ASIC Stoic: Feb 8, 2019

original document link

Hi,

As a friendly reminder in the first document of the series: “CMOS Analog Design Basics: an example

of transistor sizing when using the first time new techno ( e.g. 1um ) and we need NMOS transistor

in saturation with ID=400uA and VGS bias is 1.5V” as a title said our goal was to using these input

data:

1μ techo ( here is a Spice models of CMOS transistors ):

Alim. of Vdd =5V
Assuming bias VGS=1.5V

To achieve ID=400μA .

And in the past, for this purpose I have chosen a following schematic:

For the schematic I calculated and adjusted in simulation the NMOS transistor sizing W/L=17 and

LTspice simulation showed: IDS = approx. 398 μA and VDSsat= approx. 711mV.

Now I would like to replace R1 with a diode connected PMOS ( and calculate it’s sizing W/L ):

To keep everything the same it should be: ID2 = 400 μA and VDS1_sat= VD2 = 711mV.

=> VSD2 in saturation is equal: VSD2_sat = VDD - VDS1_sat

=> VSD2_sat should be equal: VSD2_sat = 5V - 711mv = 4,289 V

This is the transistor equation used for the sizing calculation:

μ of PMOS is from the Spice model :

Step 1: Cox is calculated using following formula:

where

is a constant:

and

( units are meters[m] ) is from Spice model of the transistor:

Here is a spreadsheet to calculate Cox:

P-mos ( techno data )		P-mos ( calculated data )
TOX[m]	UO [cm^^2/Vsec]	Cox[F/meter ^^2)	Cox[fF/micro meter ^^2)
2.00E-08	250	1.73E-03	1.73E+04

Or here is an Octave program that do the same Cox calculation:

#include <octave/oct.h>

syms ID min eox tox WdivL VGS VTH Cox min_times_Cox VDSsat R VDD

e0x = 3.9*8.854*power(10, -12) ;

tox = 200*power(10,-10) ; % units: m

min = 250 ;

VDD = 5 ;

Cox = e0x/tox ;

disp( "Cox =" ), disp( Cox ), disp( "units: F/m2" ) ;

Cox = (e0x/tox)*( power(10,15)/power(10,8) ) ;

disp( "or Cox =" ), disp( Cox ), disp( "units: fF/micro-m2" ) ;

Result of the Octave program run:

Cox =
0.0017265
units: F/m2
or Cox =
1.7265e+04
units: fF/micro-m2

Step 2: Calculating Cox * μ pmos (units: F/Vsec)

Here is a spreadsheet to calculate Cox * μ :

P-mos ( techno data )		P-mos ( calculated data )
TOX[m]	UO [cm^^2/Vsec]	Cox[F/meter ^^2)	Cox[fF/micro meter ^^2)	Cox[F/micro meter ^^2)UO [cm^^2/Vsec]= Cox[F/micro meter ^^2)UO [micro meter ^^2/Vsec]10^^4[fF/micro meter ^^2]= Cox[F/micro meter ^^2)UO [micro meter ^^2/Vsec]10^^410^^-15[F/micro meter ^^2]
2.00E-08	250	1.73E-03	1.73E+04	4.31632500E-05

Or here is an Octave program that do the same Cox * μ calculation:

#include <octave/oct.h>

syms ID min eox tox WdivL VGS VTH Cox min_times_Cox VDSsat R VDD

e0x = 3.9*8.854*power(10, -12) ;

tox = 200*power(10,-10) ; % units: m

min = 250 ;

VDD = 5 ;

Cox = e0x/tox ;

disp( "Cox =" ), disp( Cox ), disp( "units: F/m2" ) ;

Cox = (e0x/tox)*( power(10,15)/power(10,8) ) ;

disp( "or Cox =" ), disp( Cox ), disp( "units: fF/micro-m2" ) ;

min_times_Cox = min * Cox * power(10,4) * power(10,-15) ;

disp ( "min * Cox =" ), disp( min_times_Cox ), disp( "units: F/Vsec" ) ;

Result of the Octave program run:

Cox =
0.0017265
units: F/m2
or Cox =
1.7265e+04
units: fF/micro-m2
min * Cox =
4.3163e-05
units: F/Vsec

Step 3: Calculating W/L

Summary:

Reminder, this is the equation we are using:

We know that we want:

ID1 = ID2 = 400μA

We know we need M2 transistor in saturation :

VSD2_sat = 5V - 711mv = 4,289 V

From techno, meaning from SPICE model, the value of Vth[V] of a PMOS transistor is:

We already calculated Cox * μ [F/Vsec] = 1.7265*10^^4

=> W/L of M2 PMOS = 1.637 or approx. 2

Here is an Octave program that do the same W/L calculation:

