Ripple BCD Down Counter Circuit – Troubleshooting Ripple Counter Issues

proteusripple-counter

I want to implement a BCD ripple down-counter in Proteus.

The following is my circuit. The logic is simple. When the "original" 4-bit binary ripple counter reaches 1111 (15), the 2nd and 3rd flip-flops must reset to 0. To this end, I AND the outputs of all flip-flops and feed the output of the AND gate to the reset input of the 2nd and the 3rd flip flop.

But the circuit does not work properly. When I turn the circuit on, the seven-segment display shows the following sequence:

9 –> 8 –> 1 –> 0

And after 0, it goes back to 9 and continues.

By the way, if I cut the output wire of the AND gate, the circuit perfectly counts from 15 to 0.

I have tried another flip-flops such as D and another model of J-K flip-flops; But the problem did not solve.

Please help me.
Thanks

Best Answer

This happens because it is a ripple counter and ripple counters will produce glitches that will cause your intermediary reset circuit to reset even when you believe 1111 isn't present. You need to have a better circuit. When the result is settled at 8 and you clock the circuit one more time, you get a series of intermediary values like this: -

1000 = 8 (settled result waiting for next clock)
1001 at the instant the next clock comes along
1011 pretty soon after the above clock occurred
1111 a short time later and this resets the circuit producing your error
0111 <-- it never gets to 7

As you can see, for each clock, the flip-flops ripple through their results hence, why it's called a ripple counter.

Related Solutions

Electronic – In a JK Binary Counter from 0 to 9, why is the NAND gate connected to the second and fourth J-K flip flop and not the first and fourth

The circuit you cited is a ripple counter, not a synchronous counter. It actually has eleven states, 0000 through 1010, but as soon as the last state is reached, the NAND gate immediately (asychronously) resets the flip-flops to the 0000 state.

In a synchronous counter, all of the flip-flops would share a common clock, and you'd control the sequence of states by driving their J and K inputs. This would require a few additional gates, which would include decoding the 1001 state so that the counter goes directly to the 0000 state on the next clock edge.

Resetting a counter when it reaches 6, without logic gates

The resistor and LED will almost certainly not work, or be very unreliable. This is because the LED and resistor are analog components. Connecting this to 4013B's may illuminate very slightly when the output is 001/010/100, more with 011/101/110, and most with 111, but that is three "brightness levels." Digital means "OFF" or "no voltage", meaning no electricity goes anywhere, and "ON" or "voltage", meaning Vcc or the supply voltage comes out. Any component with more than these states is not digital.

This also will not follow a logical brightening sequence:

Binary Decimal Brightness
000    0       0
001    1       1
010    2       1
011    3       2
100    4       1
101    5       2
110    6       2
111    7       3

Digital circuits do not like to "drive" analog circuits. The 4013B outputs can only drive so much current into anything else, which typically limits how many other digital inputs they can be connected to. This is called "fan-out" and must be taken into consideration when connecting multiple components. Some devices are better at this than others; the NXP HEF4013B outputs can source/sink about 1mA when supplied from a 9v battery, see page 5. Even though you could connect an LED like this, it will probably not illuminate visibly with 1+1+1 = 3mA of current. This also is not recommended because it will cause the outputs to "float" somewhere other than LOW or HIGH states, as the LED is demanding more power than the 4013 can supply. Whenever an input or output is "somewhere between low or high", the device can dissipate too much power and overheat, leading to it's destruction. Digital inputs are especially sensitive to this. All unused inputs must be tied to either LOW or HIGH.

The best answer is to use some other logic gates to do the actual logic, and use a device designed to drive analog components to illuminate any LED's desired (such as 1-6, or sides of a die.) Such devices are typically called "buffers" or "line drivers" and there are literally hundreds of variants.

Best Answer

Related Solutions

Electronic – In a JK Binary Counter from 0 to 9, why is the NAND gate connected to the second and fourth J-K flip flop and not the first and fourth

Resetting a counter when it reaches 6, without logic gates

Related Topic