Analog Design Basics: NMOS

Introduction: Why do we need CMOS transistors ?

A bipolar transistor needs a base current to keep the transistor ON. And this is the main problem for bipolars in spite of the fact that its base current is small in a small signal operation.

When bipolar transistor is used in power applications, as a switch,  the base current could be really high, to keep the transistor ON, adding significantly to a power consumption ( e.g.  a need for a extended battery life in portable electronics …. ).

In CMOS technology there are two complementary types of transistors—n-channel (NMOS) and p-channel (PMOS).

NMOS: How it works ?

NMOS transistor ( n channel device ) has three terminals: gate, source and drain. NMOS conducts with a positive gate voltage, and to distinguish between source and drain terminal: a source has a lower voltage.  

As a positive gate voltage is applied, the gate attracts a negative charge forming a channel of mobile electrons connecting the drain and source regions.

• Vtn ( a threshold voltage for n channel device ) is the minimum positive voltage, when applied to a gate, that makes possible conducting electricity between the drain and source.

NMOS is OFF if Vgs < Vtn ( there is no channel formed and there is no possibility of a current flowing from drain to source  ) .

Consequently  NMOS is ON if Vgs  > Vtn ( the channel is formed and if  Vd  > Vs, there will be a current Ids .

• Veff ( effective gate-source voltage ) is: Vgs -Vtn

NMOS modes of functioning: Linear vs Saturation mode

When NMOS is ON (  Vgs  > Vtn  ) and  Vds << Veff, there is approximately  linear relationship between Vds and Ids ( the linear mode ).

NMOS in Linear mode: Ids as linear function of Vgs

But if  Vds > Veff , the channel charge concentration decreases close to the drain terminal ( the channel becomes pinched off ), making a current Ids constant (saturated ) and independent of further increase of Vds. This is the saturated mode.



NMOS in Saturation mode: Id independent of Vds because ( 1+ λ*Vds ) is approx.1 if Kn= μn * Cox *W/L then a formula of saturation Vgs = f(Id = Iref, Vds) is:

  

NMOS transistor exits linear mode and enters the saturated mode when increasing Vds reaches this point:

Vds(saturated) = Vgs - Vtn = Veff

e.g. NMOS transistor application: Current Mirror

The design of current mirror is based on "copying" current from a precisely defined reference.

In a current mirror both NMOS  transistors are in saturated mode, meaning Ids1 ( equal to Iref ) and Ids2 ( equal to Iout ) are independent of its respective Vds voltages and dependent only on its respective Vgs voltages.

• Vds1 = Vgs1= Vgs2      

Vtn +     2Id1/(Kn*( 1+*Vds1) = Vtn +     2Id2/(Kn*( 1+*Vds2)

Id1/(1+*Vds1) = Id2/(1+*Vds2)

Under assumption that: 1+*Vds1= 1+*Vds2

Id1 = Id2

 © 2011 ASIC Stoic. All rights reserved.