Lab 5 Common Emitter Amplifier Discussion The emitter amplifier

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Lab 5
Common Emitter Amplifier
Discussion
The emitter amplifier with emitter degeneration is displayed in fig. 5.1. The
choice of values for resistors RC , RE , R1 , and R2 constitutes the design of the
amplifier. The amplifier as drawn is for a.c. signals. The a.c. signal vi is coupled to the amplifier by the input capacitance Ci and the output appears after
capacitor Co . The design of the amplifier is not an exact process in the sense
that we will be able exactly to choose the right values of the components and
obtain exactly the response we want. On the other hand, using the input and
output characteristic curves from lab 4 we shall be able to make reasonable first
choices for the values of the resistors. Before we go into that design process let
us redraw the amplifier as in fig. 5.2 using the h parameters. This will enable
us to see the importance of the different components. First to be noted in the
equivalent circuit of fig. 5.2 is that this is an a.c. circuit so the bias network
R1 −R2 , as well as the capacitors have been dropped. Also notice that the input
signal vi and the output signal v0 are measured relative to ground and that the
emitter is not at ground potential. The response function of the amplifier is
called amplification and is defined by
A = v0/vi
(1)
Using Kirchoff’s laws we obtain the following relations;
about loop o : ic RC + vo = 0
(2a)
about loop i: ibhie + hre vce + ie RE = vi
emitter node: ib + hf e ib = ie
collector node: ic + io = hf eib
(2b)
(2c)
(2d)
Using equations 2 we can write for A
A = −hf e RC /(hre vce /ib + hie + (1 + hf e)RE )
(3)
This expression for A can be substantially simplified once we realize tha hf e is
a big number and usually hre and hie are small. To the extent that hf e RE >>
hie + hre vce/ib the amplification reduces to
A = −RC /RE
(4)
The negative sign is important because it tells us that there is a 180◦ phase
shift between the input and output signals.
We can now start the design analysis. This will consist in giving us the correct
resistances so that the circuit has the desired quiescent point. The quiescent
point is the steady state operaton of the amplifier when no input signal is applied. We choose the q. point so the amplifier operates in a reasonably linear
portion of the I-V characteristic of the output circuit. Attached are graphs
of a typical transistor which will be used for illustration. From graph G1 we
choose IC = 4 mA for the quiescent collector current. Suppose we decide that
the supply voltage should be +15V, and we want the amplification A = 10.
The choise of VC is usually VCC /2 because then the output swings symetrically
about VC with the least chance of clipping vo . We determine the values of RC
and RE by use of Kirchoff’s law in the output circuit.
VCC = IC RC + VCE + IE RE = VCE + IC (RC + RE )
and emitter currents are equal) (5)
so the load line equation is
IC = (VCC − VCE )/(RCC + RE )
( assuming collector
(6)
Notice that in deriving the load line equation we assumed IE is nearly equal to
IC . For VC = VCC /2 we get for the quiescent current ICQ
ICQ = VCC /(2ARE )
(7)
Hence now we can determine the desired values of the output circuit resistors.
Output Circuit Design
(A) Choose A = 10, ICQ = 4 mA,VCC = +15V, VC = +7.5V
(B) from equation (7) we obtain RE = 188 Ω
(C) from equation (4) RC = 1880 Ω
(D) from equation (6) we obtain the IC intercept when VCE = 0. This enables
us to draw the load line since the the intercept on the VCE axis occurs when
VCE = VCC and then IC = 0.
(E) Hunting for resistor in the lab turns up two resistors, RE = 176Ω and
RC = 1794Ω. These are close enough to the design values so that we can proceed to the bias network design.
