What's the Difference Between These Two Diode Limiter Problems?

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Discussion Overview

The discussion revolves around two examples of diode limiters, focusing on understanding the differences in behavior between the two circuits when subjected to positive and negative input signals. Participants explore the conditions under which the diode is forward or reverse biased and how this affects the output voltage.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that in the first example, the diode will be forward biased during the negative half of the signal, limiting the output to -0.7 V and calculates the peak output voltage.
  • Another participant questions whether the diode will be forward or reverse biased for the positive input signal in the second example and what that implies for current through the resistor.
  • A participant asserts that for the positive input, the diode is reverse biased (no current), while for the negative input, it is forward biased (current flows).
  • There is a calculation proposed for the output voltage during the negative half, suggesting it will be -10 V + 0.7 V = -9.3 V.
  • Another participant confirms that the output waveform will follow the input waveform minus the diode drop when the diode is conducting.
  • A participant seeks clarification on whether the output waveform will equal zero during the positive half and follow a specific calculation during the negative half.

Areas of Agreement / Disagreement

Participants generally agree on the behavior of the diode in terms of forward and reverse biasing, but there is uncertainty regarding the exact output voltage calculations and waveform behavior in the second example.

Contextual Notes

Some participants express uncertainty about how to calculate the output voltage in the second example, and there are unresolved questions about the output waveform during different phases of the input signal.

bibo_dvd
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hello guys

while studing the diode limitters

i found two examples i solved the first one but i couldn't with the second
1 -

ohwNJGZ.png


2-

ih6plrd.png


in the first i know that the diode while be forward biased in the negative half of the signal
so the negative half will be limitted to -0.7 v
and the V peak output will be = [(RL/(RL+R1)]*Vinput = 9.09v

but i don't know what will be the difference between the first and the second problem !

so please help me guys ...Thanks :)
 
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In the second one, for + input signal, will the diode be forward or reverse biased? What does that mean about current through the resistor?

For - input signal, ask the same questions...
 
berkeman said:
In the second one, for + input signal, will the diode be forward or reverse biased? What does that mean about current through the resistor?

For - input signal, ask the same questions...

okay in the + input the diode will be reverse biased so the current will not pass , but in the case of - input the diode will be forward biased and the current will pass

am i right ??
 
bibo_dvd said:
okay in the + input the diode will be reverse biased so the current will not pass , but in the case of - input the diode will be forward biased and the current will pass

am i right ??

Correct. And V=IR, so you know when you will have an output voltage. Good job! :smile:
 
berkeman said:
Correct. And V=IR, so you know when you will have an output voltage. Good job! :smile:

okay i know that , but i still don't know how will i calculate Vout

i think Vout will be = -10+0.7= -9.3v .. is that right ??
 
bibo_dvd said:
okay i know that , but i still don't know how will i calculate Vout

i think Vout will be = -10+0.7= -9.3v .. is that right ??

That would be the negative peak voltage. The output waveform will follow the input waveform (minus the diode drop) when the diode is conducting.
 
berkeman said:
That would be the negative peak voltage. The output waveform will follow the input waveform (minus the diode drop) when the diode is conducting.

you mean that the output waveform will equal zero in the positive half and in the negative half it will equal -10v-0.7v ?? or what ??
 
bibo_dvd said:
you mean that the output waveform will equal zero in the positive half and in the negative half it will equal -10v-0.7v ?? or what ??

In the negative half cycle, the output voltage equals the input voltage, minus the diode drop. Sketch that on top of the input waveform plot...
 

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