PHY405F 2009
EXPERIMENT 6 – SIMPLE TRANSISTOR CIRCUITS
Due Date (NOTE CHANGE): Thursday, Nov 12th @ 5 pm; Late penalty in effect!
Most active electronic devices are based on the transistor as the fundamental solid state component. We
want to help you develop some understanding of its characteristics. The most common form of transistor
is called a Bipolar Junction Transistor (BJT). There are two types of BJT, NPN and PNP, with different
circuit symbols. The letters refer to the solid state properties of the layers of semiconductor material
used to make the transistor. Most transistors used today are NPN because they are the easier type to
manufacture from silicon. The leads are labelled base (B), collector (C) and emitter (E).
The three leads which must be connected correctly in any circuit before switching on any power
supplies. A wrongly connected transistor will be damaged instantly. You will find the appropriate
locations of the three wire leads on the plastic package by consulting the spec sheet of the transistor you
use.
WHAT TO DO:
1. Testing transistors. Select an NPN and a PNP transistor pair such as the 2N3904 and the 2N3906.
You will need a basic multimeter. Set a digital multimeter to “diode test”; an analogue multimeter
to a low resistance range.
Test each pair of leads in both directions (six tests in total). The base-emitter (BE) junction
should behave like a diode and conduct one way only. The base-collector (BC) junction should
behave like a diode and conduct one way only. The collector-emitter (CE) should not conduct
either way. The diagram shows how the junctions behave in an NPN transistor. The diodes are
reversed in a PNP transistor but the same test procedure can be used. Record all your data.
2. The transistor switch. The transistor is basically an amplifier of current. A small current injected
between the base and the emitter “encourages” a larger current to flow from the collector to the
emitter. The phenomenon can be easily illustrated by constructing the circuit below.
Connect an NPN transistor into the circuit shown above which uses the transistor as a switch.
The supply voltage is not critical, anything between 5 and 12V is suitable. The LED should
light when the switch is pressed and go out when the switch is released. Repeat the experiment
using a PNP transistor in the same circuit but reverse the LED and the supply voltage. For both
experiments measure the current into the base and collector and the current out of the emitter.
Also record the base-emitter and collector-emitter voltages in the two states. Increase the value
of the 10k resistor until the LED no longer switches. Deduce the maximum current gain of the
transistor.
3. The Common Emitter AC Amplifier The basic style of a BJT common emitter amplifier is
shown in the figure below. Let us go through the design of the amplifier and the selection of
components. I used the following reference
http://www.electronicstutorials.ws/amplifier/amp_2.html
Choose the transistor and the voltage supply Vcc – I suggest 2N3904 and 12 VDC. The quiescent base
voltage (Vb) is determined by the potential divider network formed by the resistors, R1 and R2 and the
power supply voltage Vcc and is given as:
The supply voltage also determines the collector current, Ic when the transistor is fully "ON"
(saturation), Vce = 0. The base current Ib for the transistor is found from the collector current, Ic and
the DC current gain beta, β of the transistor.
The next steps require the family of characteristic curves for the transistor shown below. The curves
are determined experimentally. They consist of plots of the collector current against the collector
emitter voltage for range of values of the base current.
Lets assume a typical load resistor, RL of 1.2kΩ. We can calculate the collector current (Ic) flowing
through the load resistor when the transistor is switched fully "ON" by assuming Vce is near zero.
Given Ic, we can find an Re with a voltage drop of say 1V across it.
We can plot a load line on the characteristic curves. The point "A" on the collector current vertical
axis of the characteristic curves occurs when Vce = 0 and IC=9.2 mA.
Now when the transistor is switched fully "OFF", there is no voltage drop across either resistor Re
or RL as no current is flowing through them. Then the voltage drop across the transistor, Vce is
equal to the supply voltage, Vcc. This is point "B" on the horizontal axis of the characteristic curves.
Generally, the operating or Q-point of the amplifier is selected to be half-way along the load line so
the collector current will be given as half of 9.2mA. Therefore, at Q, IC= 4.6mA.
The DC load line produces a straight line equation whose slope is given as: -1/(RL + Re). It crosses
the vertical Ic axis at a point equal to Vcc/(RL + Re). The actual position of the Q-point on the DC
load line is determined by the mean value of Ib.
The collector current of the transistor, Ic is also equal to the product of the DC gain beta and the
base current (β x Ib). If we assume a (minimum) β value for the transistor of say 100, the base
current Ib flowing into the transistor will be given as:
Resistors, R1 and R2 are chosen to give a quiescent base current of 92uA. The current flowing
through the potential divider circuit has to be large compared to the actual base current Ib. Assume a
value of 10 times Ib flowing through the resistor R2 (rule of thumb). Then the value of R2 is given
by:
If the current flowing through resistor R2 is 10 times the value of the base current, then the current
flowing through resistor R1 must be 11 times the value of the base current at 1012uA or 1mA. The
voltage across resistor R1 is equal to Vcc - 1.7v (Vre = 0.7 for silicon transistors) which equals
10.3V, therefore R1 is given as:
So, for our example above, the preferred values of the resistors chosen to give a tolerance of 5%
are:
Then, our original Common Emitter circuit above can triumphantly be rewritten to include the
values of the components that we have just calculated. In the “old” days, hours were spent on these
kinds of calculations. Try to reproduce the results using electronic workbench.
Build the circuit and measure the quiescent (DC) operating voltages and currents. Do they differ
from those calculated? If so, what erroneous assumption may have been made? What value of C1
should therefore be selected to avoid any attenuation of signals above 100 Hz? Measure the input
impedance of the circuit? What should the value of C2 be to bypass AC signals above 100 Hz
around the resistor Re? If C2 is included in the load line calculation, is the resulting AC load line
different from the DC load line we calculated? Measure the AC gain of the circuit above 100 Hz
and the greatest sinusoidal input voltage that can be amplified without any distortion of the output
signal. Measure the effect of removing C2 on the gain of the circuit?