The transistor circuit shown in Fig. P7.58 has a constant voltage of + 5 V at th

Question

The transistor circuit shown in Fig. P7.58 has a constant voltage of + 5 V at the base. The voltage at the emitter is 4.3 V because the ON voltage of the base-emitter junction is 0.7 V. The beat for the transistor is 100. Is the transistor in the active, cutoff, or saturation region? What is the base current if a and h have no load? If a is shorted to b, what is the current that would flow in the short circuit? Draw a Thevenin or Notion circuit with a and b is the output terminals. Note: Although the transistor is a nonlinear device, here it operates in a linear region of its characteristics. Hence, we may use an equivalent circuit to represent circuit output characteristics in that linear region.

Explanation / Answer

a) Vce = 10-4.3 = 5.7 V active

b)Ie = 4.3/1k = 4.3 mA = Ic + Ib = Ib/100 + Ib = 101 Ib /100

Ib = 4.257 mA

c)Ie = 4.3 / 0.5 k = 2.15 mA

current in short ckt = 2.15/2 1.075 mA

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