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VCE Physics Electronics and Photonics The n-p-n transistor voltage amplifier
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VCE Physics Electronics and Photonics

The n-p-n transistor voltage amplifier

1. Introduction
There is a significant change in the content of the electronics component of the VCE Physics,
Unit 3, Electronics & Photonics, and this concerns the dropping of the operational amplifier and
the inclusion of (amongst other items) the n-p-n transistor in a voltage amplifier circuit. While
this is a relatively sm
as additional support and resource material to assist students and their teachers as they study this
all part of the overall Unit 3, it is not an easy topic. These notes are provided
transistor topic.

It is assumed that you are reasonably familiar with the diode. If a diode is forward biased the
p-type material (the anode) is at a higher potential (voltage) than the n-type material (the
cathode), and in this situation a significant current can flow through the diode. If the p-n junction
is reverse biased then effectively no current flows. It is also assumed that you are fairly confident
in using Kirchhoffs voltage and current laws and Ohms law, and understand the way resistors
and capacitors act in a circuit.

2. Definitions
The n-p-n transistor could be viewed as two p-n diodes joined back-to-back with a common
anode (p-type semiconductor material) region as the base of the transistor, as depicted in Figure
1. This is not really valid, but for the base, B, and emitter, E, regions (the lower part of the
transistor as it is normally drawn) this diode-like behaviour may be assumed to a fair
approximation. In this way, when forward biased, the base-emitter, B-E, junction will have a
voltage across it approximating the silicon diode cut-in voltage, V , so V
BE
= V = 0.7 V.
















Figure 1: The n-p-n transistor structure and symbol.

Note that in Figure 1 for this bi-polar junction transistor, BJT, the arrow on the emitter part of the
symbol indicates the direction of conventional current flow when the B-E junction is forward
biased; thus coinciding with the arrow head direction on the normal diode symbol.

To construct and analyse a voltage amplifier based around an n-p-n BJT is not particularly
straightforward so a few basic concepts and definitions are in order.
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The intrinsic behaviour of a BJT is that of a current amplifier where, when correctly biased,
the collector current, I
C
, is an amplified version of the base current, I
B
. The current
amplification factor is represented by (sometimes also denoted by h
fe
) such that I
C
= I
B
,
with typically in the range 50 to 200 depending on the transistor make/model. We shall
choose = 100 for the purposes of this discussion ( is always positive).

Voltage amplification arises by taking account of the fact that a current, I, passing through a
resistor of resistance R (the collector resistor, R
C
, is pertinent here) provides a voltage across
the resistor of V = IR (Ohms law).

The BJT transistor amplifier required in the VCE Physics Unit 3 is in the so-called common
emitter (CE) configuration where the input voltage, V
IN
, is between the base and the emitter
side, and the output voltage, V
OUT
, is between the collector and the emitter side. The emitter
side of the transistor is thus common to both voltages.

The n-p-n BJT CE amplifier in normal operation has the base-emitter, B-E, junction forward
biased and the collector-base, C-B, junction reverse biased.

The n-p-n BJT CE voltage amplifier is an inverting amplifier where the timevarying, AC,
output voltage, v
OUT
, is 180� out of phase (or inverted) with respect to the time-varying AC
input voltage, v
IN
. From this we can write that v
OUT
= -A
V
v
IN
, where A
V
is the gain
(amplification magnitude) of the amplifier; the subscript V simply indicate that we are talking
about voltage amplification.

It is possible to have two types of BJT CE voltage amplifiers; one based on an n-p-n
transistor and the other based on a p-n-p transistor. The n-p-n transistor is preferred here at an
introductory level as all the important DC currents and DC voltages in the analysis end up
being sensibly defined as positive quantities (they are negative in the p-n-n case).

