# Current through a resistor and diode in series

Current i is same for every element connected in series in a circuit. Let, there is a diode and a resistor connected in series . Current through the resistor should be same as current through diode . But From the graph of load line and diode current equation it seems that current through the resistor will increase as voltage increases. But it is known that current through the diode will not increase that much(a tiny increase will happen).
Isn't this a contradiction that elements connected in series shares the same current ?

A sample graph of diode i-v characteristics and load line:

Edit: voltage through the diode will not increase that much

• I don't know if this is a school assignment or what, but you've got the current-voltage relationship backward. The current through a diode will increase a lot as the voltage rises. – Dampmaskin Mar 14 '17 at 16:03
• Look up Kirchoff's current law. If the currents in series elements weren't the same (and i do mean identical) there would be a massive build up of charge at one of the junctions - what goes in must come out. – JIm Dearden Mar 14 '17 at 16:05
• "But it is known that current through the diode will not increase that much..." Voltage across the diode will not increase that much. current will increase a lot. – dannyf Mar 14 '17 at 16:17

But it is known that current through the diode will not increase that much(a tiny increase will happen).

Once the diode is turned on, a change in voltage across it will cause a large change in current, not a small one.

Isn't this a contradiction that elements connected in series shares the same current ?

No. The load line graph shows exactly what will happen. If you increase the source voltage, the green line will shift to the right. Then you will find a new intersection between the IV curve for the source+resistor and the diode.

It will not cause a large increase in the diode current because the resistor limits the current. This is because $V_d = V_s - V_r$. The diode voltage is equal to the source voltage minus the resistor voltage. So as the source voltage rises, either the diode or the resistor can take up that change. And because the diode has very low differential resistance, it's the resistor that will take up most of the added voltage.

Edit

Another way you know this is not a contradiction is that the fact the resistor and diode currents are equal was a fundamental part of drawing the load line graph.

The two curves are

$$i=I_s \exp \left(\frac{nv}{V_T}-1\right)$$

and

$$i=\frac{V_s-v}{R}$$

The $v$ on the x-axis is the voltage across the diode. The $i$ on the y-axis is the current flowing in common through all three elements in the circuit (source, resistor, and diode).

Nope.. The current increases at the same rate in the diode. It is the voltage across it that does not change that much.

As you raise the voltage, it simply means the resistor will have a greater voltage drop across it and generate more heat.