Sketching Waveforms in Circuits w/ Diodes

In summary, a 5V peak to peak input signal will cause both the normal diode and the zener diode to short out in the positive half of the cycle, and the zener will clamp the signal at 5V until it drops below the zener voltage threshold.
  • #1
dancergirlie
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0

Homework Statement



sketch the output waveforms when a 5Vp sine wave with a frequency of 100 Hz is applied to each of the following circuits **see attached diagrams**

Homework Equations





The Attempt at a Solution

 

Attachments

  • Circuit 1.png
    Circuit 1.png
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  • Circuit 2.png
    Circuit 2.png
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  • Circuit 3.png
    Circuit 3.png
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  • #2
I am only still stuck on the first one!
 
  • #3
Normal diodes conduct (ideally becoming short circuits) when they are forward biased, and are open circuits when reverse biased. Zener diodes act like regular diodes in the forward direction, but they "fail" in the reverse direction after some specified voltage -- they too ideally become short circuits after the "zener voltage" threshold is reached in the reverse direction.

So your task is to imagine this 5V peak-to-peak input signal being connected to these circuits and how the diodes are going to react as the input varies. In your first circuit, for example, both the normal diode and the zener diode are going to short out all the positive half of the voltage cycle, since they'll both be forward biased. In the negative half of the cycle the normal diode will be reverse biased, so it can't conduct (so just ignore it). The zener will also stay cut off until its zener voltage threshold is reached. Then it'll clamp the signal at that value until it drops below the threshold again.
 
  • #4
since my zener diode has a threshold of 6.3V, and since my voltage is 5V peak to peak, it will never reach that threshold, so won't it always just be zero output?
 
  • #5
dancergirlie said:
since my zener diode has a threshold of 6.3V, and since my voltage is 5V peak to peak, it will never reach that threshold, so won't it always just be zero output?

A 6.3V threshold means that it'll never conduct in the reverse direction, yes. So it'll behave like a normal diode in this case. Note how it's wired in the same direction as the normal diode beside it; they'll both just behave as normal diodes as long as the input signal is never more than 6.3V negative (which it won't do here).
 
  • #6
so would the fact that I have two "normal" diodes in the same biased in the same direction in parallel change what my output waveform would look like, or would it just resemble what it would look like if I had one diode in the circuit?
 
  • #7
You can't make a short circuit shorter! :smile: Yup, It'll behave just like a single diode.
 
  • #8
thanks so much for the help :D
 

1. How do diodes affect the waveform in a circuit?

Diodes act as one-way valves for electrical current, allowing it to flow in only one direction. This results in a distorted waveform, as the diode blocks the negative half of the alternating current signal.

2. Can you sketch a waveform with multiple diodes in a circuit?

Yes, a circuit can have multiple diodes which will each affect the waveform in their own way. The resulting waveform will depend on the specific placement and orientation of the diodes in the circuit.

3. What is the difference between a forward-biased and reverse-biased diode?

A forward-biased diode allows current to flow through it, while a reverse-biased diode blocks current flow. In a circuit, this results in a different shape of the waveform depending on the biasing of the diode.

4. How do you calculate the voltage drop across a diode in a circuit?

The voltage drop across a diode can be calculated using Ohm's law, where the voltage drop (Vd) is equal to the current (I) multiplied by the resistance of the diode (Rd): Vd = I * Rd. The resistance of a diode is typically very small, so the voltage drop is usually negligible.

5. Can diodes be used to rectify AC current into DC current?

Yes, diodes can be used in a circuit known as a rectifier to convert alternating current (AC) into direct current (DC). The diodes block the negative half of the AC signal, resulting in a pulsating DC waveform. This can then be smoothed out using capacitors to produce a more constant DC signal.

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