Electrical – How to design a binary up down counter using D flip flop’s and the ‘up’ or ‘down’ count will be controlled by another flip flop

counterflipflop

I have to design a 4bit Binary Up Down counter using D flip-flops . But here is the catch I cannot directly use a switch to control whether the counter will count up or down, I have to do that using a D flip flop. Any help on how I can solve this?

N.B. I can design an up down counter , but I need to know how to use another D flip flop to control the whether I want up or down count

Best Answer

Without knowing exactly what "using a D flip flop" means, I can provide at least two approaches.

First, instead of a single switch, you can provide two - an up/down switch and a "load control" switch. Using an NO switch operating on the open-to-close action for the control switch can be done by

schematic

^{simulate this circuit – Schematic created using CircuitLab}

But keep in mind that I have provided no debouncing on the clock switch, which you must provide.

Alternatively, if you want the control FF driven by the same clock which drives the counter, you can do it but you have to delay the clock between the control FF and the counter in order to allow a change in the control input to settle before the counter acts on it.

schematic

^{simulate this circuit}

Note that a pair of inverters makes a perfectly usable delay, and multiple pairs of inverters make for longer delays.

Related Solutions

Electronic – 3bit Gray counter using D Flip Flops and logic gates

If you can use freely the logic gates, you can do this way:

Extract from the truth table the logic function that defines each bit of the next state from the previous state: \$ b_0(n) = f(b_0(n-1), b_1(n-1), b_2(n-1)) \$
Create the logic network that implement the 3 logic functions that you have found;
Now put it in a feedback from the output to the input of 3 D flip-flops, so then at each clock edge the state will be updated;
Reset the circuit to the 000 state.

If you really know logic gates, flip flops and counters, you can easily understand how this circuit works, and the rest is a breeze.

3 bit synchronous counter design d flip flop

The key thing is to treat each bit individually.

For example, for the low-order bit, lets call it next_state[0].

state     | next_state[0]
---------------
   000    | 0
   001    | 0
   010    | 1
   011    | X
   100    | 0
   101    | 0
   110    | 0
   111    | 1

So you can write

next_state[0] = state[0] & state[1] & state[2] | ~state[0] & state[1] & ~state[2]

If you are building this with discrete logic you could simplify this to

next_state[0] = state[1] & (state[0] & state[2] | ~state[0] & ~state[2])

And you have an equation you can use to drive the flip-flop that will generate the next condition for the low-bit of the state variable.

Finding the equations for the other two state bits works the same way.

Best Answer

Related Solutions

Electronic – 3bit Gray counter using D Flip Flops and logic gates

3 bit synchronous counter design d flip flop

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