21 2 1VT VT VT2(B) INCREASING Η INCREASES I1 BECAUSE THE THRES...

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1 2 1

v

T

v

T

v

T

2

(b) Increasing η increases I

1

because the threshold is effectively reduced. The

change is:

(c) Increasing η increases I

2

because the threshold is effectively reduced. However,

the relative increase is less than that of I

1

. Solve numerically for the change in I

2

to

be a factor of 2.25, with x = 83 mV.

d) First solve for x

( ) ( )

η − − +η − − −V x V x V x− +

−

  

V x x

t

DD

DD

t

′ =  −  =  − 

v v v v

I I e e I e e

T

T

T

T

2 0

1

0

1

ds ds

 

   

(The last term is approximately unity)

( )

η η− − −

 

V xx x x

DD

=  −  ≈

1

I e e I e e

1 1

 

η− −x

 + 

= ( )

v

e e

T

v

T

x V

(Assume << )

= + ≈ ∆

x v e

∆

v

ln 1

T T

Then substitute x to find the relative currents in stacked vs. unstacked transistors:

−x x

′ =    −    =   −  

∆ −∆v v

1 1

T

T

2 1 1

′ ≈

I e

∆

2

I

− ∆2 1

′

1

(e) When DIBL is significant, we see from (d) that the current in stacked transistors

is exponentially smaller because the bottom transistor sees a small drain voltage and

thus much less DIBL than a single transistor.