| ??? 02/22/07 14:47 Modified: 02/22/07 16:29 Read: times |
#133505 - One-bit memory == DC filter ??? Responding to: ???'s previous message |
Ricahrd said:
It works either way, depending on what you want to do. You can, if you want, charge the cap through the LED. If you don't first discharge the cap, no current will flow once it's charged. If you set it up so it discharges in 10 usec, but charges for 10 msec (extreme case, but you get the idea), you can go along and discharge lots of caps during that charge time. It involves more parts and more thought, but it allows you more freedom. Thanks, Richard. I understand now. You can sort of think of the capacitor in this case as a leaky, one-bit memory device that in effect stores the LED's desired state (on or off) for some portion of the matrix refresh cycle. In a vague sort of way, adding the capacitor is sort of like going halfway along the path from a multiplexed display (with no memory) to a non-multiplexed display, where the LED state would be stored in a "real" memory device (a flip-flop, for example). Another way to think of all this is to realize that some kind of low-pass filter is needed to eliminate the perceived flicker of a multiplexed display. With a fully multiplexed display, this filter resides somewhere in the user's retina and/or the apparatus of his visual perception. That filter's time constant is relatively short, which makes it necessary to refresh the matrix fairly quickly. Adding a capacitor in series with each LED as you suggest inserts a second filter with a longer time constant into the system, ahead of the filter(s) in the user's head. Now if you go all the way and implement a non-multiplexed display, with one bit of memory for each LED, the net result is the same as if there was a filter with an infinite time constant in the system! So ... if you're so inclined, does this mean that you can think of a one-bit memory as a perfect DC filter? Seems like it. Sort of. Maybe. -- Russ |
