Rongge

I have a circuit where I want to have a constant current of 2mA through a variable resistor. I've been told that I could probably use a LM317 as a current regulator, with one resistor on the ouput. But I've read some places that the LM317 takes minimum 5-10mA load to function correctly. How can I achieve a constant current output of 2mA when I don't know the resistance of the variable resistor? The input voltage is about 2.755V. Output voltage doesn't matter, just the current. Here's an image to my feeble attempt at a schematic: The LM317 with the single series resistor between output and adjust input is actually a fixed current source, not a current limiter. You don't need the LM317 to create a current limiter, a few discrete components will do: For a limiting at 2mA you select a 330 Ω resistor for R S E N S E . If there flows 2mA through it Q2 will start to conduct and reduce the base voltage of Q1, so that its current is cut off. edit (re changed question) ...

To open the lock, buttons S1, S2, S3, and S4 must be pressed in this order. They must be pressed for more than 0.7 seconds and less than 1.3 seconds. Alarm Using 4-Buttons Circuit Diagram Reset button S5 and disable button S6 are also included with the other buttons and if the disable button is pressed, the circuit will not accept any code for 60 seconds. Each of the 3v3 zeners can be replaced with two red LEDs and this will show how you are progressing through the code. Make sure the LEDs are not visible to other users.