I have a 5K Ohm linear potentiometer built into a servo.
I want 11 light diodes (LED) in a row, indicating the potentiometer position, i.a. when the pot-meter says
0 Ohm = 1 LED lit
500 Ohm = 2 diodes lit
1000 Ohm = 3 diodes lit
o.s.v.
5000 Ohm = all 11 diodes lit
LEDs are 5 mm 2.2 volts 20 mAmp
Power supply between 12 and 14 volts dc.
Who can and want to help me with that circuit diagram?

 

If you could go with just 10 LED's, an LM3914 would work fine for you.
http://www.national.com/mpf/LM/LM3914.html

If you want 11 LED's, you could add a single comparator to the LM3914.

 

An LM3814 will get you most of the way there - http://www.national.com/mpf/LM/LM3914.html.

You can search the forums for posts relating to how to use one.

 

If the limit of 10 LEDs imposed by the LM3914 is a deal-breaker, you could cascade two LM3914's and go for 20 LEDs.

hgmjr

 

The LM3914 will not light the first LED when the input is 0V.
So you need to light the first LED with a resistor. It will always be lit.

The LM3914 steps are linear. Feed a constant current to the resistor. The LM3914 lights 10 LEDs when the resistance is 5k then one LED will light when the resistance is 500 ohms. Plus the LED that is always turned on.

 

Thank you very much for the quick and constructive answers, especially to Audioguro for the information about "the missing light" in the first diode. My problem has been solved.
/Freddy