# Transistor equation solving

Status
Not open for further replies.

Hi all,
I was reading the following paper:
https://ece.uwaterloo.ca/~cdr/pubs/maymandi/3.pdf.

I dont understand the transistor behaviour given by the relation (1) in the paper.

This equation is not for transistor in linear or saturation region. Could somebody explain this new expression and when we use it ?

In a second step the autor try to calculate the voltage Vo using the equation (3).

And he say

I applyed the laplace transform and than I applied the inverse laplace transform for the relation to calculate Vo(t). I found

The result is different. I dont know what mistake I made in my calculation.
Could somebody help ?

#### Attachments

• 2.6 KB Views: 66
• 1.8 KB Views: 66
• 6.5 KB Views: 66
• 1.5 KB Views: 66

#### erikl

##### Super Moderator
Staff member
I dont understand the transistor behaviour given by the relation (1) in the paper.

This equation is not for transistor in linear or saturation region. Could somebody explain this new expression and when we use it ?
I think this equation presents an approximation to the $(V_{GS} - V_T)^2$ dependency for the drain current in strong inversion at full velocity saturation (in saturation region), s. e.g. David M. Binkley "Tradeoffs and Optimization in Analog CMOS Design" Sect. 2.4.3 , Equs. (2.15) & (2.17), pp. 16 & 17 .

points: 2

#### Teddy

erikl is right - in case you consider large gate length devices (1um and up) - that is what you find in books. But in sub -micron technologies those equations are barely close to reality.
Good example is calculation when device is in saturation compared to simulated vdsat.
To answer the question in detail I recommend to look at BSIM 3 and higher spice models and equations for the devices.

points: 2

#### dgnani

Hi all,
I was reading the following paper:
https://ece.uwaterloo.ca/~cdr/pubs/maymandi/3.pdf.

I dont understand the transistor behaviour given by the relation (1) in the paper.

This equation is not for transistor in linear or saturation region. Could somebody explain this new expression and when we use it ?

In a second step the autor try to calculate the voltage Vo using the equation (3).

And he say

I applyed the laplace transform and than I applied the inverse laplace transform for the relation to calculate Vo(t). I found

The result is different. I dont know what mistake I made in my calculation.
Could somebody help ?
It looks like someone already found what the first equation means, note that Kn is not the usual spice level 1 parameter because it has dimension of current over voltage (rather than current over voltage squared).

As of your differential equation, remember that the general solution (for any initial condition) is the sum of the solution of the homogeneous equation (i.e. without the first K1 on the right side) which for a first order equation has one undefined parameter and a particular solution (e.g. a constant solution) of the complete equation

points: 2

#### dgnani

Hi,
Thank you so much Erikl, Teddy and dgnani.
I understand now.
Concerning the second question. The Laplace transform applied to Eq(1) gives the following transfer function:
Vo=-k/(l*k+C*s)
(Here k=K1, l=λ7 and C=CL1)
The result is given here
http://www.wolframalpha.com/input/?i=InverseLaplaceTransform[-k%2F%28k*l%2Bc*s%29%2Cs%2Ct]

When we apply the initial condition like used by the author the results is different.
Did not check on wolfram alpha but your Laplace transform is wrong:
C -> C/s
f'(t) -> sF(s)-f(0)

is all you need and you get exactly the paper result

points: 2

Yeah !
The mistake was the f(0).
But C is a constant so it remain as it is. If I change it by C/S the exponential part will not appear in the Vo expression.
The result obtained is Vo(t)=Vdd.exp(-t/τ).

Last edited:

#### dgnani

By C->C/s I mean any constant transforms accordingly, specifically in the case of your equation
K->K/s
constants transform as Heavyside (step) functions
-C dV/dt = K + Kλ V
becomes, using F(s) as lapl transform of V(t):
-C[s F-V(t=0)]=K/s + Kλ F

Which gives you the correct result

You're right. If I use the laplace transform that you provide I get correct result.
Thank you so much dgnani.
Thank you all for helping me.
Cheers,

#### elgiovo

##### Junior Member level 1
I assume kn = mun * Cox; if this is the case, the expression for the short channel saturation current is wrong, since it has not to be divided by 2L. (or it is not a current, dimensionally). Moreover, remember that it's just an approximation valid for L->0.

Status
Not open for further replies.