counters and registers

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7-4 Asynchronous (Ripple) down counter

7-4 Asynchronous (Ripple) down counter

23=8 ---> MOD-8 Down Counter

7-5 Propagation Delay in Ripple Counters


Ripple counters are the simplest type of counters.

Although they are simple, they have a major drawback which is the propagation delay caused by their operation principle.

Ripple counters are the simplest type of counters.

Although they are simple, they have a major drawback which is the propagation delay caused by their operation principle.

To avoid this problem, we should make sure of the following:

N = number of FFs Tpd = propagation delay time

To avoid this problem, we should make sure of the following:

N = number of FFs Tpd = propagation delay time



€

TClock ≥ N × t pd

fmax =1

N × tpd

ExampleExample

A 4-bit Ripple counter with tpHL= 16 nS and tpLH= 24 nS, using 74LS112 J-K FF. find the maximum frequency for a proper operation of the counter. Assume a worst case scenario

A 4-bit Ripple counter with tpHL= 16 nS and tpLH= 24 nS, using 74LS112 J-K FF. find the maximum frequency for a proper operation of the counter. Assume a worst case scenario

€

fmax =1

N × tpd=

1

4 × 24nS=10.4MHz

€

fmax =1

N × tpd=

1

6 × 24nS= 6.9MHz

For 6-bit Ripple counter that has 6 FFs. For 6-bit Ripple counter that has 6 FFs.

7-6 SYNCHRONOUS (PARALLEL) COUNTERS


The synchronous counters have all of the FF’s triggered simultaneously. That is, all the CLK inputs are connected together

Synchronous counters require more circuitry than the asynchronous counters.

Synchronous counters needs extra logic gates to be added.

The synchronous counters have all of the FF’s triggered simultaneously. That is, all the CLK inputs are connected together

Synchronous counters require more circuitry than the asynchronous counters.

Synchronous counters needs extra logic gates to be added.



Each FF should have its j and k inputs connected such that they are HIGH only when the outputs of ALL lower-order FFs are in the HIGH state



Each FF is clocked by the NGT of the clock input signal so that all the FF transitions occur at the same time.

Only A has its J-K inputs permanently at the HIGH level.

A changes at each NGT clock (A FF toggles)

B changes when A=1 and a NGT clock occurs

C changes when A=B=1 and a NGT clock occurs

D changes when A=B=C=1 and a NGT clock occurs

Each FF is clocked by the NGT of the clock input signal so that all the FF transitions occur at the same time.

Only A has its J-K inputs permanently at the HIGH level.

A changes at each NGT clock (A FF toggles)

B changes when A=1 and a NGT clock occurs

C changes when A=B=1 and a NGT clock occurs

D changes when A=B=C=1 and a NGT clock occurs

Advantage of synchronous Counters over AsynchronousAdvantage of synchronous

Counters over Asynchronous In a parallel counter, all FF will change simultaneously. Propagation delay of FF do not add together to produce

the overall delay.

Total delay = FF tpd + AND tpd

Fmax = FF Fmax+AND Fmax

The total delay is the same no matter how many FF are used.

A synchronous counter can operate at much higher frequency, but the circuitry is more complex than that of the asynchronous counter.

In a parallel counter, all FF will change simultaneously. Propagation delay of FF do not add together to produce

the overall delay.

Total delay = FF tpd + AND tpd

Fmax = FF Fmax+AND Fmax

The total delay is the same no matter how many FF are used.

A synchronous counter can operate at much higher frequency, but the circuitry is more complex than that of the asynchronous counter.

ExampleExample

Determine fmax for the synchronous MOD-16 counter if tpd=50ns for each FF and tpd=20ns for each AND gate.

Determine fmax for the synchronous MOD-16 counter if tpd=50ns for each FF and tpd=20ns for each AND gate.

€

fmax =1

50 + 20=14.3MHz

Determine the equivalent fmax for asynchronous MOD-16 counter

Determine the equivalent fmax for asynchronous MOD-16 counter

€

fmax =1

4 × 50= 5MHz

ExampleExample

What must be done to convert this counter to MOD-32 parallel counter? A Fifth Flip-Flop must be added “25=32”

Determine fmax for the MOD-32 parallel (synchronous) counter

For a MOD-32 ripple counter

What must be done to convert this counter to MOD-32 parallel counter? A Fifth Flip-Flop must be added “25=32”

Determine fmax for the MOD-32 parallel (synchronous) counter

For a MOD-32 ripple counter

€

fmax =1

50 + 20=14.3MHz

€

fmax =1

5 × 50= 4MHz

Unchanged

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