ASIC Stoic: Feb 6, 2019

https://docs.google.com/document/d/1T7STY4RCtczVMoJTGfcYT43lXyCiwTUO5hNv8xqMIIc/edit?usp=sharing

Hi,

I just got into new 1um CMOS techno so here is its transistors SPICE models:

You can download the SPICE model here: cmosedu_models.txt .

I would like to do sizing ( e.g. finding W/L ) of an NMOS transistor, so this transistor is in saturation
with ID = 400uA and biased with VGS=1.5V .

This is the transistor equation used for the sizing calculation:

μ is from 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:

N-mos ( techno data )		N-mos ( calculated data )
TOX[m]	UO [cm^^2/Vsec]	Cox[F/meter ^^2)	Cox[fF/micro meter ^^2)
2.00E-08	650	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 = 650 ;

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 (units: F/Vsec)

Here is a spreadsheet to calculate Cox * μ :

N-mos ( techno data )		N-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	650	1.73E-03	1.73E+04	1.12224450E-04

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 = 650 ;

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 =
1.1222e-04
units: F/Vsec

Step 3: Calculating W/L

Summary:

Reminder, this is the equation we are using:

We know that we want:

ID = 400μA

We know we will bias the transistor with :

VGS = 1.5V

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

We already calculated Cox * μ [F/Vsec] : 1.1222 * 10^^-4

=> W/L = 14.548 or approx. 15

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 = 650 ;

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) ), 1.5, 0.8]);

% 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 =
  1.1222e-04
units: F/Vsec
f = (sym)

                          2
       7⋅WdivL⋅(VGS - VTH)
-ID + ────────────────────
              124750

result = (sym)

4990
────
343

ans = 14.548

Step 4: Verifying that when we have an NMOS ( in the techno used ) with W/L = 15 and

we bias it with VGS=1.5V and alim. for this techno is VDD=5V,

then the transistor should be in saturation with ID = 400μA

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

Here is how I calculated R1 = 10.77K in this schematic:

Step 4.1: to calculate R1 we would need VDSsat because R1= (VDD -VDSsat)/IDS

R1= (5V -VDSsat)/400μA

A friendly reminder: we are still using the same equation:

We know that we want:

ID = 400μA

We already calculated Cox * μ [F/Vsec] : 1.1222 * 10^^-4

We already calculated ( under assumption that VGS = 1.5 V and

knowing from the techno, meaning from SPICE model,
that the value of Vth[V] is:

=> W/L = approx. 15

Note*: We expect VDSsat minimum to be VGS - Vth = 1.5 - 0.8 = 0.7V

Here is a Octave program to calculate VDSsat:

#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 = 650 ;

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) ), 1.5, 0.8]);

% 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) ), 15 ]);

% 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:

4990
────
343

ans = 14.548

f = (sym)

               2
       7⋅VDSsat ⋅WdivL
-ID + ───────────────
            124750

result = (sym 2×1 matrix)

⎡-√20958 ⎤
⎢────────⎥
⎢ 210   ⎥
⎢       ⎥
⎢ √20958 ⎥
⎢ ────── ⎥
⎣ 210   ⎦

ans =

-0.68938
  0.68938

Source code of Octave program is here: octave_program

Step 4.2:

A friendly reminder from previous step: R1= (5V -VDSsat)/400μA

=> R1 = (5V -0.68938)/400μA

=> R1 = 1.0777e+04 = approx. 10.78KΩ

LTspice simulation showed: IDS = 354 μA (expected 400 μA ) and VDSsat= 1.18V ( expected result: 689mV )

Name: m1

Model: n_1u

Id: 3.54e-04

Vgs: 1.50e+00

Vds: 1.18e+00

Vbs: 0.00e+00

Vth: 8.39e-01

Vdsat: 5.59e-01

Finally, by fine adjusting of W/L from 15 to 17, LTspice simulation showed: IDS = approx. 398 μA (expected 400 μA ) and VDSsat= approx. 711mV ( expected result: 689mV )

Name: m1

Model: n_1u

Id: 3.98e-04

Vgs: 1.50e+00

Vds: 7.11e-01

Vbs: 0.00e+00

Vth: 8.39e-01

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Wednesday, February 6, 2019

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

https://docs.google.com/document/d/1T7STY4RCtczVMoJTGfcYT43lXyCiwTUO5hNv8xqMIIc/edit?usp=sharing

Step 1: Cox is calculated using following formula:

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

Step 3: Calculating W/L

Step 4.1: to calculate R1 we would need VDSsat because R1= (VDD -VDSsat)/IDS

R1= (5V -VDSsat)/400μA

Step 4.2: