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Common-Emitter Amplifier
The circuit diagram of a common-emitter (CE) amplifier is shown in Fig. 1 (a). The capacitor CB is used to couple the input signal to the input port of the amplifier, and CC is used to couple the amplifier output to the load resistor RL. We are interested in the bias currents and voltages, mid-band gain, and input and output resistances of the amplifier.
amplifier
(a) (b)
R 1
R 2 RE^ CE
RC
CC CB vO vS
R 1
R 2
VCC
RE
RC
VCC RL
Figure 1: Common-emitter amplifier: (a) circuit diagram, (b) circuit for DC bias calculation.
Bias computation
The term “bias” refers to the DC conditions (currents and voltages) inside the amplifier circuit. The capacitors CB , CE , and CC are replaced with open circuits under DC conditions, and the circuit reduces to that shown in Fig. 1 (b). If the transistor β is assumed to be large (β → ∞), the base current can be neglected, and the R 1 - R 2 network is then simply a voltage divider, giving
VB = (^) RR^2 1 +^ R 2
VCC. (1)
For the circuit to operate as an amplifier, it is designed such that the BJT operates in its active region, with the B-E junction under forward bias and the B-C junction under reverse bias. The B-E voltage drop (VBE = VB − VE ) is about 0.7 V for a silicon BJT, and that gives us VE as
VE = VB − 0 .7 = (^) RR^2 1 +^ R 2
VCC − 0. 7. (2)
The emitter current IE is then obtained as IE = VE /RE , and IC = (^) β β+ 1 IE ≈ IE since we have
assumed β to be large. The DC collector-emitter voltage is
VCE = VC − VE = VCC − IC RC − IE RE ≈ VCC − IC (RC + RE ). (3)
The above procedure gives a good estimate of the DC bias quantities. If the base current IB is to be taken into account in the bias computation, the Thevenin equivalent circuit shown in Fig. 2 can be used. KVL in the B-E loop gives
R 2 R 2
R 1
R 1
VCC
VCC
IB IB
RC RC RC
IE
IC IC IC
VCE VCE VCE
VTh
VCC RTh VCC
Figure 2: Bias computation for the common-emitter amplifier with finite base current.
VTh = IB RTh + VBE + (β + 1) IB RE. (4)
The collector current IC is then given by
IC = βIB = β (^) R VTh^ −^ VBE Th + (β^ + 1)RE
where RTh = (R 1 ‖ R 2 ), and VTh = (^) RR^2 1 +^ R 2
VCC.
AC representation of an amplifier
amplifier
vi
Ro
AV 0 vi
Ri vo RL
Rs
vs
Figure 3: AC representation of an amplifier.
An amplifier can be represented by the AC equivalent circuit enclosed by the box in Fig. 3. Note that the signal source (voltage Vs with a series resistance Rs) and the load resistance RL are external to the amplifier. The coupling capacitors (CB and CC ) are not shown in the AC circuit since their impedances are negligibly small in the “mid-band” region (see Fig. 4). The amplifier equivalent circuit is characterised by the input resistance Ri (ideally infinite), output
(a)
(b)
BJT
CE amplifier
R 1
vs
RC
vo^ RL
R 1 R 2 RC RL
CB CC
RE CE
R 2
vs vo
vbe
gmvbe
vbe
E
B C
ro
rπ gmvbe
ro
B C
E
rπ
AC ground
Figure 5: (a) Small-signal equivalent circuit of a CE amplifier, (b) simplified circuit after replac- ing the coupling and bypass capacitors with short circuits.
To measure AV L and AV 0 , we apply a sinusoidal input voltage^1 (vs in Fig. 1 (a)) and measure vo with RL in place and with RL → ∞ (i.e., open circuit), respectively.
