555 digital timer

Thread Starter

innernerd

Joined Nov 15, 2008
5
I'm working on a project for school, a 555 based one minute timer, and I'm stuck trying to figure out how to activate the second digit. It keeps counting 0-9 and starting over, while the tens digit just sits there. I'm using a 555 to drive a pair of 7490 chips that feed 7447 chips, output to two 7 segment LEDs. The Qd of the first 7490 feeds the CLK-A on the second 7490. Any help will be greatly appreciated.

I did find some other threads regarding similar projects, but none that helped.

innernerd
 

bertus

Joined Apr 5, 2008
22,278
Hello,

Can you post a schematic of what you have upto now?
It can be that the reset signals are not correct of the second 7490.

Greetings,
Bertus
 

bertus

Joined Apr 5, 2008
22,278
Hello,

Until what value do you want the second 7490 to count?
If it is until 9 you can also put pins 2 and 3 to ground.
When they are not connected (floting) they will probably go high and the counter will stop counting.

Greetings,
Bertus
 

Thread Starter

innernerd

Joined Nov 15, 2008
5
The circuit is supposed to count to 59 (60) and reset, thus being a repeating one minute timer. Pins 2 and 3 are grounded on the first chip, but are left open with ??? by them on the schematic for the second chip. If I ground them, both digits count by ones and I get 11, 22, 33 etc for a display. It needs to count to 9 and then place a 1 on the MSD, while the LSD resets and starts counting from 0 again.
 
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SgtWookie

Joined Jul 17, 2007
22,230
As Bertus implied, if pins 2 and 3 are not grounded (or held below 0.8v), the J-K flip-flops within the IC will be constantly reset, with their outputs at a logic low level.

If you wish to add a "reset" function to your counters, you could use a NC (Normally Closed) pushbutton switch. Connect all of the 7490's pins 2 and 3 to one switch terminal, and ground to the other switch terminal.
 

SgtWookie

Joined Jul 17, 2007
22,230
The circuit is supposed to count to 59 (60) and reset, thus being a repeating one minute timer.
You then need an AND gate.
A decimal 6 is 0110. So, when the MSD's 7490 Qb and Qc (changed from Q1 and Q2) are HIGH, you need to raise both pins 2 and 3 on the MSD's counter to reset it.

Pins 2 and 3 are grounded on the first chip, but are left open with ??? by them on the schematic. If I ground them, both digits count by ones and I get 11, 22, 33 etc for a display. It needs to count to 9 and then place a 1 on the MSD, while the LSD resets and starts counting from 0 again.
You must have connected the output of the 555 timer to pin 14 of both of the 7490's - either that, or you have not used bypass capacitors across the VCC/Gnd pins of each of the IC's. Bypass capacitors are typically 0.1uF (100nF) ceramic or tantalum capacitors.

555 timers also need to have a bypass capacitor across their VCC/GND pins, but it needs to be larger (1uF-10uF) because when the timer output toggles states, the internal transistors temporarily create a short between VCC and GND.
 
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Thread Starter

innernerd

Joined Nov 15, 2008
5
Thanks for the tips, I grounded pins 2 and 3 on the second chip (as I had at 3 a.m. this morning!) and now it is counting as it should. Not sure what I did different but the time of day may have played a part. Now I just have to get it to reset on the minute, preferably on its own. Progress!:)
 

SgtWookie

Joined Jul 17, 2007
22,230
You could use a 74x08 for an AND function. Each 74x08 contains four 2-input AND gates.

You could also use a 74x09, which is similar - but you would have to use a 470 Ohm pull-up resistor connected to +5v on the output of the gate, because it has open-collector outputs.

Open-collector outputs are very handy for controlling items that are outside of TTL levels; but they can "bite" you if you forget the pull-up resistors when interfacing with other TTL IC's.
 
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