Design Technic

14
1 Super Buffer Design Driving Large Capacitive Loads prepared by Mr. KOTHANDAPANI R ME-VLSI DESIGN

Transcript of Design Technic

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1

Super Buffer Design

Driving Large Capacitive Loads

prepared

by

Mr. KOTHANDAPANI R

ME-VLSI DESIGN

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Given a large capacitance load C load How many stages are needed to minimize the delay?

How to size the inverters?

2

Supper Buffer

C load

1 2 N1  

C g C dC g

C gC d 2 C d N C g N 

C dC load

Equiv INV 

N:  number of inverter stages

:  optimal stage scale factor  

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  where C  g: the input capacitance of the first stage inverter. C d : the drain capacitance of the first stage inverter. Each inverter is scaled up by a factor of per stage. C load  =  N+1C  g   All inverters have identical delay of

0(C d +  C  g )/(C d +C  g )  which 0 is per gate delay forEquiv INV in ring oscillator circuit with load

capacitance = C  g+C d

3

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Consider N  stages, each inverter has same delay

0(C d +  C  g )/(C d +C  g ).

Therefore,  

  

 

C C 

C C  N 

 g d 

 g d total 

      01

4

1 2 N1  

C g C d C g 2 C gC d 2 C d N C g N C d C load

Equiv INV 

dd dd

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Goal: Choose and N  to minimize total .  By C load  =  N+1C  g , we have

Plug the above equation into total , we get

To minimize total :

 

 

 

 

  

  

C C 

C C C C 

d    g 

 g d  g 

load 

total 

  

     0

ln

ln

 ln

ln

  

 

  C 

 N   g 

load 

5

0

ln1

ln

1

ln20  

  

  

  

  

  

  

C C C 

C C C C 

C C 

d    g 

 g 

d    g 

 g d 

 g 

load total 

  

   

  

 g 

d opt opt    1ln  

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For the special case C d =0 ln(  opt)=0  opt = e. However, inreality the drain parasitics cannot be ignored.

Example: For C d =0.5 fF, C  g=1 fF, determine opt and N  for C load  =50 pF.

opt (ln opt -1) = 0.5  opt = 3.18

The Super Buffer Design which minimizes total  for C load  = 50 pF is N=7  Equiv INV stages, and opt = 3.18

36.6

118.3ln/101/1050ln

1ln//ln

ln

/ln1

1412

 

 

opt  g load 

opt 

 g load 

C C  N 

C C  N 

6

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Oscillation period T  is equal to

T = PHL1+ PLH1+ PHL2+ PLH2+ PHL3+ PLL3 =2  p+2  p+2  p =3·2  p=6  p

For arbitrary odd number (n) of cascade-connected

invertes, we have  f=1/T=1/(2·n·  p )  Also, we can write

  p=1/(2·n·f)

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V 1

C load,1 C load,2 C load,3

V 2 V 31   2   3  

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τ  PHL2

V out

V OH

V 50%

t  

V OL

τ  PLH3 τ  PHL1 τ  PLH2 τ  PHL3 τ  PLH1

V 2 V 1 V 3 V 2 V 1 V 3

T  

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Figure 8.8 CMOS inverter circuit

High speed design can be obtain fromstudying the characteristic delay throughinverter

n p   L

W r 

 L

W  

  

 

 

  

 

1''

 p

n

 p

n

k k r 

  

)(

1

T  DD

 pnV V 

 R R R

  

]1[)(   / t 

 DDout    eV t V   

 /)(   t 

 DDout    eV t V   

)(  L FET out    C C  R RC     

 L s   C t t        0

)(

1

T  DD   V V  R

  

 

)( GpGnox

GpGnin

 A AC C C C 

Gn

nox

 pnoxin

C r 

 LW C r 

W W  LC C 

)1(

))(1(

)(

)2/(  DD M    V V   

(8.42)

(8.43)

(8.44)

(8.50)(8.45)

(8.48)

(8.49)

(8.46)

(8.47)

(8.51)

(p-network pre-charge function)

(n-network dis-charge function)

(assume ts = tr = t f )

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Figure 8.9, since the load capacitance is the same

as the gate’s own input capacitance , we call this a

unit load value

in L   C C    1

in s   C t t        01

       S '

 R R '

   '

 L s   C S 

t t     

  

    

0

nn   SW W    '

inin   SC C   '

(unit load) (8.52)

(switching time)

(When C L >> C in , using S > 1)

(new switching time)

(8.53)

(8.54)

(8.55)

(8.52)

(8.53)

(8.54)

(8.55)

(When C L=S C in , using S > 1)

Figure 8.9 Concept of a unit load

Figure 8.10 Driving a largeinput capacitance gate

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In figure 8.11. To drive the large load capacitance,let the 1-th be the unit gate

Figure 8.11 Inverter chainanalysis

 N  N                  1321   ...

12         S 

23         S 

 j j   S        1

12        S 

1

2

23           S S   

1

3

34           S 

1

)1(          j

 j   S 

1

)1( C S C    j

 j

)1(  

 j j

 R R

1   j j j   C  R 

Figure 8.12 Characteristics of atypical stage in the chain

(8.60)

(8.61)

(8.62)

(8.63)

(8.64)

(8.65)

(8.66)

(8.67)

cetanctranscondudevice

cetanresis FET  R

etanccapaciinput C 

1

1

1

  

(1-th stage parameters)

(assume C  j+1 >> C FET,j)

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 L N  N  N 

 N  N d 

C  R R RC  RC  RC  R  

1433221

1321

.. .

...         

1

1

C S 

C C 

 N 

 N  L

 

11

11

1

2

11

3

2

11

2111   .. .   C S 

 RC S 

 RC S 

 RC S 

 RSC  R   N 

 N 

 N 

 N d   

   

)(

.. .

11

1111111111

C SR N 

C SRC SRC SRC SRC SRd 

 

r d    NS     

1C S C    N 

 L  

)ln(ln)ln(1

S  N C 

C S    L N    

  

)ln(

ln1

 N 

 L

 

  

 

 

 

 

 

)ln(ln1   S 

C  Lr d      

0)ln(

S S 

d  

1)ln(   S or 

eS  

 

  

 

 

  

 

1

1ln

)ln(

ln

 N    L

 L

 L

d  C 

e       

 

 

 

1ln

Figure 8.13 Time constants in the cascade

(8.68)

(8.69)

(8.70)

(8.71)

(8.72)

(8.73)

(8.74)

(8.75)

(8.77)

(8.78)

(8.79)

(8.80)

(8.81)

(8.76)

(8.82)

(N stages)

11,   C  Rwhere r   

0)][ln()ln(

12 

S S 

(e = 2.71…) 

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Figure 8.14 shows the  j-th  stage circuit with the parasitic FET

capacitance C  Fj included at the output S  > e (in physical design)

Figure 8.14 Driver chain withinternal FET capacitance

)( 1,     j j F  j j   C C  R 

1,

)1(

,   F 

 j

 j F    C S C   

)(...)()( ,32,221,1   L N  F  N  F  F d    C C  RC C  RC C  R    

)( 111,1   C SR N C  NR  F d     

  

  

  

  

1

ln)ln()ln(   C 

 Lr 

 xd     

  

1,1   F  x   C  Rwhere    

 xS S  

 1)ln(

(8.68)

(8.69)

(8.70)

(8.71)

(8.72)

(8.73)

(8.74)

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