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You could do this with a comparator, comparing the input voltage to a voltage reference of 2.25 volts.
The output would go from 0v to Vcc as the input goes from 2.2 to 2.3 volts.
The circuit would need to be powered from something like 5 volts.
Alternatively you could use a 2.25 volt (ish) voltage detect chip and a LED with a Vf < 2.3v
The output would go from 0v to Vcc as the input goes from 2.2 to 2.3 volts.
The circuit would need to be powered from something like 5 volts.
Alternatively you could use a 2.25 volt (ish) voltage detect chip and a LED with a Vf < 2.3v
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What's the power supply, and what's the desired current draw before lighting the LED?
Has this something to do with your other thread?
Has this something to do with your other thread?
yes, i am building a circuit that will charge the capacitor, the LED is used as a status light for charge complete. this is what i have so far. is there a better way?
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That's pretty impressive.
I would guess that the only component which wouldn't be damaged or destroyed would be the 100 ohm resistor.
You're going to have to be more forthcoming on why you're charging the capacitor, where the 5V is coming from, and what you intend to do with the charged capacitor.
That way we can do something more than come up with literal answers to your question which don't appear to be very satisfying for you.
And please don't start another thread if you don't like the answer...
The 5v is coming from a LM317 that has input from an unregulated dc voltage.
The charged capacitor is used to power a load that will be directly connected to the capacitor through a dc step up voltage converter.
The load is going to be turned off while the capacitor is charging (is that necessary?).
Set up one LM317 as a constant current source (pick a suitable value based on the capacitor and your power supply (i.e. do not exceed the capacitor ripple current or the maximum power supply current).
Use this to feed another LM317 set up as a 2.3V voltage regulator.
This will charge the capacitor up to 2.3V as quickly as safely possible.
You may be able to place a LED across 3 or 4 series diodes somewhere before the voltage regulator. This will go off when the current drops to zero this should be a good indication that the capacitor is charged.
I see no reason why the load needs to be removed, but not knowing the nature of the load I can't be certain.
Use this to feed another LM317 set up as a 2.3V voltage regulator.
This will charge the capacitor up to 2.3V as quickly as safely possible.
You may be able to place a LED across 3 or 4 series diodes somewhere before the voltage regulator. This will go off when the current drops to zero this should be a good indication that the capacitor is charged.
I see no reason why the load needs to be removed, but not knowing the nature of the load I can't be certain.
Can you explain what your circuit is supposed to do? In simulation it doesn't seem to do anything.
This will do pretty much what you require.
C1 is the filter capacitor of your unregulated supply.
D1, 2, 3 serve to create a voltage drop while current is flowing that will allow a red LED (D4) to light during charging.
U1 is an LM317 with R1 determining the max current. This should be set to the lesser of the ripple current for your capacitor and the max current from your power supply.
U2 is another LM317 with R2/R3 determining the output voltage (it should be set to 2.2V.
Ignoring esoteria like switchmode regulators, this will charge your capacitor to 2.2V as quickly and safely as is possible given your capacitor and power supply.
If charge time is less of a consideration, removing U1 and R1 and replacing them with a simple resistor would also be OK.
In this configuration your input voltage needs to be at least 6 volts higher than your capacitor voltage.
This will do pretty much what you require.
C1 is the filter capacitor of your unregulated supply.
D1, 2, 3 serve to create a voltage drop while current is flowing that will allow a red LED (D4) to light during charging.
U1 is an LM317 with R1 determining the max current. This should be set to the lesser of the ripple current for your capacitor and the max current from your power supply.
U2 is another LM317 with R2/R3 determining the output voltage (it should be set to 2.2V.
Ignoring esoteria like switchmode regulators, this will charge your capacitor to 2.2V as quickly and safely as is possible given your capacitor and power supply.
If charge time is less of a consideration, removing U1 and R1 and replacing them with a simple resistor would also be OK.
In this configuration your input voltage needs to be at least 6 volts higher than your capacitor voltage.
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here is the same schematic but with a the input voltage is from the LM317, resistor added, for some reason the simulator charges the capacitor very fast.
the point of this schematic is to charge the capacitor and then run a load off the capacitor
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the simulation charges the capacitor, while its charging the red led is on. when the capacitor is charged another led is lit by the capacitor indicating that the charge is complete.
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Won't the diode limit the voltage across the capacitor?
Can both the power supply and capacitor handle a 1.5A charge?
*how* does the transistor get turned on and off?
I have serious doubts about this working in practice.
If you increase the 1 ohm resistor to 3.3 ohms, and remove everything except the resistor, the battery and the capacitor, then you have something that will charge the battery.
Presuming you use an LM317 to get the 2.3V then you need to look at my response above where I've given you a circuit that gets far closer to doing what you want.
Can both the power supply and capacitor handle a 1.5A charge?
*how* does the transistor get turned on and off?
I have serious doubts about this working in practice.
If you increase the 1 ohm resistor to 3.3 ohms, and remove everything except the resistor, the battery and the capacitor, then you have something that will charge the battery.
Presuming you use an LM317 to get the 2.3V then you need to look at my response above where I've given you a circuit that gets far closer to doing what you want.
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