3. Single-Stage Amplifiers
3.3 Common-Source Stage
Small-signal analysis gives Av = -gmRD. If take into consideration the short-channel effect:
Av = -gm(rO//RD) = -gmrORD/(rO + RD).
gm increases as Vin increases to the boundary point of saturation and triode. To maximize gain, we set Vin – VTH close to Vout while leaving certain room for output voltage swing. The gain can be increased by increasing W/L, which results in greater device capacitance.
For a diode-connected load (which behaves like a [1/(gm + gmb)]//rO resistor),
Av = -gm1/(gm2 + gmb2 + 1/rO1 + 1/rO2),
where gm1 and gm2 are the transconductance of the amplifying transistor and the load transistor, respectively. If we ignore body effect and channel-length modulation and assume equal current from M2 (PMOS) to M1 (NMOS),
Av = -gm1/gm2 = √[[μn(W/L)1]/[μp(W/L)2]],
indicating that |VGS2 – VTH2|/(VGS1 – VTH1) is proportional to the gain. We can add a current source in parallel to the diode-connected load to relieve the ratio and provide more output voltage swing.
For a current-source load (a PMOS with fixed bias voltage on its gate), we have
Av = -gm1(rO1//rO2).
The voltage |VDS2,min| = |VGS2 – VTH2| can be reduced to less than a hundred millivolts by simply increasing the width of M2.
rO at a given ID can be scaled by changing the channel length, i.e., to the first order, λ ∝ 1/L, and hence rO ∝ L/ID. With the given gain equation, we may surmise that longer transistor yield a higher voltage gain. For M1, W1 needs to be scaled proportionally with L1 in order for overdrive voltage and gm1 to stay constant. Note that λ depends more strongly on L than gm does, and gmrO decreases as ID increases.
For an active load M2 (PMOS) that takes on the same gate voltage as M1, the configuration is simply an inverter, and
Av = -(gm1 + gm2)(rO1//rO2).
The circuit exhibits the same output resistance as CS with current-source load, but a higher transconductance. The critical issues with this amplifier is that the bias current is a strong function of PVT, and the circuit amplifies supply noise.
In CS stage with diode-connected load, drain current is relatively linear to overdrive voltage. Alternatively, this can be accomplished by placing a degeneration resistor in series with the source terminal. Small-signal analysis gives
Av = -gmRDrO/(rO + RS + gmrORS + RD) ≈ gmRD/(1 + gmRS);
Gm = Iout/Vin = gmrO/[Rs + rO + (gm + gmb)RsrO].
As the overdrive and therefore gm increase, the effect of degeneration, 1 + gmRs, becomes more significant. For large values of Vin, ID is approximately a linear function of Vin and Gm approaches 1/RS. The output resistance is given by:
Rout = rO + RS + gmrORS.
