Find effective resistance and current?

Question

The diode, a very common two-terminal nonlinear device, can be modeled using the following current-voltage relationship:

I = 10^-9 * ((e^39V) - 1))

1) what is the effective resistance of the diode at V = 0.55 V?
2) at what current does the diode have a resistance of 1 Ohm?

answer 1 = 266 mΩ
answer 2 =514 mA

how????

Answer 1

R= V/I = 0.55/10^-9(e^39*0.55 -1)=0.2659 ohm
As e^39V is much greater than 1 as a first approximation you can take
1= V*10^9/e^39V so e^39V =10^9*V which by trial and error gives you 514 mV which as R =1 ohm correspond to 514 mA.
The equation is not solvable by algebraic means.
I solved it with a scientific HP calculater

Answer 2

That is the Schottky equation which I didn't know until this question came up. The funny thing is, I am learning more than the person who asked the question. I'll just try and work it out. Check back in 10 minutes.

Answer 1.
I=0.000000001*(2.71828183^(0.55*39)-1)
I=2.07
From Ohm's law R=V/I=0.55/2.07 = 0.266Ω or 266mΩ

Answer 2.
Ah, its too late. I'm off to bed. I couldn't work it out due to my poor maths skills and goal seek on Excel went nowhere.

ps. muddypupuk. The more homework I do, the cleverer I get and the less likely this guy is going to be coming after my job.

Answer 3

a million. there isn't any longer adequate counsel right here however the respond they are finding for is possibly specific. in maximum situations it rather is real, yet no longer all. 2. No. while in comparison with maximum metals Tungsten has a severe resistance and a severe melting factor, which makes its useful in incandescent easy bulbs.

Answer 4

2.928635478382834 ohms