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# Tutorial 7

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Tutorial 7
IC
RB
IB
VCE
VBB
VCC
VBE
IE
Fig. 1(a)
1.
Figure 1(a) shows an npn-transistor is biased in Common-Emitter connection. The transistor
curves are shown in Figure 1(b).
a) In normal operation of a transistor, the emitter-junction is (forward, reverse)* biased and
the collector-junction is (forward, reverse)* biased.
b) What is the relation between IE, IC and IB?
c) Why IB is very much smaller than IC?
 dc
d) Define dc and dc of a transistor and show that  dc 
.
1   dc
e) Calculate the value of dc and dc at point A, B and C.
f) Define ac and ac of the transistor and calculate the value of ac and ac for the change
from point B to point C.
g) The collector current, IC is controlled by (IB, VCE)*. Explained why the transistor can be
considered as a controlled current source.
h) What is VCE,sat? What is ‘cutoff’ point?
i) There is extremely small collector current when IB is zero, where is the current come
from and what is it called?
j) In transistor curves, what does ‘active regions’ means?
Note: *(select the right answer)
For all problems, the transistor is silicon with VBE = 0.7V.
+15V
RB
4k
600
Fig. 3
2. In the amplifier shown in Figure 3, βdc = 60. Draw the dc equivalent circuit for the amplifier
and find the value of RB required to set VCE to 8V. Draw the dc load line and show the Q
point, (IC,VCE).
+VCC
+30V
RC
20k
4k
RB
dc = 50
10k
Fig. 4
5k
Fig. 5
3. For the voltage-divider bias circuit shown in Figure 4, determine the Q-point (IB, IC, VCE).
Draw the dc load-line for the circuit.
By assuming IC = IE, (IB  0), again determine the Q-point (IC, VCE). Are there any significant
differences?
4. Circuit in Figure 5 shown the transistor is biased using collector-feedback. What is the
V
advantage of collector-feedback biased? What is the condition for VCE  CC ?
2
If RC = 2k, RB = 330k and the transistor has dc = 150, what are the values of IC and VCE.
5. The biased circuit in Figure 6 is called ‘emitter-bias’.
Show that the collector current is almost independent
on dc.
Show that VCE  VCC - ICRC.
If VCC = +20V, RC = 2.2k, RE = 3.9k, RB =
5.1k, find IC and VCE.
+VCC
RC
RB
-VEE
Fig. 6
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