Electronic – Need help understanding simple diode clamp & voltage divide combination

diodesvoltage divider

I am trying to understand Art of Electronics 2nd Edition page 50. The bit that I don't understand is that the impedance looking into the divider has to be small compared with with R (see first picture). Can any one explain this? I know how to get the Thevenin equivalent shown in the second picture, and I understand why voltage dividers' output can drop when a load is attached (especially for relatively low resistance load).

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Best Answer

The goal of the circuit is to clamp the output voltage to just over 5 V.

To make it easier to analyze the circuit, imagine swapping the location of the diode and the equivalent resistor:


simulate this circuit – Schematic created using CircuitLab

This doesn't affect the voltage that appears at the output, but it makes it easy to see you can analyze the circuit as a voltage divider between the input node and an approximate 5.7 V source at the diode's anode.

Say R is 20 kohms. Then if the input voltage is, say, 10 V, the output voltage will be held to roughly 5.8 V (assuming an idealized 0.7 V drop across the diode).

If R is 10 ohms, on the other hand, most of the voltage drop ends up across the 667-ohm equivalent resistor, giving an output voltage of 9.96 V. Which means the circuit has done effectively nothing to limit the output voltage.