2. Small-Signal Operation and Models
• Conceptual circuit • NMOS biased using constant VGS
• Input signal to be amplified vgs
• Total voltage at gate is:
• vGS = VGS + vgs
• Note nomenclature
• vGS – ac + dc signal
• VGS – dc or constant signal
• Vgs – ac or varying signal
• Output voltage taken at drain
thus Vo = VDS
• Drain current iD
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11. Small-Signal Equivalent-Circuit Models
• From a signal point of view, the FET behaves as a voltage-controlled current
source.
• It accepts a signal vgs between gate and source and provides a current gmvgs
at the drain terminal.
• The input resistance of this controlled source is very high—ideally, infinite.
• The output resistance—that is, the resistance looking into the drain—also
is high, and we have assumed it to be infinite thus far.
• Putting all of this together, we arrive at the circuit in Fig. 7.13(a), which
represents the small-signal operation of the MOSFET and is thus a small-
signal model or a small-signal equivalent circuit.
• Dependence of id on vds (channel length modulation) is represented by
including ro in parallel with the controlled source in Fig. 7.13(b).
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13. Small-Signal Equivalent-Circuit Models
• In the analysis of a MOSFET amplifier circuit, the transistor can be replaced
by the equivalent-circuit model shown in Fig. 7.13(a).
• The rest of the circuit remains unchanged except that ideal constant dc
voltage sources are replaced by short circuits.
• This is a result of the fact that the voltage across an ideal constant dc
voltage source does not change, and thus there will always be a zero
voltage signal across a constant dc voltage source.
• A dual statement applies for constant dc current sources; namely, the
signal current of an ideal constant dc current source will always be zero,
and thus an ideal constant dc current source can be replaced by an open
circuit in the small-signal equivalent circuit of the amplifier.
• The circuit resulting can then be used to perform any required signal
analysis, such as calculating voltage gain.
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18. Example: DC circuit
• We first determine the dc
operating point. For this
purpose, we eliminate the input
signal vi, and open-circuit the
two coupling capacitors (since
they block dc currents). The
result is the circuit shown.
• We note that since IG = 0, the dc
voltage drop across RG will be
zero, and
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20. Example: Small signal analysis
• Next we proceed with the small-signal analysis of the amplifier.
Toward that end we replace the MOSFET with its small-signal model
to obtain the small-signal equivalent circuit of the amplifier, shown in
Fig. 7.15(c).
• Observe that we have replaced the coupling capacitors with short
circuits.
• Recall impedance of capacitor Zc = 1/(jωC), for high ω Zc is small
hence the short circuit
• The dc voltage supply VDD has also been replaced with a short circuit
to ground.
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