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Let's consider this simple circuit

schematic

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?

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Simply apply kirchoffs current law - Is = IL

The zener is, for all practical purposes, removed the circuit since Iz = 0 and the circuit reduces to a potential divider where Vout = Vin * RL/(RL + Rs)

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  • \$\begingroup\$ 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\$? \$\endgroup\$ – BowPark Feb 1 '17 at 16:33
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    \$\begingroup\$ As long as the the voltage across it is below the knee of the Zener voltage (determined by the resistor values) the Zener diode acts like any normal reverse biased diode. There may be a tiny leakage current but nothing significant. As RL decreases beyond a critical value (Is =IL) the output voltage will fall and regulation will fail. That's why you design with extra current through the Zener at no load so that when you take current from the output it always allows some current to flow through the Zener to maintain regulation. \$\endgroup\$ – JIm Dearden Feb 1 '17 at 16:54

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