Download The transistor amplifier

VCE Physics 12: The transistor amplifier
[Revised Apr 2005. Note to teachers – you could delete blue answers to make worksheet]]
The circuit shown is a typical transistor amplifier circuit. Signal
voltages at Vin are amplified by ‘transistor action’ (see the text)
and appear at Vout.
Here we look at the various parts of this circuit in turn.
Vcc is the ‘supply voltage’, that is, the voltage supplied by a
battery, or by a ‘power supply’ circuit in the device being
considered. It needs to be a constant DC supply with little or no
ripple (this is achieved by diodes and a capacitor as discussed
earlier). Vcc is usually fixed somewhere between 5 and 15 volts.
Note that the lower line in diagrams such as this is assumed to be
connected to ‘earth’ and therefore at 0 V.
The transistor itself is explained in the text, but the basic idea is
that small currents (Ib around 10 A) going into the base (b) have a large effect on the current Ic and Ie (a few
mA) flowing through the collector and emitter connections. Ic and Ie and almost the same, Ie being just a little
more because Ie = Ic + Ib. The aim is to have the transistor ‘half on’, that is, arrange the base current so that the
transistor is half way between being turned off because Vb is too low (less than about 0.6 V) and fully on
because Vb is too high (greater than about 0.8 V) Silicon transistors are in this state when the base voltage is
about 0.7 V. Then small increases in Vb (and hence Ib) will cause Ic to increase considerably, while decreases
will cause Ic to decrease. Setting Vb correctly to about 0.7 V is called ‘biasing’ the transistor. In this circuit it is
achieved by the ‘voltage divider’ made up of R1 and R2.
The resistors R1 and R2 are a voltage divider which sets the ‘bias voltage’, Vb for the base of the transistor.
(Resistors can be represented by boxes like these or by zig-zag lines as in the text.) In this context the voltage
divider equation becomes: Vb = Vcc R2/(R1 + R2). It can also be convenient to express this as either:
R2 = R1 Vb/(Vcc – Vb) or R1 = R2 (Vcc – Vb)/Vb [This can be seen as simply the ratio of the ΔV’s = R ratios]
It is important to remember that the current through these resistors should be much greater than Ib so that Ib does
not itself affect the bias voltage. (Remember that Ib is very small so this is not a problem.) Typical values for the
resistors would be a few k, with R2 being several times greater than R1 (as Vb needs to be less than 1 V). This
means that the current through the resistors would typically be around V cc/Rtot  10 V/10 k  1 mA, compared to
Ib which will be a few microamps.
Sometimes the current through the transistor (Ic) is used to drive an output (such as a loudspeaker), but usually
the desired output is a varying amplified output voltage. This is where the collector resistor Rc (sometimes
called a load resistor) comes into the scene. As the current through it changes, so does the voltage across it. The
voltage at the top is always Vcc which is the supply voltage. Thus the voltage at the bottom, which is Vc, is less
than Vcc by the drop across Rc, or: Vc = Vcc – IcRc.
When the transistor is fully ‘turned on’, that is Vb (and hence Ib) are high, the collector current, Ic is high and so
Vc will be low. For this reason this sort of amplifier is called an inverting amplifier – input V goes up, output V
goes down. But remember that small increases in Vin lead to large decreases in Vout hence the signal is
amplified.
As explained previously, the capacitors Cin and Cout are to allow the signal (changing, or AC) voltage through
without allowing any steady (DC) component of the voltage through. Cin allows the small rapidly changing
input voltages (from a microphone or other amplifier stage etc.) to change Vb up and down, but only around the
steady bias voltage already determined by R1 and R2. Note that these changes will typically be measured in a
small number of millivolts, they are not large changes. At most, the signal voltage will only be changing V b
between about 0.6 and 0.8 volts. C2 performs a similar function at the output. If, for example, the output (Vc)
was connected directly to a loudspeaker, a steady current would flow from the supply through Rc and the
speaker. This current would both upset the level of Vc as well as wasting power by sending a useless current
through the speaker (and possibly damaging it). Capacitor Cout blocks the steady current but lets through the
changing signal voltage to drive the speaker.
