# Ohmic Resistance

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1. May 10, 2017

### Faisal Moshiur

How am I supposed to calculate ohmic resistance from a non-linear electronic device component?
Say, for example, I need to calculate ohmic resistance of a pn junction of bipolar junction transistor from its forward I-V characteristics curves. (It can be either Base-Emitter junction or Base-collector junction.)

2. May 10, 2017

### LvW

A non-linear device (like a pn junction) has no "ohmic resistance" per definition.
For such a non-linear V-I characteristic we can define only (a) a static resistance or (b) a dynamic/differential resistance.
Both resistances depend (vary) on (with) the corresponding DC operating point (Q point).

3. May 10, 2017

### scottdave

I am not sure exactly what you are asking. Perhaps the curve is fairly linear during certain operating ranges, therefore it could be approximated as a resistor for certain calculations.

4. May 10, 2017

### Staff: Mentor

It depends on your definition. The slope of the V-I curve at any point is a resistance. But the resistance changes for every different point on the curve.

Similarly, pick any two points on the V-I curve, draw a straight line between them. The slope of the line is the average resistance in that region.

That may not be useful, but it is one way to define it.

If that is not what you mean, then tell us how you want to define ohmic resistance.

5. May 10, 2017

### LvW

Yes - that´s what we call differential resistance r=dV/dI..[/QUOTE]
More than that, pick any point on the V-I curve and draw a straight line between this point and the origin.
This gives you the static resistance R=Vo/Io (but it is not an "ohmic resistor" because this value depends on the selected point Vo,Io- the value R is not a constant as required for an ohmic characteristic).

6. May 10, 2017

### Baluncore

Pass a small current such as 10 uA through the forward biassed PN junction. Measure the voltage drop across the junction, plotting the voltage across the junction against the Log10 of the current as you go. Increase the current in steps to 100 uA, 1 mA, 10 mA … At the start you should get a straight line, but at higher currents the measured voltage will begin rise faster than the extrapolated straight line. That extra voltage above the straight line is due to the ohmic resistance of the junction. You know the current flowing, and you know the extra voltage above the straight line, ohms law will then give you the ohmic resistance.
Keep the junction at a stable temperature. Apply higher currents for shorter periods and watch out for heating effects which will change the junction voltage.