Help me on RC circuit

[Jerson]

Charge the capacitor via a constant current source. This will give you a linear curve.

[Johan]

Mack: I am welcoming any other solution
The circuit is to dampen sudden voltage changes of a potentiometer ( voltage divider ), what important is the linearity of the output


Jerson: could you please tell me how to make constant current charging schematic that suit my need?

[Ioannis]

I really cannot understand what will be the difference if the charge/discharge will be linear or exponential relative to the pot circuit. As long as the pot is at a position the capacitor will charge and stay at that charge. A new pot position will give little or more change at the output but then again it will stabilize at the new level.

A current source will charge the capacitor linearly (Q=C*V, Q=I*t, t=C*V/I).

But as is your example, I don't know how it can be accomplished or what the result would be. The Current source will charge the capacitor and then will stop function because there will be no current to flow untill you will change the pot to a higher level. For a lower level a second curent source would be needed to discharge.

Ioannis

[mackrackit]

Does the voltage on pin#3 in your drawing change or is the change only from the pot position?
Either way a ADC looks like the way to go. The capacitor in your drawing will serve to "dampen" any voltage change (in its range) making Vout smooth.

If you have not used ADC before, take a look here.
http://www.rentron.com/PICX2.htm
http://www.rentron.com/PicBasic/LM34.htm
http://www.rentron.com/serial.htm

If you are expecting large spikes you would set up the voltage divider so that the maximum voltage in would still be a maximum of 5 volts out of the divider.

The ADC is linear. With an 8 bit setup, 0 volts will read 0, 2.5 volts will read 127 and 5 volts will read 255. 10 bit will have a 1024 step range.

If the voltage is never more than 5, say 3 volts. The reference voltage can be set for this and you will still have full scale.

[Johan]

Thanks for your responses

Well , this is not a simple circuit indeed. I have found the explanation about it on Nutsvolt , February 2007,p20

Thanks again