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Dayang Nur Salmi Dharmiza
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Transistor is a VCCS (voltagecontrolled current source)
To cause changes in current, VBE is
changed.
Base or emitter terminals can be usedas input terminal
Collector is not used as an inputterminal because collector voltage hasnegligible effect on terminal currents
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Large changes in Ic or Iecreate large voltage dropsacross collector and emitterresistors
Collector or emitter can beused as ouput terminal
Base is not used as outputterminal due to small Ib
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Constraints for signal injection and extractionyield three families of amplifiers
Type of amplifier Input Output
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Steps for using small-signal models: Determine DC operating point of BJT
Decide the collector current
Calculate small-signal model parameters for this
DC operating point Eliminate DC sources
Replace DC voltage sources with short circuits
Replace DC current sources with open circuits
Replace BJT with an equivalent small-signal model Choose most convenient one depending on
surrounding circuitry
Analyze
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Redraw network by replacing capacitors with shortcircuit and dc voltage to ground.
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Substitute with re
model
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Zi: the figure clearly reveals that
= ohms
usually RB is greater than re. So,
RB>10 ohms
Zo: output impedance isdetermined when Vi=0.
= ohms
usually ro is greater than RC. So,
ro>10 ohms
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Av: resistors ro and RC are in parallel, and since
= = =-
=
, substitute into =
Thus, =
=
If 10, =
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Ai: apply current divider rule to the input and output circuits
Therefore,
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Since = , and =
,
so =
+and = -
(+ )
Replace in Ai above, then
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Negative sign in the resulting equation for Avreveals that a 180o phase shift occurs betweeninput and output signals
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Find: (a) re, with ro=, determine (b)Zi, (c)Zo, (d)Av, (e)Ai,
Repeat parts (c)-(e) including ro=50 in allcalculations and compare results
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Which is only slightly different due to the accuracycarried through the calculations
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Biased using four-resisternetwork (voltage divider)
AC input is injected to
base through a couplingcapacitor
AC output is taken fromcollector
Emitter is grounded
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All capacitors open circuits, disconnecting Vi, R1 and R3from circuit
Q-point can be found from dc equivalent circuit by using DCmodel for BJT
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Replace all capacitors by shortcircuits, inductors by opencircuits
Set all independent DC sources
to 0.
Replace transistor by small-signal model
Use small-signal AC equivalentto analyze AC characteristics ofamplifier
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Ifro>10 , ro>10 ohms
Av: Since RC and ro are in parallel,
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Ifro>10 ,
Since the network is so similar with fixed bias CEexcept for the fact that R=R1||R2=RB, theequation will have the same format
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Since = , and =
,
so =
+and = -
(+ )
Replace in Ai above, then
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Find: (a) re, (b)Zi with ro=, determine (c)Zo, (d)Av, (e)Ai,
Repeat parts (b)-(e) if ro=1/hoe=50k and compare results
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-10-
-10 mins break-
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CE bias configuration Emitter-follower configuration
CB configuration
Collector DC feedback configuration
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Apply Kirchhoffs law:
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Magnitude of BR is often too closeto Zb to permit the approximationIb=Ic. Applying current divider ruleto the input will result in
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The output is taken fromemitter terminal
Output voltage is always
slightly less than the inputsignal due to the dropfrom base to emitter
Voltage is in phase withinput signal follower
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Zi (same with CE emitter-bias):
Zo: Vi is set to 0V,
Ai: apply voltage divider rule
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Zi:
Zo:
Av:
Ai: assuming
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=
=
=
= 1
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Employs feedback path from collector to base toincrease stability
Quite difficult to analyze network
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Determine re, Zi, Zo, Av, Ai Determine again with ro=20k and compare
results.
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DC feedback resistor for increased stability Capacitor C3 shifts portions of the feedback
resistance to the input and output
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Determine: