Transcript cont`d

TUTORIAL 4 – Questions & Answers
Q1. The power amplifier in Figure 1 is required to deliver 10 W to the load
resistance RL at a single frequency sinusoidal input signal, vI. Both QN and
QP are matched transistors and have the following parameters;
ISQ = 1.45 x 10-12 A and  = 50
D1 and D2 are identical diodes with ISD = 1.45 x 10-12 A . The diode current is
assumed to follow the exponential law;
I D  I SD eVD / VT
and the transistor collector current is assumed to follow the exponential law;
I C  I SQeVBE /VT
where;
VT  26 mV
TUTORIAL 4 – Questions & Answers
Q1. (cont’d)
(a) Determine the value of Ibias to ensure the diode current iD  1 mA
at any time.
(b) Calculate the collector currents of QN and QP when vO = 0 .
You may assume that iBN = iBP = 0 under this condition.
(c) Determine the required amplitude of the input voltage, vI.
TUTORIAL 4 – Questions & Answers
Q1. (cont’d)
Figure 1
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(a) The value of Ibias to ensure the diode current iD  1 mA at any
time.
The r.m.s value of output
voltage is given by;
vO rms 
R
2
 P  10 W
vO rms   10  8 W
 8.944 V
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(a) (cont’d)
The peak value is;
vO p   2  8.944
 12.65 V
The peak load current is;
iL  p  
VO p 
RL
12.65

8
 1.581 A
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(a) (cont’d)
This load current is supplied by QN
i.e. iN. Hence;
iN p   iCN p   1.581 A
and
iBN p  
iCN p 

1.581

 31.6 mA
50
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(a) (cont’d)
To guarantee a minimum of 1 mA
diode current, select
I bias  33 mA
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(b) The collector currents of QN and QP when vO = 0.
When vO = 0, the whole of Ibias
flows through the diodes
(assuming iBN is negligible).
ID  ISe
VD / VT
or;
 ID 
VD  VT ln  
 IS 
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(b) (cont’d)
Substituting values;
 33  10 3 

VD  26 ln 
12 
 1.45  10 
 620 mV
Since the diodes are identical,
VBB  2VD  1.24 V
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(b) (cont’d)
and;
VBEN  VEBP
VBB

 620 mV
2
iCN 0   iCP 0 
 I SQeVBE / VT
 1.45 10 12 e 620/ 26
 32.9 mA
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(c) The required amplitude of the input voltage, vI.
From (a), the amplitude of vO is
12.65 V and the corresponding
base current iBn is 31.6 mA. The
diode current is therefore;
iD  I bias  iBN
 33  31.6
 1.4 mA
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(c) (cont’d)
And at this value of diode current;
 ID 

VD  VT ln 
 I SD 
 1.4 10 3 

 26 ln 
12 
 1.45 10 
 538 mV
VBB  2VD  1.075 V
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(c) (cont’d)
The peak load current occurs at
the peak value of the output
voltage which corresponds to the
peak value of the input voltage.
This current approximately equals
the emitter current iN. Hence;
iL  iN  1.581 A
and
iCN  1.581

 1
50
 1.581
 1.55 A
50  1
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(c) (cont’d)
Using the equation
vBEN
I C  I S eVBE / VT
 iCN 

 VT ln 
I 
 SQ 
 1.55

 26 ln 
12 
 1.45 10 
 720 mV
TUTORIAL 4 – Questions & Answers
Q1 – Solution
(c) (cont’d)
vI  VBB  vBEN  vO
Substituting values;
vI  1.075  0.72  12.65
vI  12.3 V
TUTORIAL 4 – Questions & Answers
Q2. Class-AB power amplifier in Figure 2 employs input buffer transistors
Q1 and Q2 to establish the required quiescent bias current. Q1, Q2, Q3
and Q4 are matched transistors. Derive an expression for the
approximate current gain, Ai where;
iO
Ai 
iI
State any assumption made in your derivation.
TUTORIAL 4 – Questions & Answers
Q2. (cont’d)
Figure 2
TUTORIAL 4 – Questions & Answers
Q2 – Solution
Applying KCL at the input node;
iI  iB 2  iB1
Assuming iB3 is negligibly
small compared to iE1;
V   VEB1  vI
iB1 
1  1 R
TUTORIAL 4 – Questions & Answers
Q2 – Solution
Assuming iB4 is negligibly
small compared to iE2;
iB 2
vI  VBE 2  V 

1   2 R
TUTORIAL 4 – Questions & Answers
Q2 – Solution
Hence;
iI  iB 2  iB1
vI  VBE 2  V  V   VEB1  vI


1   2 R
1  1 R
Since all the transistors are matched;
1   2   ;
vI  VBE  V   V   VEB  vI
iI 
1   R

2vI  V  V

1   R

VBE  VEB
V   V 
TUTORIAL 4 – Questions & Answers
Q2 – Solution
Also, since
V   V 
2vI  V   V 
2vI
iI 

1   R
1   R
V   V 
TUTORIAL 4 – Questions & Answers
Q2 – Solution
Since the voltage gain of
the amplifier is assumed
unity,
vO  vI
and;
vO
vI
iO 

RL RL
V   V 
TUTORIAL 4 – Questions & Answers
Q2 – Solution
The current gain is;
iO v I
2v I

