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8 Problems with Solutions on Electronics I - Assignment 8 | ECE 340, Assignments of Basic Electronics

University of Illinois - ChicagoBasic Electronics

Material Type: Assignment; Class: Electronics I; Subject: Electrical and Computer Engr; University: University of Illinois - Chicago; Term: Spring 2004;

Typology: Assignments

2011/2012

Uploaded on 05/18/2012

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2004 University of Illinois at Chicago ECE 340 V. Goncharoff
Web code:
Check your lab meeting time:
Tuesday
Thursday
8 AM
11 AM
2 PM
Last name:
First name:
Signature:
ECE 340 - Electronics I
Homework #8: due Wednesday 4/21/04.
(Clearly draw the circuits that you have designed and show all work on pages stapled to this cover sheet.)
In each of the problems assume that
β
> 100 and ro ≈ 100 KΩ.
vout
+
−
vin
+
−
Your Circuit
Figure 8.1
1. Design a common-emitter NPN BJT amplifier circuit having 25 dB voltage
gain ( inoutvvvA /
= as shown in Figure 8.1). Use a single +5 volt d.c. power
supply, and make the value of C
I= 1.0 mA.
2. Repeat Problem 1, this time using dual ±5 volt d.c. power supplies and a
current-mirror transistor biasing scheme.
3. Design a common-base NPN BJT amplifier circuit having 40 dB voltage gain
(inoutvvvA /
= as shown in Figure 8.1). Use a single +10 volt d.c. power
supply, and make the value of C
I= 2.0 mA.
4. Design a common-collector NPN BJT amplifier circuit having approximately
unity voltage gain ( inoutvvvA /
= as shown in Figure 8.1). Use a single +6 volt
d.c. power supply, and make the value of 02.0
=
m
g mhos.
pf3
pf4
pf5

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Download 8 Problems with Solutions on Electronics I - Assignment 8 | ECE 340 and more Assignments Basic Electronics in PDF only on Docsity!

Web code:

Check your lab meeting time:  Tuesday  Thursday

 8 AM  11 AM  2 PM

Last name:

First name:

Signature:

ECE 340 - Electronics I

Homework #8: due Wednesday 4/21/04. (Clearly draw the circuits that you have designed and show all work on pages stapled to this cover sheet.) In each of the problems assume that β > 100 and ro ≈ 100 KΩ.

vout

−

v in

−

Your Circuit

Figure 8.

  1. Design a common-emitter NPN BJT amplifier circuit having 25 dB voltage gain ( (^) Av = vout / vin as shown in Figure 8.1). Use a single +5 volt d.c. power supply, and make the value of I (^) C = 1.0 mA.
  2. Repeat Problem 1, this time using dual ±5 volt d.c. power supplies and a current-mirror transistor biasing scheme.
  3. Design a common-base NPN BJT amplifier circuit having 40 dB voltage gain ( Av (^) = vout / vin as shown in Figure 8.1). Use a single +10 volt d.c. power supply, and make the value of I (^) C = 2.0 mA.
  4. Design a common-collector NPN BJT amplifier circuit having approximately unity voltage gain ( Av = vout / vin as shown in Figure 8.1). Use a single +6 volt d.c. power supply, and make the value of g (^) m = 0. 02 mhos.

vout

−

Your Circuit

R S= 50 Ω

vgen

−

Figure 8.

  1. Design a common-emitter NPN BJT amplifier circuit having 40 dB voltage gain ( Av = vout / vgen as shown in Figure 8.2). Use a single +15 volt d.c. power supply, and make the value of I (^) C = 1.0 mA.
  2. Repeat Problem 5, this time using a PNP transistor.
  3. Design a common-base NPN BJT amplifier circuit having 20 dB voltage gain ( Av = vout / vgen as shown in Figure 8.2). Make the value of re = 50 Ω. Use a +10 volt d.c. power supply.

R S = 50 Ω

R L = 50 Ω vout

−

v gen

−

Your Circuit

Figure 8.

  1. Referring to Figure 8.3, design a common-base NPN BJT amplifier circuit having the value of g (^) m = 0. 02 mhos. What is the highest voltage gain that your circuit may produce ( (^) Av = (^) vout / vgen )?

when designing the biasing network. This also makes I (^) C fairly insensitive to β (see Homework 7 solution).

