01 ram a krishna david devanand comments

8/8/2019 01 Ram a Krishna David Devanand Comments

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BASICS OF VLSI

LAB ASSIGNMENT 1

BY

DAVID JOSE (M100388EC)N RAMAKRISHNA (M100410EC)

DEVANAND T (M100404EC)


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Exercise 1 Extract threshold voltage, drain induced barrier lowering, body effect,channel length modulation for different lengths. ( Lmin, 2Lmin, 3Lmin, 4Lmin, 5Lmin)

These exercises are done with W / L ratio = 2.

V DS is kept constant at 1V and I D vsV GS graphs are used to determine thethreshold voltage.

I D versus V GS graphs are plotted at twodifferent V DS to obtain the change inthreshold voltage.

Lambda is evaluated from I D versusV DS graph by noting down the values of I D at two different values of V DS .

Length(nm) Threshold voltage(mV)180 405.12360 481.72540 456.78720 431.36900 416.1

Length(nm) DIBL180 -0.03906360 -0.01938540 -0.02034720 -0.01356900 -0.02

Length(nm) Lambda180 0.167360 0.094540 0.078720 0.059900 0.048

100 200 300 400 500 600 700 800 900405

410415420425430435440445450455460465470475480485

Threshold voltage - length

Vt(mV))

V t ( m

V )

100 200 300 400 500 600 700 800 900

0.040

0.050

0.060

0.070

0.080

0.0900.100

0.110

0.120

0.130

0.140

0.150

0.160

0.170

Lambda - Length

Lambda

Length(nm)

L a m

b d a

100 200 300 400 500 600 700 800 900

-0.04000

-0.03750

-0.03500

-0.03250

-0.03000

-0.02750

-0.02500

-0.02250

-0.02000

-0.01750

-0.01500

DIBL-Length

DIBL

Length(nm)

D I B L


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Gamma is evaluated by obtaining the values of V T with V BS =0V and V BS =-0.5V.

Exercise 2 Use the extracted values to plot I D vs V GS

These plots are drawn by fixing V DS at 1 Volt and varying V GS from 0 to 1.8V

0 0.25 0.5 0.75 1 1.25 1.5 1.75 21.00E-013

1.00E-012

1.00E-011

1.00E-010

1.00E-009

1.00E-008

1.00E-007

1.00E-006

1.00E-005

1.00E-004

1.00E-003

log Id Vs Vgs

L=180nm

L=360nm

L=540nm

L = 720 nm

length(nm) Gamma180 0.1542360 0.2606540 0.3016720 0.3303900 0.3375

100 200 300 400 500 600 700 800 900

0.1400

0.1600

0.1800

0.2000

0.2200

0.2400

0.2600

0.2800

0.3000

0.3200

0.3400

Gamma-Length

Gamma

length(nm)

G a m m a


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Exercise 3. Report the minimum and maximum error in percentage(in linearand saturation regions)

Length 180nm 360 nm 540nm 720nm 900nm

% error in I DS

max min max min max min max min max min

Linear 29.1 0.78 18.74 1.22 27 0.12 33.06 7.35 37.25 13.8

Saturation 17.75 0.37 11.84 0.64 7.99 0.6 16.33 7.87 22.47 14.3

These values are obtained by comparing the the graphs plotted between V DS and I DS

(by maintaining V GS at 1 Volt.) using the theoretical expression and cadence simulator.V DS is varied from 0V to 1.8V to obtain the I DS versus V DS graph.

Exercise 4. List the reasons for the error.

Reasons for error in Drain current :

In saturation region ,the ideal drain current equation is

I D SAT = 1 /2 n C ox W / L V GS V TH 2

In linear region drain current equation is

I DLIN = n C ox W / L V GS V TH V DS V DS 2 /2

The graph plotted using the above theoretical equations doesn't consider the effects of channel length modulation , velocity saturation etc. The graph plotted using the cadencesimulator takes into account the above factors. The theoretical graphs are plotted with amobility value of 250 cm

2 /Vs . But practically the value of surface mobility is about 10to 20 cm

2 /Vs .The simulator also takes into account the factors like dependence of bulk depletion chargeon channel voltage and sub threshold conduction.


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Exercise 5. Analyze the reasons for the fashion in which results vary withlength.

Reasons for variation of extracted parameters with length.

Threshold voltage ( V T )Threshold voltage increases slightly with increase in channel length. This is because

gate induced bulk depletion charge increase with increase in channel length. Short-channeleffect leads to the decrease of V T with decrease in L.

Drain Induced Barrier Lowering ( DIBL )

DIBL = - V T / V DS V T decreases with increase in V DS .Hence (- V T ) is a measure of barrier

lowering. In the graph,we have plotted V T / V DS .These values are negative. Thevalues become less negative as L increases .Hence barrier lowering decreases with increasein L. This is because ,with lower L, drain-body depletion region makes a major share incharge balance. At higher values of L,the drain-body depletion region width is smallercompared with L. Hence barrier lowering is low.

Body bias coefficient ( )

= V T V T0 / 2 F V SB 2 F

is found to increase initially with increase in L and later it remains constant.But ideally, should have remained constant throughout. It is given as

= 2 qN A Si /C ox .This initial variation is due to short channel effects etc.

Channel length modulation coefficient ( )

As V DS increases, channel pinch off occurs at the drain end. This change inlength L is proportional to V DS . Mathematically, = L/ Leff V DS . Thus ,

for a given V DS , change in channel length ( L ) with respect to original channellength (L) will be larger for a smaller value of L . Hence lambda decreases with increase inchannel length.


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Exercise 6. Comment on the transition frequency for L=.8Lmin to 5Lmin

The transition frequency f T is given as : f T = g m /2 C gs whereC gs= WLC ox .Here C ox = ox /t ox

In order to compute f T , g m value is obtained from the slope of I D V GS characteristics.

f T is a measure of the time taken by a charge carrier to move from source todrain. Hence as channel length (L) increases, time taken for transition increases. So f T decreases. The transition frequency f T in terms of channel length L is given as :

f T = 3 n V DSAT / 4 L2

*******************************

0 200 400 600 800 1000

0

5

10

15

20

25

30

35

40

45

50

55

60

Transition frequency vs channel length

Transitionfrequency

length(nm)

T r a n s i t

i o n

f r e q u e n c y

( G H z )

Length(nm) Transition frequency180 57.06360 19.42540 9.96720 6.16

900 4.3

01 ram a krishna david devanand comments

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