# Voltage across Zener diode

Let's consider this simple circuit

simulate this circuit – Schematic created using CircuitLab

with

$$V_S = 15 \ V$$ $$R_S = 500 \ \Omega$$ $$V_Z = 5.1 \ V$$ $$R_L = 1 \ k\Omega$$

It can be described by the system of equations

$$V_S = R_S (I_Z + I_L) + V_Z$$ $$V_S = R_S (I_Z + I_L) + R_L I_L$$

where $V_S, R_S, V_Z, R_L$ are constant and only $I_L, I_Z$ are unknown.

Supposing that the Zener diode is inversely biased, voltage at node A will always be the Zener voltage $V_Z = 5.1 \ V$ and with these values $I_L = 5.1 \ mA$ and $I_Z = 14.7 \ mA$. If the resistance $R_L$ is decreased, the value of $I_L$ will raise and $I_Z$ will be lower. The limit condition is when $R_L = R_L^*$ is so small that it requires $I_L = I_S$ and the Zener branch has no current.

What does happen if $R_L$ is lowered beneath $R_L^*$ ?

What assumptions should be followed to write new equations? How would the Zener diode behave and how can it be considered?

• So, if no current can flow through the Zener, regardless of the voltage across it, it simply turns off, becomes an open circuit and stops forcing $V_A$ being equal to $5.1 \ V$? – BowPark Feb 1 '17 at 16:33