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seragnoor252
modified 6 months ago

Zero Volt

6
10
352
04:40:31
36 LED by Zero Volt. Why?
published 1 year ago
Redstone_guy
1 year ago
It’s not 0V, but 100mV that’s powering the LEDs. But why 100mV works, I don’t know
seragnoor252
1 year ago
👍
billoute720
1 year ago
This is normal. At 10 GHz, the intrinsic capacitance of the led (5pF) makes the diode passing. The 200 mV of the generator is enough to create a current of 20 mA that lights led.
seragnoor252
1 year ago
Well done! But can this be used practically?
seragnoor252
1 year ago
Can it be used to save energy?
EricsElectronicAdventures
1 year ago
I assume you are refering to the 0V in the upper righ hand of the circuit. This wire is tied to the CIRCUIT GROUND which is by deffinition 0V. Plot that node and it will read 0V. Move the ground to say the bottom leg of the generator and plot the same node as before. It will no longer be 0V. The circuit ground is the refrence from which all other voltages are measured. Try this, stack up several DC sources in series, measure the voltages across them, move the ground then measure again. Frist do you get say 0V, 5V, 10V. Next do you get -5V, 0V, 5V, Or -10, -5 0? It is all relative to ground (defined to be 0V)
seragnoor252
1 year ago
Zero volt is an approximate percentage, dear, and the real voltage is 200 mV
seragnoor252
1 year ago
Does 200 millivolts light up the LED?
giomix
1 year ago
Led Is working in current, not in voltage. To enlight It you have to flow from 5 to 20 mA , depending on ite color. It Easy to demonstrate that ONLY in simulator and not in real Life , increasing the frequency of the sinusoidal generator the current increase up to enlight leds. Decreasing the frequency current decrease and all led are turned off.
billoute720
1 year ago
Here is the calculation for Vmax=200 mV and C=5 pF and F=10 GHz: I=V/Z with V=Vmax/1.414 and Z=1/(2 x 3.14 x F x C) => I(mA) =Vmax(mV) x 2 x 3.14 x F x C / 1.414 => I=200 x 2 x 3.14 x 10 x 10^9 x 5 x 10^(-12) / 1.414 => I=44 mA
seragnoor252
1 year ago
Well done everyone, I have benefited a lot from this discussion.

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