11
(Saturated) MOSFET Small-Signal Model Transconductance
■ Concept: find an equivalent circuit which interrelates the incremental changes The small-signal drain current due to vgs is therefore given by in iD, vGS, vDS, etc. for the MOSFET in saturation id = gm vgs.
vGS = VGS + vgs , iD = ID + id -- we want to find id = (?) vgs
D iD = ID + id
G We have the functional dependence of the total drain current in saturation: + + B VDS = 4 V µ 2 λ vgs _ iD = n Cox (W/2L) (vGS - VTn ) (1 + nvDS) = iD(vGS, vDS) _ + S Solution: do a Taylor expansion around the DC operating point (also called the VGS = 3 V_ quiescent point or Q point) defined by the DC voltages Q(VGS, VDS):
2 ∂ ∂ 600 iD 1iD 2 i =I ++()v --- ()v +… D D ∂ gs 2 gs 500 vGS 2∂ i v = V + v Q v GS d GS GS gs Q iD 400 Q vGS = VGS = 3 V (µA) 300 If the small-signal voltage is really “small,” then we can neglect all everything past gm = id / vgs the linear term -- 200
100 ∂ iD i ==I + ()v I+ gv D D ∂ gs D mgs 1 2 3 456 vGS Q VDS (V)
where the partial derivative is defined as the transconductance, gm.
EE 105 Fall 1998 EE 105 Fall 1998 Lecture 11 Lecture 11 Another View of gm Transconductance (cont.)
* Plot the drain current as a function of the gate-source voltage, so that ■ Evaluating the partial derivative: the slope can be identified with the transconductance:
W g =µ C ----- ()V – V ()1+ λV m n oxL GS Tn nDS D iD = ID + id
G + ■ In order to find a simple expression that highlights the dependence of gm on the + B VDS = 4 V DC drain current, we neglect the (usually) small error in writing: vgs _ _ + S VGS = 3 V_ 2I µW D gm==2nCox----- ID ------L VGS – VTn iD(vGS, VDS = 4 V) 600 µ µ µ -2 500 For typical values (W/L) = 10, ID = 100 A, and nCox = 50 AV we find that id gm = id / vgs iD 400 Q (µA) 300 µ -1 gm = 320 AV = 0.32 mS 200
100
123456 vGS (V) vGS = VGS = 3 V vGS = VGS + vgs
EE 105 Fall 1998 EE 105 Fall 1998 Lecture 11 Lecture 11 (Partial) Small-Signal Circuit Model Output Conductance/Resistance
■ How do we make a circuit which expresses id = gm vgs ? Since the current is not ■ We can also find the change in drain current due to an increment in the drain- across its “controlling” voltage, we need a voltage-controlled current source: source voltage: ∂i D µ W ()2λ≅λ go ==∂------n C ox------VGS – VTn nnID vDS 2L Q
gate id The output resistance is the inverse of the output conductance + drain g v vgs m gs _ 11 source ro==----- λ------_ gon I D
The (partial) small-signal circuit model with ro added looks like:
id = gm vgs + (1/ro)vds
gate drain + + id gmvgs r vgs o vds _ source _
EE 105 Fall 1998 EE 105 Fall 1998 Lecture 11 Lecture 11 MOSFET Capacitances in Saturation Complete Small-Signal Model fringe electric field lines gate ■ All these capacitances are “patched” onto the small-signal circuit schematic source drain