Need help understanding simple diode clamp & voltage divide combination
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 &ndash; 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.