  • This design problem is greatly simplified by the fact that RS = 0. As a result the values of RB (^) 1 , RB (^) 2 and r π have no effect on voltage gain.
  • Since RC || ro ≈ RC , we have ignored the effects of ro on voltage gain.
  • VC = 5 − (1 mA)(462 Ω) = 4.54 volts > VB = 2.5 volts, so the BJT is not in saturation.
  • The peak output signal amplitude is 4.54 − 2.5 ≈ 2.0 volts to stay out of saturation, so therefore v (^) in (max) = 2.0/17.87 ≈ 100 mV peak. However, remembering the small-signal linearity assumption that | vbe |(max)< VT , we should keep v (^) in (max) ≈ 25 mV peak.
  1. Repeat Problem 1, this time using dual ±5 volt d.c. power supplies and a current-mirror transistor biasing scheme.

Solution: Replace RE with a current source, and the only other change has been to replace RB (^) 1 and RB (^) 2 with a single resistor R (^) B = 10 K Ω(to provide a d.c. path for I (^) B ):

462 Ω

  • 5 V

10 KΩ

vin

vout

− 5 V

4.3 KΩ

  1. Design a common-base NPN BJT amplifier circuit having 40 dB voltage gain ( Av = vout / vin as shown in Figure 8.1). Use a single +10 volt d.c. power supply, and make the value of I (^) C = 2.0 mA.

Solution: 40 dB= 20 log 10 ( 100 ), so Av = gmRC = 100. I (^) C = 2.0 mA = gm VT , so gm = 2 mA / 26 mV= 1 / 13 mhos. Thus R (^) C = 100 / gm = 100 × 13 Ω= 1. 3 K Ω:

Comments about Problem 3:

  • VC = 10 − (2 mA)(1.3 KΩ) = 7.4 volts > VB = 5.0 volts, so the BJT is not in saturation.
  • The peak output signal amplitude is 7.4 − 5.0 = 2.4 volts to stay out of saturation, so therefore v (^) in (max) = 2.4/100 = 24 mV peak. This also satisfies the small-signal linearity assumption that (^) | vbe (^) |(max)< VT.
  1. Design a common-collector NPN BJT amplifier circuit having approximately unity voltage gain ( Av = vout / vin as shown in Figure 8.1). Use a single +6 volt d.c. power supply, and make the value of g (^) m = 0. 02 mhos.

Solution: g (^) m = IC / VT = 0. 02 mhos, so I (^) C = VTgm = 26 mV×0.02mhos= 0.52 mA. Thus RE ≈ VE / 0. 52 mA= 4.4 KΩ when VB = VCC / 2 = 3 volts. Note that a collector resistor is no longer required, and tying the BJT collector directly to VCC saves us one large bypass capacitor.

2.15 K Ω

2.5 K Ω 1.3 K Ω

  • 10 V

2.5 K Ω

vin

vout

vout

−

Your Circuit

R S= 50^ Ω

vgen

−

Figure 8.

  1. Design a common-emitter NPN BJT amplifier circuit having 40 dB voltage gain ( Av = vout / vgen as shown in Figure 8.2). Use a single +15 volt d.c. power supply, and make the value of I^ C = 1.0 mA.

Solution: Let RB (^) 1 (^) = RB 2 =7.5 KΩ, so that RB (^) 1 (^) || RB 2 = 3.75 KΩ. Call this value RB = 3.75 KΩ. From the amplifier specs., 40 dB= 20 log 10 ( 100 )so Av = (( r π || RB )/(( r π|| RB )+ RS ))(− gmRC )=− 100. The value of g (^) m = IC / VT = 1 / 26 mhos, so r π ≈ β/ gm = 100 × 26 Ω = 2.6 KΩ. The ratio r π /( r π + RS )= 2600 /( 50 + 2600 )= 0. 98 ≈ 1 , so the 50 Ω source resistance doesn't affect Av much: Av ≈ − gmRC =− 100 , yielding RC = 2. 6 KΩ:

6.8 K Ω

7.5 K Ω 2.6 K Ω

  • 15 V

7.5 K Ω

vgen

vout 50 Ω