On Mon, Oct 5, 2009 at 7:19 AM, Marechiare wrote: > Mark Rages wrote: >> >> When you get back, >> http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem > > And what is your answer to the next: > > *** > For the circuit: > > "perfect 0 to 5V square wave followed by the 2K,3.9K ohm divider, > driving some capacitive load" (the capacitive load being connected > between the middle of the divider and ground). > > We decrease divider's bottom resistor 3.9K to 3.85K, the top divider's > resistor 2K remains the same. The impedance as per Olin > (R1*R2)/(R1+R2) will decrease. > > How will change the time to rise the signal on the capacitor from 0V to 2.9V ? > > Will it decrease as per Olin's statement that - the lower impedance - > the higher speed of signal rising on that capacitor. > *** > > Just "Yes, the time will decrease" or "No, the time will increase". > This should not be a problem to you if you are sure you are > referencing me to a relevant web link. > You are changing the voltage the circuit is dividing to. Rise time is always defined relative to "full" voltage, not some absolute 2.9V. So relative to the new full voltage of 3.29V, the time will decrease. Anyway, the subject is 5V to 3.3V conversion. Suggesting conversion to other levels is not very interesting. Regards, Mark markrages@gmail -- Mark Rages, Engineer Midwest Telecine LLC markrages@midwesttelecine.com -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist