# P-N Junction Diode - Charge question

1. Mar 18, 2013

### bsr25

We have been looking at P-N junction diodes in our lab session this week and investigating the current through the circuit and how this relates to the Voltage.
We are using the approximation:

I = (I0)*exp(qV/kT)

We measured the voltage and the current for a few different voltages and this could be plotted into a graph. A regression analysis gave the reverse saturation current as I0 = 3.2 x 10^-6 A, and the effective charge carrier charge as q = 4.78 x 10^-21 C.

The diode being used is the 1N4148 junction diode.

Im trying to understand why the carrier charge is not equal to 1.6*10^-19 C. We were told that it has something to do with the data-sheet for this diode. But so far Im not sure what on the sheet I should be looking for and even how this would related to solving this problem.

Any help would be appreciated,

Thanks Ben.

2. Mar 18, 2013

### CWatters

Is that a bit high?

3. Mar 18, 2013

### rude man

That makes no sense. q = charge of an electron. If you came up with any other value then you have operated the diode outside the region where i = i0 exp(qV/kT) applies. A 1N4148's 'good' region is around i = 0.01 mA to 10 mA.

I agree the value of i0 is totally unrealistic. It should be in the pico to femtoampere region. Are you running the diode at or near room temperature?

4. Mar 18, 2013

### bsr25

Yeah. This is my problem too. I expected the q value we got to at least be on the order of magnitude of the elementery charge. And from research I had done, expected I0 to be less.

However, the lab manual states: "Finally, if all goes well your value for the electron charge, q, will probabbly not be what you expect. At this point you should go back and look carefully at the datasheet for a PN-junction diode, and the values for 1N4148 in particular and try and reconcile the results".

This has me baffled.

5. Mar 18, 2013

### bsr25

Oh right so I just looked at my data, and the currents we calculated were within the range:

0.08 mA to 14 mA.

And the diode is just at room temperature.

6. Mar 18, 2013

### rude man

Will get back later on this. The problem seems to lie with i0 which does not behave like a constant over voltage.
Obviously q = 1.6e-19 C, you would not want to come up with a model that says anything else ...

7. Mar 19, 2013

### ehild

I guess you measured on the forward-biased diode. The original equation for the junction diode is I=Io(exp(qV/kT)-1), and Io is usually determined by measuring on reverse-biased diode (negative V).

Even that equation is quite bad approximation for the relation between the measured voltage and current at high currents, when the current is limited by the parasitic resistances and internal resistances of the meters.

ehild

8. Mar 19, 2013

### CWatters

Was the voltmeter (or whatever you used to measure the voltage) connected when you were measuring the reverse current?

9. Mar 19, 2013

### CWatters

10. Mar 20, 2013

### bsr25

A square wave generator was connected to a diode and resistor that were in series. Leads going to an oscilloscope were placed over the diode to measure that voltage drop. Leads were than placed over both the diode and resistor to get the voltage of the source. The oscilloscope measured both peak voltages and from that we calculated the resistance over the resistor (i.e. the source voltage minus the diode voltage). From this we got the current in the circuit. For a number of different source voltages we did this and got the current.

This was plotted as a graph (I against V across diode). The fit gave us values for I0 and q.
Is there an error in this setup? Perhaps where we are measuring something extra or something?

I dont know really. Shooting blanks here.

Thanks for the help so far!
Ben