Don't understand clipper circuit

i've only drawn the necessary stuff to show half a cycle.

i don't understand how the battery acts to limit the voltage at V_out. how does battery counteract the part of the cycle where V_in > V_battery? for this part of the cycle shouldn't it be V_in - V_b?

jim hardy
Gold Member
Dearly Missed
Write Kirchoff's Voltage Law around the loop Vout-diode-battery....

NascentOxygen
Staff Emeritus
As Jim said. When the diode is conducting, what is the sum of the voltages : Vbattery + Vdiode?

what about the outer loop? what is happening to V_source?

and is the resistor absolutely necessary?

and is the resistor absolutely necessary?
What do you think, what's going on with current when diode conducts and the resistor value is R=0 Ω ?

Your output voltage should be the rightmost one. In the diagram, the one V(out) you have written in black.

When V(in) < V(b) then the diode won't be forward biased and won't conduct. So the output voltage is V(in). There is no voltage drop across the resistor, since no current is flowing across it.

When V(in)>V(b) then the diode will be forward biased and will conduct. So the output voltage will be V(b). When forward biased the drop across diode is around 0.4 or 0.7V.

The battery does counteract the input voltage, which is reflected across the resistance. Think of it as cold water flowing from V(input). There is a thermometer at Resistance and hot source of water from V(b). The difference of temperature is reflected at the thermometer but the temperature of either source of water won't change. So if you place another thermometer across the cold source of water, you will get the cold water's termperature.