Counters Questions and Answers

Answer: A

The maximum operating frequency is 4 MHz.

A MOD-32 counter needs 5 flip-flops, since \(2^5 = 32\). In a ripple counter each flip-flop is clocked by the one before it, so in the worst case the delays add up before the count settles:

\[ \begin{aligned} t_{total} &= 5 \times 50\ \text{ns} = 250\ \text{ns} \\ f_{max} &= \frac{1}{250\ \text{ns}} = 4\ \text{MHz} \end{aligned} \]

4 GHz is the same digits with the wrong prefix. 3 MHz and 2 MHz match no sensible count of flip-flops: 2 MHz would need ten flip-flops' delay, which is a MOD-1024 counter.

Answer: D

The output frequency is 8 Hz.

Each stage of a binary divider (one flip-flop) halves the frequency, so 14 stages divide by \(2^{14}\):

\[ \begin{aligned} 2^{14} &= 16384 \\ f_{o} &= \frac{131072}{16384} = 8\ \text{Hz} \end{aligned} \]

131072 is \(2^{17}\), so the result is \(2^{3}\) exactly. 1 kHz or 100 Hz would need a decimal (divide-by-10) chain, and 13 Hz does not come from dividing by any power of two.

Answer: B

The output at \(Q_0\) is 25 kHz with 25% duty cycle.

A 4-bit ring counter loaded with 1000 circulates the single 1 through its four stages, so each output is high for one clock period out of every four. The output frequency and duty cycle are

\[ \begin{aligned} f &= \frac{100\ \text{kHz}}{4} = 25\ \text{kHz} \\ D &= \frac{1}{4} = 25\% \end{aligned} \]

50 kHz would come from dividing by 2, as a single toggle flip-flop does, and a 50% duty cycle would need the output high for two of the four clock periods, as in a Johnson counter.

Answer: A

A ripple counter with \(n\) flip-flops has a maximum modulus of \(2^{n}\).

Each flip-flop stores one bit, so \(n\) flip-flops hold \(n\) bits and can represent \(2^{n}\) distinct states. A straight binary ripple counter steps through all of them, 0 to \(2^{n} - 1\), before returning to 0, so it divides the clock by \(2^{n}\). For example, a 4-bit ripple counter counts 0 to 15 and is a mod-16 counter.

\(2^{n} - 1\) is the largest count reached, not the number of states, which is the common slip. \(n\) is the number of states of a ring counter with \(n\) flip-flops, and \(n^{2}\) has no meaning here.

Answer: D

The total delay is 48 ns.

In a ripple (asynchronous) counter, each flip-flop is clocked by the output of the one before it. Going from 1111 to 0000, every bit toggles, and each toggle waits for the previous one, so the delays add:

\[ t = 4 \times 12\ \text{ns} = 48\ \text{ns} \]

12 ns would be the delay of a synchronous counter, where all flip-flops are clocked together. 24 and 36 ns count only two or three of the four stages.

Answer: C

A 4-bit ripple counter needs 16 clock pulses for a complete cycle.

An n-bit binary counter has \(2^{n}\) states. For n = 4 it counts 0000 to 1111, 0 to 15, which is 16 states, so 16 clock pulses take it through every state and back to 0000. Its modulus is 16.

4 is the number of flip-flops, not the number of states. 8 is the number of states of a 3-bit counter, and 2 that of a single flip-flop.

Answer: B

A modulo-6 counter needs 3 flip-flops.

n flip-flops give up to \(2^{n}\) states, so the smallest n with \(2^{n} \ge 6\) is needed: \(2^{2} = 4\) is too few and \(2^{3} = 8\) is enough. A 3-bit counter is made to reset after state 5 (101), giving the six states 0 to 5.

Two flip-flops give only four states. Four flip-flops would work but are more than needed. Six flip-flops would be used only in a ring counter, which needs one flip-flop per state, not in a binary modulo counter.

Answer: D

The answer is None of these: the counter needs 4 flip-flops.

The counter steps through 0, 1, 6, 10, 4 and back. Its flip-flops must hold each of these values as a binary number, so the number of flip-flops is set by the largest value, 10:

\[ 10_{10} = 1010_{2} \Rightarrow 4\ \text{bits} \]

Three flip-flops can represent only 0 to 7, so 3 is too few, even though there are only five states. 5 and 6 are more than needed: 4 flip-flops already cover 0 to 15. Since the correct count, 4, is not listed, the answer is None of these. The unused states (2, 3, 5, 7, 8, 9 and 11 to 15) are treated as don't-cares when the next-state logic is designed.

Answer: C

A decade counter skips binary states 1010 to 1111.

A decade (mod-10) counter counts through ten states, 0000 to 1001 (0 to 9), and then resets to 0000. Of the 16 states a 4-bit counter can have, the six from 1010 (10) to 1111 (15) are never used.

The states 1000 and 1001 (8 and 9) and 0000 to 0011 (0 to 3) are counted, so the options skipping them are wrong, and a 4-bit counter has no states above 1111.

Answer: C

A 3-bit binary counter has 8 states.

Each bit has two values, so n bits give \(2^n\) combinations. For 3 bits, \(2^3 = 8\) states: 000 to 111, that is 0 to 7.

3 is the number of bits, 6 is 2 times 3 rather than 2 to the power 3, and 16 is the count for 4 bits.

A counter with n bits counts from 0 to \(2^n - 1\) and then wraps round, so a 3-bit counter is also called a mod-8 counter.