Input resistance Ri
The input resistance of an amplifier can be found by applying a voltage vs and measuring by some means the current^2 iin shown in Fig. 6 (a) to obtain Ri = vs/iin. From the AC equivalent circuit of Fig. 5 (b), we can see that the input resistance is
Ri = (R 1 ‖ R 2 ‖ rπ), where rπ = β gm
= βVT IC
with IC being the bias (DC) value of the collector current. A simple way for experimental measurement of Ri is shown in Fig. 6 (b). We connect the input voltage source to the amplifier^3 through a variable resistance (pot) Rs. Keeping (^1) vs must be sufficiently small to ensure that the output voltage is purely sinusoidal. (^2) Note that vs and iin are AC quantities. (^3) The coupling capacitor CB is not shown explicitly in Fig. 6 (b), but it must be connected so that the bias
(a) (b)
vs^ vi
iin Ro
AV 0 vi
Ri vo RL
Ri
vs^ vi
Ro
AV 0 vi
Ri vo RL
Ri
Rs
Figure 6: (a) Theoretical interpretation of Ri, (b) practical technique to measure Ri.
vs constant, we then vary Rs and measure vo. If v o′ corresponds to Rs = 0 Ω, then the input resistance Ri is equal to the value of Rs which gives vo = v o′/2.
Output resistance Ro
(a) (b)
R 1 R 2 RC gmvbe
vbe (^) ro
Ro Ro
vi Ri
Ro
AV vi rπ E
B C
Figure 7: (a) AC equivalent circuit of an amplifier with vs = 0, (b) AC equivalent circuit of the CE amplifier with vs = 0.
If we apply vs = 0 to the generic amplifier shown in Fig. 3, we obtain the circuit shown in Fig. 7 (a). Since vi = 0, the dependent voltage source gets replaced by a short circuit, and looking at the circuit from the output port, we only see Ro. If we do that for the AC equivalent circuit of the CE amplifier (Fig. 7 (b)), we get
Ro = ro ‖ RC ≈ RC , (10)
since ro of a BJT is typically much larger than RC. To experimentally measure Ro of the CE amplifier, we can use a procedure similar to that discussed for Ri. We connect a variable load resistance RL (through a suitably large coupling capacitor) as shown in Fig. 8. Keeping vs constant, we first measure vo ≡ v o′ with RL → ∞ (open circuit). Ro is given by the value of RL which gives vo = v′ o/2.
values are not disturbed.
RE with (RE 1 + RE 2 ). For computing the AC quantities of interest (gain, Ri, Ro), we use the circuit shown in Fig. 9 (b). Since ie = (β + 1) ib, the resistance RE 1 appears as (β + 1)RE 1 as seen from the base, and we can write
vs = ib [rπ + (β + 1)RE 1 ]. (12)
The output voltage is vo = −β ib × (RC ‖ RL), (13)
and the gain with load AV L is therefore
AV L = v vo s
= − (^) r β^ (RC^ ‖^ RL) π + (β^ + 1)RE 1
If (β+1)RE 1 ≫ rπ, AV L → − (RC^ ‖^ RL) RE 1
, and the open-circuit gain AV 0 = AV L|RL→∞ = − RC RE 1
Note that the gain of the CE amplifier with partial bypass is less than that of the CE amplifier (compare Eqs. 7 and 14) as we would expect from an amplifier with negative feedback^4. The input resistance, by inspection of Fig. 9 (b) is Ri = R 1 ‖ R 2 ‖ (rπ + (β + 1)RE 1 ), (15)
and the output resistance is Ro ≈ RC , assuming ro of the BJT to be large. An important point to note is that the base-emitter small-signal voltage vbe in this case is much smaller than vs (see Fig. 9 (b)):
vbe vs^ =^
rπib rπib + (β + 1)RE 1 ib^ =^
rπ rπ + (β + 1)RE 1.^ (16)
As a result, the small-signal condition vbe ≪ VT means that vs^ rπ rπ + (β + 1)RE 1
≪ VT or
vs ≪ VT
( 1 + (β^ + 1)RE^1 rπ
) , i.e., a larger vs can be applied (as compared to the CE amplifier)
without causing distortion in the output voltage.
References
- A.S. Sedra and K.C. Smith and A.N. Chandorkar, Microelectronic Circuits Theory and Applications. New Delhi: Oxford University Press, 2009.
- P.R. Gray and R.G Meyer, Analysis and Design of Analog Integrated Circuits. Singapore: John Wiley and Sons, 1995.
- M.B. Patil, Basic Electronic Devices and Circuits. Prentice-Hall India: Delhi, 2013.
(^4) The feedback involved in the CE amplifier with partial bypass is of the series-series type. On the output side, the output current ic causes a voltage drop RE 1 ie ≈ RE 1 ic across RE 1 , and this voltage drop gets subtracted from the input voltage vs.
