Why does a clamper circuit shift the input voltage upward by 20V?

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Abdulwahab Hajar
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Hello everyone, I hope you could help me with the analysis of this circuit.
the output voltage is supposed to be similar to the input but shifted upwards by 20V, so basically it's between 0 and 40V... but why?
I mean when the input is 20V, the diode is reverse biased... and the capacitor charges... our output voltage depends on the ratio of the resistance to the capacitive reactance... at least until the capacitor fully charges... but then what happens?

Thank you
 
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  • Figure P2.39(a): clamper circuit with input v_I, series capacitor C, shunt diode and resistor R to ground, output v_O. Beside it, a square-wave input swinging between 20 V and -20 V, with 0 V marked.
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Abdulwahab Hajar said:
I mean when the input is 20V, the diode is reverse biased... and the capacitor charges...

no. They told you the time constant is large, so the capacotor does NOT charge through the resistor.

Your thought tool should be this: "Time constant is large, eh? If i take that to the extreme how would the circuit behave? Limit would be infinite time constant... so what would go on if that resistor were nearly ∞ or not even there ? "

In your mind snip out that resistor and repeat your thought experiment , considering both Vt transitions.
 
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jim hardy said:
no. They told you the time constant is large, so the capacotor does NOT charge through the resistor.

Your thought tool should be this: "Time constant is large, eh? If i take that to the extreme how would the circuit behave? Limit would be infinite time constant... so what would go on if that resistor were nearly ∞ or not even there ? "

In your mind snip out that resistor and repeat your thought experiment , considering both Vt transitions.
I really can't see where this is going, a bit more insight would be highly appreciated...
I mean a very high resistance would mean almost 0 current given that the voltage supply is constant... but that doesn't really get me anywhere..
Also sir, they said that the time constant is large... how would we know if that indicates a large resistance or a large capacitance.. since the time constant is RC and dependent on both...

Thank you!
 
Abdulwahab Hajar said:
I really can't see where this is going, a bit more insight would be highly appreciated...
I mean a very high resistance would mean almost 0 current

What about the diode ?
 
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Abdulwahab Hajar said:
the voltage supply is constant
??
Square-wave voltage vs. time graph alternating between 20 V and -20 V, with a dashed continuation after the second high level, time axis labeled t.

looks to me like voltage goes from +20 to -20 then back to +20, maybe repeating that forever they didn't say .

You are not operating the circuit in your mind.
 
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  • Square-wave voltage vs. time graph alternating between 20 V and -20 V, with a dashed continuation after the second high level, time axis labeled t.
    upload_2018-3-10_15-31-31.png
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jim hardy said:
What about the diode ?
Got it sir, I forgot that the diode would determine the voltage drop Vo...
Thank you!
 
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