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431/531 Class Notes 6
Following our discussion last time of the basic transistor switch and emitter follower, we will likewise intro duce the basic relations for two other transistor circuit con gurations: the current source and the common-emitter ampli er. We will then return to the issue of input and output imp edance so that we can build realistic circuits using these con gurations.
5.4 Transistor Current Source
Figure 20 illustrates the basic con guration for a single-transistor current source. VCC is a constant p ositive voltage from a DC p ower supply. Hence, the base voltage VB is also a constant, with VB = VCC R 2 =(R 1 + R 2 ). RL represents a load which we intend to p ower with a current which is approximately indep endent of the sp eci c value of RL.
R 1
R 2
IB VB
I
IE
Vcc
R
RE
L L
Figure 20: Basic transistor current source.
When the transistor is on, we have IE = ( +1)IB. In addition, we have VE = VB 0 :6; and VE = IE RE = ( + 1)IB RE. Solving for IB in this last equation gives IB = VE =(( + 1)RE ). We can combine these to solve for the current which passes through RL :
IL = IC = IB =
VE
( + 1)RE
VB 0 : 6
RE
VB 0 : 6
RE
Hence, we see that indeed IL is indep endent of RL. Of course, there are limitations to the range of RL for which the current source b ehavior is reasonable. Recall that the transistor will shut down if VB VE or if VCE is less than 0 : 2 V. These criteria determine the compliance of the current source, that is its useful op erating range. So, for example, if we have VCC = 15 V and VB = 5 V in our circuit ab ove, then VE = 5 0 : 6 = 4 : 4 V, and the range of compliance for the collector voltage VC will b e approximately 4 : 6 V to 15 V.
5.5 Common-emitter Ampli er
Figure 21 represents the basic con guration of the common-emitter ampli er. To determine the output for this circuit, we assume at this p oint that the input is a sum of a DC o set voltage V 0 and a time-varying signal vin , as discussed last time. (In the next section we will discuss how to achieve these.) V 0 provides the transistor \bias", so that VB > VE , and the signal of interest is vin.
RC
RE
Vcc
Vout
Vin
Figure 21: Basic common-emitter ampli er.
The incoming signal shows up on the emitter: vin = (VE + 0 :6) = VE vE. And by Ohm's Law, iE = vE =RE = vB =RE. As we found previously, iE = iC + iB iC. Now, the voltage at the output is Vout = VC = VCC IC RC. And therefore, Vout vout = iC RC. Putting all of this together, vout = iC RC iE RC = (vB =RE ) RC , giving the voltage gain G: G vout =vin = RC =RE (26)
5.6 Circuit Biasing and Input
Now we need to gure out how to provide inputs to our basic circuits. In Fig. 22 b elow we show the input network for a common-emitter ampli er. The same considerations we apply here apply equally to the input of an emitter follower. The idea is that the voltage divider R 1 and R 2 provide the DC bias voltage (V 0 in our discussion ab ove), and the time varying signal is input through the capacitor (which blo cks the DC). We need to gure out what design criteria should b e applied to this design. We need to make sure that our input circuit do es not load the ampli er, C is chosen to give a reasonable RC cuto , and that the gain of the ampli er is what we want. We will start by designing the DC comp onent of the input network, that is cho osing R 1 and R 2. It is helpful when designing the input network to consider the equivalent circuit shown in Fig. 23. The dio de and resistor lab elled Zin represent the transistor input: the voltage drop across the base-emitter \dio de" and the input imp edance from Eqn. 23. RTH is the Thenenin equivalent resistance for the DC input network. So our design pro cedure can b e as follows:
