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I came across some examples to calculate the current in a circuit with diode and resistance in series. Consider the circuit below:

enter image description here

Current was calculated as follows:

I = (9 - 0.7)/R1

Does that mean that no matter what the resistance R1 is, the voltage drop across it is always 9 - 0.7 Volts? Or in other words, diodes decide the drop across a resistor by varying current through it unlike how in a circuit with only resistors, increasing value of resistance also increased the voltage drop across it?

Also, if the above assumption is true (i.e. taking higher resistance will lower the current value in the circuit but the drop across diode is 0.7v only), but doesn't that violate the V-I graph of a diode where at a specific voltage, a specific current valule should be there while we see different values of current for same value of voltage drop (0.7v)? (based on assumption)

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The forward voltage drop of a silicon diode is only approximately 0.7 V. The current through a diode depends exponentially on the voltage across it (i.e. the current is a very strong function of voltage). This means the voltage is a weak function of current, so for currents that you are likely to encounter when the diode is conducting (maybe tens of mA, depending on the application), the forward voltage of the diode typically doesn't deviate much from a fixed value (about 0.7 V). So clearly diodes are different from resistors in this respect.

The constant forward voltage drop approximation is equivalent to the I-V relation of the diode being approximated by I = 0 for V < 0.7 V, and I = ∞ for V > 0.7 V. This means that as long as the diode is "turned on" (its voltage is about 0.7 V), its current could be anything positive depending on the circuit the diode is connected to: in this case the battery and the resistor.

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  • \$\begingroup\$ Thank you for explaining Puk. You mentioned that "as long as diode is ON", true, but how can we predict it will be ON or OFF (I'm talking about voltage, assuming it is forward-biased). I mean if a diode is connected to some other components and a power source greater than 0.7v in series, can we always say it will be ON? What if two diodes are connected in series with a resistor and power source 1v? My point is, how can we be sure that a diode will get enough voltage to be ON unless we know the voltage divided in circuit in first place which assumes diode is ON to calculate it, deadlock? \$\endgroup\$
    – J J
    Jun 27 '20 at 5:22
  • \$\begingroup\$ In complex circuits, it may not be easy to tell if a diode will be on. So you can assume it will be on: if you find a negative current through the diode as a result, this tells you that you made the wrong assumption and that the diode should be off. More often than not you can tell by inspection if a diode will be on. For example, you can find the direction of current that would flow if there were no diode (if it were shorted). If the polarity of the diode is such that the diode allows current in that direction, chances are the diode will be on. \$\endgroup\$
    – Puk
    Jun 27 '20 at 5:29
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    \$\begingroup\$ Nice explanation, tysm. But what about two diodes in series with 1v? That confuses me since they require 1.4v drop but have only 1v available \$\endgroup\$
    – J J
    Jun 27 '20 at 5:30
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As already mentioned, the foward voltage drop diode model is the most basic model for simple approximate calculations. This forward voltage drop also depends on the diode model. Now if you know the I-V characteristic diagram of the diode you can make a more accurate calculation. Alternatively you can use the shockley equation. A similar question was answered in more detail already Current through a resistor with diode

Regarding calculations with the simple voltage drop model in multiple diode circuits, diodes in series just add up their forward voltage drop. This means, that if the supply voltage is lower than the total forward voltages of the diodes, there will be no current and thus no voltage across the resistor. If the supply voltage is higher than the Forward voltage, the remainder of the supply voltage will drop across the resistor and therefore set the current.

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  • \$\begingroup\$ Thank you Lars for such a good answer. This helps me. \$\endgroup\$
    – J J
    Jun 27 '20 at 6:22

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