Resistance of LED circuit

[Exemplaris]

I have device that has two 1.5Vdc batteries in series giving me 3.162Vdc when measured. The circuit using this power source contains a switch and an LED. If I remove the batteries and check the resistance at the contacts I get an open, with the switch open or closed. If I use the diode test at this same point I get 1.884Vdc with the switch closed.

I need to change the power source from batteries to a power supply that can be supplied from a permanent source, probably 24Vdc but it doesn't really matter as I have multiple options. I also need to add a relay to this circuit that will activate at the same time as the LED and provide a contact for a 24Vdc signal.

From what I have read it seems that I could use a 12Vdc supply and a resistor to drop the excess voltage.

My questions are:
How do I determine the resistance of the circuit as it is now?
Is it possible that the LED does not require a resistor now because the voltage is only 3.162Vdc?
If that is the case then a 450Ω resistor should dissipate the extra 9Vdc, assuming 20mA LED, and a 12Vdc supply correct?
If that is correct how will a 12Vdc relay's coil affect the circuit?
Am I missing a simpler solution?

 

[Hero999]

It sounds like you've not connected a resistor in series with the LED so the supply voltage is clamped to the LED's forward voltage which will destroy the LED, if the battery has a low internal resistance and it's connected for too long.

You should always use a series resistor unless you're powering the LED from tiny button cells which have a high resistance.

 

[MikeMl]

With only 3.1V applied to the LED, it is just slightly beyond the point at which the LED is forward biased. When converting the circuit to operate on a higher voltage, you MUST add the resistor. To compute the resistance, R = (Vin-3.1)/If, where Vin is the power supply voltage, and If is the desired current (in A), so for 12V and 0.02A, R = (12-3.1)/0.02 = 445Ω The power dissipation in the resistor is I*E = I*(Vin-3.1) = 0.02*(12-3.1) = 0.178W, so a 1/2W resistor is right.

 

[Hero999]

It depends on the LED.

1.88V is only 59% of 3.16V.

If the battery has a series resistance of any less than 64Ω a typical LED will fry if it's left connected for too long.

 

[Exemplaris]

Thanks MikeMl and Hero999!

Hero999, I am modifying this working device and all I can see is what I have described. The batteries are standard 1.5Vdc AA. so I do not know if they have the resistance that you mention. I can say that the LED will only be lit for a couple of minutes at most at any given time. Don't know if that matters.

MikeMl, so I was on the right track, if I get a 12Vdc power supply and as close to 445Ω resistor my original circuit should be OK. What affect will the coil of a 12Vdc relay have on this circuit? I know there is inductive reactance in the coil. I assume that I would need to find that value and subtract it from the resistor's value to keep the circuit "balanced" so to speak, correct?

 

[MikeMl]

Thanks MikeMl and Hero999!

Hero999, I am modifying this working device and all I can see is what I have described. The batteries are standard 1.5Vdc AA. so I do not know if they have the resistance that you mention. I can say that the LED will only be lit for a couple of minutes at most at any given time. Don't know if that matters.

MikeMl, so I was on the right track, if I get a 12Vdc power supply and as close to 445Ω resistor my original circuit should be OK. What affect will the coil of a 12Vdc relay have on this circuit? I know there is inductive reactance in the coil. I assume that I would need to find that value and subtract it from the resistor's value to keep the circuit "balanced" so to speak, correct?

What does the relay have to do with lighting the LED? Are you using the relay's contacts to turn on the LED? Is the relay COIL operated on the same (12V?) voltage as the LED circuit?. If yes, you need to put a snubber diode across the relay coil, cathode to +.

A circuit diagram would help.

 

[Exemplaris]

Yes, the relay COIL will be operated on the same voltage as the LED circuit. Yes thank you I almost forgot the snubber diode. The circuit is very basic: right now just the power source (3.1Vdc battery supplied) a switch and the LED. I want the relay COIL to energize at the same time as the LED so I assumed that they would be in parallel with each other and in series with the resistor. I just don't know how the coil will affect the value of the resistor.

 

[MikeMl]

If the relay coil is in PARALLEL with the [resistor+LED], then the math doesn't change. The Snubber is VERY important; without it, you will blow-up the LED!!!

 

[audioguru]

If you connect the relay coil parallel to the LEDs and both in series with a current-limiting resistor then the voltage to the relay might be too low to acivate the relay. Do you have a relay that works from only 3V or less? The lowest I have seen is a 5V relay.

Don't forget that two 1.5V AA cells make 3V only when they are brand new. As they are used their voltage quickly drops to 2.4V then drops slower to 2.0V when they are dead.

 

[Exemplaris]

So, the LED and the resistor would be in SERIES and the COIL would be in PARALLEL with both of them? What would happen if the LED and the COIL were in PARALLEL together and then in SERIES with the resistor?

 

[Exemplaris]

Thanks audioguru, I posted my question as your post came in. That answers my question.

I think I have a handle on it now and will do some experimenting.

Thanks everyone!

Sorry, the relay is 12Vdc.