Contents

1 Problem 1.61 2

2 Problem 1.67 3

3 Problem 1.71 6

1

1 Problem 1.61

For the circuit in Figure 1, show that

vcvb

=βRl

rπ + (β + 1)Re(1)

and

vevb

=Re

Re + [rπ/ (β + 1)](2)

Figure 1: Circuit for P1.61

Solution:

Start by finding the equation to describe the base-emitter mesh

ib(rπ +Re)− vb + βibRe = 0 (3)

Solving (3) for vb gives

vb = ib(rπ +Re) + βibRe

vb = ib (rπ +Re + βRe)

vb = ib [rπ + (1 + β)Re] (4)

2

Now, by inspection, the relation for vc is

vc = −βibRl (5)

Utilizing (4) and (5) the following ratio can be deduced

vcvb

=−βibRl

ib [rπ + (1 + β)Re]

vcvb

=−βRl

rπ + (1 + β)ReX (6)

To prove (2), start by finding the relationship for the emitter voltage

ve = (ib + βib)Re

ve = ib(1 + β)Re (7)

Now, utilizing (4) and (5), the following relationship can be formulated

vevb

= ib(1 + β)Re

ib [rπ + (1 + β)Re]

vevb

=(1 + β)Re

rπ + (1 + β)Re(8)

Factoring out a (1 + β) from the denominator of (8) provides

vevb

=(1 + β)Re

(1 + β)[

rπ(1+β) +Re

]vevb

=Re[

rπ(1+β)

]+Re

X (9)

2 Problem 1.67

For the circuit shown in Figure 2, find the transfer function T (s) = Vo(s)/Vi(s), and arrange it in the appropriate

standard form from Table 1. Is this a high-pass or a low-pass network? What is the transmission at very high

frequencies? [Estimate this directly, as well as by letting s → ∞ in your expression for T (s).] What is the corner

frequency ω0? For R1 = 10 kΩ, R2 = 40 kΩ, and C = 0.1 µF, find f0. What is the value of |T (jω0)|?

3

Table 1: Table for P1.67

Figure 2: Circuit for P1.67

Solution:

The following transfer function is found by using a voltage divider

Vo(s) =R2

R1 +R2 + 1sC

Vi(s)

Vo(s)

Vi(s)=

sCR2

sC(R1 +R2) + 1

Vo(s)

Vi(s)=

s(

R2

R1+R2

)s+ 1

C(R1+R2)

. (10)

Using Table 1, it can be concluded that the above s-domain transfer function describes a high-pass filter. The corner

4

frequency is

ω0 =1

C(R1 +R2)(11)

=1

0.1 µF(10 kΩ + 40 kΩ

=1

5e 3

∴ ω0 = 200 rad/s (12)

Noting the following relationship

f0 =ω0

2π(13)

allows us to say that the corner frequency is

f0 =200 rad/s

2π

f0 = 31.8309 Hz (14)

Letting s→∞ gives

T (∞) = lims→∞

s(

R2

R1+R2

)s+ 1

C(R1+R2)

= lims→∞

(R2

R1+R2

)1 + 1

sC(R1+R2)

=

(R2

R1+R2

)1

=R2

R1 +R2

=40 kΩ

10 kΩ + 40 kΩ

∴ T (∞) = 0.8V

V(15)

Finally, to find the magnitude to the transfer function, note that

5

T (jω) =jω(

R2

R1+R2

)jω + 1

C(R1+R2)

. (16)

Therefore,

|T (jω)| =ω(

R2

R1+R2

)√ω2 +

(1

C(R1+R2)

)2

=ω(

40 kΩ10 kΩ+40 kΩ

)√ω2 +

(1

C(10 kΩ+40 kΩ)

)2

=0.8ω√

ω2 + 2002

∴ |T (jω)| = 0.8ω√ω2 + 4e4

(17)

3 Problem 1.71

The unity-gain voltage amplifiers in the circuit of Figure 3 have infinite input resistance and zero output resistances

and thus this function as perfect buffers. Convince yourself that the overall gain Vo/Vi will drop by 3 dB below the

value at dc at the frequency for which the gain of each RC circuit is 1.0 dB down. What is that frequency in terms of

CR?

Figure 3: Circuit for P1.71

The RC combination in each stage acts as a low pass filter, so the transfer function is

T (s) =1

1 + sω0

(18)

where ω0 is the cut-off frequency defined as

6

ω0 =1

RC(19)

T (jω) =1

1 + jωω0

(20)

Thus the magnitude is

|T (jω)| = 1√12 +

(ωω0

)2(21)

Now, if we take the log of both sides, we get

20 log(|T (jω)|) = 20 log

1√12 +

(ωω0

)2

−1 = 20 log

1√12 +

(ωω0

)2

−1

20= log

1√12 +

(ωω0

)2

(22)

After using MathCAD, we get

ω

ω0=√

0.2588

ω = ω0(0.5087) (23)

Now, substituting (19) into (23) gives

ω =1

RC(0.5087)

∴ ω =0.5087

RC(24)

7