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I am experimenting with a circuit for adding two numbers represented by two digital counters and two cascaded 4 bits adders (74HC283). I tested separately with cascaded 74HC163 and with asynchronous 74HC4040 counters and I ended up with the same results.

enter image description here

I feed one of the counters with a 1 kHz com signal and the second counter with 512 Hz signals. All signals are well formed square signals.

I use a 16Hz crystal-controlled time base. The counters count during logic high of the time base that is 1/32 second or about 31 milliseconds. The counters are reset in the second half of the logic-low of the time base.

When I verify the outputs of the first counter, my oscilloscope shows pulses at some of the outputs and I consider them as logic high. The output of the first counter is binary 00011111 to represent the 1kHz clk input, almost as expected. The output of the second counter is binary 00001111 to represent the 512 Hz clk input also during the 1/32 sec. So far so good because I can read the output of the counters correctly.

I feed input a[] of the cascaded adders by the outputs of the first counter and input b[] of the adders by outputs of the second counter. I expect to see the same pulsing output pins of the adders to show the correct sum. My expectation is to see the correct sum as the binary number 00101110 at output s[] of the 8 bits adder, the same way that I see the correct binary numbers at the pulsing pins of the two counters.

Instead, I see binary 00111111 as the sum.

If I change the clk input of the first counter to 2kHz and I leave the clk input of the second counter as it was 512 Hz, I see binary number 01111111 instead of the correct sum 01001110 as sum at outputs s[] of the adder.

If I change the clk input of the first counter to 512kHz like the second counter, I see the correct output of 00011110 as sum at outputs s[] of the adder.

I understand that as the counter is cycling through the addition is performed. My thoughts are that the total propagation delay of the counters and the adders is in the order of hundreds of nanoseconds but the counters are not counting for many milliseconds before they are reset.

My expectation is to see the correct sum at output s[] of the adder at their pulsing output pins while the counters are counting during the logic high of the time base.

I mean, I expect to see the right output of the adders at their pulsing output pins just like I can see the right output of the counters at their pulsing output pins. Is my expectation justified while counters periodically pause counting? If so, why do I get the wrong sum?

Please note that it is not about what kind of counters I am using. It is about the wrong sum no matter which counter I use.

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  • \$\begingroup\$ I apologize that the schematic which I included does not look standard - what prevents you from drawing it? There is even an embedded schematic editor here. \$\endgroup\$ – Eugene Sh. Apr 22 at 17:06
  • \$\begingroup\$ "Hz" or "hertz" but not "hz". See SI standards. \$\endgroup\$ – Transistor Apr 22 at 20:56
  • \$\begingroup\$ Please see my new schematic and submit your answer if you have. Thank you. \$\endgroup\$ – Shawn-HR Apr 22 at 21:39
  • \$\begingroup\$ "my oscilloscope realizes pulses at some of the outputs and I consider them as logic high" Do you really consider "pulses of 1" as the value "1"? It does not work that way, during the time a line is high, its value is "1", but when it's low, its value is "0". So a train of pulses is a sequence of ones and zeroes on the line. \$\endgroup\$ – the busybee Apr 23 at 6:03
  • \$\begingroup\$ A minor point. Although a kilobyte is (typically) 1024 bytes, 1kHz is 1000 Hz. So half of 1 kHz is not 512 Hz. That might be important if your frequencies are important to you. \$\endgroup\$ – Math Keeps Me Busy Apr 25 at 3:19
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Assuming the counter "a" bit is the least significant bit, it looks like you have the Carry Out of the high adder feeding the Carry In of the low adder. This is backwards probably

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  • \$\begingroup\$ Please see the new schematic. \$\endgroup\$ – Shawn-HR Apr 22 at 21:31
  • \$\begingroup\$ how can I reply to the answer? \$\endgroup\$ – Shawn-HR Apr 23 at 23:49
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    \$\begingroup\$ @Shawn-HR You just did. \$\endgroup\$ – Elliot Alderson Apr 24 at 0:20

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