# Why does a voltage drive an infinite current?

This is a follow up to my previous question, where I asked why a generator will generate inconsistent voltage, leading to fluctuations in the current that might burn out an LED, unless we introduce resistors into the circuit. A user answered my question by saying that a resistor is required to achieve the desired current because any voltage will drive an infinite current. So why exactly is that? and as an additional side question: can voltage and current be mathematically related without resistance being part of the equation?

• Even when forward biased the led has a finite resistance. If the voltage is such that the current through the led exceeds the rated value (noting that this current is not infinite) the led will be destroyed. – Farcher Jun 30 '18 at 6:39

A user answered my question by saying that a resistor is required to achieve the desired current because any voltage will drive an infinite current.

That was probably a hyperbole used to make a point that increasing the forward voltage across an LED could lead to unexpectedly high currents.

When I say "unexpectedly high", I mean that, in LED's (as in other diodes) the current increases disproportionally to the increase of the voltage, i.e., LED's, unlike more familiar resistors, do not abide by Ohms's law.

Instead, the forward current in diodes grows exponentially with the forward voltage, i.e., for a given voltage increment, the current is multiplied by a certain factor.

One of the commonly used equations approximating this exponential relationship is Shockley diode equation:

$I=I_s(e^{\frac V {nV_T}}-1)$

If we assume $n=1$ (ideal diode) and $V_T=26mV$ (room temperature), we can calculate that, for every $60mV$ increment of the diode voltage, its current will increase by about $10$x.
For example, if the current of a diode at $0.7V$ was $10mA$, at $0.76V$ it would increase to $100mA$ and at $0.82V$, it would jump to $1A$. As a comparison, the current in a resistor, due to the same voltage changes, would increase to about $11mA$ and $12mA$, i.e., proportionally to the applied voltage.