Figure 1. Single diode -3 to 3 V current-voltage characteristic.
Plot the absolute current against voltage on a semi-log plot (Log-I vs V). You will use this plot to extract parameters from the diode’s electrical (I-V) characteristic. Paste the plot in here.
Copy in diode semi-log I-V Plot
QUESTIONS/TASKS
1. What is the current-rectification ratio at ± 3 V (I at +3 V divided by I at -3 V)?
Ans.
2. What is the saturation current I0? (I at zero bias)
Ans.
3. Estimate the series resistance Rs of the diode by extrapolating the linear portion of the I-V curve to the I = 1 mA line and evaluating Rs = ΔVd/I where ΔVd is the voltage difference between the linear extrapolation and the measured curve at 1 mA. Mark the extrapolation from the linear region of I-V data on your MS Excel graph using a simple line. (You could plot the linear region only and use the linear fit function as an alternative)
Ans.
4. Series resistance causes power loss in the diode but is inevitable. Where do you think the series resistance comes from?
Ans.
5. In an ideal semiconductor diode, the current is given by I = I0(exp(qV/kt)-1). No real diodes are ideal. Non-ideality is conveniently incorporated into the preceding equation by introducing η so that I = I0(exp(qV/ηkt)-1)
This equation can be rearranged so that η = qΔVlog(e)/kT = ΔV/0.05783 at room temperature. ΔV here is the change in junction voltage per decade of current measured from the linear part of the forward bias data on the semi-log plot. Fig. 5 shows extraction of series resistance and ideality factor from a diode current-voltage characteristic:
Figure 5. Extraction of series resistance and ideality factor from semi-log diode I-V data (taken from Microwave Mixers by S A Maas)
Estimate the ideality factor of the 1N4148 diode (to two significant figures) using the above approach.
Ans.
Data:
V2.V
I(AM1.nB)
-3
-2.52E-09
-2.999
-2.52E-09
-2.998
-2.52E-09
-2.997
-2.52E-09
-2.996
-2.52E-09
-2.995
-2.52E-09
-2.994
-2.52E-09
-2.993
-2.52E-09
-2.992
-2.52E-09
-2.991
-2.52E-09
-2.99
-2.52E-09
-2.989
-2.52E-09
-2.988
-2.52E-09
-2.987
-2.52E-09
-2.986
-2.52E-09
-2.985
-2.52E-09
-2.984
-2.52E-09
-2.983
-2.52E-09
-2.982
-2.52E-09
-2.981
-2.52E-09
-2.98
-2.52E-09
-2.979
-2.52E-09
-2.978
-2.52E-09
-2.977
-2.52E-09
-2.976
-2.52E-09
-2.975
-2.52E-09
-2.974
-2.52E-09
-2.973
-2.52E-09
-2.972
-2.52E-09
-2.971
-2.52E-09
-2.97
-2.52E-09
-2.969
-2.52E-09
-2.968
-2.52E-09
-2.967
-2.52E-09
-2.966
-2.52E-09
-2.965
-2.52E-09
-2.964
-2.52E-09
-2.963
-2.52E-09
-2.962
-2.52E-09
-2.961
-2.52E-09
-2.96
-2.52E-09
-2.959
-2.52E-09
-2.958
-2.52E-09
-2.957
-2.52E-09
-2.956
-2.52E-09
-2.955
-2.52E-09
-2.954
-2.52E-09
-2.953
-2.52E-09
-2.952
-2.52E-09
-2.951
-2.52E-09
-2.95
-2.52E-09
-2.949
-2.52E-09
-2.948
-2.52E-09
-2.947
-2.52E-09
-2.946
-2.52E-09
-2.945
-2.52E-09
-2.944
-2.52E-09
-2.943
-2.52E-09
-2.942
-2.52E-09
-2.941
-2.52E-09
-2.94
-2.52E-09
-2.939
-2.52E-09
-2.938
-2.52E-09
-2.937
-2.52E-09
-2.936
-2.52.
Figure 1. Single diode -3 to 3 V current-voltage characteristi.docx
1. Figure 1. Single diode -3 to 3 V current-voltage characteristic.
Plot the absolute current against voltage on a semi-log plot
(Log-I vs V). You will use this plot to extract parameters from
the diode’s electrical (I-V) characteristic. Paste the plot in here.
Copy in diode semi-log I-V Plot
QUESTIONS/TASKS
1. What is the current-rectification ratio at ± 3 V (I at +3 V
divided by I at -3 V)?
Ans.
2. What is the saturation current I0? (I at zero bias)
Ans.
3. Estimate the series resistance Rs of the diode by
extrapolating the linear portion of the I-V curve to the I = 1 mA
line and evaluating Rs = ΔVd/I where ΔVd is the voltage
difference between the linear extrapolation and the measured
curve at 1 mA. Mark the extrapolation from the linear region of
I-V data on your MS Excel graph using a simple line. (You
could plot the linear region only and use the linear fit function
as an alternative)
Ans.
4. Series resistance causes power loss in the diode but is
inevitable. Where do you think the series resistance comes
2. from?
Ans.
5. In an ideal semiconductor diode, the current is given by I =
I0(exp(qV/kt)-1). No real diodes are ideal. Non-ideality is
conveniently incorporated into the preceding equation by
introducing η so that I = I0(exp(qV/ηkt)-1)
This equation can be rearranged so that η = qΔVlog(e)/kT =
ΔV/0.05783 at room temperature. ΔV here is the change in
junction voltage per decade of current measured from the linear
part of the forward bias data on the semi-log plot. Fig. 5 shows
extraction of series resistance and ideality factor from a diode
current-voltage characteristic:
Figure 5. Extraction of series resistance and ideality factor from
semi-log diode I-V data (taken from Microwave Mixers by S A
Maas)
Estimate the ideality factor of the 1N4148 diode (to two
significant figures) using the above approach.
Ans.
