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RC Delay for Active-Lo Enable

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cwrasmussen

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I need to make a simple circuit that will go low after about 200ms to work as a delayed active-lo enable line (specifically for the 74ls540/541 chip). I tried the RC circuit in the embedded image, but I realized that won't exactly pull low because there is still a resistance to the ground. Sure enough, the signal goes to 2V (5v system). That signal goes to 6 541 chips. Should I take a second look to see if something else is causing the issue? Any other suggestions you guys have would be great!

56_1207706478.jpg


P.S. I know I should know this stuff, so feel free to give me hell for my lack of education before I entered the workforce. :p
 

Hi,
You may use a two input OR gate with an RC delayed signal to one input and the signal direct to the other input, and use the output of the OR gate.. In this case you have to make sure that the high period of the signal is long enough to charge the capacitor to high level even for the first time.

An alternative method is to trigger a monoshot using the negative edge of the signal and connect the signal and the +ve pulse output of the monoshot to an OR gate. A small RC filter may be added to the output of the OR gate to remove any glitch at the negative edge of the signal due to the delay in triggering of the monoshot.

Regards,
Laktronics
 

if you are looking for something reliable then rc is not a good option ..
there is a lot of cheap 3-pin, 4-pin, and 8-pin por circuits that have been specially designed to provide reliable start-up conditions ..

so, have a look at, for example, ds1232/max1232, max809/810/811, ... and see how they can be easily implemented into your circuit ..

rgds,
ianp
 

cwrasmussen,
if you need exact RC reset circuit... well
check this

**broken link removed**

about 10 years ago I had same problem ;)
anyway in this appnote intel advise reset IC to be used
good luck
 

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