电子信息类专业英语(第三版) 第5章

The variation in base emitter voltage could be produced by the AC signal VS. This might require signal amplitude of perhaps ±10 mV. If VS=±10 mV produces VO=±2.5 V, the signal may be said to be amplified by a factor of VO/VS=2.5 V/10 mV=250, or circuit amplification is 250.
Assume that RB is selected to give a base current of IB=20 μA. Also, let the DC current gain factor of the transistor be
β=50. Then IC≈βIB=50×20×10-6=1mA
IC≈βIB=50×25×10-6=1.25 mA The voltage drop across RL is IC·RL=1.25 mA×10 kΩ=12.5 V, and VCE= VCC-(IC·RL) =20 V-12.5 V=7.5 V.
When IB is 20 μA, VCE=10 V, and when IB is 25 μA, VCE=7.5 V.
1. Principles of Common Emitter Circuit An NPN transistor is shown in Figure 3.1 with a load resistor (RL=10 kΩ) in series with the collector terminal. A collector supply voltage (VCC=20 V) is provided with a polarity that reverse biases the collector base junction. A base current IB is also provided via RB, and this is results in a forward bias (VBE) at the base emitter junction.
Figure 5.1 Common Emitter Circuit
A signal voltage VS having a source resistance RS is capacitor coupled via C1 to the transistor base. The output is derived via another capacitor C2 connected to the transistor collector. Both capacitors are open circuit to direct currents, but offer very low impedance to AC signals. If the signal source were direct connected instead of capacitor coupled, there would be a low resistance path from the base to the negative supply line, and this would affect the circuit bias conditions. Similarly, an external load directly connected to the transistor collector might alter the collector voltage.
Unit ５ Electrical Technique
Passage A Analog Circuit Passage B Binary System and Logic Systems Passage C Magnitude Locked Loop
Passage A Analog Circuit
The voltage drop across RL is IC·RL=1 mA×10 kΩ=10 V, and the collector to emitter voltage VCE is VCC-(IC·RL) =20 V-10 V=10 V.
The circuit DC conditions have been established as IB=20 μA, IC=1 mA, VCE=10 V, VCC=20 V. If VBE is increased until IB=25μA, then
The transistor current and voltage variations have no effect
on the supply voltage (VCC).So, when assessing the AC performance of the circuit; VCC can be treated as a short circuit. The coupling capacitor C1 also becomes a short circuit to AC signals. Redrawing the circuit of Figure 3.1 with VCC and C1 shorted gives the AC equivalent circuit shown in Figure 5.2.
Hence, for an increase in IB of 5 μA, VCE decreased by 2.5V (i.e., VCE changed by the same amount as the voltage change across RL).
Similarly, if VBE is decreased until IB is 15 μA, IC becomes 50×15×10-6=0.ቤተ መጻሕፍቲ ባይዱ5 mA and IC·RL=0.75 mA×10 kΩ=7.5 V. Thus, VCE=20 V-7.5 V=12.5 V.Therefore, for a 5 μA decrease in IB, VCE increases by 2.5 V.