# Linearizing tunnel diode equation

1. Oct 1, 2008

### rjspurling

1. The problem statement, all variables and given/known data
Hi, I need to produce a linearized equation of the following,

I = D. V. exp(-V/B)
eq(1)​

I is the current
V is the voltage
D is a constant
B is a constant

Data was collected in an experiment designed to investigate the characteristics of a tunnel diode. I didn't do the experiment myself, I just have to find the linearized form of eq(1) and determine the constants D and B. I have a set of I values and a set of V values given to me to allow me to calculate the constants.

2. Relevant equations
I = D. V. exp(-V/B)
eq(1)​

3. The attempt at a solution
The problem I am having is the V term before the exponential term.

Taking logs of both sides,
Ln(I) = Ln(D) + Ln(V) - V/B
eq(2)​

I'm going to use excel to determine B and D. I'm just not sure how to graph it. If there was no V term before the exponential term then it would be,

Ln(I) = Ln(D) - V/B
eq(3)​

and i could simply graph Ln(I) vs V, and 1/B would be my gradient and Ln(D) would be my y-intercept. I can't do this for eq(2) though because of the Ln(V) term.

If I graphed Ln(I) vs Ln(v) - V, i can't determine B. I just can't seem to find a way to do it. I'm sure there's a simply way. I just need to work out how to bring the V's together.

Any help would be greatly appreciated.

2. Oct 1, 2008

### tiny-tim

Welcome to PF!

Hi rjspurling! Welcome to PF!
Stop trying to keep I and V separate …

you're allowed to mix-and-match!

Hint: you want an equation with B on the LHS, and a fraction on the RHS.

3. Oct 1, 2008

### rjspurling

argh mixing, of course!

Ln(I) = Ln(D) + Ln(V) - V/B

Ln(I) - Ln(V) = Ln(D) - V/B

Ln(I/V) = Ln(D) - V/B

Ln(I/V) vs V

Ln(D) is the y-intercept.

-1/B is the gradient.

I think thats it.

Thanks tiny-tim! much appreciated.