LABORATORY WRITE-UP TRANSISTOR CHARACTERISTICS

parkagendaElectronics - Devices

Nov 2, 2013 (3 years and 1 month ago)

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LABORATORY WRITE
-
UP

TRANSISTOR
CHARACTERISTICS


AUTHOR’S NAME GOES H
ERE

STUDENT NUMBER: 111
-
22
-
3333












2

TRANSI STOR CHARACTER
I STI CS


1. PURPOSE



The transistor ranks as one of the greatest inventions of 20th century technology. It finds
applica
tion in virtually all electronic devices from radios to computers. Integrated circuits
typically contain millions of transistors, formed on a single tiny chip of silicon. Two of the
basic uses of a transistor, which will be explored in this experiment, a
re as an amplifier and
as a switch.




Fig. 1 a) The pnp transistor b) Circuit symbol c) Common emitter amplifier circuit


The
pnp

transistor, shown in Fig. 1a) contains three distinct regions, a p
-
type "emitter", an n
-
type "base" and a p
-
type "collect
or", which together form two
pn

junctions. In a typical
amplifier circuit, voltages are supplied so that the emitter
-
base junction is forward
-
biased and
the collector
-
base junction is reverse
-
biased. This means that V
CE

> V
BE
. Fig. 1c illustrates a
"com
mon emitter" circuit, so called because the emitter is common to the input circuit on
the left and the output circuit on the right.


Consider first the forward
-
biased emitter
-
base junction. The doping of the emitter is made
much heavier than that of the b
ase so that positive holes from the emitter form almost all of
the current, I
E
, from emitter to base. The base, being lightly doped, does not have many
electrons available for recombination with these holes to form neutral atoms. It is also very
narrow (
< 1

m) making it easy for a large fraction,

, of the holes to diffuse across to the
collector
-
base junction where the junction voltage accelerates them into the collector region
to form the collector current, I
C
.

Thus, I
C

=

I
E

.......................(1
)


The remaining fraction, (1
-

), of holes leave the base through the external connection to
form the base current, I
B
, where



3

I
B

= (1
-

)I
E

...................(2)

The "current gain",

, of the transistor is defined by




= I
C
/I
B

.....................(3)



=

/(1
-

) .


For typical transistors,

~ 0.9 to 0.995, giving values of


~ 10 to 200. Thus we have a
"current amplifier", in that a small change in I
B

will cause a large change in I
C
. The "voltage
gain
", A
V
, is the ratio of the voltage drop, I
C
R
C
, across the output resistor, R
C
, to the voltage,
V
BB
, of the input source:

A
V

= I
C
R
C
/V
BB
.


Applying the loop theorem to the input circuit in Fig. 1c), and assuming I
E

~ I
C

=


I
B
, it is
easy to show that

A
V

=


R
C
/(R
B
+

r
b
)………………….4 .


where r
b

is the resistance of the emitter
-
base junction. By a suitable choice of resistors, an
appreciable voltage amplification may be obtained.




Fig. 2 a) The AC amplifier

b) Equivalent circuit


Fig. 2a shows how the transistor may be used as an AC amplifier to amplify a small signal
from a signal generator (or, say, from the play
-
back head of a tape deck). Now the two
batteries in the circuit behave like large ca
pacitors with impedances (1/

C) ~ 0, so that the
equivalent circuit is as shown in Fig. 2b. Once again the voltage gain is given by equation
(4). However, this is a simplified situation. In reality, the transistor junctions possess
capacitance, and the
corresponding reactances are frequency dependent. Thus we can
expect A
V

to be a function of frequency.




4

2. PROCEDURE


Preliminary
Set up the circuit shown in Fig. 1c using the power supply outputs for the
voltages V
BB

and V
CC
. Note the symbols e, b and
c denoting the transistor connections. Use
a 3000


resistor for R
B

and a 220


resistor for R
C
. Turn the supply outputs to zero then
turn on the unit. Set one of the digital meters to the 20 V DC range and connect it to
measure V
CC

(+ lead to ground on

the transistor board). Adjust V
CC

to approximately 15 V.
Reconnect the meter to measure V
CE
. This should also read 15 V, indicating I
C

= 0.
Connect the second meter to measure V
BB
, also with the + lead to ground. Slowly increase
V
BB

up to 2 V and not
e V
CE

decreasing, indicating an increasing I
C
. Your amplifier is now
working.


DC operation


Set V
BB

to 0.75 V. Reconnect the meter to measure V
BE

and calculate I
B

[= (V
BB
-
V
BE
)/R
B
].
Now reconnect the meter to measure V
CC
. (The first meter should still
be measuring V
CE
.)
Adjust V
CC

to 1, 3, 5, ....15 V and calculate corresponding values of I
C

[=(V
CC
-
V
CE
)/R
C
].


Repeat with V
BB

= 0.90 and 1.05 V. Plot on a single graph I
C

(ordinate) vs V
CE

(abscissa) for
each value of I
B
. The "operating region" of the t
ransistor is where the curves level off.
Determine


for the middle curve at V
CE

= 3 V. Is


generally constant?


Transistor as a switch

Reconnect the meters to measure V
CC

and V
BB
, set V
CC

to 6 V and decrease V
BB

to zero.
Replace R
C

with the light bulb
. Slowly increase VBB until the bulb is at maximum intensity.
Connect the two
-
way switch as shown in Fig. 3. Note the effect of operating the switch.
Measure V
BE

to determine the very small current I
B

that you are turning on and off to
control the much

larger current (~ 1A) through the light bulb.


Fig. 3 Switch connection

AC amplifier

Reconnect the circuit of Fig 1c with the meters to measure V
CC
and V
CE
. Increase V
CC

to 15
V and turn V
BB

down to zero. Increase V
BB

until V
CE

= 7.5 V. Connect the si
gnal generator
in series with V
BB

as shown in Fig. 2a and adjust it to 100 Hz. Connect the oscilloscope also
as in Fig. 2a
WITH THE BLACK LEADS TO GROUND ON THE TRANSISTOR
BOARD FOR BOTH CONNECTIONS
. Observe the input and output waveforms. Note
that adj
usting V
BB

causes distortion of the output waveform. Can you explain this? From
the ratio of peak
-
to
-
peak voltages, determine A
V

for frequencies of 100 Hz, 1 kHz, 10 kHz


5

and 20kHz. Would this amplifier be good as an audio
-
amplifier? Replace RC with a 1
20



resistor and note the effect on AV. Is this an expected result?