Example 6.3 For the circuit in Fig. 6.22, it is required to determine the value of the voltage Vag that results in the transistor operating (a) in the active mode with Veg =5 V (b) at the edge of saturation (c) deep in saturation with Brced = 10 For simplicity, assume that Vag remains constant at 0.7 V. The transistor ß is specified to be 50. Vec= 10V Ict Rc = 1 kn = 10 kn VCR VRE Figure 6.22 Circuit for Example 6.3.

Introductory Circuit Analysis (13th Edition)
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Example 6.3
For the circuit in Fig. 6.22, it is required to determine the value of the voltage Vg that results in the
transistor operating
(a) in the active mode with Veg =5 V
(b) at the edge of saturation
(c) deep in saturation with Berced = 10
For simplicity, assume that Vg remains constant at 0.7 V. The transistor B is specified to be 50.
Vcc = 10V
V BB
Icv Ž Rc = 1 kfN
R
= 10 kN°
VCE
VBE
Figure 6.22 Circuit for Example 6.3.
Transcribed Image Text:Example 6.3 For the circuit in Fig. 6.22, it is required to determine the value of the voltage Vg that results in the transistor operating (a) in the active mode with Veg =5 V (b) at the edge of saturation (c) deep in saturation with Berced = 10 For simplicity, assume that Vg remains constant at 0.7 V. The transistor B is specified to be 50. Vcc = 10V V BB Icv Ž Rc = 1 kfN R = 10 kN° VCE VBE Figure 6.22 Circuit for Example 6.3.
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