- #1
Sketching Waveforms in Circuits w/ Diodes
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The discussion focuses on analyzing the output waveforms when a 5Vp sine wave is applied to circuits with normal and Zener diodes. Normal diodes conduct during the positive half of the cycle and block during the negative half, while Zener diodes only conduct in reverse after reaching their specified threshold. In this case, with a Zener threshold of 6.3V and a 5V peak-to-peak input, the Zener will not conduct, resulting in zero output. The presence of two normal diodes in parallel does not change the output waveform, as they will behave like a single diode. The overall conclusion is that the output will remain at zero due to the Zener diode's threshold not being met.
- #2
dancergirlie
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I am only still stuck on the first one!
- #3
gneill
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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.
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
dancergirlie
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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
gneill
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dancergirlie said:
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
dancergirlie
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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
gneill
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You can't make a short circuit shorter! 
Yup, It'll behave just like a single diode.
Yup, It'll behave just like a single diode.- #8
dancergirlie
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thanks so much for the help :D