THE COMMON EMITTER AMPLIFIER - A More Advanced Look
The amplifier shown is the most commonly used small signal (low power) amplifier
using discrete transistors. It has long been replaced by the integrated circuit operational
amplifier (op amp) for small signal applications. The reason for examining this amplifier
is to understand what goes on inside an IC amplifier since people tend to forget that the
op amp and other ICs contain transistor circuitry based on the common emitter amplifier
and other transistor amplifier circuits. There are also plenty of these discrete circuits still
widely used in older equipment. Let us now examine this circuit.
To analyze the circuit we use a technique called “superposition”. Before you panic at this
exotic term, all it means is that we first analyze the circuit for DC by setting the AC
signal source Vs to 0V. We then set the DC source (Vcc) to 0V and analyze the circuit
for the AC values resulting when Vs is applied. The DC and AC values are then added
together to get the totals (the DC and AC values are superimposed on one another).
To start with the DC analysis, which you have all ready done when you did bias circuit
calculations, we regard all capacitors as open circuits (recall from AC that Xc = 1/2πfC
and f= 0Hz at DC so Xc = ∞……an open circuit). Our circuit now looks like this:
This is simply the DC bias circuit:
VB = (5.6K/32.6K)X 10V = 1.7V
VE = 1.7V – 0.7V
= 1.0V
= 1.0ma
Ic ≈ IE = 1.0V/1K
Vc = 10V – (1.0ma)(4.7K) = 5.3V
These are the DC bias values.
AC ANALYSIS
This is the more difficult part of the circuit analysis and we start by assuming that the
capacitors are approximate short circuits at the normal operating frequencies of the
amplifier, For example, at 5KHZ we get
Xce = 1/(2π X5000X25µf) = 1.3Ω
Since Ce is in parallel with Re2 it effectively short circuits it for AC. Similarly, C1 and
C2 are 32Ω each and are negligibly small compared to the resistors connected to them.
Calling the capacitors short circuits to AC will change the appearance of the circuit
considerably and we will introduce an AC equivalent circuit of the amplifier for our
analysis. There are two other items however that must be introduced. The Vcc terminal is
0VAC. There is no AC component at the Vcc terminal. Any point in a circuit that is
always 0V is the same as if it was grounded even if not directly connected to ground!
The Vcc terminal, for AC analysis is at (AC) ground.
The transistor is replaced by a mathematical model which is simplified here by ignoring
the internal capacitances of the transistor which only come into play at very high
frequencies, much higher frequencies than we deal with in this course.
The only internal resistance of the transistor
Of interest here is the internal resistance of the emitter
Region which is called
re’
From some calculus and the curves relating IE and VBE
The value of re ' is generally approximated as:
25mv
re'=
IEdc
ohms
IEdc is the bias value of IE used in the circuit. The AC equivalent circuit will look like
this:
re ' = 25mv/1ma =25Ω
You don’t need to know the derivations for the formulas which follow but they are
presented here for those few students who want to know “where they came from”. It is
customary to use lower case letters for AC quantities (v and i)
Voltage Gain is arrived at by starting with the basic
Av = vout/vin
From the AC circuit vout = icRc and vin = vB
But note here that vB is also equal to the voltage across re’ + Re1 which can be found
as:
vB = ie(re' + Re1 ) and recall that ic ≈ ie
Thus the voltage gain, Av = icRc/ic(re ' + Re1 ) which simplifies to Av = Rc/( re ' + Re1)
For our circuit then, we have
Av = 4.7KΩ/(25 + 100)Ω = 37.6
If vin, for example, is 100mv then vout will be 37.6 X 100mv = 3.76v
As discussed earlier in class, we need to know the input resistance of the amplifier so we
can determine how much loading of the signal source will occur when the amplifier is
connected. Similarly we need to know the Rout of the amplifier so that we can see how
connecting a load affects the output voltage.
Input Resistance is found as follows:
We first need the resistance looking into the base terminal
and we will call this rb
rb = vB / ib = ie(re'+ Re1)/ib
but ie ≈ βib so rb = βib(re ' + Re1)/ib = β(re ' + Re1)
Now, the total Rin is simply R1 ‫ ׀׀‬R2 ‫ ׀׀‬rb = R1 ‫ ׀׀‬R2 ‫ ׀׀‬β(re' + Re1)
For this circuit, if β = 100 then Rin = 27K ‫ ׀׀‬5.6K ‫ ׀׀‬100(25 +100) = 3.4KΩ
The output resistance of the amplifier is much simpler, it is simply Rc = 4.7KΩ
As noted earlier you don’t need to know the derivations of the formulas and here they are
in summary:
Av = Rc/(re' + Re1)
Rin = R1 ‫ ׀׀‬R2 ‫ ׀׀‬β(re' + Re1) Rout = Rc
How are these applied?
Let us go back to the original circuit and look at the effects of adding a load and a signal
source.
If we connect the signal source
with Vs = 100mv we will get,
at Vin an input voltage of:
Vin = Rin/(Rin + Rs) X Vs
= 3.4K/ (3.4K + 100) X 100mv
= 97.1 mv
This is a loading error of (100 – 97.1)/100 X 100% = 2.9% (not too bad)
With no load connected our Vout would be 97.1mv X 37.6 = 3.65V
With the 10KΩ load connected Vout will be Voutno load X RL/(Rout + RL)
Vout = 3.65V X 10K/(4.7K + 10K) = 2.48V
This is a loading error of (3.65V – 2.48V)/3.65V X 100% = 32% (ouch)
The total amplifier gain with the source and load connected is found as
Av = Voutloaded/ Vs = 2.48V/100mv = 24.8
The gain with no load and no signal source loading was 37.6 for a total drop of
( 37.6 -24.8 )/37.6 X 100% = 34%
It should be evident from this example that we want the amplifier Rin to be as high as
possible relative to the source resistance, and we want Rout as low as possible relative to
the load resistance. The common emitter amplifier has the highest voltage gain of the
various amplifier configurations but only a moderate Rin (to perhaps 10KΩ or so ) and a
high Rout since Rout must be high if voltage gain is to be high (look back at the gain
formula). It turns out that the common collector amplifier which we will examine later
has a high Rin and a very low Rout but only has a voltage gain of X1. The two amplifers
can work together to combine their best features!
Finally you may have wondered why we used two separate emitter resistors and only
bypassed one with a capacitor. Read on!
If we used a single Re and bypassed it fully with a capacitor, the voltage gain would be:
Av = Rc/re' = 4.7K/25 = 188
This is very impressive, however re' is approximate and is subject to some slight
variations due to temperature changes and variations in Ie. This means that although the
gain is very high, it is not stable. By using a partial emitter bypass as in our circuit, the
gain is lower but much more stable as the unbypassed 100Ω is much bigger than re' and
tends to “swamp out” the variations in re'. The choice of Re1, the unbypassed portion of
the emitter resistance is a compromise between high voltage gain and stability and will
depend upon the requirements of the application.
An assignment will follow.
FORMULA SUMMARY:
Voltage gain from the base to the collector with no load connected:
Av = Rc/(re' + RE )
where RE is any unbypassed RE
Input resistance of the amplifier:
Rin = R1 ‫ ׀׀‬R2 ‫ ׀׀‬β(re' + RE )
where RE is any unbypassed RE
Output resistance of the amplifier :
Rout ≈ Rc
When a signal source with an internal resistance Rs is connected the voltage into the base
input becomes:
Vin = Rin/(Rs + Rin) X Vs
When a load RL is connected to the amplifier output, the unloaded Vout becomes:
Voutloaded =
RL/(Rc + RL) X Voutopen
D Fuller March 2005