Lecture Notes for Analog Electronics - The Electronic Universe ...

5.1.2 Transistor Switch and Saturation
From the preceding discussion, the most straightforward way to turn the transistor “on” or
“off” is by controlling VBE. This is illustrated by the circuit below which was introduced in
Lab 2. We will follow the lab steps again here.
R
+5 V
33
LED
2N2222A
Figure 18: A transistor switch.
First, let R = 10 kΩ. When the switch is open, IC = βIB = 0, of course. When the switch
is closed, then VBE becomes positive and VB = VE +0.6=0.6V.IB =(5−0.6)/10 4 =0.44
mA. Hence, IC = βIB = 44 mA. Then, assuming negligible voltage drop across the LED,
VC =5−33 × 0.044 = 3.5 V.So,VCE > 0andVCB > 0. So this should work just fine.
Substituting R =1kΩgivesIB =4.4mAandβIB = 440 mA. Setting this equal to
IC would give VC = −9.5 V. This is not possible. In order to stay in operation VCE must
be positive, and depending upon the transistor species, usually can only go as low as ≈ 0.2
V. (Appendix K of the text, pages 1066-1067, gives data for a typical model.) Hence, IC is
limited to a maximum value of IC =(5−0.2)/33 ≈ 150 mA. So, effectively, the current gain
has been reduced to β = IC/IB = 150/4.4 = 34. In this mode of operation, the transistor is
said to be saturated. It turns out that for high-speed switching applications, for example in
computers, the transistors are generally operated in a partially saturated mode, for reasons
discussed in Section 2.02 of the text.
5.2 Notation
We will now look at some other typical transistor configurations, including the emitter
follower, the current source, and the common-emitter amplifier. But first we need to set
some notation. We will often be considering voltages or currents which consist of a time
varying signal superposed with a constant DC value. That is,
V (t) =V0+v(t); I(t)=IO+i(t)
where V0 and I0 are the DC quantities, and v or i represent time-varying signals. Hence,
∆V = v ; ∆I = i
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