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becks098
modified 5 years ago

Please Advise....

0
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01:48:46
Should this circuit not act as a voltage indicator? i.e full voltage, all LED on, low voltage, just the one red one? Ive gotten it to work in real life but this app doesnt seem to act the same way?
published 5 years ago
giomix
5 years ago
No, it cannot acts as voltage indicator, all led an resistor are in parallel to the source, then all element are at the same potential.
becks098
5 years ago
Ao in order for them to act as a voltage indicator, this set up needs to be in series?
giomix
5 years ago
Do you gotten it to work in real.? Surely not this circuit you post here.
becks098
5 years ago
Yup this exact one. Its from a youtube video, followed it on a breadboard and its fine. Put it on here and no dice :(
becks098
5 years ago
Mind you i made a simple oscillator which went on infinitely, it didnt trail off as expected....
lmccoig
5 years ago
Disconnect all the resistors and the circuit still lights the same. Analogue volt meter uses resistors to have a certain voltage across a scale. You need circuit to select voltage level and output to correct indicator lamp.
tranist
5 years ago
Connect each resistor in series with its respective LED, and that resistor should be grounded. That way you'll have 4 led and resistor networks in parallel. Make the red led have the lowest resistance, the blue have the highest.
giomix
5 years ago
An example http://everycircuit.com/circuit/6185575642824704
wyoelk
5 years ago
The reason it would work in real life is each led has a different voltage and amperage rating by color. In this app all leds are 2v and 20ma no matter what color.
mlewus
5 years ago
LEDs have a characteristic forward voltage, just like diodes. This based on the material they are made of . In general red LEDs have a forward voltage drop of around 1.5 volts , where is Blue LEDs have a forward drop of more like 3.5 volts. Orange and green or somewhere in the middle. In real life the red LED would be on because it has the lowest forward voltage. That would hold the voltage too low to light the other LEDs.

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