1   R
Ai  


iI
RL 1   R
2 RL
In deriving the above expression, the following assumptions were made:
(a) All transistors are perfectly matched;
(b) The voltage gain of the system is unity;
and
(c) The base currents of Q3 and Q4 are negligibly small as
compared to the emitter currents of Q1 and Q2 respectively.
TUTORIAL 4 – Questions & Answers
Q3. All the transistors in class-AB power amplifier shown in Figure 3 are
matched. If  = 40, VBE(npn) = VEB(pnp) = 0.7 V and V+ = V = 12 V;
(a) determine iE1, iE2, iB1 and iB2 when vI = 0;
(b) determine iO, iE1, iE2, iB1, iB2 and iI when vI = 5 V;
(c) use the results of part (b) to determine the current gain,
Ai where;
iO
Ai 
iI
TUTORIAL 4 – Questions & Answers
Q3. (cont’d)
Figure 3
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(a) iE1, iE2, iB1 and iB2 when vI = 0;
Since all transistors are perfectly
matched, all their respective
quiescent currents (when vI = 0)
are equal i.e.
iB1  iB 2  iB 3  iB 4
iC1  iC 2  iC 3  iC 4
i E 1  iE 2  iE 3  iE 4
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(a) (cont’d)
V   vEB1
iR 
R
12  0.7

250
 45.2 mA
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(a) (cont’d)
iR  iE 1  iB 3
iE 3
 iE1 
1 
iE1
 iE1 
1 

1 

 iE1 1 
 1  
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(a) (cont’d)
iE 2  iE 1
 1  

 iR 
2 
 1  40 
 45.2

 2  40 
 44.1 mA
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(a) (cont’d)
i B 2  i B1
 iE1 

 
1  
 44.1 


 1  40 
 1.08 mA
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(b) iO, iE1, iE2, iB1, iB2 and iI when vI = 5 V;
vO  vI  5 V
vO
iO 
RL
5
  0.625 A
8
iE 3  iO  0.625 A
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(b) (cont’d)
Consider the Q1 – Q3 pair;
iE 3
iB 3 
1 
0.625

 15.2 mA
1  40
V   vEB1  vI
iR 
R
12  0.7  5

 25.2 mA
250
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(b) (cont’d)
i E1  i R  i B 3
 25.2  15.2  10 mA
iE1
iB1 
1 
10

 0.244 mA
1  40
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(b) (cont’d)
Consider the Q2 – Q4 pair;
vI  vBE 2  V 
iR 
R
5  0.7  12

250
 65.2 mA
Neglecting iB4;
iE 2  iR  65.2 mA
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(b) (cont’d)
iE 2
1 
65.2

 1.59 mA
1  40
iB 2 
iI  iB 2  iB1
 1.59  0.244  1.35 mA
TUTORIAL 4 – Questions & Answers
Q3 – Solution
(c) The current gain, Ai;
Ai 
iO
iI

0.625
 463  53.3 dB
0.00135
Comparing with the
expression of approximate
gain;
Ai

1   R

2 RL

1  40250  641  56.1 dB
28
TUTORIAL 4 – Questions & Answers
Q4. All the transistors in class-AB power amplifier shown in Figure 4 are
matched. The parameters of the transistors are:  = 60 and
IS = 5 x 10-13 A . Let V+ = 10 V and V = 10 V.
(a) Determine the quiescent collector currents in the four transistors
when vI = vO = 0.
(b) If RL = 200  and vO(peak) = 6 V, determine the current gain,
Ai and the voltage gain, Av of the circuit where;
iO
Ai 
iI
and
vO
Av 
vI
TUTORIAL 4 – Questions & Answers
Q4. (cont’d)
Figure 4
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(a) The quiescent collector currents
in the four transistors when
vI = vO = 0.
iE1  iB 3  3 mA
iE 3
iE1 
 3 mA
1 
Because the transistors are
matched;
iE1  iE 3  iE
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(a) (cont’d)
Hence;

1 
iE 1 
  3 mA
 1  
 1  

iE  3
2 
 1  60 
 3

 2  60 
 2.952 mA
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(a) (cont’d)
iE
1   
2.952

 48.4 μA
1  60
iB 
iC  iB
 60  48.4 μA  2.903 mA
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) The current gain and voltage gains
when RL = 200  and vO(peak) = 6 V.
vO
6
iO 

 30 mA
RL 200
iE 3  iO peak   30 mA
iB 3
iE 3

1 
30

 0.492 mA
1  60
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) (cont’d)
iE 1  3  i B 3
 3  0.492  2.51 mA
iE1
1 
2.51

 41.1 μA
1  60
iB1 
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) (cont’d)
Neglecting iB4;
iE 2  3 mA
iE 2
iB 2 
1 
3

 49.2 μA
1  60
iI  iB 2  iB1
 49.2  41.1  8.1 μA
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) (cont’d)
iO
30
Ai  
 3704
iI 0.0081
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) (cont’d)
VBE 3
 iE 3 
 VT ln  
 iS 
 30  10 3 

 0.026 ln 
13 
 5  10 
VBE3  0.6453 V
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) (cont’d)
 iE 1 
VEB1  VT ln  
 iS 
 2.508  10 3 

 0.026 ln 
13
 5  10

VEB1  0.5807 V
TUTORIAL 4 – Questions & Answers
Q4 – Solution
(b) (cont’d)
vI  vO  VBE 3  VEB1
 6  0.6453  0.5807
 6.0646 V
Av 
vO
vI
6

6.0646
 0.989