# How to find R,without knowing diode resistance

I can't able to find R,because I dont know diode resistance

• en.wikipedia.org/wiki/Shockley_diode_equation The simplified model suffices. You may even go with a linearization of the exponential function. – Janka Aug 12 at 7:29
• Why do you need diode resistance? Why can’t you assume a fixed forward voltage? – winny Aug 12 at 7:55
• Given Voltage across diodes is 2.4,but it has to be 2.1 if we consider forward voltage – raghu ram Aug 12 at 8:08
• Please crop and rotate your image. – Marcus Müller Aug 12 at 8:53
• The diodes have a drop of 0.7V at 1mA as stated in the question, so you have to work out what current you need to get a drop of 0.8V – Finbarr Aug 12 at 9:05

You do not need to know the diode resistance in this problem.

First, let’s remind how a diode works in forward mode: It is a current source controlled by voltage. So, the tension of the diode varies little but the current of the diode varies a lot. Consequently, increasing the tension of the diode output more current and vice versa. This is why the diode is an important component of electronic. There are several mathematical model of the diode, some are very simple and others are more complex. In your problem, we will choose the one using the Shockley diode equation: The diode is modeled by this equation:

• I is the diode current,
• Is is the reverse bias saturation current (or scale current),
• U is the voltage across the diode,
• Vt is the thermal voltage (Boltzmann constant times temperature divided by electron charge). We will consider that Vt = 25mV = 0.025V (at 27°C)

Now, you can see that, we do not know Is. To find this value, we have to use the fact that “the diodes have a voltage drop of 0.7Volts at 1.0mA” (as the hint in your problem says).

So, we know have:

The model of the diode is entirely known.

Back to the problem, the idea of this circuit is the following: by tuning the resistor R, the current crossing the diodes will change, consequently, the tensions of the diodes will vary too. We know that if the resistor R imposes a current of 1mA, each diode will have a tension of 0.7V, so the output voltage will be 3 times 0.7V (because there are 3 diodes in series) so Vo = 2.1V.

However, we want Vo = 2.4V (meaning 2.4/3 volts for each diodes, so 0.8 volts), so the current crossing the diodes is too low, we want to increase it. The first thing to do is to find the current crossing the diodes, we have to use the equation above for that:

We can see that 55mA is far more important than the previous 1mA and it seem to respect what we said earlier (“the tension of the diode varies little but the current of the diode varies a lot”).

You know that you want a current of 55mA, you now have to choose the right value for your resistor. This is the easy part, the current crossing the resistor is 55mA, and the tension of the resistor is 10V-2.4V=7.6Volts. You can use the ohm law:

Note that, to be precise, you have to say that the output current is zero or negligible in comparaison of the other current in the circuit (otherwise the current crossing the resistor would be different from the one crossing the diodes).

EDIT: My bad, the current I on the schematic should be downward facing, the same way the diodes are.