We want to analyze the circuit of Fig. 6.28(a) to determine the voltages at all nodes and the currents through all branches. The minimum value of ß is specified to be 30. 10 ΚΩ www +5 V www 1 ΚΩ 10 kn -5 V 5- (VB+,0.7) 1 2/B=V₂/10 7lc 10 ΚΩ www +5 V (a) 6.28 Example 6.9: (a) circuit; (b) analysis with steps numbered. VI VB V₁+0.5-(-5) -- 10. 1 ΚΩ (b) OV=V₂ +0.7 3 + VEC = 0.2 V -O Vc VB +0.5 6 10 k -5. V Solution quick glance at this circuit reveals that the transistor will be either active or saturated. Assuming e-mode operation and neglecting the base current, we see that the base voltage will be approximately volts, the emitter voltage will be approximately +0.7 V, and the emitter current will be approximately mA. Since the maximum current that the collector can support while the transistor remains in the active. de is approximately 0.5 mA, it follows that the transistor is definitely saturated.

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Can someone explain problem 6.9 step by step please.
The circuit of Fig. 6.27(a) is to be fabricated using a transistor type whose ß is specified to be in the
range of 50 to 150. That is, individual units of this same transistor type can have 8 values anywhere
in this range. Redesign the circuit by selecting a new value for Re so that all fabricated circuits
are guaranteed to be in the active mode. What is the range of collector voltages that the fabricated
circuits may exhibit?
Ans. Rc=1.5 ks; Vc=0.3 V to 6.8 V
ple 6.9
We want to analyze the circuit of Fig. 6.28(a) to determine the voltages at all nodes and the currents through
all branches. The minimum value of ß is specified to be 30.
10 ΚΩ
www
+5 V
ΣικΩ
10 kQ
-5 V
4 lg =
5- (VB+,0.7)
2 B = VB/10
7 Ic=
10 ΚΩ
www
VB
+5 V
VB+0.5-(-5)
10.
(a)
6.28 Example 6.9: (a) circuit; (b) analysis with steps numbered.
1k0
-OVE = VB + 0.7 3
+
VECat = 0.2 V S
(b)
-O Vc VB +0.5 6
≥ 10 ΚΩ
-5 V
Solution
quick glance at this circuit reveals that the transistor will be either active or saturated. Assuming
e-mode operation and neglecting the base current, we see that the base voltage will be approximately
volts, the emitter voltage will be approximately +0.7 V, and the emitter current will be approximately
mA. Since the maximum current that the collector can support while the transistor remains in the active
de is approximately 0.5 mA, it follows that the transistor is definitely saturated.
Transcribed Image Text:The circuit of Fig. 6.27(a) is to be fabricated using a transistor type whose ß is specified to be in the range of 50 to 150. That is, individual units of this same transistor type can have 8 values anywhere in this range. Redesign the circuit by selecting a new value for Re so that all fabricated circuits are guaranteed to be in the active mode. What is the range of collector voltages that the fabricated circuits may exhibit? Ans. Rc=1.5 ks; Vc=0.3 V to 6.8 V ple 6.9 We want to analyze the circuit of Fig. 6.28(a) to determine the voltages at all nodes and the currents through all branches. The minimum value of ß is specified to be 30. 10 ΚΩ www +5 V ΣικΩ 10 kQ -5 V 4 lg = 5- (VB+,0.7) 2 B = VB/10 7 Ic= 10 ΚΩ www VB +5 V VB+0.5-(-5) 10. (a) 6.28 Example 6.9: (a) circuit; (b) analysis with steps numbered. 1k0 -OVE = VB + 0.7 3 + VECat = 0.2 V S (b) -O Vc VB +0.5 6 ≥ 10 ΚΩ -5 V Solution quick glance at this circuit reveals that the transistor will be either active or saturated. Assuming e-mode operation and neglecting the base current, we see that the base voltage will be approximately volts, the emitter voltage will be approximately +0.7 V, and the emitter current will be approximately mA. Since the maximum current that the collector can support while the transistor remains in the active de is approximately 0.5 mA, it follows that the transistor is definitely saturated.
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