# How does this voltage clamping circuit work? I'm reading a book which states this circuit prevents the input voltage from exceeding 5.6 V (0.6V is the forward drop of the diode), and I'm having a hard time understanding it.

The diode is forward biased and acts as a closed switch at any voltage above 0.6, so I'm having trouble understanding why it clamps voltage above 5.6 volts.

• Voltage is always between two points. Oct 12, 2021 at 4:29

If voltage at In terminal is less than 5.6v the diode is not active, so current flows from In to Out only. If the voltage at In is more than 5.6v some aditional current starts to flow thru diode to 5v rail. Since the Vf of diode is "always" 0.6v the voltage at Out cannot be more than 5.6v, the rest of voltage (voltage above 5v6) will be dropped on 1k.

It's 0.6V from the diode PLUS 5V from the power supply.

So it's clamping "out" at 5+0.6=5.6V ABOVE GROUND

The 'above ground' part may be what you're missing.
When not explicitly stating the voltage reference, it's generally assumed one is referencing ground (i.e. 0 volts)

Any voltage that "in" gets higher than 5.6V will be dropped in the resistor. SO if "in" is 8V, "out" is still clamped to 5.6, then the voltage ACROSS the resistor (not referenced to ground) would be 8-5.6=2.4V simulate this circuit – Schematic created using CircuitLab

Just looking at the peak voltage, can you see these are the same?

A better Simulation

For any voltage at input terminal less than 5.6 V, the diode does not conduct. So the same voltage will be seen at the output. But when input voltage exceeds 5.6V, the diode starts to conduct and current flows from input to 5V via resistor and diode. This current creats a voltage drop accros input resistor so you will always see maximum 5.6V at the output.