Convert 0 5V PWM signal to +2.5V -2.5V signal

Thread Starter

hengis

Joined Sep 9, 2011
3
I wish to take a 1kHz PWM signal and create a symmetrical 2.5V signal from it. I do not want to have polarisation effects altering my measurements. The frequency if that is the correct term is about 1kHz Any help would be appreciated
 

SgtWookie

Joined Jul 17, 2007
22,210
Do you have +2.5v and -2.5v supplies available?

If not, do you have at least a negative supply available that is <= -2.5v?
 

Thread Starter

hengis

Joined Sep 9, 2011
3
Thanks for your quick reply. No to both questions. I would have to add a circuit to create such a voltage.
 

Thread Starter

hengis

Joined Sep 9, 2011
3
I want to measure water content (wetness) in plant pots. A dc signal causes polarisation about the electrodes. Using an alternating +2.5V pulse followed by a -2,5V pulse gets over this problem. Using a MCU is more practical in a greenhouse particularly as I am reluctant to "play" with mains voltages in a greenhouse with water and dampness all around.
 

ErnieM

Joined Apr 24, 2011
8,007
Just insert a series capacitor. That keeps out the DC but for the small leakage current of the cap itself.
 

John P

Joined Oct 14, 2008
1,763
But if it's PWM you'll have more charge transfer in one direction than the other, unless the duty cycle is exactly 50%, i.e. a square wave.
 

John P

Joined Oct 14, 2008
1,763
I don't see how it can not be true. Suppose you have +2.5V feeding through a resistance to Gnd for 9 seconds, followed by -2.5V through the same resistance to Gnd for 1 second, i.e. a very slow PWM signal. Won't the ratio of the amount of charge transfered be the same as the ratio of time intervals, 9:1? Note that I'm talking about "charge" rather than "current" because while it's flowing, the current is constant. And I assume it's integrated charge that affects the electrodes chemically.
 

ErnieM

Joined Apr 24, 2011
8,007
As a capacitor is incapable of passing a DC current then the " integrated charge that affects the electrodes chemically" is zero irregardless of the driving signal.

Aside: While it is possible to get a constant current to flow thru a capacitor, you need to drive it in a very special way to accomplist this, or in other words, the current in a cap is only a constant in very special circumstances. A linearly changing voltage can accomplish this.
 
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