Name: Ogwuafi Ineme-Awaji
Group No.: 2
Experiment No.: 3
Title of Experiment: Measurements of small currents
Purpose of Experiment:
1. We will measure small currents produced by a photodiode
2. We will study the relationship between the luminance and the output voltage across the diode and the current
through the diode.
Brief theory:
A photodiode is simply a diode based on the PN- junction sensitivity to light. Once a photon strikes on its surface and it
has energy enough to overcome the material bandgap, an electron is free to circulate in the circuit creating a photo-
current ( current that is light-dependent). The current produced by a photodiode is rather too small to measure, hence
we will amplify it with an operational amplifier. The photodiode acts as a current source when operated in reverse bias
mode.
Figure 1: reverse bias connection
Figure 2: IV-characteristic of a diode
Where VR , VF : reversed bias and forward bias voltages respectively, I R, IF - reversed bias and forward bias currents
respectively
Figure 3: photodiode; junction properties at illumination
The equivalent circuit of a photodiode is shown below
Figure 4: The equivalent circuit of a photodiode
Apparatus:
1. Voltage source (0 -12V) 6. Operational amplifier
2. Digital multimeter 7. Connecting cables
3. LED light source
4. UA-962 UYIGAO lightmeter
5. Photodiode
Experimental setup:
Figure 5: Experimental setup
Procedure:
Task1.
Connect the photodiode to a voltmeter and measure the voltage produced by the Photodiode vs. the illuminance. How
is the dependence of the voltage vs. illuminance? Why?
Figure 6: experimental setup for task 1
Task2.
Instead of the voltmeter, connect the ammeter to the photodiode and repeat the experiment. In this case, you will
measure the dependence of the current vs. the illuminance. How does it differ from the curve of the voltage vs.
illuminance? Why?
Figure 7: experimental setup for task 2
Task3.
The current that you will have measured in point 2 is small, you need to amplify it. You can use a current to voltage
converter which takes the current and converts it to a voltage using a 100,000 multiplying factor (i.e. if you have a
current of 10 μA you obtain at the output a voltage of 1 V). This is the schematic:
Figure 8: experimental setup for task 3
Task4:
Then, if it suffices to inject the current in a 100 kΩ resistor to multiply it by 100,000 times, why don’t you connect
directly the photodiode to the resistor?
Figure 9: experimental setup for task 4
Measured Data:
Luminance [ lux ] Voltage
[ mV]
1.0 146.42
50.8 239.04
101.1 258.19
153.7 269.68
201.8 277.41
250.9 283.47
301.9 288.81
350.0 293.11
502.0 303.88
1007.0 326.21
1516.0 338.51
Table 1. table of measured values for task1
Luminance [ lux ] Current [ μA ]
6.2 0
252.4 2.4
502.9 4.8
753.2 7.3
1016 9.8
1262 12.2
1519 14.7
Table 2. table of measured values for task2
Luminance [ lux ] Voltage [mV]
0.0 0.03
252.0 236.22
506.4 475.2
752.6 704.5
1000.0 942.2
1246.0 1162.3
1500.0 1402.9
Table 3. table of measured values for task3
Graphs and computations:
From task 3, involving the operational amplifier circuit the current can be calculated by dividing the voltage by the
resistor value.
For resistance R = 100 kΩ, and voltage = 236.22 mV ; the current I = voltage / resistance = 2.3622 μA .
And subsequent current values were calculated from voltage values given in table 3, and are tabulated below
Luminance [ lux ] Voltage [ mV ] Current [ μA ]
0 0.03 0.300
252 236.22 2.362
506.4 475.20 4.752
752.6 704.50 7.045
1000 942.20 9.422
1246 1162.29 11.623
1500 1402.90 14.029
Table 4. table depicting current values calculated from voltage values
Figure 10: Graph of voltage against luminance.
16
graph of luminance against current
14
12
current [ μA]
10
8
6
4
2
0
0 200 400 600 800 1000 1200 1400 1600
luminance [ lux ]
Figure 11: Graph of luminance against current
graph of luminance against voltage
1600
1400
1200
1000
voltage [mV]
800
600
400
200
0
0 200 400 600 800 1000 1200 1400 1600
Luminance [lux]
Figure 12: graph of luminance against voltage with operational amplifier circuit
Discussions and Conclusion:
In task1, when we measure the voltage with a voltmeter, the ideal property of a voltmeter implies that the load
resistance is very high compared to the shunt resistance of the diode( from figure 4). The generated photocurrent flows
through the shunt resistor causing a voltage drop across the diode. This voltage opposes the bandgap potential of the
photodiode junction, forward biasing it. The value of Rd drops exponentially as the illumination increases. Thus the
photo-generated voltage is a logarithmic function of incident light intensity, this also produces a similar result with task
4.
In task2 and task 3., the ammeter and the operational amplifier both act as short circuits, hence the shunt resistance is
very high and the generated current flows through the load resistor, if we replace the load resistor with an ammeter, the
current values can be read and these values can be seen to increase linearly with illumination (figure 11). A current to
voltage converter amplifier can be used to convert the photodiode current to a voltage and keep the diode voltage at
zero. The resultant voltage is therefore linearly dependent on the incident radiation level. Since the operational
amplifier has low load resistance, an amplified output voltage can be obtained by feeding the photocurrent to an
operational amplifier and the voltage will have a linear behavior as can be seen in figure 12.
