# Simple digital clock

Discussion in 'General Electronics Discussion' started by Karthik rajagopal, May 9, 2017.

1. ### Karthik rajagopal

234
7
May 9, 2016
Hi all,
I decided to make a 24 hrs clock with IC 4026 and few logic gates. I drew the circuit diagram for that and have posted it below. Here I have given the output of the IC to the logic ICs so that I will be able to reset the min and hours after it reaches 60 and 24 respectively. The problem was that when the output of the IC 4026 was connected to to the logic IC (for example OR gate in the circuit) , the pins were left floating when there was no output from the 4026 ic which gave a wrong output. So I tried connecting a pull down resistor to the inputs, but adding these pull down resistos made the segments of the LED display ( segments that are connected to logic gate IC)glow dimmer than the others . So ,I need a solution to have my logic gates functioning properly without compromising the diplay's brightness. I have connected the output of the logic gates to the resemt pin of the ic 4026.Please help me in solving this problem lem. If you need any clarifications in the circuit, I will answer to those questions in the comments. Thanks in advance.
X and y are written to just denote those led displays segments with their pin.
I have not drawn the diplay's pin in order.

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,489
2,830
Jan 21, 2010
Use a higher value pull-down. 10k should be fine.

3. ### Harald KappModeratorModerator

11,434
2,624
Nov 17, 2011
Under which condition should there be no output from the 4026? Afaik teh 4026 has a push-pull output stage, so there should always be a well defined high or low signal.

4. ### Karthik rajagopal

234
7
May 9, 2016
For example let's say that we need a reset in the ic when the 2 and segmet from the right reaches six. So what I have done here is I have selected the output pins that ar unique for six alone. So the pin mentioned as y that is the b segmet will be low and the pin mentioned as X that is the e segment will be high will be high while displaying the number 6. So I have taken this unique output for each number as my logic input so as to make my circuit simple. There is a correction to be made for that display which I don't mark there. That X and y will change

Last edited: May 9, 2017
5. ### Karthik rajagopal

234
7
May 9, 2016
Thank you Steve. It works.

6. ### AnalogKid

2,496
718
Jun 10, 2015
Or 100K. In theory, even 1M will work, but that might leave the gate inputs susceptible to noise.

ak

7. ### Harald KappModeratorModerator

11,434
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Nov 17, 2011
Sorry, I made my self not clear: the outputs of the 4026 are totem-pole or push-pull. This means they will actively drive either a high voltage or a low voltage on the ouput pins. There is no need for pull-up or pull-down resistors. I see no condition where any output of the 4026 will be floating.

I see, however, another issue with your circuit. YOu're using common cathode displays with one single current limiting resistor per display. This will lead to current sharing between the LEDs and the brightness of any single segment within one display will depend on the number of active segments within the display (say: a 1 will have 2 bright segments, an 8 will have 7 dim segments).
I recommend one resistor per segment:

8. ### Karthik rajagopal

234
7
May 9, 2016
Yes, I thought the same but it's not working that way. When the outputs are not high it leaves the pin floating. I say that the pin are floating as I didn't get the actual output from the logic gate when output of ic 4026 was given as the input. So, I had to necessarily connect these pull down resistors to have my logic connected to ground when the output was low.
And about the second thing you mentioned, there was no problem about the current sharing between the segments in my circuit but I think your idea of connecting a resistor to each segment would be better. I will connect each segment with a resistor. Thank you for your guidance and help.

9. ### Karthik rajagopal

234
7
May 9, 2016
Why do you say that 1 M resistor might leave the gate inputs susceptible to noise? Please explain.
Thanks.

10. ### Harald KappModeratorModerator

11,434
2,624
Nov 17, 2011
The higher the resistance, the smaller the noise current required to produce sufficient voltage drop to generate a logic high instead of the required logic low.

11. ### Karthik rajagopal

234
7
May 9, 2016
Thanks for the explanation.

12. ### AnalogKid

2,496
718
Jun 10, 2015
1 M is a very high input impedance. It would take very little radiated energy to impress a couple of volts across it, enough to cause the gate to change state.

ak