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enter image description hereSo I have a circuit using the 4510 and 4511 chips as well as 3 seven segment displays on a PCB (designed and created). What it is meant to do is count up to 9.99 in 1's, 10's and 100's by cascading and manual input. It can also count up and down as well as being able to reset. I have used Push-to-make switches, push-to-break switches and a rocker switch (display ON/OFF). When powering the circuit everything works as expected - when I count 2 out of the 3 seven segment displays can show up to a 9. However, the 10's seven segment display only counts to 8 and then goes straight to 0, thus cancelling the cascading effect onto the 100's seven segment. It displays all digits on the seven segment display except 9! However this only happens when counting up as when counting down it seems to be displaying a 9 on the display.

I have no idea how to troubleshoot this problem - I have replaced the chip and have also checked the continuity of the track and the same thing happens. Any ideas on how to solve this problem (I know I haven't explained this well)?

All help appreciated.

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    \$\begingroup\$ Checked for accidental shorts? \$\endgroup\$ – immibis Feb 27 '17 at 22:56
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    \$\begingroup\$ Could you add a schematic (even hand drawn) of your connections? \$\endgroup\$ – Danny Staple Feb 27 '17 at 23:06
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    \$\begingroup\$ ^^ .... But preferably using the built in schematic editor so it's a familiar format. \$\endgroup\$ – Asmyldof Feb 27 '17 at 23:07
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    \$\begingroup\$ Do you have power supply decoupling capacitors across each IC? \$\endgroup\$ – Bruce Abbott Feb 27 '17 at 23:08
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    \$\begingroup\$ You have a lot of reading to do \$\endgroup\$ – Sunnyskyguy EE75 Feb 27 '17 at 23:23
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It's pretty clear that you don't understand how to use your counters. The fact that you have your carry in (pin 5) grounded on all chips establishes this. So.

1) Connect the clock to all 3 chips. I suspect, although your fuzzy layout gives me a headache when I try to look at it, that you are running your Q3 to the clock of the next chip. Don't.

2) Connect the up/down to all 3 chips.

3) Ground the carry in on the 1's digit only.

4) Connect the carry out of the 1's to the carry in of the 10's. Connect the carry out of the 10's to the carry in of the 100's.

5) Ground all of the Din pins of each counter.

6) Put a 0.1 uF ceramic cap from pin 8 to pin 16 of each chip, and do it with very short leads.

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  • \$\begingroup\$ Pretty sure that pin 5 should be grounded... look at any 4510 data sheet... \$\endgroup\$ – Kronixion Feb 27 '17 at 23:46
  • \$\begingroup\$ And if what you are saying is true, then my circuit shouldn't work at all but it does. I have ONE issue with ONE digit on ONE display - I clearly haven't gone very wrong. \$\endgroup\$ – Kronixion Feb 27 '17 at 23:47
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    \$\begingroup\$ @Kronixion - Oh no, I suspect your circuit will sort of work, although I can't figure the details without the schematic which you have not produced. I'm pretty sure you've taken a sychronous counter chain and converted it to a ripple counter. Like I say, you've missed the point of the IC design. \$\endgroup\$ – WhatRoughBeast Feb 28 '17 at 0:24

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