NAND astable timer

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madmikejt12

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hi, i am designing an electronic alarm, i origionaly had 3x 2 input or gates feeding into a NOT gate which fed into a 555 timer which then went to a siren gen to a speaker... i was told it had too many IC's in it (need a PIC for a seperate keypad circuit aswel) so i tried using various logic gated to try and eliminate 1 chip.... i did it in the end by using 3x or gates and 3 NAND gates (1 to use as a not gate and 2 for the latch to replace the timer) i then found out that it was illegal because it can only go off for a certain length of time so i thaught instead of doing a logic latch with 2 NAND gates, i could do an astable timer with them, but i cant find how to do it

i want it so that when a switch is pressesd (or released, either is good) the alarm sounds for a set time even if that switch returns to its original state so i need a constant power source AND an input and a reset aswell.
 
Who invented the stupid term, "NOT gate"? It isn't a gate. A logic inverter is a logic inverter, isn't it?

Go to any manufacturer's site and search for a "hex NOT gate", you won't find it. They have hex inverters instead.
 
hi, sorry i do not have a schematic, i have just used an alternative.... i have used 3 or gates, an inverter, a 555 timer and a siren gen and i am not using the keypad (i am using a key switch instead) so i have cut out one chip.

Thanks
Mike
 
Since you won't sketch a simple schematic then I guess you won't get any help.

I use Microsoft Paint program to make schematics. Straight lines are made with the Shift key down. I copy and paste parts of other schematics or from datasheets into the sketch. Why can't you do it?
 
I have the choice of computer software (circuit maker 6 student, its free), paint, or by hand. I nearly always do them by hand, possibly not the neatest, but much quicker (to me anyhow. the software often has parts missing too). A scanner or decent camera should get it onto your computer.
 
hi, i wasnt going to use that circuit but i have changed my mind again, i want to use it so there is a time delay when the front door is opened (i will not need another chip if i can do this using 3 or less NAND gates) as i am now using a keyswitch, it should be easy to de-activate it.

Here is a schematic...

(p.s. it doesnt work, the output Q wont change state... that is all i want, a simple on/off time delay with a set button)
 

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You made a schematic with a BMP file type that uses more than 500k of memory space. Its space is so huge that you had to zip it to attach it.
Next time, save it as a PNG file type and it will be 1/100th the size and won't need zipping. Here is my corrected PNG of your schematic.

Do you want a time delay so the door switch can open but the alarm is delayed from sounding until the person keys in a correct code?
Your circuit won't do anything when the switch is opened because the logic is backwards. Swap the 100k resistor and switch positions.

A time delay is roughly equal to the resistance times the capacitance. So your delay is only 1k x 1nF= 1us. If you want 10 seconds then use a 4.7uF film capacitor (an electrolytic has too high a leakage current) and a 2.2M resistor. Also, add a 22M resistor in series with the input of gate C to isolate its input protection diode.
 

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Thanks for the tip

I am not using a keypad now, i am using a key switch to de-activate the alarm (i know it is less secure but our teacher said it will be too complicated with too many chips)

i just put those values in to see if the circuit would work.

I will try your sugestion now, thanks
Mike
 
Sorry, my circuit doesn't make a delay. Instead it makes a 10 seconds long output pulse. Re-arrange its parts to make a delay.
 
hi, i thaught that was the monostable circuit??

i did that circuit on crocodile tech. and nothing happened (Q still didnt change state)
 
In my circuit, when the switch is closed then it makes the input of gate B high. The 2.2M resistor makes the input of gate C low so its output is high. Gate B then has both inputs high so its output is low and the capacitor remains uncharged. Because the output of gate C is high then the output of gate D is low.

When the switch is opened then the input of gate B goes low, therefore its output goes high. Therefore the input of gate C also goes high due to coupling from the uncharged capacitor. With the input of gate C high, then its output goes low which makes the input of gate B also low and also makes the output of gate D go high.

The capacitor charges through the 2.2M resistor until the input of gate C is low, causing its output to go high and also the inputs of gates B and D. Therefore the output of gate D goes low.
 
This circuit has a low output which goes high about 10 seconds after the switch is opened and stays high even if the switch becomes closed.
Reset it by shorting the capacitor with a switch.
 

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hi, tried this circuit but the output still doesnt change state!!

The top input to the first NAND gate stays low constantly, and the other one goes high when the switch is pressed, so the output of that gate stays high
 
madmikejt12 said:
hi, tried this circuit but the output still doesnt change state!!

The top input to the first NAND gate stays low constantly, and the other one goes high when the switch is pressed, so the output of that gate stays high
Click to expand...
If you are talking about my last circuit, then the uncharged capacitor holds the input of the gate C low, so its output is high. Therefore the top input of gate B is also high. Maybe your diode is backwards. See this sketch:
 

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madmikejt12 said:
here is what happens with the logic, the 1st gate doesnt go high
Click to expand...
You forgot to label what happens in your sketch.

Oh, oh. You re-numbered the gates so my previous explanation needs a translation.

The gates in a CD4011 are NAND. The output of a NAND gate goes high if any input is low. When the switch is open then the 100k rsistor makes the input of gate A low, so its output must be high. Then the 22M resistor can slowly charge the capacitor.
 
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