Resetting a counter when it reaches 6, without logic gates

counterflipflopresistorsripple-counter

I have a 3 bit ripple counter (using t flip flops) and I want the counter to reset to one when it reaches 7 i.e. the counter counts from 1 to 6. My idea was that I could hook the three outputs from the counter together and place a resistor and led in series such that when the counter reaches 111 (6 in binary) that it will be able to overcome the resistor and light up the led (the led is just to test if it works, then I can hook the led up to the necessary reset and set pins of the counter). When the number is 001 or 011 for example, there will not be enough power from the outputs to over come the resistor.

I was just wondering if this theory is possible, or if there is another way to do this without using logic gates. I'm currently using a 9v battery and the cmos dual d-type flip flops are 4013be components.

Also i'm pretty new to electronics so sorry if I'm missing anything obvious!

Regards,
David

Best Answer

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.

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