Constructing a Circuit to Limit Voltage of 20V Peak-to-Peak Sine Wave

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Discussion Overview

The discussion focuses on constructing a circuit to limit the voltage of a 20V peak-to-peak sine wave, specifically targeting a positive limit of 5.6V and a negative limit of -2.5V. The conversation explores circuit design, including diode clipping and scaling considerations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the construction of a circuit to limit the sine wave voltage, mentioning a 1 k ohm resistor and a forward-biased diode.
  • Another participant questions the available voltage rails and whether the output waveform should be a scaled version of the input or simply clipped, seeking clarification on the input characteristics.
  • A participant describes their attempt to create a diode clipper circuit that outputs 5V, expressing uncertainty about whether scaling is necessary before clipping.
  • Further elaboration on the diode clipper circuit is provided, with a suggestion to draw the positive rail at the top of the schematic and the negative rail at the bottom, along with a note on the standard conventions for schematic representation.
  • Discussion includes the feasibility of clamping at 5.6V and the challenges associated with clamping at -2.5V, with a mention of using a transistor and resistors for achieving non-standard voltage drops.

Areas of Agreement / Disagreement

Participants express differing views on whether the output should be scaled or simply clipped, and there is no consensus on the specifics of the voltage rails available for the circuit.

Contextual Notes

Unresolved questions include the exact input characteristics and the implications of using non-standard voltage rails for clamping. The discussion reflects uncertainty about the design requirements and assumptions regarding scaling.

laminatedevildoll
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How do I construct a circuit that will limit the positive half of a 20V peak-to-peak sine wave to 5.6V and the negative half to -2.5V?

So far, I have a 1 k ohm resistor and the diode is forward biased leading to VL and Vout.
 
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What voltage rails do you have available? Is the resulting waveform supposed to be a scaled version of the input waveform, or just clipped on top and bottom to the voltages you mention? Can it be scaled and clipped, or is it not supposed to be scaled before clipping? What are the input characteristics of whatever you are feeding your output waveform into?
 
For instance, I attempted to draw the diode clipper circuit which gives out a 5V instead. The input graph is suppposed to be a sine wave and the output is supposed to be a cut off at the top of the wave.

I am not sure if it is supposed to be scaled or not, but I am thinking no.
 

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laminatedevildoll said:
For instance, I attempted to draw the diode clipper circuit which gives out a 5V instead. The input graph is suppposed to be a sine wave and the output is supposed to be a cut off at the top of the wave.
I am not sure if it is supposed to be scaled or not, but I am thinking no.
That looks right for the positive side clipper. BTW, the usual convention is to draw the positive rails at the upper side of the schematic, and the negative rails at the bottom part of the schematic. So in your sketch, draw the positive catch diode pointing up, with its cathode connected to +5V at the top of the page.

Now you just need the negative clipper. That's why I asked about what voltage rails you can assume that you have. Clamping at 5.6V is pretty straightforward, since that's just about a diode drop above a 5V rail. But clamping at -2.5V is a little weird, since having a -1.9V rail to clamp to would be pretty non-standard. You can make a diode clamp of any voltage drop with a transistor and two resistors, so maybe that's what they are looking for.
 

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