Download Diodes and Transistors

Diodes and Transistors©98
EXPERIMENT 12
Objective: To examine the characteristics of simple diodes and transistors, and to
build a transistor amplifier.
DISCUSSION:
Positive
Negative
A diode is a crystalline device that permits
Region
Region
the flow of charge in only one direction. There are
two parts to the device, one called p (for positive)
_
and the other n (for negative). These are shown
+
schematically in Fig. 1. Silicon (or germanium) is
the principal element in both parts. However in
the p-region, a small amount of gallium (or boron,
Figure 1: The Diode.
or aluminum, or indium) has been added to the
crystalline lattice formed by silicon. In the n-region a small amount of arsenic (or
phosphorus, or antimony) has been added. Relative to the p-region, the n-region
has an excess of electrons, and so, if an external electric field is applied across the
crystal so as to force electrons from the n-region to the p-region, they move easily
across the p-n junction (that is, across the p-n surface). However if the electric
field is applied in the opposite direction so as to force electrons from the p-region
to the n-region, they do not move easily. Thus, when a diode is placed in a
circuit, it permits the flow of charge in only one direction. Positive current flows
from the p-region to the n-region. The symbol for the diode, indicates the
direction of the positive current in the diode.
Collector Base Emitter
Collector Base Emitter
A transistor is a crystalline
N
P
N
P
N
P
device consisting of three regions in
tandem, in the order of either n-p-n,
Collector
Collector
or p-n-p, as shown in Fig. 2. The
region at one end of the transistor is
Base
Base
called the collector and the other end
the emitter. The central region is
called the base. The essential
Emitter
Emitter
electrical property of the transistor is
that the base may act as a kind of
Figure 2: The Transistor
electrical valve in permitting the flow
of electricity through the device. Fig. 2 also shows the circuit diagrams for these
two kinds of transistors. The arrows indicate the direction of the positive current.
We first examine the n-p-n transistor. If we apply a relatively large
voltage Vce across the collector and emitter to try to cause a current from collector
to emitter, the presence of the positively biased base prevents an actual flow of
charge. However, if a very small voltage Vbe is applied across the base and
emitter so as to cause a small current Ib from base to emitter, that voltage opens
12-1
the valve, so to speak, and permits a large current Ic from collector to emitter.
The larger the small current from base to emitter, the greater the current from
collector to emitter.
The p-n-p transistor works in much the same way, except that the principal
current is from emitter to collector. We first apply a relatively large voltage
across the emitter and collector to try to cause positive charge to flow from
emitter to collector. Very little, if any charge does flow until the "valve" is
"opened" at the base. This opening is effected by applying a small voltage from
emitter to base, causing a small amount of positive charge to flow from emitter to
base. The larger this voltage, the greater the current from emitter to collector.
Fig. 3 is a schematic diagram of the n-p-n
transistor in operation. The current from the
emitter Ie is the sum of the currents from the
base and the collector. That is
Ie  Ic  Ib.
Ic Collector
Ib
(1)
+
_ Vce
Base
+
Ie
The most important property of the transistor
_
Emitter
is the effect the base current, Ib, has on the
collector current, Ic. This effect is measured
Vbe
by considering the ratio of the change Ic
Figure 3: An electrical diagram of a transistor.
that is produced in Ic by a change Ib in Ib.
Specifically, we define the current amplification factor, , of the transistor to be

I
I
c
(2)
.
b
The amplification factor of a transistor is ideally constant but in fact it depends
somewhat on the voltages Vce and Vbe. This final result, that the transistor is a
current amplifier, is the essence of transistor behavior.
V
+
+
A
+
_
_
100W
EXERCISES:
1. Build the circuit shown in Fig. 4. The
function of the resistor (R=100W) is to
limit the current through the diode. Excess
current generates heat which can damage
the crystal. The rheostat R provides a
source of variable voltage V.
(a) With the diode oriented as shown in Fig.
4, determine the current I in the diode as a
function of the voltage V across the diode.
Record a set of data for I and V for
increments in V of 0.2 volts.
_
R
+
Figure 4. The diode circuit.
12-2
(b) Reverse the orientation of the diode and note the current I as a function of
the voltage V.
2. Build the circuit shown in Fig. 5. The power supply provides a constant
voltage drop across the rheostat R1 and across R2=180W, Rc, across the
transistor, and Re. The rheostat is used to provide a variable voltage across
Rb, the base-emitter junction of the transistor, and Re.
R2=180
Vc
Rc
Vb
+
_
R1
Collector
Base
Rb
27k
Emitter
Re
Figure 5: The transistor circuit. One voltmeter measures the voltage drop V c across Rc.
The other voltmeter measures the voltage drop Vb across Rb. These voltmeter readings
can be used in combination with the corresponding resistance values to give the currents
Ic and Ib provided the resistances Rc and Rb are known. The transistor and its
corresponding resistors are already wired on the board for you.
From Ohm’s law we can write
I 
c
I 
b
V
c
R
(3)
c
V
b
R
(4)
b
(a) Vary the base voltage Vb by moving the sliding contact or wiper along the
rheostat Vb . Increment the base voltage so that the collector voltage
increases by 0.33 volts for each step. For each step in Vc, record the values
of Vc and Vb. Continue stepping the voltage until the collector voltage does
not change. At this point you are saturating the transistor gain and no more
increase in Vc can be achieved, even though you increase Vb.
(b) Using Eqs.(3) and (4), find the corresponding values of Ic and Ib.
12-3
(c) Calculate the increments Ic and Ib which occur when Vc is changed in
steps of 0.33 volts. From these values of Ic and Ib, calculate the value
of  corresponding to each value of Vb. Note whether  remains constant
over the range of Vb.
(d) Graph Ic versus Ib. Calculate  by finding the slope of the line.
12-4