In the circuit shown, assume that: RB = 50 kilo-ohms, RC = 1 kilo-ohms, VCC = 5

Question

In the circuit shown, assume that:
RB = 50 kilo-ohms, RC = 1 kilo-ohms,
VCC = 5 volts, Vin = 2.096 volts.

The BJTs parameter are:
VBEON = 0.7 volts, VCESAT = 0.2 volts, beta = 88.

Two BJTs (T1) and (T2) are connected as shown on the left, in a configuration known as a Darlington Pair. The reason for using a darlington pair is to produce a much higher current gain. In the circuit shown, assume that: RB = 50 kilo-ohms, RC = 1 kilo-ohms, VCC = 5 volts, Vin = 2.096 volts. The BJTs parameter are: VBEON = 0.7 volts, VCESAT = 0.2 volts, beta = 88. What is the value of VBE1? (in volts) What is the value of VBE2? (in volts) What is the value of VB? (in volts) What is the value of the base current iB? (in MicroAmps) What is the value of the collector current iC1? (in MilliAmps) What is the value of the current iE? (in MilliAmps)

Explanation / Answer

The base emitter voltage of T1 and T2 will be the same VBE1 = 0.7V VBE2 = 0.7V VB = VBE1 + VBE2 VB = 0.7 + 0.7 VB = 1.4V Applying KVL to the input sections Vin = IB RB + VBE1 + VBE2 2.096 = 50000 IB + 0.7 + 0.7 50000 IB = 1.5 IB = 30µA IC1 = ß IB IC1 = (88) ( 30 *10-6 ) IC1 = 2.64mA IE = IB + IC1 IE = (30*10-6) + (2.64*10-3) IE = 2.67mA IE = 2.67mA

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