r/beneater • u/Icy-Departure7752 • 7d ago
Help Needed Can someone explain
I am new to electronics and was recently watching the series on building a digital clock. On this circuit I probed the diode on the left (the one highlighted with blue) and resistor on the right. In both the scopes the waveform of current is as expected only in the positive cycle but I don't quite get the waveform of voltage of diode why does it show a voltage spike in the negative cycle and a flat line on the positive cycle.
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u/flatfinger 7d ago
In any loop, the sum of all the voltages between points on the loop must equal zero.
If at some moment in time the right side of the AC supply is 10 volts higher than the left, then the right half of the diode will be about 0.7 volts below the left side, and the bottom of the resistor will be about 9.3 volts below the top.
If at some other moment in time, the right side of the AC supply is 10 volts lower than the left, then the left half of the diode would be lower than the right side while the diode passed almost no current. Since the resistor would have almost no current passing through it, there would be almost no voltage between the top and bottom of the resistor. The only way to make the voltage going around the loop sum to zero would be for the left half of the diode to be about 10 volts below the right side.
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u/Educational-Pop-2939 7d ago
Thank you for the question. I teach electronics and we are working on diodes right bow as they apply to altering sine waves and I showed them your post and we got to talk about it and i even demonstrated limiting and clamping. It was a good class.
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u/HydroPage 6d ago
Ohm’s law says there will only be a voltage difference across the resistor if current is flowing through it.
When the source swings positive, it matches the orientation of the diode and lets current flow into the resistor and build up a voltage across it.
When the source swings negative, the diode doesn’t allow current to flow. If the diode wasn’t there, you’d see a negative voltage on the resistor, because a current flows in the opposite direction as it did before.
However, since there is a diode, the opposite-direction current is not allowed to flow at all: no current flows through the resistor, and V = I*R, I = 0, so V = 0
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u/The8BitEnthusiast 7d ago
There is something unique about a diode which is that it only starts conducting current after the voltage across its pins reaches a certain minimum, and once it conducts, that voltage remains relatively constant. That's the flat line you see on the positive half of the cycle, which the scope reports as '593.199 mV', or just short of 0.6V. When voltage drops below that, the diode cuts off and behaves like an open switch, so the voltage across its pins will follow the supply voltage. That's what you see in the negative half of the cycle.
I've just watched this video from Mr. Carlson's Lab, which was published today. It actually explains this concept in great detail, and also shows a practical use for it. Highly recommended if you don't mind another 15 minutes on Youtube! ;-)