EXAMPLE 4-2: Determine Ig, Ic, Ig, VBE. VCE, and VcB in the circuit of Figure 4-7. The transistor has a Bpc = 150. FIGURE 4-7 Rc 100N
EXAMPLE 4-2: Determine Ig, Ic, Ig, VBE. VCE, and VcB in the circuit of Figure 4-7. The transistor has a Bpc = 150. FIGURE 4-7 Rc 100N
Introductory Circuit Analysis (13th Edition)
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![EXAMPLE 4-2: Determine Ig, Ic, Ig, VBE. VCE, and VCB in the circuit of Figure 4-7. The transistor
has a Bpc = 150.
FIGURE 4-7
Rc
100 Ω
Solution From Equation 4-3, VBE = 0.7 v.
Calculate the base, collector, and emitter
Vcc
10 V
10kN
VB
5 V
currents as follows:
VBB – VBE 5 V – 0.7 V
- 430 μΑ
RB
10 kn
Ic = Bocls = (150)(430 µA) = 64.5 mA
Ig = Ic + Ig = 64.5 mA + 430 µA = 64.9 mA
Solve for VCE and VCB-
VCE = Vcc - lc Rc = 10 V – (64.5 mA)(100 N) = 10 V – 6.45 V = 3.55 V
VCB = VCE - VBE = 3.55 V – 0.7 V = 2.85 V
!!
%3D
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Transcribed Image Text:EXAMPLE 4-2: Determine Ig, Ic, Ig, VBE. VCE, and VCB in the circuit of Figure 4-7. The transistor
has a Bpc = 150.
FIGURE 4-7
Rc
100 Ω
Solution From Equation 4-3, VBE = 0.7 v.
Calculate the base, collector, and emitter
Vcc
10 V
10kN
VB
5 V
currents as follows:
VBB – VBE 5 V – 0.7 V
- 430 μΑ
RB
10 kn
Ic = Bocls = (150)(430 µA) = 64.5 mA
Ig = Ic + Ig = 64.5 mA + 430 µA = 64.9 mA
Solve for VCE and VCB-
VCE = Vcc - lc Rc = 10 V – (64.5 mA)(100 N) = 10 V – 6.45 V = 3.55 V
VCB = VCE - VBE = 3.55 V – 0.7 V = 2.85 V
!!
%3D
%3D
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