Achi Ech

ECEN 326 LAB 10 Design of a BJT Shunt-Series Feedback Amplifier
1 Circuit Topology

Circuit schematic of the feedback amplifier to be designed in this lab is shown in Fig. 1.
VCC RC1 RS Vs Rin CE RE1 RE2 CB Q1 RF Q2 RC2 Vo CC RL Rout

Figure 1: Feedback amplifier.

1.1 DC Biasing
First, the emitter DC bias voltage of Q1 (VE1 ) needs to be determined. Since only IB1 flows through the resistorRF , the DC voltage drop on RF can be assumed to be negligible. Thus, VE2 and VC1 can be expressed as VE2 ≈ VE1 + 0.7 VC1 ≈ VE1 + 1.4 (1) (2)

To maximize the voltage swing at the output, load-line analysis is needed for the second stage. The DC equation including IC2 and VCE2 can be written as VCC ≈ RC2 IC2 + VCE2 + VE2 Figure 2 shows the AC load line to obtain the maximum output swing.
ic2(3)

DC bias point IC2

0 VCE,sat

VCE2 2VCE2−VCE,sat

vce2 For maximum symmetrical swing

Figure 2: AC load line. From the AC equivalent of Fig. 1, slope of the AC load line can be determined as ∆ic2 =− ∆vce2 (RC2 1 1 RL ) + RE2 (4)

Using the slope and the DC bias point (ic2 , vce2 ) = (IC2 , VCE2 ), the load line equation can be obtained as ic2 − IC2 =− vce2 − VCE2 (RC2 1 RL ) +RE2 (5)

Evaluating Eq. (5) at the point (ic2 , vce2 ) = (0, 2VCE2 − VCE,sat ), IC2 ((RC2 RL ) + RE2 ) = VCE2 − VCE,sat (6)

Solving Eqs. (3) and (6), the optimum IC2 to obtain the maximum symmetrical swing can be found as IC2 = VCC − 2VE2 − VCE,sat RC2 + (RC2 RL ) (7)

After determining IC2 , 0-to-peak voltage swing at the output can be calculated as Vsw = IC2 (RC2 RL ) = VCC − 2VE2 −VCE,sat VCC − 2VE2 − VCE,sat = RC2 RC2 1+ 2+ RC2 RL RL (8)

Since VE1 and VC1 are already determined, IC1 can be chosen based on other specifications. The remaining components can be calculated as RE1 = RC1 RE2 VE1 IC1 VCC − VC1 = IC1 VE2 = IC2 (9) (10) (11)

1.2 Feedback Analysis and Mid-band Frequency Response
AC equivalent of the amplifier in the mid-band frequency range is shown in Fig. 3.
ioRS vs ii ifb iε Q2 Q1 RC1 io RF RE2 RL RC2 vo

Feedback network

Figure 3: AC equivalent circuit. The input port of the feedback network in Fig. 3 is not directly connected to the output node (vo ). Therefore, the sampled output signal is not a voltage. Defining the output current as io = − vo RL RC2 (12)

it can be concluded that io is sampled by the feedback network. At the amplifier’s input,subtraction of the feedback signal is performed in the current domain, ii − if b = iε (13)

2

Therefore, the type of feedback is shunt-series. The next step is to obtain the g parameters of the feedback network as shown in Fig. 4.
i1 v1 RF RE2 i2 v2

g11f = g22f
v2

i1 v1

=
i2 =0

1 RF + RE2 RF

i1 v1 1 g11f g12f i2

g22f g21f v1

i2

v2 = i2 i1 i2

= RE2
v1 =0g12f =

=−
v1 =0

RE2 RE2 + RF

Figure 4: Calculation of the g-parameters of the feedback network. Replacing the feedback network with its two-port equivalent and converting the input source into current, the amplifier circuit can be arranged as in Fig. 5.
Ri2 vb2 iε vs =i RS s RS vε RF+RE2 Q1 RC1 RF ||RE2 io Q2 vo RL||RC2

f io

io

Feedback network

Figure 5: Amplifier with the idealfeedback network. From Fig. 5, assuming ro1 and ro2 are large, the parameters a and f can be obtained as follows vε = iε (RS (RF + RE2 ) rπ1 ) vb2 RC1 Ri2 =− vε re1 Ri2 = (β + 1)(re2 + (RF RE2 )) vo RL RC2 =− vb2 re2 + (RF RE2 ) vo = −io (RL RC2 ) io (RS (RF + RE2 ) rπ1 )(RC1 a= =− iε re1 (re2 + (RF RE2 )) RE2 f = g12f = − RE2 + RF The current-mode close-loop amplifier parameters are io a = is 1 +af zi Zi = 1 + af Zo = (1 + af )zo Ai = 3 (21) (22) (23) (14) (15) (16) (17) (18) Ri2 ) (19) (20)

where zi = R S (RF + RE2 ) rπ1 (24) (25) (26) rπ2 zo = RT + ro2 + gm2 ro2 RT rπ2 + (ro1 RC1 ) RT = RF RE2 [rπ2 + (ro1 RC1 )] Figure 6 shows the current-mode equivalent model of the amplifier.
io is Zi Ai is Zo RC2 RL

Figure 6: Current-mode equivalent model of the amplifier. As the final step,...