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Circuit diagram2I was doing an experiment to plot the $I-V$ characteristics of a diode. I connected a diode and a $47\Omega$ resistance in series, and measured the voltage and current across the diode. I recorded the values in a table. When I changed the resistance to $660\Omega$ ohms, I got a different set of diode voltages for the corresponding diode current. Why should the diode voltage change?

$I_d(mA)$ $\space$ $V_d(V)(47\Omega)$$\space$ $V_d(V)(660\Omega)$

15   0.746     0.749
17   0.753     0.758
19   0.760     0.765
23   0.771     0.773

EDIT 1:

I tried the experiment again. The voltmeter resistance showed infinite and ammeter resistance showed 0. I got the same values for $47\Omega$. I then waited for 15 mins to allow the diode to cool down and return to normal. I got a different set of values:

$I_d(mA)$ $\space$ $V_d(V)(47\Omega)$$\space$ $V_d(V)(660\Omega)$

15   0.746     0.758
17   0.752     0.767
19   0.760     0.780
23   0.772     0.794

Any ideas?

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  • $\begingroup$ Welcome to Physics Stack Exchange. All questions about circuits should come with a complete circuit diagram. Complete means that the diagram shows each element in the circuit and the connections of the measurement devices. All elements should be labeled. Without this diagram, we can't tell what you actually did. $\endgroup$ – DanielSank Sep 17 '16 at 14:12
  • $\begingroup$ Please include the units of the measured quantities. Doing so will help us judge whether akhmeteli's answer could apply. $\endgroup$ – DanielSank Sep 17 '16 at 14:38
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I am not sure which of the following may be the cause, but I suspect one of the following is worth considering: the voltage source resistance and amp. meter resistance are not zero, and voltmeter resistance is not infinite.

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  • $\begingroup$ But I have used the same ammeter, voltmeter and voltage source for both cases $\endgroup$ – user3442005 Sep 17 '16 at 14:35
  • $\begingroup$ Given that the voltages are different by less than 1%, I'd say akhmeteli is probably right on track. @user3442005 should add the resistance of the ammeter or of the volt meter to the circuit diagram and see how that would affect the predictions. $\endgroup$ – DanielSank Sep 17 '16 at 14:49
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The change in resistance will change the current in the circuit and given the specific I-V characteristics of the potential drop over the diode will settle at a different voltage. The wiki page on diodes discusses how this can be modeled using the Shockley diode equation which relates the voltage over a diode to its current: https://en.wikipedia.org/wiki/Diode_modelling.

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The current registered on the ammeter is the current passing through the diode and the voltmeter.
Given that the meters and the diode were not changed the fact that there were different readings for the two experiments was probably not due to the fact that the voltmeter had a finite resistance.

With a larger resistor present to obtain the same current the power supply must be set at a higher voltage.
This might mean that the adjustment of the voltage of the power supply to obtain a given current could be done more accurately.
That is the control knob would have to be turned further between readings.

Noting that the currents were only give to about $5\%$ it could be that the reading on the ammeter was either low or high compared with the actual current depending on which way the voltage control knob on the power supply was turned.

It could be that the temperature of the diode changed between the two experiments.

All in all this is a good example why repeated readings should be taken and in this case when the anomaly was noted then another set of results with the $47 \; \Omega$ resistor again in the circuit should have been taken.

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protected by Qmechanic Sep 17 '16 at 15:57

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