|Radio and Electronics (DED Philippinen, 66 p.)|
Let us suppose the amplifier we are looking at has to amplify a sinusoidal signal as shown in 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.
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.
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)
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.
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.
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.
To reduce this problem, it is prefered to bias with voltage-dividers like shown in 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.
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.
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.