As discussed in the first part of this chapter, a transistor is an excellent amplifier. The picture below shows an example.
The input signal is connected to the amplifier via C1. C1 prevents a DC current flow in R1 and the input signal source, e.g. a microphone. DC currents may destroy the microphone (unless it's an electret; a type of microphone with a built-in amplifier).
The characteristics of transistor T1 are: hFE=100 and VBE=0.6V.
Assume we want to connect an end amplifier with a 10k input resistance to the OUT terminal. For maximum power transfer, the output resistance of our amplifier must be equal to the input resistance of the end amplifier. The output impedance of an amplifier is defined as vOUT/iOUT. In our case vOUT = vRC and iOUT = iRC. So the output impedance of this amplifier is vRC/iRC = RC. So RC = 10k. For stability reasons VRE must be VS/5. Since VS = 9V, VRE must be 1.8V. This means that the voltage at the OUT terminal can vary between 1.8 and 9V. So the maximum AC output voltage (vOUT,max) is 9-1.8=7.2Vtt. Obviously, this can only be arranged if the quiescent output voltage (when vIN=0) is exactly between 1.8 and 9V. This means VOUT=1.8+(9-1.8)/2=5.4V. We already know that RC=10k, so IC = (9-5.4V)/10k = 0.36mA. IE will also be 0.36mA, so RE=VRE/IE=1.8/0.36m=5k.
VB = VBE + VRE. VBE is always 0.6V, so vRE = vB. (Remember: AC voltages and currents are written in lower case letters.) When C1 is large enough, vIN = vB = vRE. VOUT = VS - VRC = 9V - VRC => vOUT = -vRC.
Knowing this, we can calculate the gain A of the amplifier, which is defined as: A = vOUT/vIN = -vRC/vRE=-(iC∙RC)/(iE∙RE). Since iC=iE (hFE is large enough to neglect iB), A=-(iC∙RC)/(iC∙RE) = -RC/RE. This means our amplifier's gain is -10k/5k =-2.
VR1 = VS-VBE-VRE = 9-0.6-1.8 = 6.6V. IR1=IC/hFE=0.36mA/100=3.6μA. R1 = 6.6V/3.6μA = 1.8M.
Unfortunately, transistors with the same type designation can have a wide range of hFE. For example, the hFE of a 2N3904 transistor ranges from 100 thru 300. The question is: will our amplifier still function properly if hFE = 300? Let's see...
IB = (VS-VBE-VRE)/R1. VRE = IC∙RC = hFE∙IR1∙RE = hFE∙RE∙(VS-VBE-VRE)/R1 = hFE∙RE∙(VS-VBE)/R1 - hFE∙RE∙VRE/R1 =>
VRE+(hFE∙RE/R1)∙VRE = hFE∙RE∙(VS-VBE)/R1 => (1+300∙5k/1.8M)∙VRE = 300∙5k∙8.4/1.8M => 1.833∙VRE=7 => VRE = 7/1.833 = 3.8V.
IB = (8.4-3.8)/1.8M = 2.6μA. IC = 300∙2.6μA = 0.767mA. However, IC,max = VS/(RC+RE)=9/15k=0.6mA. So hFE∙IB > IC,max, which means that the transistor is saturated and thus acts like a closed switch!
Solution: add an extra resistor which makes VRE (and therefore IC) independent of hFE:
Make sure IR1 >> IB => IR1≈IR2. Let's estimate proper values for R1 and R2.
VR2 = VBE+VRE = 0.6+1.8V = 2.4V. VR1 = VS-VR2 = 9-2.4 = 6.6V. So R1:R2=6.6:2.4. E.g. R1=33k and R2=12k. In that case IR1(=IR2) = 6.6V/33k = 0.2mA, which is much larger than IB.
As mentioned before, the voltage gain of this amplifier is just (-)2. In many cases that will not be enough. You can easily increase the gain by adding an extra resistor and capacitor as shown in the picture below: the most common transistor amplifier.
Capacitor C2 shorts RE2 for AC voltages. So for DC signals, RE = RE1 + RE2, and for AC signals, RE = RE1. If RE1 = 500Ω and RE2 = 4.5k, we have an amplifier with the same characteristics as above, but the gain is 10k/500=20.
The impedance of C2 must be much smaller than RE1:
1/(2∙π∙fmin∙C2) « RE1 => C2 » 1/(2∙π∙fmin∙RE1) where fmin is the lowest frequency the amplifier must be able to handle.
For example: if fmin = 20Hz, C2 » 1/(2∙π∙20∙500) = 16μF. 47 or 100μF is a good choice.
The AC input resistance of the amplifier is approximately R1//R2 = 8.8k. So the impedance of C1 must be much less than 8.8k => C1 » 1/(2∙π∙20∙8.8k) = 0.9μF. 10μF is a good choose. The positive terminal of C1 must be connected to the amplifier, unless the input signal's DC component is larger than 2.4V.