#include <octave/oct.h>

syms ID min eox tox WdivL VGS VTH Cox min_times_Cox VDSsat R VDD

e0x = 3.9*8.854*power(10, -12) ;

tox = 200*power(10,-10) ; % units: m

min = 250 ;

VDD = 5 ;

Cox = e0x/tox ;

disp( "Cox =" ), disp( Cox ), disp( "units: F/m2" ) ;

Cox = (e0x/tox)*( power(10,15)/power(10,8) ) ;

disp( "or Cox =" ), disp( Cox ), disp( "units: fF/micro-m2" ) ;

min_times_Cox = min * Cox * power(10,4) * power(10,-15) ;

disp ( "min * Cox =" ), disp( min_times_Cox ), disp( "units: F/Vsec" ) ;

f = -ID + (1/2)*min_times_Cox*(WdivL)*power((VGS-VTH),2)

% Substitute in values that are known

newf = subs(f, [ID VGS VTH], [( 400*power(10, -6) ), (VDD - 0.711), 0.9]);

% Solve the resulting symbolic expression for x

result = solve(newf == 0, WdivL)

% And if you need a numeric (rather than symbolic) result

double(result)

Result of the Octave program run:

Cox =
0.0017265
units: F/m2
or Cox =
  1.7265e+04
units: fF/micro-m2
min * Cox =
  4.3163e-05
units: F/Vsec
f = (sym)

result = (sym)


ans = 1.6137

As a sanity check let’s verify do we really have VSD2_sat = 5V - 711mv = 4,289 V when using M2 PMOS sizing W/L = 2 and after calculations result is VSD2_sat = 3.0442V ( expecting 4.289V )

Here is an Octave program that do the same VSD2_sat calculation:

#include <octave/oct.h>

syms ID min eox tox WdivL VGS VTH Cox min_times_Cox VDSsat R VDD

e0x = 3.9*8.854*power(10, -12) ;

tox = 200*power(10,-10) ; % units: m

min = 250 ;

VDD = 5 ;

Cox = e0x/tox ;

disp( "Cox =" ), disp( Cox ), disp( "units: F/m2" ) ;

Cox = (e0x/tox)*( power(10,15)/power(10,8) ) ;

disp( "or Cox =" ), disp( Cox ), disp( "units: fF/micro-m2" ) ;

min_times_Cox = min * Cox * power(10,4) * power(10,-15) ;

disp ( "min * Cox =" ), disp( min_times_Cox ), disp( "units: F/Vsec" ) ;

f = -ID + (1/2)*min_times_Cox*(WdivL)*power((VGS-VTH),2)

% Substitute in values that are known

newf = subs(f, [ID VGS VTH], [( 400*power(10, -6) ), (VDD - 0.711), 0.9]);

% Solve the resulting symbolic expression for x

result = solve(newf == 0, WdivL)

% And if you need a numeric (rather than symbolic) result

double(result)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

f = -ID + (1/2)*min_times_Cox*(WdivL)*power(VDSsat,2)

% Substitute in values that are known

newf = subs(f, [ID WdivL] , [( 400*power(10, -6) ), 2 ]);

% Solve the resulting symbolic expression for x

result = solve(newf == 0,VDSsat)

% And if you need a numeric (rather than symbolic) result

double(result)

Result of the Octave program run:

Cox =
0.0017265
units: F/m2
or Cox =
  1.7265e+04
units: fF/micro-m2
min * Cox =
  4.3163e-05
units: F/Vsec
f = (sym)

result = (sym)

              ans = 1.6137
f = (sym)

       result = (sym 2×1 matrix)

⎡-√45409 ⎤
⎢────────⎥
⎢ 70 ⎥
⎢       ⎥
⎢ √45409 ⎥
⎢ ────── ⎥
⎣ 70 ⎦

ans =

-3.0442
  3.0442

Source code of Octave program is here: octave program

Step 4: Verifying that when we have an PMOS ( in the techno used ) with W/L = 2 and and we want VSD2_sat = 4,289 V and alim. for this techno is VDD=5V, then the transistor M2 PMOS should be in saturation with ID = 400μA

For this verification I will use LTSpice simulation of following schematic:

Note*: We expect VDS1_sat minimum to be VGS1 - Vth nmos = 1.5 - 0.8 = 0.7V

=> we expect that in this schematic VDS2_sat max. = VDD - VDS1_sat min. = 4.3V

LTspice simulation showed: ID2=ID1 = approx. 276 μA (expected 400 μA ) and VDS1_sat= 0.244V ( expected result: 711mV ) which is too low so M1 is not even in saturation.