Don't understand clipper circuit

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Discussion Overview

The discussion revolves around understanding the operation of a clipper circuit, specifically how the battery influences the output voltage (V_out) in relation to the input voltage (V_in) during different phases of the input cycle. Participants explore the roles of components such as diodes and resistors in the circuit.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how the battery limits V_out when V_in exceeds V_battery, suggesting that V_out should be V_in - V_b.
  • Another participant suggests applying Kirchhoff's Voltage Law around the loop involving V_out, the diode, and the battery.
  • It is noted that when the diode conducts, the sum of the voltages is V_battery + V_diode.
  • Concerns are raised about the role of the outer loop and the necessity of the resistor in the circuit.
  • Another participant reiterates the question about the necessity of the resistor and inquires about the current behavior when the diode conducts with a resistor value of R=0 Ω.
  • A participant explains that when V_in is less than V_b, the diode does not conduct, and V_out equals V_in, with no current flowing through the resistor.
  • When V_in exceeds V_b, the diode conducts, and V_out equals V_b, with a forward voltage drop across the diode noted to be around 0.4 or 0.7V.
  • A metaphor involving cold and hot water is used to describe how the battery counters the input voltage, reflecting the difference across the resistance.

Areas of Agreement / Disagreement

Participants express differing views on the role of the resistor and the behavior of the circuit under various conditions, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants have not reached consensus on the necessity of the resistor or the implications of having R=0 Ω, and there are unresolved questions regarding the behavior of the circuit in different voltage conditions.

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

fhTC4SV.png


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?
 
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Write Kirchoff's Voltage Law around the loop Vout-diode-battery...
 
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?
 
iScience said:
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.
 

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