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(b) The Probe feature of PSpice is used to plot iS , iL , nS , and nL as displayed by Fig. 3-18. Notice the 1808 phase shift between input and output quantities.
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Fig. 3-18
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Find the value of the emitter resistor RE that, when added to the Si transistor circuit of Fig. 3-17, would bias for operation about VCEQ 5 V. Let ICEO 0; 80; RF 220 k; RC 2 k, and VCC 12 V.
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Application of KVL around the transistor terminals yields IBQ VCEQ VBEQ 5 0:7 19:545 A RF 220 103
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Since leakage current is zero, (3.1) and (3.2) give IEQ 1 ICQ ; thus KVL around the collector circuit gives IBQ IBQ RC 1 IBQ RE VCC VCEQ VCC VCEQ 1 IBQ RC 12 5 80 1 19:545 10 6 2 103 2:42 k 1 IBQ 80 1 19:545 10 6
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In the circuit of Fig. 3-12, IBQ 30 A; RE 1 k; VCC 15 V, and 80. Find the minimum value of RC that will maintain the transistor quiescent point at saturation, if VCEsat 0:2 V, is constant, and leakage current is negligible.
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We rst nd 80 0:9876 1 81
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[CHAP. 3
Then the use of (3.2) and (3.1) with negligible leakage current yields ICQ IBQ 80 30 10 6 2:4 mA and IEQ ICQ 2:4 10 3 2:43 mA 0:9876
Now KVL around the collector circuit leads to the minimum value of RC to ensure saturation: RC VCC VCEsat IEQ RE 15 0:2 2:43 1 5:154 k ICQ 2:4 10 3
The Si transistor of Fig. 3-19 has 50 and negligible leakage current. Let VCC 18 V, VEE 4 V; RE 200 , and RC 4 k. (a) Find RB so that ICQ 2 mA. (b) Determine the value of VCEQ for VB of part (a).
+ VCC RC
RB G RE _V
Fig. 3-19 (a) KVL around the base-emitter-ground loop gives VEE IBQ RB VBEQ IEQ RE Also, from (3.1) and (3.2), IEQ 1 ICQ 2 1
Now, using (3.2) and (2) in (1) and solving for RB yields RB VEE VBEQ 50 4 0:7 1 RE 50 1 200 72:3 k ICQ 2 10 3
(b) KVL around the collector-emitter-ground loop gives   1 VCEQ VCC VEE RC RE ICQ   50 1 200 2 10 3 13:59 V 18 4 4 103 50
The dc current source IS 10 A of Fig. 3-19 is connected from G to node B. The Si transistor has negligible leakage current and 50. If RB 75 k, RE 200 , and RC 4 k, nd the dc current-gain ratio ICQ =IS for (a) VCC 18 V and VEE 4 V, and (b) VCC 22 V and VEE 0 V.
CHAP. 3]
CHARACTERISTICS OF BIPOLAR JUNCTION TRANSISTORS
(a) A Thevenin equivalent for the network to the left of terminals B; G has VTh RB IS and RTh RB . With the Thevenin equivalent circuit in place, KVL around the base-emitter loop yields RB IS VEE IBQ RB VBEQ IEQ RE 1
Using (3.2) and (2) of Problem 3.17 in (1), solving for ICQ , and then dividing by IS results in the desired ratio: ICQ RB IS VEE VBEQ 75 103 10 10 6 4 0:7 ! 237:67   3 RB 1 IS 6 75 10 50 1 200 RE IS 10 10 50 50 2
Note that the value of VCC must be large enough so that cuto does not occur, but otherwise it does not a ect the value of ICQ . (b) VEE 0 in (2) directly gives ICQ IS 75 103 10 10 6 0:7 10 10 6 75 103 50 1 200 50 50 ! 2:93
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