EGR220 Than & Bhavin Lab #1
Page 1
Introduction and Objectives
Since Diodes are one of the fundamental electronic
devices, which has various applications, it is important
to know their characteristics and how they work.
Therefore, in this lab, we were instructed to measure
and analyze the static characteristic of diodes. The
primary objectives of this lab are:
1. To analyze and understand the nature of I-V
curve of the diode
2. To understand about the revers saturation
current Is, and the ideality factor n of the
diode.
3. To be able to collect data and to plot them on
semi-log scale
4. To be able to calculate current through and
voltage across the diode using piece-wise
linear model
Equipments and Components used
In this lab, the equipments and components we used
are:- diodes: 1N914 (x3), 1N 60 (x3); resistors: 100Ω
@ 1W (or more) (x1), 1KΩ(x2); a breadboard, a
waveform generator, ±20V power supply, a multi-
meter, an Oscilloscope to capture the I-V curve, wires
and cords.
Procedures
Procedure 1: Capturing I-V curve of the Diode
Figure 1
In order to capture the I-V curve of the Diode on
Oscilloscope, we used time varying voltage source
( ±5V Sine Wave with frequency of 1kHz) to trace
forward and reverse characteristic of the diode. One
oscilloscope probe was placed across the resistor to
measure the diode current (Vr /R) and the other was
placed across the diode to measure diode voltage.
However, because both probes have a common
ground, we could not get a proper I-V curve.
Therefore, we re-designed the circuit as follows by
using a resistor with low resistance (3Ώ) (to avoid
unnecessary voltage drop).
Figure 2: Circuit Design for capturing I-V curve of
Diode
We built the above circuit and captured the
oscilloscope image of the I-V curve of 1N914 and
1N60, and figured out the values of VDO and RD, by
moving the cursors of oscilloscope display. We found
that VDO for 1N914 is about 0.6±0.5V, RD = (VD-VDO) /
ID =35Ώ and VDO for 1N60 is about 0.3V, RD = 11Ω.
Figure 3: I-V characteristic of Si Diode
Procedure 2: Measuring and Plotting of Diode
Current Voltage Point by Point
By using the circuit in figure 1, we measured the diode
EGR220 Than & Bhavin Lab #1
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voltage and current of 1N914 and 1N60, by giving the
source voltage, ranging from -10V to 10V using 0.5
voltage increment.
In order to calculate the ideality factor n, we used the
following formulas [1]
:
I2 / I1 = e(V2 – V1)/nVT
(1)
n = (V2 – V1) / VT ln(I2-I1) (2)
where VT = 25mV at room temperature. By plugging in
the two diode currents and voltage, we calculated that
the value of n for 1N914 is 2.2 ± 0.09 and that of 1N60
is 2.3± 0.01. From our calculation, we concluded that
the value of n must be 2 for both 1N914 and 1N60. By
using n = 2, we, then, calculated the Is Value for both
1N914 and 1N60, using the following formulas [1]
.
I = Is (e V/nVT
– 1) (3)
I ≈ Ise V/nVT
(4)
Is ≈ Ie –V/nVT
(5)
Our calculation shows that the value of Is for 1N914 is
2.88±0.003 ×10-7
A, and that of 1N60 is 1.0±0.3 × 10-3
A. According the formula (3), in the reverse bias
region, the reverse current would be approximated
with negative Is as the exponential term would
disappear. However, in the lab, due to systematic errors
(because of measurement methods and faulty devices),
we were not able to detect the reverse current. Then,
we plotted the data we measured, using both
semi-log and linear scale.
From the linear graphs, we tried to find the VDO
and then calculated for RD for both 1N914 and
1N60, [1]
by
RD = (VDO – VD) / ID (6)
-9.5
-7.9
9
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Linear Vs Expoential I-V Curve (1N914)
Voltage (V)
Cu
rre
nt (
mA
)
EGR220 Than & Bhavin Lab #1
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From our linear graphs, we found that VDO for
1N914 is approximately 0.65V and VDO for 1N60
is approximately 0.3V. Then we calculated RD for
both 1N914 and 1N60. RD for 1N914 is
approximately 34.29Ω and that of1N60 is 7.92
Ω.
Procedure 3:Finding diode currents with
estimations and comparing those results with
the actual measurements
Figure 4
By using the values of RD and VDO, we calculated
the currents through the diodes by using piece-
wise linear model.
For 1N914, in circuit 4 (a) we got ID1 = 0A and
ID2 = 2.13 mA, in circuit 4(b), we found both ID1
and ID2 = 0A because the voltage at the junction
is less than VDO and so, no current cannot pass
through the diodes. For 1N60, in circuit 4 (a) we
got ID1 = 0A and ID2 = 2.37 mA, in circuit 4(b),
we also found both ID1 and ID2 = 0A.
From our measurements, for 1N914, we got ID1
= 0 A and ID2 = 1.48 mA for circuit 4(a). and ID1
and ID2 = 0 A for circuit 4(b). For 1N60, we got
ID1 = 0A and ID2 = 1.67 mA for circuit 4(a) and
ID1 and ID2 = 0A for circuit 4(b).
Although the errors are within the range of 40%:
2.13 – 1.48 = 0.65 × 10-3
and 2.37 – 1.67 = 0.7
× 10-3
, considering the fact that our calculation
was based on piece-wise linear model, not the
actual exponential model, and there might be
errors accumulated from measurements and
calculation of VDO and RD, we assumed that the
results are within the acceptable range.
Discussion
Since didoes are non-linear devices, their
characteristic are much different from that of
resistors, which are linear devices. The
relationship between voltage and current for
diodes are theoretically modeled with a complex
exponential function, which depends on many
factors, such as temperature, ideality factor etc.
However, for some applications, which do not
need so much accuracy, piece-wise linear model,
could be used instead of complex model. In our
case, the error was 40% of actual observations.
Obviously, piece-wise linear model can give us a
faster and easier method to calculate, it is not
useful enough to cope with accuracy demanding
applications.
References [1] Sedra, Adel S., and Smith. Kenneth C.
“Microelectronics Circuits”. 5th. New York: Oxford
University Press, 2004.
-9.5 -8
-6.5 -5
-3.5 -2
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Linear Vs. Exponential I-V Curve (1N60)
Voltage
Curr
ent (m
A)
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