Bias Network Design
(A) We want the quiescent value of VE to be VEQ = ICQ ∗RE = 4mA∗176Ω = 0.7V
(B) In general for a silicon transistor the base-emitter voltage difference is on
the order of 0.6V ( a forward bias pn junction) . So VB = 1.3V . Actually, we
can make a better estimate of VB by using the IC vs VBE curve that we measured for the Ebers-Moll model. Investigating graph G2, fig. 5.G2, we see that
for a collector current of 4 mA we need VBE = 0.64V . Thus we must ensure
that VBE = VEQ + 0.64V = 1.34V
(C) If the output impedance of the bias network were very small compared
to the input impedance of the transistor looking into the base then we could
simply choose R1 , R2 from the voltage divider equation. It turns out that the
input impedance of the the transistor can be quite large and is on the order of
hf e RE . You can obtain this by analysing fig. 5.2. Let us suppose that the input
impedance of the transistor is in fact Rin = hf e RE . The equivalent impedance
of the bias network is obtained from Thevenin’s theorem, Req = R1 ||R2 ( parallel combination). The equivalent circuits are in fig.5.3a and fig.5.3b. A sensible choice for Req = Rin/10. Since hf e is about 200 from fig.5.G1 we get
Rin = 35KΩ so choose Req = 3.5KΩ. The value of VB in fig.3b is then
VB = Veq Rin /(Rin + Req )
(8)
So then Veq = 1.47V
(D) Knowing Veq and Req we calculate that R1 = 35.7KΩ, R2 = 3.9KΩ.
Hunting about the lab we find resistors that give us R2 = 3901Ω and R1 =
30.40K + 5.873K = 36.3KΩ. (Here we added two resistors to get the desired
value of R1.
The circuit in fig. 5.1 was assembled with these components and the quiescent point values are shown in table 1.
Table 1- Quiescent point values
item
design
actual
VCC
+15V
+14.99V
VC
7.5 V
7.67 V
VE
0.7 V
0.723V
IB
23.7µA
RC
1880Ω
1794Ω
RE
188Ω
176Ω
R1
35.7KΩ
36.3KΩ
R2
3900Ω
3901Ω
Procedure
(1) Build your amplifier according to your design results. Make a table like
Table 1. You can determine IB by placing a small resistor in place of the conductor from points A to B in fig. 5.1 and measuring the potential drop across
it.
(2) Choose Co , Ci of about 0.15µF . Place an a.c. signal across vi and measure
the amplification A = vo /vi. You will need to reduce the sine wave generator
using a voltage divider because the sine wave signal is too large on its own.
Choose three frequencies that span the range of the sine wave generator and
measure the amplification at these frequencies. Consider in your observations
both the absolute value of A and the phase difference between vi and vo .
(3) The amplifier we built is called the common emitter amplifier with emitter
degeneration. Despite this rather pejorative name it has very good properties.
You can see from eqn. 4 that the amplification depends mainly on the ratio of
two external resistors. These resistances do not change much when the temperature changes. In addition the amplifier has a farily flat frequency response and
constant phase shift. This ensure that the output signal will be a reasonably
faithful replica of the input signal except for a change in magnitude and an
overall change in sign. The amplifier is a linear device. Recall from Fourier
analysis that if the Fourier components of the input signal get treated differently by the network then the ouput shape will be different from the input shape.
In contrast, the transistor h parameters are sensitively dependent on temperature. If you were to design an amplifier that depends on the exact values of
the h parameters you would have a bad design. In our design we have used the
h parameters in only a qualitative way, namely is parameter x much bigger or
smaller than parameter y.
If you must have the largest possible amplification and do not care about the
linearity of the amplifier you place a bypass capacitor across resistor RE , that is
in parallel, as in fig.5.4. Choose a value of CE of about 0.22µF (This does not
affect the quiescent point. Why?) With the bypass capacitor in place we see
from fig. 5.2 that the impedance bewteen E and G is now frequency dependent.
In fact, at high enough frequencies the impedance ZEG approaches zero and RE
is essentially by passed (for signals !). From eqn. (3) the amplification is then
A = −hf e RC /hie
(9).
(4) Choose two frequencies of the sine wave generator such that the output
signal does not get clipped. Observe the amplification and phase shift at both
these frequencies and record their values.
Figure 1: Measured amplification for 123AP npn transistor vs theory. data (circles), theory(x’s
and triangles)
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