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Obviously, VEE strongly controls the dc current gain of this ampli er.
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In the circuit of Fig. 3-20, VCC 12 V, VS 2 V; RC 4 k, and RS 100 k. The Ge transistor is characterized by 50; ICEO 0, and VCEsat 0:2 V. Find the value of RB that just results in saturation if (a) the capacitor is present, and (b) the capacitor is replaced with a short circuit.
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(a) Application of KVL around the collector loop gives the collector current at the onset of saturation as ICQ VCC VCEsat 12 0:2 2:95 mA RC 4 103
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With C blocking, IS 0; hence the use of KVL leads to RB VCC VBEQ VCC VBEQ 12 0:3 198:3 k IBQ ICQ = 2:95 10 3 =50
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+VCC IRB RB IS + VS _ RS C RC +
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Fig. 3-20
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CHARACTERISTICS OF BIPOLAR JUNCTION TRANSISTORS
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[CHAP. 3
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(b) With C shorted, the application of (3.2), KCL, and KVL results in IBQ so that ICQ VS VBEQ VCC VBEQ IS IRB RS RB VCC VBEQ 12 0:3 RB 278:6 k ICQ VS VBEQ 2:95 10 3 2 0:3 RS 50 100 103
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The Si Darlington transistor pair of Fig. 3-21 has negligible leakage current, and 1 2 50. Let VCC 12 V; RE 1 k, and R2 ! 1. (a) Find the value of R1 needed to bias the circuit so that VCEQ2 6 V. (b) with R1 as found in part a, nd VCEQ1 .
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+ VCC R1 IR1 a T1 T2 RE b
R2 IR2
Fig. 3-21 (a) Since R2 ! 1; IR2 0 and IBQ1 IR1 . IEQ2 Now and IBQ2 IR1 By KVL,
VCC VCEQ2 12 6 6 mA RE 1 103 IEQ2 IEQ1 2 1 IEQ1 IEQ2 6 10 3 2:31 A IBQ1 1 1 1 1 2 1 50 1 50 1
By KVL (around a path that includes R1 , both transistors, and RE ) and Ohm s law, R1 VR1 VCC VBEQ1 VBEQ2 IEQ2 RE 12 0:7 0:7 6 10 3 1 103 1:99 M IR1 IR1 2:31 10 6 VCEQ1 VCC VBEQ2 IEQ2 RE 12 0:7 6 10 3 1 103 5:3 V
(b) Applying KVL around a path including both transistors and RE , we have
The Si Darlington transistor pair of Fig. 3-21 has negligible leakage current, and 1 2 60. Let R1 R2 1 M; RE 500 , and VCC 12 V. Find (a) IEQ2 , b VCEQ2 , and (c) ICQ1 .
(a) A Thevenin equivalent for the circuit to the left of terminals a; b has VTh and RTh R2 1 106 V 12 6 V R1 R2 CC 1 106 1 106 R1 R2 1 106 1 106 500 k R1 R2 1 106 1 106
CHAP. 3]
CHARACTERISTICS OF BIPOLAR JUNCTION TRANSISTORS
With the Thevenin circuit in place, KVL gives VTh IBQ1 RTh VBEQ1 VBEQ2 IEQ2 RE Realizing that IEQ2 2 1 IBQ2 2 1 1 1 IBQ1 we can substitute for IBQ1 in (1) and solve for IEQ2 , obtaining 1 1 2 1 VTh VBEQ1 VBEQ2 60 1 60 1 6 0:7 0:7 7:25 mA RTh 1 1 2 1 RE 500 103 60 1 60 1 500 1
IEQ2
(b) By KVL, VCEQ2 VCC IEQ2 RE 12 7:25 10 3 500 8:375 V (c) From (3.1) and (3.2), ICQ1 IEQ2 1 1 1 60 7:25 10 3 I I 116:9 A 1 1 EQ1 1 1 BQ2 1 1 2 1 60 1 60 1
The Si transistors in the di erential ampli er circuit of Fig. 3-22 have negligible leakage current, and 1 2 60. Also, RC 6:8 k, RB 10 k, and VCC VEE 15 V. Find the value of RE needed to bias the ampli er such that VCEQ1 VCEQ2 8 V.
5 + VCC RC
RC 4 + 1 + T1 3 iE
6 + 7 T2 +
RE 2 _V
Fig. 3-22 By symmetry, IEQ1 IEQ2 . Then, by KCL, iE IEQ1 IEQ2 2IEQ1 1
Using (1) and (2) of Problem 3.17 (which apply to the T1 circuit here), along with KVL around the left collector loop, gives VCC VEE 1 I R VCEQ1 2IEQ1 RE 1 1 EQ1 C 2
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