The remaining resistor and capacitor in the emitter lead of the transistor are used to reduce the voltage gain
of the circuit. They are not always used, but have the effect of making the circuit more stable. Re would be a
relatively small value, perhaps a few hundred ohm. With about 1 mA flowing through it there would only be
less than 1 V across it. The actual Vbe voltage which controls the current flow in the transistor is Vb – Ve . Now
as Vbe increases, and the Ic current therefore increases, so does Ve . This has the effect of somewhat decreasing
Vbe and thus reducing the increase in Ic . This may seem a little counter productive, but in fact it has the effect of
making the circuit more stable. For example, if the transistor gets hot, the current will increase because electrons
are ‘released’ from their bonds by the heat. Thus the voltage across Re will increase, but this has the effect of
decreasing the Vbe (base-emitter voltage) and thus reducing the current. This is an example of ‘negative
feedback’, a technique used frequently in electronics to ensure stability of amplifying circuits.
(A similar situation applies when using a microphone near a loudspeaker. If the gain is too high, positive
feedback can make a horrible howl. The same sort of thing can occur in a circuit with too much voltage gain.)
Putting numbers into it
When working with actual values in circuits like these you should always work in milliamps and kilohms as
the currents and resistors in these circuits are mostly in these ranges. It is useful to note that if you use V = IR
with I in mA and R in k, the voltage will be in volts. This is because 1 mA  1 k = 0.001 A  1000  = 1 V,
that is, the milli’s cancel the kilo’s. Remember to express resistors such as 200  as 0.2 k and currents such as
10 A as 0.01 mA however.
The actual gain of the transistor is expressed as a or a current gain:
or voltage gain: (lower case v represents changing voltages)
AI = Ic/Ib
AV = Vout/Vin or = vout/vin
The current gain is typically 100 or more, that is, a current of, say, 10 A in the base will produce an Ic of 1 mA.
(Note that zero current in the base results in virtually zero Ic.)
The voltage gain depends on the actual values of the various resistors, but for our purposes it can normally be
approximated to AV = –Rc/Re (Don’t ask why this is the case, it is just one of those ‘rules of thumb’.)
As explained in the text, there are some ‘rules of thumb’ which designers use for circuits like this. They are:
1. Vc should be about ½Vcc. (So that Vc can move up and down around the normal value.)
2. Ve should be about 1/10th of Vcc, (that is, usually about 1 V) (This makes the circuit more stable.)
3. R2 should be about but not more than 10 Re. (So that the current in the divider is considerably greater than Ib.)
Let’s say that we have a transistor with a current gain of 200 which operates at a typical ‘midway’ current of Ic
= 5 mA. We will work out suitable values for the various resistors for a power supply voltage (Vcc) of 12 V.
First, let’s determine a suitable load resistor (Rc). This needs to be such that the output voltage is around 7 V for
our operating current of 5 mA:
That is, VRc = 12 - 7 = 5 and so Rc = 5/5 = 1.0 k
To find Re we first need to determine Ve - now Ve should be about 1.2 V from the ‘thumb’ rule. So
As Ie = 5 mA we need Re = 1.2/5 = 0.24 k = 240 
To find R1 and R2 we need to ensure that the current running through them is considerably more than the base
current. What will be the normal base current in this transistor?
Ib = Ic/200 = 5/200 = 0.025 mA = 25 A
Use thumb rule 3 to pick a value for R2 :
R2 = 10  Re = 2.4 k
Now find the appropriate value for R1 :
R1 = R2 (Vcc – Vb)/Vb and Vb needs to be Ve + 0.7 = 1.2 + 0.7 = 1.9 V
So R1 = 2.4 (12 – 1.9)/1.9 = 12.8 k
What is the current through the divider and is it suitable?
ID = Vcc/Rtot = 12/(2.4 + 12.8) = 0.79 mA = 790 A which is a lot more than the 25 A in Ib and so is suitable.
What is the approximate voltage gain for this amplifier? And what output voltage would a 20 mV signal
produce?
AV = Vout/Vin = –Rc/Re = –1 /0.24 = – 4.2 and so if vin = 20mV, vout = 84mV
KB April 2005