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in the calculations of the AC gains This dual role served by coupling and bypass capacitors in transistor ampli ers is of fundamental importance in their practical operation Figure 109 depicts the path taken by AC and DC currents in the circuit of Figure 107 Example 102 further explains the use of coupling capacitors
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Figure 109 Effect of coupling capacitors on DC and AC current paths
EXAMPLE 102 Computing the Value of the AC Coupling Capacitor for Audio-Range Ampli er Operation
Problem
Determine the value of the coupling capacitor CC in Figure 109 that will permit ampli er operation in the audio range
Solution
Known Quantities: Audio-frequency range Find: Value of CC such that the series impedance of the capacitors is high at low frequencies and low at frequencies in the audio range Schematics, Diagrams, Circuits, and Given Data: Audio-frequency range:
40 40,000 (20 to 20,000 Hz)
Assumptions: The input resistance of the ampli er is in the range of 1 k Analysis: If the input resistance of the ampli er is expected to be around 1 k , the impedance of a series capacitor to be used for AC coupling in the ampli er of Figure 109 should be signi cantly smaller in the frequency range of interest, say, 10 The
10
Transistor Ampli ers and Switches
expression for the impedance of the capacitor is: ZC = 1 j C
thus, the capacitor impedance will be larger at the lower frequencies, and we should require the magnitude of the above impedance to be less than or equal to the desired value of 10 at the frequency = 40 We therefore require that |ZC | = or C= 1 1 = = 2,190 F |ZC | 400 1 = 10 C at = 40
Comments: The coupling capacitor offers in nite resistance to DC currents (ie, for
= 0) The capacitor value calculated in this example is fairly large; in practice it would be reasonable to select a somewhat smaller value A standard value would be 470 F
VCC R2
(a) DC circuit
The ampli er of Figure 107 employs a single DC supply, VCC , as does the DC self-bias circuit of Example 96 The resistors R1 and R2 , in effect, act as a voltage divider that provides a suitable bias for the BE junction This effect is most readily understood if separate DC and AC equivalent circuits for the commonemitter ampli er are portrayed as in Figure 1010 To properly interpret the DC and AC equivalent circuits of Figure 1010, a few comments are in order Consider, rst, the DC circuit As far as DC currents are concerned, the two coupling capacitors and the emitter bypass capacitor are open circuits Further, note that the supply voltage, VCC , appears across two branches, the rst consisting of the emitter and collector resistors and of the CE junction, the second of the base resistors This DC equivalent circuit is used in determining the Q point of the ampli er that is, the quantities IBQ , VCEQ , and ICQ In drawing the AC equivalent circuit, each of the capacitors has been replaced by a short circuit, as has the DC supply The effect of the latter substitution (which applies only to AC signals) is to create a direct path to ground for the resistors R1 and RC Thus, R1 appears in parallel with R2 Note, also, that in the AC equivalent circuit, the collector resistance RC appears in parallel with the load This will have an important effect on the overall gain of the ampli er
DC Analysis of the Common-Emitter Ampli er
RS RC RL
vS(t) + R1 || R2 ~ _
We redraw the DC circuit of Figure 1010 in a slightly different form, recognizing that the Th venin equivalent circuit seen by the base consists of the equivalent e voltage VBB = R2 VCC R1 + R 2 (1014)
(b) AC circuit
Figure 1010 DC and AC circuits for the common-emitter ampli er
and of the equivalent resistance RB = R1 R2 (1015)
Part II
Electronics
The resulting circuit is sketched in Figure 1011 Application of KVL around the base and collector-emitter circuits yields the following equations, the solution of which determines the Q point of the transistor ampli er: VCEQ = VCC ICQ RC +1 ICQ RE +1 ICQ RE (1016)
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