Tutorial work - digital circuits tutorial sheet 1 and 2 with solutions PDF

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 1

FEEG1004 Electrical and Electronic Systems Electronics tutorial sheet 1

1. An ideal diode is defined as having a voltage-current characteristic as shown in Figure 1. i Reverse bias

Forward bias

v

Figure 1: An ideal diode voltage-current characteristic

Figures 2(a)-(d) show circuits containing ideal diodes. In each case, determine the values of I and Vout.

(a)

(b) 2.5 kΩ I

+5V

(c) -5V

2.5 kΩ Vout

2.5 kΩ

(d) I

Vout

-5V

Figure 2

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I

+5V

Vout

2.5 kΩ I

Vout

FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 1

2. (a)

In the circuit shown in Figure 3, assume vin to be sinusoidal, and the diode to be ideal. Sketch the transfer characteristic of the circuit, i.e. vout (y-axis) verses vin (x-axis).

vD

vin

vout

~

Figure 3

(b)

Again, for the circuit shown in Figure 3, sketch the diode voltage vD verses time. Include vin on your sketch .

3. Assuming the diodes in Figure 4 to be ideal, describe the transfer characteristic of the circuit.

10kΩ

vin

5V

~

10kΩ

5V 10kΩ

Figure 4

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vout

FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 1

4. An AC voltmeter can be modelled by the circuit within the dashed box in Figure 5. The diode is assumed to be ideal. The moving coil meter gives a fullscale reading when the average current flowing through it is 1 mA. The meter has a resistance of 50 Ω. Determine the value of R which will result in a full-scale reading for a sinusoidal test voltage of 20 V peak-to-peak. (Hint: the average value of a half-cycle of a sine wave with amplitude A is 2A ).

π

vtest

R

~

meter

Figure 5

5. The input and output voltages for a particular half-wave rectifier are shown in Figure 6. The rectifier itself is a circuit similar to that shown Figure 3, except the diode is not ideal in this case, and has a forward-biased voltage drop across it of 0.7V. If the input signal is sinusoidal with a peak-to-peak value of 20V, determine the total angle per cycle that the diode is conducting (conduction angle).

Volts

vout 0 time

vin Figure 6

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 1

6. A full-wave rectifier with smoothing capacitor is shown in Figure 7. The diodes have forward-biased voltage drops each of 0.7 V. Vout is required to have a value of 10 V with a ripple of 0.2 V in order to maintain an average load current I = 5 mA. If the frequency of vin = 50 Hz (a) what is the value of the smoothing capacitor? (b) what should be the peak-to-peak value of vin?

I vin

~ C

Figure 7

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vout

R

FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 1

FEEG1004 Electrical and Electronic Systems Electronics tutorial sheet 1 - SOLUTIONS 1. An ideal diode behaves like an open circuit in the reverse direction and a short circuit in the forward direction. Therefore:

(a) Vout = 0V; I = Vin/R = 5/2500 = 2mA (b) Vout = 5V; I = 0A (c) Vout = -5V; I = 0A (d) Vout = 0V; I = Vin/R = -5/2500 = -2mA

2. (a) (ideal) half-wave rectifier: vout 1 1

vin

(b) vD = vin - vout Volts

vin 0 time

VD

3. -5V < vin < +5V

vout = vin

vin < -5V

vout = vin/2 - 2.5V

vin > +5V

vout = vin/2 + 2.5V

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 1

4. The circuit is similar to a ha lf-wave rectifier. Therefore the voltage is 0 over half of every cycle. The average voltage is therefore the average over a half cycle divided by 2. Vave = 20/2π = 3.183V Total resistance = Vave/Iave = 3.183/1mA = 3183Ω Therefore R = 3183 - 50 = 3133 Ω 5. When diode begins to conduct, the cycle of the sine wave has progressed an angle of sin-1 [0.7/amplitude] = sin-1 [0.7/10] = ~4 degrees. Therefore ~8 degrees lost per half cycle. The other half cycle no conduction. Therefore total conduction angle = 360 – 180 – 8 degrees = 172 degrees Equivalently 2π – π – 0.14 rads = 3 rads

6. (a) From the course notes page 18 iDC 5mA   250 F 2fΔV (2)(50)(0.2) (b) peak output voltage = 10 + ripple/2 = 10.1V. C 

peak voltage + 2 diode voltage drops = 10.1 + 0.7 + 0.7 = 11.5 V. therefore peak-to-peak vin = 23 V.

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

FEEG1004 Electrical and Electronic Systems Electronics tutorial sheet 2 1. For the operational amplifier circuit shown below, determine the circuit transfer function.

100 kΩ

25 kΩ

50 kΩ

100 kΩ

V1 25 kΩ

V2 50 kΩ

150 kΩ V out

2. Prove using truth tables, the following Boolean identities: (a) (A.B )  ( A.B)  A  B (b) A.(B  C)  (A.B)  (A.C)

3. (a) Obtain a Boolean expression for the output F for the logic circuit diagram shown below. (b) Show that F  A.B.C by using Boolean algebra only.

A B F

C

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

4. Use a Karnaugh map to simplify the following ‘sum of products’ expressions. (a) F  A.B.C  A.B.C  A.B.C  A.B.C (b) F  A.B.C  C.D  A.B.C.D  B.C.D

5. The input resistance of an inverting operational amplifier is limited by the value of input resistor R1. If a high input resistance is required then for large gains the value of the feedback resistor (RF in the course notes) must be much higher still. The stability of these high value resistors can be poor. The ‘T’ feedback network, shown below, overcomes this problem by increasing the gain of the amplifier. By taking into consideration the ‘golden rules for op-amp circuit analysis’ show that the transfer function of the circuit is given by:

Vout R  R R   2 1  4  4 Vin R1  R2 R3

  

Vx I2

I 4 R4

R2 I3

I1

R3

R1

Vin

+

V1

Vout

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

FEEG1004 Electrical and Electronic Systems Electronics tutorial sheet 2 - SOLUTIONS 1.

V1'

100 kΩ

25 kΩ

V3

50 kΩ

100 kΩ

V1 25 kΩ

V1'

V2

150 kΩ 50 kΩ

A

B

C

Vout

D

A: Potential divider C: Summing amplifier B: Voltage follower D: Non-inverting amplifier

Section A

V1 ' 

25kΩ 1 V1  V1 (100  25)kΩ 5

Section B

V1 '  V1 ' Section C

50kΩ  50kΩ  V 3   V1 '  V2  100kΩ   25kΩ  50 1  5  V3   . V1   V2   25 5  10  1 2 V3   V1  V2 2 5 Section D

 150kΩ  V out  1  V3  4V3 50kΩ   1 John Mills 2014

FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

1  2  Vout  4  V1  V2  5 2   Circuit transfer function:

8 Vout   V1  2V2 5 2. (a)

B

A

AB

0 0 1 1

0 1 0 1

0 1 1 0

A.B 0 1 0 0

(A.B )  ( A.B) 0 1 1 0

A.B 0 0 1 0

(A.B )  ( A.B)  A  B

(b) C

B

A

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

A.B

A.C

(A.B)  ( A.C)

B C

0 0 0 1 0 0 0 1

0 0 0 0 0 1 0 1

0 0 0 1 0 1 0 1

0 0 1 1 1 1 1 1

A.(B  C)  (A.B)  ( A.C)

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A.(B  C)

0 0 0 1 0 1 0 1

FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

3.

(a) A

A.B

B

F  A.B  (B  C)

C B C

(b) F  A.B  (B  C)  ( A.B).(B  C)

 (A.B).(B.C  B.C)

(Algebra helped with identities given on page 74 of the course notes)

 (A.B).( B.C ).(B.C )  (A.B).( B  C).(B C )  ( A.B).( B.B  C.B  B.C  CC )  ( A.B).(C.B  B.C )

 A.B.C.B  A.B.B.C

 A.B.C

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

4. (a)

F  A.B.C  A.B.C  A.B.C  A.B.C

AB C 0

00 1

01 0

11 0

10 1

1

1

0

0

1

Therefore F  B (b)

F  A.B.C  C.D  A.B.C.D  B.C.D

AB CD 00

00 1

01 1

11 1

10 1

01

1

1

1

1

11

0

0

0

0

10

0

0

0

0

Therefore F  C

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FEEG1004 - Diodes, Transistors, Op-amps and Digital Electronics TUTORIAL SHEET 2

5.

Vx I2

I 4 R4

R2 I3

R3

R1

I1

-

I1 

+

V1

Vin

Vout

V in V1 V in  R1 R1

I 2  I1 

Vin R1

V x  V1  I 2R 2  0 

I3 

Vin R R2   2 Vin R1 R1

0 V x R2 Vin  R3 R1R3

I4  I2  I3 

V in R2 Vin  R1 R 1R 3

V out  V x - I 4R 4  

 V R2 R V in   in  2 Vin  R4 R1   R1 R1 R3

R Vout R  R   2  4 1  2 Vin R3  R1 R1 

Vout R  2 Vin R1

  

 R4 R4   1    R2 R3 

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