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How to light an LED when a capacitor is full?

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W-Unit

New Member
Hey guys! I'm just getting into the world of electronics, been fiddling around with various stuff, mixing parts from various old devices and so forth, and it's been great fun!

Just a quick question here: how can I make a simple circuit with a battery, capacitor, and LED in which the LED will light only when the capacitor is fully charged? To test I'll also add some method of discharging the capacitor... probably I'll just use a momentary DPDT switch that, when pressed, will change the current from flowing between the battery and the capacitor to flowing between the capacitor and the ground.

Thanks and cheers! :D
 

Hero999

Banned
Well technically a capacitor never fully charges so it's more sensible to turn on the LED when it's 95% charged.

With a comparator; that's how.

Comparator - Wikipedia, the free encyclopedia

The problem is comparators generally don't work with the inputs near the positive supply so you need to measure the voltage with a potential divider.
 

audioguru

Well-Known Member
Most Helpful Member
You could measure the charging current of the capacitor. When the current is low then the capacitor is almost fully charged.

Some opamps have inputs that work at the positive supply voltage (TL07x and TL08x) and others work when their inputs are near 0V (MC3317x, MC3407x, LM358 and LM324).
 

Hero999

Banned
You could measure the charging current of the capacitor. When the current is low then the capacitor is almost fully charged.
That's another way of doing it.

The disadvantage of that idea is that it normally involves using a current sense resistor which increases the ESR of the capacitor. If the comparator can't take negative input voltages without being damaged then you'll need to add a Schottky diode across the sense resistor to protect the comparator.

Some opamps have inputs that work at the positive supply voltage (TL07x and TL08x) and others work when their inputs are near 0V (MC3317x, MC3407x, LM358 and LM324).
I wouldn't recommend using an op-amp.

I would recommend using a comparator such as the LM311 (my favourite if only one is required) or LM393.
 
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bountyhunter

Well-Known Member
You can use a FET input op-amp as a comparator with a high impedance resistive divider from the + side of the cap to ground. The advantage of the FET input is it takes zero bias current and so you can use very large divider resistors. Set the op amp up for untiy gain positive and connect the center tap of the divider to the pos input. Connect the neg input from a reference like an LM336 and adjust the values of the resistive divider so that the voltage of the center tap goes above the reference voltage when the cap's voltage is high enough to consider it fully charged (like maybe 95% of the supply voltage). The output of the O/A will go higjh when the cap reaches the threshold.
 
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Hero999

Banned
I agree with all of that apart from the voltage reference idea. I think a potential divider would be a better option since the supply voltage can vary. Using a voltage reference the % charge of the capacitor will depend on the power supply, using a potential divider, the % charge will remain constant regardless of the supply voltage.
 

bountyhunter

Well-Known Member
I agree with all of that apart from the voltage reference idea. I think a potential divider would be a better option since the supply voltage can vary. Using a voltage reference the % charge of the capacitor will depend on the power supply, using a potential divider, the % charge will remain constant regardless of the supply voltage.
That's true. Use a divider down from the rail as a reference point so they will track as the supply changes.
 
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