# Reverse saturation current for 1N4004

Discussion in 'Homework Help' started by danielwbbr, Oct 11, 2013.

1. ### danielwbbr Thread Starter New Member

Oct 11, 2013
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0
Hi all!

I've started playing with diodes for the first time but I've run into a little problem. I'm currently attempting to plot the V-I characteristics of a 1N4004 silicon diode using a approximation of Shockley's equation namely:

I ≈ (${I}s$)($e^{40*{V}D}$)​

where ${V}D$ is the operating voltage of the diode and I is the operating current.

${I}s$ is the reverse saturation voltage from what I've read but I have no idea where to find this value for the 1N4004.
Is the reverse saturation current something that is known or is it a dynamic value that will have to be independently calculated?

Thanks all in advance for any input as I'm really confused at the moment.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,939
1,089
You can do two things:
Firs is to find a spice model for a diode and find that
Is =14.11nA a

Id = Is*e^(VF/Vd)

Or you need to measure it on the bench. You need to do four measurements:
First measure a diode forward current Id1 and diode VF1 for that Id1 current.
Next you increase diode current to the value Id2 and measure Id2 an VF2.

And then we can find VD thermal voltage

$\Large V_{D}=\frac{V_{F2} - V_{F1}}{ln \frac{I_{d2}}{I_{d1}}}$

and

$\Large Is = I_{d2} *e^{\frac{-V_{F2}}{{V_D}}}$

For example for
Id1 = 50mA, VF1 = 0.768V
Id2 = 100mA , VF2 = 0.792V

We have

Vd = (0.782V - 0.768V)/( ln(100mA/50mA)) = 34.624mV

and

Is = 100mA*e^(-0.792V/34.624mV) = 11.64pA

Last edited: Oct 11, 2013
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3. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
Calculating an actual number for Is is seldom done and seldom worth it (now, for some occupations/applications, this isn't the case).

Instead, what you generally want is a reference point -- what is Vd at a particular Id, say Vref @ Iref -- and then a convenient way to find Vd for different values of Id (or Id for different values of Vd). This is easily done for the simplified diode equation you are working with. I'll let you play with that notion awhile.

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4. ### danielwbbr Thread Starter New Member

Oct 11, 2013
2
0
Thanks for the feedback!!

I think I have a handle on whats going on now!

Mar 31, 2012
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