2/23: Op Amps II

in which Julia investigates current topics in op amps

Lab 8-5: Current source

Op amp current source diagram (Electronics pdf); we used a 220 Ω resistor for V- to ground.

This circuit provides a constant current through the ammeter.
The expected current was that corresponding to 0.75 V applied over 220 Ω of resistance.  Ohm's Law tells us that V = I*R so I = V/R; thus we expect (0.75/220 = ) 0.0034 A current, or 3.4 mA.  This was in good agreement with our experimentally measured value: 3.4 mA.

When the load resistance is varied, we would expect the current to drop, because increasing the load makes the voltage drop bigger, and voltage drop is capped at 12 V.  Assuming that the op amp can output 11 V, the maximum resistance for 3.4 mA of current is:

11 V = (3.4 mA)*(R)

R = 11 / 0.0034 = 3200 Ω

We confirmed this using a resistance decade box to vary the load; the current decayed an appreciable amount at 3.1 kΩ but not at 3 kΩ.

Our experimental setup.  Left: breadboard with op amp and +/- 12 V connections.  Middle: ammeter.  Right: resistance decade box (connected to op amp's output and directly to the ammeter).  Left foreground: the discarded potentiometer initially used to vary load resistance.

The main limitations of this current source are that it requires a "floating" load, where neither side is connected to ground, and its limited speed; a more complicated transistor circuit can begin to solve these problems.

Lab 8-6: Current to Voltage Converter

Part a) Photodiode

The photodiode circuit in H&H: we used an LED running backwards instead of an LPT-100.

We used an LED wired to an op amp to measure incident light on our circuit.

The photodiode's changing voltage response to the 60-Hz flicker of fluorescent lights.

The signal was clear when AC coupled, confirming that the peaks were caused by the fluorescent lights, as all outlets in the lab are coupled.

The photodiode's output decreases when it is covered (exposed to less light).

We added a capacitor (0.1  μF) in parallel with the 10 MΩ resistor to try to eliminate fuzz.

Adding the capacitor smoothed the output signal and decreased fuzz.

When we added wires to attach a capacitor, the signal smoothed before we added the capacitor.  This result was quite puzzling until we looked closely at our breadboard--we had put two wires directly adjacent in the breadboard, and we think this made a small region of the breadboard act as a capacitor (by charging two parallel strips of the breadboard connections).  This effect was less prominent than that of the 0.1 μF capacitor.

The capacitor douses high-frequency oscillations because it is a low-pass filter.  At high frequencies, most of the current would go through capacitor, which has a small impedance so this current turns into a small output voltage signal.  At low frequencies, current instead travels through the 10 MΩ resistor, which converts a small amount of current into a large voltage signal.  Thus, high-frequencies become tiny voltage oscillations obscured by the large gain of low-frequency signals, and the output curve is smoothed.

Measuring at "X" showed a voltage of 0 V, as expected because the op amp is in negative feedback, so V- should equal V+, and V+ is at ground (0 V).

The average DC output level was approximately -82 mV (-0.082 V).

The percentage modulation was appxoimately 68% (there was a 56 mV difference from middle to top).

Calculating the input photocurrent for this output:

Vout = (Iphot)*(10MΩ)

0.082 V = (Iphot)*(10e6 Ω)

0.082/10e6 = Iphot = 8.2 x 10^-9 A so 0.00082 μA

Our photodiode isn't multiplying anything; there are 1/(1.6 x 10^-19) Coulombs per electron or 6.25 x 10^18 C/e-, so ((8.2 x 10^-9)*(6.25 x 10^18) = ) 5 x 10^10 photons/s, which is less than a high-powered laser and so seems plausible.

This op amp circuit is superior to the alternative simple resistor circuit shown in H&H, because if the latter circuit were connected to 1 MΩ oscilloscope, current would flow through the scope and not through the resistor, which would artificially decrease the voltage signal.

Part B) Phototransistor

While the signal was the same shape as in the photodiode circuit, the average DC output level was approximately -1.24 V; the peaks occurred at -80 mV, so modulation was (1.16/1.24 = ) 94%.  While this is much greater than the photodiode circuit, it is difficult to compare the two as the resistor values (and thus their gain) are different.

For this circuit, the photocurrent is 1.24 V / 100 kΩ resistor, so by Ohm's Law:

(1.24 V)/(100 kΩ) = 1.24 x 10^-5 A

The current is now 1.24 x 10^-5 A, which is nearly four orders of magnitude greater than that of the photodiode--as expected, because the phototransistor releases many more electrons per incident photon than the photodiode (which has a one photon:one electron ratio).

Lab 8-7: Summing Amplifier

The summing amplifier circuit from H&H; the arrow denotes a variable resistor (potentiometer here).

Changing the potentiometer's resistance here changes the offset voltage.  This circuit successfully adds a variable DC offset to an input signal!  To subtract voltage, we could first run one signal through an inverting op amp circuit (with a gain of unity).

Alternatively, we could attempt to add the offset voltage after the op amp (adding to output)--but that would greatly increase output impedance, so this could only be useful if it were the last step in a circuit.