Find Ic and Vec for the pnp transistor circuit in Figure 5–15. 5-4 FIGURE 5-15 +10 V RE 10 kn 1.0 kn Boc 150 R 22 kn Rc 2.2 kn Solution This circuit has the configuration of Figures 5-14(b) and (c). Apply Thevenin's theorem. R1 VEE R1 + R2 22 kn VTH 10V (0.688)10 V = 6.88 V %3D 22 kN + 10kn, R;R2 R+ R 22 kn + 10kN (22 kN)(10 kn) RTH 6.88 kl %3D Use Equation 5-8 to determine /p. VTH + VBE - VEE 6.88 V + 0.7 V – 10V -2.42 V RE + RTH/BDc -2.31 mA 1.0kN + 45.9 2 1.0459 kn The negative sign on Ig indicates that the assumed current direction in the Kirchhoff's analysis is opposite from the actual current direction. From Ig, you can determine Ie and VEc as follows: Ic = E = 2.31 mA Vc = ICRC = (2.31 mA)(2.2 kN) = 5.08 V %3D VE = VEE - IRE = 10 V - (2.31 mA)(1.0 kn) = 7.68 V VEC = VE - Ve = 7.68 V – 5.08 V = 2.6 V %3D Related Problem Determine RIN(BASE) for Figure 5-15.

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Solve the related problem. Determine R in (base).
VEE =22 kN + 10k9 10 V = (0.,688)10 V = 6.88 V
E 5–4
Find Ic and VEc for the pnp transistor circuit in Figure 5–15.
FIGURE 5-15
VEE
+10 V
R
10 kn
1.0 kn
)Boc - 150
RI
$ 22 k.
2.2 kf
Solution This circuit has the configuration of Figures 5–14(b) and (c). Apply Thevenin's
theorem.
R1
22 ka
VTH =
R + R2.
R,R2
(22 kM)(10KN)
RTH =
= 6,88 kN
R + R 22 kN + 10 k.
Use Equation 5–8 to determine IẸ.
VTH + VBE – VEE 6.88 V + 0.7 V – 10 V
-2,42 V
= -2.31 mA
RE + RTH/Bpc
1.0kn + 45.9 N
1.0459 kN
The negative sign on Iɛ indicates that the assumed current direction in the Kirchhoff's
analysis is opposite from the actual current direction. From Iµ, you can determine Ie
and VEc as follows:
Ic = l = 2.31 mA
Vc = IcRc - (2.31 mA)(2.2 kM) = 5.08 V
VE = VEE - ĘRE = 10 V – (2.31 mA)(1.0 kM) = 7.68 V
VEC = VE – Vc = 7.68 V – 5.08 V = 2.6 V
Related Problem Determine RiN(BASÐ for Figure 5–15.
Transcribed Image Text:VEE =22 kN + 10k9 10 V = (0.,688)10 V = 6.88 V E 5–4 Find Ic and VEc for the pnp transistor circuit in Figure 5–15. FIGURE 5-15 VEE +10 V R 10 kn 1.0 kn )Boc - 150 RI $ 22 k. 2.2 kf Solution This circuit has the configuration of Figures 5–14(b) and (c). Apply Thevenin's theorem. R1 22 ka VTH = R + R2. R,R2 (22 kM)(10KN) RTH = = 6,88 kN R + R 22 kN + 10 k. Use Equation 5–8 to determine IẸ. VTH + VBE – VEE 6.88 V + 0.7 V – 10 V -2,42 V = -2.31 mA RE + RTH/Bpc 1.0kn + 45.9 N 1.0459 kN The negative sign on Iɛ indicates that the assumed current direction in the Kirchhoff's analysis is opposite from the actual current direction. From Iµ, you can determine Ie and VEc as follows: Ic = l = 2.31 mA Vc = IcRc - (2.31 mA)(2.2 kM) = 5.08 V VE = VEE - ĘRE = 10 V – (2.31 mA)(1.0 kM) = 7.68 V VEC = VE – Vc = 7.68 V – 5.08 V = 2.6 V Related Problem Determine RiN(BASÐ for Figure 5–15.
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