The BJT CE voltage amplifier will only amplify time-varying input voltage signals above a
certain frequency; they are not able to amplify DC voltages like an operational amplifier, and
because of this we need to distinguish carefully between the AC and DC components of all
relevant currents and voltages. This tends to be a troublesome issue for many, but we get
around this here by using the following convention: upper case I and V for DC (time
invariant, constant) currents and voltages, and lower case i and v for AC (time-varying, and
in general we shall only consider sinusoidal) currents and voltages. To these symbols we shall
add subscripts that describe the point at which the current or voltage exists (as we have
already done above in some instances).

There are a few simple texts that discuss this type of amplifier and many start with the
specific transistor current-voltage characteristics, but this approach is expressly denied in the
VCE Physics Study Design. These texts then often move on to measuring the transistor
characteristics using two DC batteries or power supplies, and then on to practical circuits.
The notes here treat a real n-p-n BJT CE voltage amplifier, commencing with a realistic
circuit and analyse this using simple models of diodes, resistors and capacitors that you
should be familiar with.

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3. Amplifier circuit
We start with the complete amplifier circuit shown in Figure 2, where all the relevant DC currents
and are indicated using a sensible convention as to direction and polarity. So all currents, I
B
, I
E

and I
C
, go in a direction towards the common/ground connection. All voltages are measured with
respect to this common/ground line at the foot of the circuit diagram.



















Figure 2: The basic n-p-n CE voltage amplifier circuit.

Typical component values are as follows, and the circuit will work if you use these: R
1
= 5 k,
R
2
= 25 k, R
C
= 4 k, R
E
= 1 k, C
IN
= 10 �F, C
E
= 47 �F, while the DC power supply is
V
CC
= 20 V.

We will now analyse this circuit in as simple a way as possible, section-by-section, using only the
model parameters defined along with Kirchhoffs laws and Ohms law. In doing this we will see
how the AC voltage input, v
IN
, is amplified to the AC voltage output, v
OUT
. However, to do this
we will analyse the circuit in two stages; first by only considering the DC currents and voltages to
set up the conditions for amplification, then second, by considering just the AC current and
voltage time-varying fluctuations. We shall look at each circuit area indicated in Figure 1

4. The input circuit
This section comprises only the AC input voltage, v
IN
, and the input capacitor, C
IN
(see Figure 2).
The rest of the circuit can be represented by a single resistor we will denote by R
B
(you will have
to take my word for this!) with R
B
1 k.

You know that a capacitor will not pass DC current (it is an open circuit), but may pass AC
current depending on the current frequency and the time constant, , of the resistor-capacitor
(here R
B
and C
IN
) combination. The time constant is given by = R
B
C
IN
. The circuit design has
to ensure that C
IN
is chosen so that the time constant is comparable to the lowest period, T, of a
sinusoidal input voltage signal that you wish to amplify. So, for example, if you wish to amplify
voltages with frequencies above 500 Hz (so T < 2 ms) we use the condition = 5 R
B
C
IN
= 0.01 s,
and with R
B
= 1 k we end up with C
IN
= 10 �F as suggested in the previous section. The factor
of 5 in the time constant expression is to allows the full smoothing effect of the R-C combination.
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The capacitor, C
IN
, has another very important function and that is to block any DC current
component from v
IN
from disturbing the DC set-up of the overall circuit. The only change to the
base current we want is from the AC input signal. This current is given by i
B
= v
IN
/R
B
.

5. Voltage divider circuit
This part of the circuit comprises the two resistors R
1
and R
2
and the DC power supply, so we are
only interested in the DC currents and voltages at this stage. The purpose of this voltage divider is
to set the DC voltage of the base, B, of the transistor to ensure the B-E junction is forward biased.
If the transistor were to be removed completely then V
B
= (V
CC
R
1
)/(R
1
+ R
2
).

In a practical amplifier circuit this expression for V
B
is not correct as the transistor loads the
divider circuit, but the result is not too far out, so we would calculate the base voltage to be
approximately V
B
= (20 � 5)/(5 + 25) = 3.3 V. This DC voltage is adequate to ensure the B-E
junction is forward biased and that V
BE
= 0.7 V (approximately). We have already said that the
transistor can be thought to present a resistance of around R
B
= 1 k so there now will be a DC
an