Sorry for this novice question, I just want the circuit to be safe.
I have a 12V indicator that I would like to use as the "power on" indicator in a system powered by 27V. The indicator uses a bulb (not LED) but is a sealed unit, i.e. bulb not replaceable. It's rated at 60mA.
I tried the voltage divider method using a 510 and 390 resistor giving ~11.7V which seemed to work, however the resistor got hot. Is that safe, and will it fail in the long term? Is there a better method using just resistors?
Thanks
I have a 12V indicator that I would like to use as the "power on" indicator in a system powered by 27V. The indicator uses a bulb (not LED) but is a sealed unit, i.e. bulb not replaceable. It's rated at 60mA.
I tried the voltage divider method using a 510 and 390 resistor giving ~11.7V which seemed to work, however the resistor got hot. Is that safe, and will it fail in the long term? Is there a better method using just resistors?
Thanks
You're going to burn (27-12)V x 0.06A = 0.9W of power somehow.
It may as well as be in one series resistor of (27-12)V / 0.06A = 250 Ohms.
An "indicator assembly".
http://www.mouser.com/ds/2/423/2110A 2111A 2112A 2113A AND 2195A Series Non-Relam-552440.pdf
http://www.mouser.com/ds/2/423/2110A 2111A 2112A 2113A AND 2195A Series Non-Relam-552440.pdf
Didn't check the price but wouldn't be surprised if just replacing the full indicator assembly is cheaper than getting the suitable lamp, which would need its own screw or bayonet base , and indicator body would also need its own screw/bayonet body.
All require metal, punching and forming it, assembly, etc.
Seen that way, making just a glass bulb with 2 pigtails and injecting some plastic around it becomes intere$$$$$ting
All require metal, punching and forming it, assembly, etc.
Seen that way, making just a glass bulb with 2 pigtails and injecting some plastic around it becomes intere$$$$$ting
Rayma did the Math on post #2 and calculated 0.9W so 1 W will be fine, but in general we like to work with some safety margin (Murphy's Law , sh*t happens , etc.) so if possible use a 2W resistor there.
It will dissipate exact same amount of heat as before, but surface temperature will be lower (because same heat distributes over a larger surface) which is good.
and calculated 0.9W so 1 W will be fine, but in general we like to work with some safety margin (Murphy's Law , sh*t happens , etc.) so if possible use a 2W resistor there.It will dissipate exact same amount of heat as before, but surface temperature will be lower (because same heat distributes over a larger surface) which is good.
Good info, thanks everyone. I love this forum.
All my resistors are 1/4W, except a handful of 47K ones which are rated at 2W. Perhaps I could make a 13.5W voltage divider with two 47K, then a 25 Ohm 1/4W series resistor to the lamp. Am I right in thinking the bulk of the excess power will be dissipated through the voltage divider resistors and only 90mW through the 25 Ohm resistor?
It may seem a little excessive, but I happen to have those resistors to hand and don't mind over complicating things a bit!
All my resistors are 1/4W, except a handful of 47K ones which are rated at 2W. Perhaps I could make a 13.5W voltage divider with two 47K, then a 25 Ohm 1/4W series resistor to the lamp. Am I right in thinking the bulk of the excess power will be dissipated through the voltage divider resistors and only 90mW through the 25 Ohm resistor?
It may seem a little excessive, but I happen to have those resistors to hand and don't mind over complicating things a bit!
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.