Operating Point Of The Bipolar Junction Transistor
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Operating Point Of The Bipolar Junction Transistor Figure 122.4 illustrates a Bipolar Junction Transistor (BJT) self-bias DC Circuit. Determine the operating point of the transistor in the circuit given the following circuit values:
R1 = 100 kΩ; R2 = 50 kΩ; RC = 5 kΩ; RE = 3 kΩ;
VCC = 15 V; offset voltage, Vγ = VBE = 0.7;
Current gain, β = IC/IB = 100.

The strings: S7P5A51 (Physical Change).

The math:
Pj Problem of Interest is of type change (physical change). Transistors are primarily used for signal amplification and switching. Both are change problems. The operating point (also called the Q point and the quiescent) of a device is the steady state DC currents and voltages present at the terminals of the device.

Operating Point of interest: IBQ; ICQ; VCEQ;
Where the Q indicates value is a Q point.

Formulas of interest:
IBQ = (VBB - VBE)/[RB + (β + 1)RE]------(1)
Where VBB= base voltage; RB = base resistor.

ICQ = βIBQ -------(2)

VCEQ = VCC - ICQ[RC + ((β + 1)/β)RE]------(3).
Equation (3) is the load-line equation. The intersection of the load-line equation and the IC - VCE characteristic curve for a given IB defines the operating point.

Now, VBB = [R2/(R2 + R2 )] = [50/(100 + 50)]15 = 5 V

Base resistance, RB = R1||RB = (100x50)106/(100 + 50)103 = 33.3 kΩ

So, by equation (1), IBQ = 12.8 μA

So, by equation (2), ICQ = 1.28 mA.

So, by equation (3), VCEQ = 4.78 V.

So, the Q point (operating point) of transistor is given by:
VCEQ = 4.78 V; ICQ = 1.28 mA; IBQ = 12.8 μA

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