12.1. STRUCTURE OF A CLASS A AMPLIFIER
INTRODUCTION
Let us suppose the amplifier we are looking at has to amplify a
sinusoidal signal as shown in fig. 140.
fig. 140
Applying our knowledge about transistors you probably would come
to suggest a circuit similar like shown in fig. 141. where a tiny signal source
supplies a base-emitter-junction and the big collector-current is originated
from the power-supply. Here is mentioned a NPN-transistor, but the principle is
exactly the same with a PNP-transistor.
fig. 141
WHAT IS THE COLLECTOR RESISTOR NECESSARY FOR?
The circuit shown in fig. 141 would be in the theoretical sense
of the word already an amplifier.
But the sinusoidal current flowing here in the so called
collector-circuit, is not useful anywhere.
Neither is it operating anything (like a loudspeaker) nor is it
producing a signal at the output which could be transmitted to another amplifier
stage.
This is the reason why we find in all amplifiers at the output
either a component which will use the signal current itself or at least cause a
voltage drop. This part of the amplifier is called the collector resistor Rc, as
shown in fig. 142a.
fig. 142a
If there is flowing now a collector current Ic it will cause a
proportional voltage-drop at this collector resistor.
WHAT IS THE POINT OF “QUIESCENCE”.
Referring to the example in the last chapter we can easily state
that:
VRC = VS -
VCE
With the formula in the last chapter we can now derive that the
voltage VRC can vary between the two extreme values.
1. Ic = 0 then VRc = 0
2. Ic =
Maximum VRc = Vs
(whereby IC maximum =
VS/RC)
fig. 142b
fig. 142c
At the output of the amplifier, we want to find an exact
sinusoidal signal of exactly the same frequency again. This can only be achieved
if the output-voltage can oscillate, which means: it must be free to increase
and to decrease, from the level which it has at an input-voltage of 0 Volts
- a status which will be called from now on:
QUIESCENCE. BIASING WHAT FOR?
If the input-signal would be directly connected to the base of
our amplifier transistor as shown in fig. 142d we would find only a change of
the output voltage during the period between point 1 and 2 as long as the input
signal is above 0.6 Volts. This would not be fitting to the precondition
developed for a class A amplifier in the last chapter.
fig. 142d
To set the output to the so called quiescent-voltage we have to
make sure that at quiescence at the input, there is already flowing a
collector-current which must be about half of the maximum collector current,
therefore we have to let flow an input-current (base-current) even if there is
not yet any input-signal.
fig. 142e
We call this method BIASING our amplifier circuit. The simplest
circuit to achieve this is shown in fig. 143. Here a base-current is allowed to
flow which is big enough to keep a collector current flowing which is half of
its maximum. But if you have a short look back to the input-characteristics of a
transistor, you can easily see, that a tiny change of the input voltage would
cause a tremendous change in base-current, which would cause again a tremendous
change of collector-current, and therefore too a tremendous change of the
output-voltage. This means, that the quiescence-voltage is not stabil. As will
be seen later, this would have very big disadvantages for the working of our
amplifier.
fig. 143
To reduce this problem, it is prefered to bias with
voltage-dividers like shown in fig. 144.
fig. 144
REPRESENTATION OF THE QUIESCENT-POINT IN A
FOUR-QUADRANT-CHARACTERISTICS OF THE AMPLIFIER TRANSISTOR.
We can find the so called quiescence-point as well in our
four-quadrant-characteristics.
Therefore we have to enter into the first quadrant the loadline
of the collector-resistor. The transistor should have the characteristics given
in fig. 143. The quiescent-voltage should be in this case about 4.5 Volts.
To achieve this, we have to supply the transistor with a
base-current of about 0.14 mA. And therefore we need a base-emitter-voltage of
0.75 V. As you see in fig. 144a.
fig. 144a
HOW TO INJECT NOW THE INPUT SIGNAL TO SUCH A AMPLIFIER CIRCUIT?
If we could connect a signal-source directly to the
input-terminals of the amplifier as shown in fig. 144 the signal-source would
pass a dc-current through the lower resistor.
But the signal-source is always a very weak energy-source. So by
such a dc-current the signal could be distorted very heavily. The problem can be
solved by the following idea:
We know, the input-signal, is always an ac-voltage (at least in
radio-technology). Therefore we have to make sure that only an ac-current is
allowed to enter the input terminals of the amplifier-circuit.
As you know from chapter 8.2 the capacitor is the component
which we need here.
If we connect such a capacitor in series there can only flow an
ac-current through the inputcircuit.
fig. 145
HOW TO GET THE SIGNAL OUT OF THE STAGE?
As long as the amplifier is not the last one in a row of stages,
its signal has to be entered to another stage.
The next stage will be constructed similar like the stage we
came to know just now.
If we would connect another stage directly without a capacitor
between the stages, there would again flow a dc current from one stage to the
other, and this would mean again distortion.
Therefore: at the output we will find another capacitor. Between
two stages, very often the input-capacitor is at the same time the
output-capacitor. Both capacitors are called: COUPLING CAPACITORS.
fig.
146
