[cadence virtuoso practice record (1)_simulated pipe parameters, μCox, λ]

1 First scan the I/V characteristic curve of the NMOS/PMOS tube

1.1 I/V curve of PMOS tube

1.1.1 PMOS tube simulation topology

 

1.1.2.ADE window parameter setting





1.1.3.ADE/Tools/Parametric Analysis set scan variable, step size

1.1.4. Run the simulation to get the I/V curve of the pmos tube

1.2 IV curve of NMOS

Same steps as above

2 Simulation pipe parameters

2.1 Look at the NMOS tube first

Set W/L to 220n/180n

2.1.2 Scan, Vds: 0-2v, Vgs: 0.6-1.2v, step size is set to 0.2

2.1.3 Record the Id value under different Vds and Vgs

	Vds=1V	vds=1.5V
Vgs=0.8V	25.83uA	27.23uA
Vgs=1V	52.36uA	54.65uA
Vgs=1.2V	81.31uA	84.51uA

2.1.4 Calculate μCox, λ according to the formula

$I_{DS}$ = $\frac{1}{2}$ $m_n$ $C_{ox}$ $\frac{W}{L}$ ($V_{GS}$ - $V_{T{H}_n}$)${}^2$ ( 1 +$l_n$$V_{DS}$)

令K _n = $m_n$ $C_{ox}$

得：$I_{DS}$ = $\frac{1}{2}$ $K_n$ $\frac{W}{L}$ ($V_{GS}$ - $V_{T{H}_n}$)${}^2$ ( 1 +$l_n$$V_{DS}$)

(1) Substitute the first row of data in the table

① Substitute into $\frac{W}{L}$ = $\frac{220n}{180n}$ , $I_{DS}$= 25.83uA , Vgs=0.8V , Vds=1V
② Substitute into $I_{DS}$ = 25.83uA ,Vgs=0.8V ，Vds=1.5V
 
25.83 x 10^-6 = $\frac{1}{2}$ $K_n$ $\frac{220n}{180n}$ (0.8 - $V_{T{H}_n}$)${}^2$ ( 1 +$l_n$ x 1)     ①

27.23 x 10^-6 = $\frac{1}{2}$ $K_n$ $\frac{220n}{180n}$ (0.8 - $V_{T{H}_n}$)${}^2$ ( 1 +$l_n$ x 1.5)   ②

Directly divide the two expressions $\frac{①}{②}$, get $l_n$ = 0.121

(2) Substitute the data in the first column of the table

① Substitute into $\frac{W}{L}$ = $\frac{220n}{180n}$ , $I_{DS}$= 25.83uA , Vgs=0.8V , Vds=1V
② Substitute into $I_{DS}$ = 52.36uA ,Vgs=1V ，Vds=1V

25.83 x 10^-6 = $\frac{1}{2}$ $K_n$ $\frac{220n}{180n}$ (0.8 - $V_{T{H}_n}$)${}^2$ ( 1 +$l_n$ x 1)     ①

52.36 x 10^-6 = $\frac{1}{2}$ $K_n$ $\frac{220n}{180n}$ (1 - $V_{T{H}_n}$)${}^2$ ( 1 +$l_n$ x 1)   ②

Directly divide the two expressions $\frac{①}{②}$ ，$V_{T{H}_n}$= 0.329V

（3）将 $\frac{W}{L}$ = $\frac{220n}{180n}$, $l_n$ = 0.121，$V_{T{H}_n}$= 0.329V, substitute into formula ①

$K_n$= 1.704 x 10^-4 A/V²

_{2.1.5 But the Vth n} = 0.483V obtained from the actual simulation data

2.2 Look at the PMOS tube again

2.2.1 The graph connection method of PMOS is slightly different from that of NMOS.
(Stuck here for a while, to understand the working conditions of PMOS)
Set W/L to 220n/180n first

2.2.2 Voltage setting of each point

Scan simulation settings, the steps are the same as NMOS

2.2.3 Scan results

Record the Id value under different Vds and Vgs.
For the convenience of calculation, Vsd and Vsg are used below

	Vsd=1V	Vsd=1.5V
Vsg=0.8V	-6.53uA	-7.10uA
Vsg=1V	-15.02uA	-16.09uA
Vsg=1.2V	-25.55uA	-27.22uA

2.2.4 Calculation parameters

$I_{DS}$ = $\frac{1}{2}$ $m_p$ $C_{ox}$ $\frac{W}{L}$( $V_{SG}$ - $|V_{T{H}_p}|$)${}^2$ ( 1 +$l_p$$V_{SD}$)

令K _p = $m_p$ $C_{ox}$

得：$I_{DS}$ = $\frac{1}{2}$ K_p $\frac{W}{L}$( $V_{SG}$ - $|V_{T{H}_p}|$)${}^2$ ( 1 +$l_p$$V_{SD}$)


(1) Substitute the first row of data in the table

① Substitute into $\frac{W}{L}$ = $\frac{220n}{180n}$ , $I_{DS}$= 6.53uA , Vsg=0.8V , Vsd=1V
② Substitute into $I_{DS}$ = 7.10uA ,Vsg=0.8V ，Vsd=1.5V
 
6.53 x 10^-6 = $\frac{1}{2}$ $K_p$ $\frac{220n}{180n}$ (0.8 - $V_{T{H}_p}$)${}^2$ ( 1 +$l_p$ x 1)     ①

7.10 x 10^-6 = $\frac{1}{2}$ $K_p$ $\frac{220n}{180n}$ (0.8 - $V_{T{H}_p}$)${}^2$ ( 1 +$l_p$ x 1.5)   ②

Directly divide the two expressions $\frac{①}{②}$, get   $l_p$ = 0.21

(2) Substitute the data in the first column of the table

① Substitute into $\frac{W}{L}$ = $\frac{220n}{180n}$ , $I_{DS}$= 6.53uA , Vsg=0.8V , Vsd=1V
② Substitute into $I_{DS}$ = 15.02uA ,Vsg=1V ，Vsd=1V
 
6.53 x 10^-6 = $\frac{1}{2}$ $K_p$ $\frac{220n}{180n}$ (0.8 - $V_{T{H}_p}$)${}^2$ ( 1 +$l_p$ x 1)     ①

15.02 x 10^-6 = $\frac{1}{2}$ $K_p$ $\frac{220n}{180n}$ (1 - $V_{T{H}_p}$)${}^2$ ( 1 +$l_p$ x 1)     ①

Directly divide the two expressions $\frac{①}{②}$ ，$|V_{T{H}_n}|$ = 0.412V

（3）将 $\frac{W}{L}$ = $\frac{220n}{180n}$, $l_n$ = 0.21，$|V_{T{H}_p}|$ = 0.412V, substituted into formula ①

$K_p$= 5.876 x 10 ^-5 A/V ²


2.2.5 But the actual simulation data obtained Vth _p = -0.487V

Self summary:

1. The calculation process needs to be careful, the formula is not difficult, do not miss items, and pay attention to the unit.
2. Although there is a gap between the actual simulation and theoretical calculation data, it still feels different if you do it yourself.
3. I learned some syntax for typing formulas. It turns out that Latex formulas and word can be shared.
4. Beginner Xiaobai, self-study records, if there are any mistakes, please feel free to enlighten me.
5. The whole process refers to the records of the following bloggers, thanks!

  [1] https://blog.csdn.net/weixin_44115643/article/details/119062516
  [2] https://blog.csdn.net/kexuedalao/article/details/122540451