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The Common-Base Amplifier

Basic Circuit

Fig. 1 shows the circuit diagram of a single stage common-base amplifier. The object is to solvefor the small-signal voltage gain, input resistance, and output resistance.

Figure 1: Common-base amplifier.

DC Solution

(a) Replace the capacitors with open circuits. Look out of the 3 BJT terminals and make Thvenin

equivalent circuits as shown in Fig. 2.

VBB =V+R2 + V

R1R1 + R2

RBB = R1kR2

VEE = V REE = RE VCC = V

+ RCC = RC

(b) Make an educated guess for VBE. Write the loop equation between the VBB and the VEEnodes. To solve for IC, this equation is

VBB VEE = IBRBB + VBE + IEREE =ICRBB + VBE +

ICREE

(c) Solve the loop equation for the currents.

IC = IE = IB =VBB VEE VBERBB/+ REE/

(d) Verify that VCB > 0 for the active mode.

VCB = VC VB = (VCC ICRCC) (VBB IBRBB) = VCC VBB IC (RCC RBB/)

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Figure 2: DC bias circuit.

Small-Signal or AC Solutions

(a) Redraw the circuit with V+ = V

= 0 and all capacitors replaced with short circuits as shownin Fig. 3.

Figure 3: Signal circuit.

(b) Calculate gm, r, re, and r0 from the DC solution..

gm =ICVT

r =VTIB

re =VTIE

r0 =VA + VCE

IC

(c) Replace the circuits looking out of the base and emitter with Thvenin equivalent circuitsas shown in Fig. 4.

vtb = 0 Rtb = 0 vte = vsRE

Rs + RERte = RskRE

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Figure 4: Signal circuit with Thvenin emitter circuit.

Exact Solution

(a) Replace the BJT in Fig. 4 with the Thvenin emitter circuit and the Norton collector circuitas shown in Fig. 5.

Figure 5: Emitter and collector equivalent circuits.

(b) Solve for ic(sc).

ic(sc) = Gmevte = GmevsRE

Rs + RE

Gme =1

Rte + r0ekr0

r0 + r0

e

r0 + r0er

0

e =rx

1 + + re

(c) Solve for vo.

vo = ic(sc)rickRCkRL = GmevsRE

Rs + RErickRCkRL

ric =r0 + r

0

ekRte1 Rte/ (r0e + Rte)

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(d) Solve for the voltage gain.

Av =vovs

=RE

Rs + REGmerickRCkRL

(e) Solve for rin.

rin = R1kR2krie rie = r0

e

r0 + Rtcr0e + r0 + Rtc/ (1 + )

(f) Solve for rout.rout = rickRC

Example 1 For the CB amplifier in Fig. 1, it is given thatRs = 100, R1 = 120 k, R2 = 100 k,RC = 4.3 k, RE = 5.6 k, R3 = 100, RL = 20 k, V

+ = 15V, V = 15 V, VBE = 0.65 V,= 99, = 0.99, rx = 20, VA = 100 V andVT = 0.025V. Solve forAv, rin, and rout.

Solution. Because the dc bias circuit is the same as for the common-emitter amplifier example,the dc bias values, re, gm, r, and r0 are the same.

In the signal circuit, the Thvenin voltage and resistance seen looking out of the emitter aregiven by

vte = RE

Rs + REvs = 0.9825vs Rte = RskRE = 98.25

The Thvenin resistances seen looking out of the base and the collector are

Rtb = 0 Rtc = RCkRL = 3.539k

Next, we calculate r0e, Gme, ric, and rie.

r0e =Rtb + rx

1 + + re = 12.03 Gme =

1

Rte + r0ekr0

r0 + r0

e

r0 + r0e=

1

111.4S

ric =

r0 + r0

ekRte

1 Rte/ (r0e + Rte) = 442.3 k

rie = r0

e

r0 + Rtc

r0e + r0 + Rtc/ (1 + ) = 12.83

The output voltage is given by

vo = Gme (rickRtc) vte = Gme (rickRtc)RE

Rs + REvs = 30.97vs

Thus the voltage gain isAv = 30.97

The input and output resistances are

rin = R1kR2krib = 12.81 rout = rickRC = 4.259k

Approximate Solutions

These solutions assume that r0 = except in calculating ric. In this case, ic(sc) = i0

c = i0

e = ib.

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Figure 6: Simplified T model circuit.

Simplifi

ed T Model Solution(a) After making the Thvenin equivalent circuits looking out of the base and emitter, replace theBJT with the simplified T model as shown in Fig. 6.

(b) Solve for i0c and ric.

0 vte = i0

e

r0e + Rte

=

i0c

r0e + Rte

= i0c = vte

r0e + Rte

ric =r0 + r

0


(c) Solve for vo.

vo = i0

crickRCkRL = vte

r0e + RterickRCkRL = vs

RERs + RE

r0e + Rte

rickRCkRL


Av =vovs

=Rs

Rs + RE

r0e + RterickRCkRL

(e) Solve for rie and rin.

0 ve = i0

er0

e = i0

e = ver0e

rie =ve

i0

e

= r0e

rin = r0

ekRE


Example 2 For Example 1, use the simplified T-model solutions to calculate the values of Av, rin,and rout.

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Av = 0.9825

8.978 103

3.511 103

= 30.97

rin = 12 rout = 4.259k

Model Solution

(a) After making the Thvenin equivalent circuits looking out of the base and emitter, replace theBJT with the model as shown in Fig. 7.

Figure 7: Hybrid- model circuit.


0 vte = ibrx + v + i0

eRte =

i0crx +

i0cgm +

i0cRte = i

0

c =

vte

rx

+ 1gm

+ Rte

ric =r0 + r

0


(c) Solve for vo.

vo = i0

crickRCkRL =vte

rx

+1

gm+

Rte

rickRCkRL = vsRE

Rs + RE

1rx

+1

gm+

Rte

rickRCkRL


Av =vovs

=RE

Rs + RE

1rx

+1

gm+

Rte

rickRCkRL

(e) Solve for rout.rout = rickRC

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(f) Solve for rie and rin.

0 ve = ib (rx + r) =i0e

1 + (rx + r) = i

0

e = ve1 +

rx + r

rie =vei0e

=rx + r1 +

rin = riekRE

Example 3 For Example 1, use the-model solutions to calculate the values ofAv, rin, and rout.

Av = 0.9825

8.978 103

3.539 103

= 30.97

rin = 12 rout = 4.259k

T Model Solution

(a) After making the Thvenin equivalent circuits looking out of the base and emitter, replace theBJT with the T model as shown in Fig.??.

Figure 8: T model circuit.


0 vte = ibrx + i0

e (re + Rte) =i0crx +

i0c

(re + Rte) = i0

c =vte

rx

+re + Rte

ric =r0 + r

0


(c) Solve for vo.

vo = i0

crickRCkRL =vte

rx

+re + Rte

rickRCkRL = vsRE

Rs + RE

1rx

+re + Rte

rickRCkRL

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Av =vovs

=RE

Rs + RE

1rx

+re + Rte

rickRCkRL

(e) Solve for rie and rin.

0

ve = ibrx + i

0

ere =

i0e

1 + rx + i

0

ere = i

0

e rx

1 + + re

=

i

0

e =

ve

rx1 +

+ re

rie =vei0e

=rx

1 + + re

rin = REkrie


Example 4 For Example 1, use the T-model solutions to calculate the values of Av, rin, androut.

Av = 0.9825

8.978 103

3.539 103

= 30.97

rin = 12 rout = 4.259k

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