Question

A common-source small-signal NMOS transistor amplifier is DC biased at VₛG = 1 V. Determine its small-signal parameters gₘ and rₒ in this state, if V₋DD = 1.8 V, RD = 8kΩ, and the device parameters are Vₜ = 0.4 V, kₙ = 0.4 mA/V², Vₐ = 5 V.

Answer 1

To determine the NMOS transistor's small-signal parameters gm and ro, the quiescent drain current (ID) is first calculated, and then gm and ro are derived using ID along with the given device parameters.

Step-by-step explanation:

To find the small-signal parameters gm (transconductance) and ro (output resistance) for a common-source NMOS transistor amplifier, we use the given DC bias VSG = 1 V, device parameters Vt = 0.4 V, kn = 0.4 mA/V², and Va = 5 V, along with the supply voltage VDD = 1.8 V and the drain resistance RD = 8kΩ.

The transconductance gm is defined as:

gm = √(2*kn*ID)

First, we must find the quiescent drain current ID. Since VSG > Vt, the transistor is in saturation and ID can be calculated using:

ID = kn*(VSG - Vt)^2

ID = 0.4 mA/V² * (1 V - 0.4 V)^2

ID = 0.144 mA

Now we calculate gm:

gm = √(2*0.4 mA/V²*0.144 mA)

gm = 2.4 mS

Next, we find ro using:

ro = Va/ID

ro = 5 V / 0.144 mA

ro = 34.72 kΩ