Hi, I want to make single LED PSU power ON indicator circuit (red, green, yellow 5mm LEDS etc.) to use on various power supplies as indicator that the PSU is on.
Can I use LM317L and a resistor? I think it will not work for low voltage levels like 1 or 2 volts etc?
My Requirements:
-As few components as possible.
-Small size of the whole circuit
-Needs to operate from 1 volt to 35volts (the wider range the better.)
Will the input to the 317 be fixed in your project? As stated in the previous reply, the LED should be on the non-variable side. Otherwise the LED brightness will change with any change in voltage. Most likely, the current limiting resistor you would need to protect the LED at 35V would make the LED very dim or not even visible at 1V.
Will the input to the 317 be fixed in your project? As stated in the previous reply, the LED should be on the non-variable side. Otherwise the LED brightness will change with any change in voltage. Most likely, the current limiting resistor you would need to protect the LED at 35V would make the LED very dim or not even visible at 1V.
I try to make universal circuit, as one indicators will be on 5V, others on 12V, some on 9V. I can't just use a resistor as motor speed will change and LED needs to be the same brightness all the time etc.
That is why a simple resitor will not work. I think LM317L would work with a resistor as stable current supply of 20mA.
The trouble is it will not work at low voltages.
The LEDs will be at a variable side of the power supply as they are add-on's to see that certain parts are switched on.
1. Motor 12V (variable voltage)
2. 2 Detectors 9V
3. Luxeon 3W LED Light 9V (variable current)
Every component can be switched on and off with it's own switch. The PSU will have a dedicated switch too.
The ON/OFF indicators have to be connected in parallel to each part of the circuit the switch will turn either on or off.
I as thinking to make simple circuit I can use everywhere I need an indicator.
Perhaps LM3419 would work but I need only 1 LED and how do program these?
Are there other IC's that have lower voltage drop compared to LM317L?
An LM317 current regulator needs 2V for the IC plus 1.25V for the resistor.
Then if you use a 2.0V red LED the minimum input voltage must be 5.25V or the LED will look dim.
You might be lucky and find an LM317 with a dropout voltage of only 1.5V so the minimum supply can be as low as 4.75V.
The LED indicators for the two nine volt supplies are trivial; just one resistor and one LED each.
For the motor circuit, can you steal power from either of the other two supplies to power the LED? If yes, then use an NPN transistor with a base-current limiting resistor to sense the motor voltage (the transistor will be on with any input voltage greater than about 0.7V). Wire the LED and its current-limiting resistor in the transistor collector to 9V.
I don't think I can steal power from any of my circuits
The double pole switch seems ok but but I got my switches and they single pole.
Now when I think about it the LM317L circuit will work OK if I manage to get some simple cheap IC that will increase voltage. This IC would need to make at least 6V from even as low as 1V to make LM317L and LED work at low voltages.
The LM317L is rated at 40V input so the IC that increases voltage should be rated the same?
I need something like **broken link removed**
But with input as low as 1V and as high as say 25V also to be cheaper.
If the indicators must be powered off variable voltages... can't each LED be driven by the 5 V supply, via a NPN lowside transistor switch?, the appropriate bases conencted via resistors to each variable supply
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