With a sine wave, we now want to see when Vout begins to drop appreciably.
Picture by Hannah
From 1 MHz to 2 MHz, we observe a huge drop. In the picture above, the 1 kHz Vmax is lined up with the grid line. We can see that at 1.5 MHz, the Vmax has dropped quite a bit.This makes sense. The slew is around 1 V/ 10 µs so when the period of the wave is larger than 10 microseconds and the frequency is around 1 MHz, the OpAmp starts having troubles.
Now we can make an integrator with the OpAmp
Driving with a 1 kHz 1 Vpp square wave...
Picture by Hannah
Note that Vout = -1/(RC) ∫ Vin dtR = 100 kOhm and C= 0.01 µF
∫ Vin dt = 0.5 V * 500 µs (Considering only half the period)
So Vout = -250 mV
Our experimental Vout is also -250 mV!
Note that Vout is also offset because of the small DC offset. The gain for DC is 100, so a small offset in Vin means a huge offset in Vout.
Picture by Hannah
Now with a Vin of 2 Vpp and 500 Hz, we get a Vout both experimentally and calculated of -1VYou may be wondering why there is a 10 MOhm resistor in the circuit. It is preventing the DC offset from entering the integral. The DC current goes through this bypass resistor instead of building on the capacitor. If the resistor were not there, the integral would drift to a rail.
Now a microphone amplifier!
The video is not uploading correctly here, but it is up on Hannah's blog.
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