Exemplaris
New Member
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?
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?