Operating Point Of A Bipolar Junction Transistor
Strings (SiPjAjk) = S7P5A51 Base Sequence = 12735 String Sequence = 12735 - 5 - 51
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.
S7P5A51 (Physical Change).
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 quuiescent) 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
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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