EGR220 Than & Bhavin Lab #1

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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)

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Linear Vs Expoential I-V Curve (1N914)

Voltage (V)

Cu

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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.

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