Digital Circuits
Counters
34 questions By Tony R. Kuphaldt
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Question 19 of 34
Consider the following four-bit binary counter integrated circuit (IC). When clocked by the square wave signal generator, it counts from 0000 to 1111 in sixteen steps and then “recycles” back to 0000 again in a single step:

There are many applications, though, where we do not wish the counter circuit to count all the way up to full count (1111), but rather recycle at some lesser terminal count value. Take for instance the application of BCD counting: from 0000 to 1001 and back again. Here is one way to truncate the counting sequence of a binary counter so that it becomes a BCD counter:

Explain how the NAND gate forces this counter to recycle after an output of 1001 instead of counting all the way up to 1111. (Hint: the reset function of this IC is assumed to be asynchronous, meaning the counter output resets to 0000 immediately when the [RST] terminal goes low.)
Also, show how you would modify this circuit to do the same count sequence (BCD) assuming the IC has a synchronous reset function, meaning the counter resets to 0000 if [RST] is low and the clock input sees a pulse.
Reveal answerA timing diagram is probably the best way to answer this question! As for the synchronous-reset BCD counter circuit, the only change necessary is a simple wire move (from output Q1 to Q0):

Notes:Although both circuits achieve a BCD count sequence, the synchronous-reset circuit is preferred because it completely avoids spurious (“ripple-like”) false outputs when recycling. Be sure to emphasize that the difference between an asynchronous and a synchronous reset function is internal to the IC, and not something the user (you) can change. For an example of two otherwise identical counters with different reset functions, compare the 74HCT161 (asynchronous) and 74HCT163 (synchronous) four-bit binary counters.
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Question 20 of 34
Suppose you had an astable multivibrator circuit that output a very precise 1 Hz square-wave signal, but you had an application which requires a pulse once every minute rather than once every second. Knowing that there are 60 seconds in a minute, can you think of a way to use digital counters to act as a “frequency divider” so that every 60 multivibrator pulses equates to 1 output pulse?
You don’t have a divide-by-60 counter available, but you do have several divide-by-10 (“decade”) counters at your disposal. Engineer a solution using these counter units:

Note: assume these counter ICs have asynchronous resets.
Reveal answerCascade two decade counters together, with a NAND gate to decode when the output is equal to 60:

Follow-up question: why can’t we take the divide-by-60 pulse from the RCO output of the second counter, as we could with the divide-by-10 pulse from the first counter?
Challenge question: re-design this circuit so that the output is a square wave with a duty cycle of 50% (“high” for 30 seconds, then “low” for 30 seconds), rather than a narrow pulse every 60 seconds.
Notes:Tell your students that counter circuits are quite often used as frequency dividers. Discuss the challenge question with them, letting them propose and discuss multiple solutions to the problem.
The “note” in the question about the asynchronous nature of the counter reset inputs is very important, as synchronous-reset counter ICs would not behave the same. Discuss this with your students, showing them how counters with synchronous reset inputs would yield a divide-by-61 ratio.
Incidentally, a divide-by-60 counter circuit is precisely what we would need to arrive at a 1 Hz pulse waveform from a 60 Hz powerline frequency signal, which is a neat “trick” for obtaining a low-speed clock of relatively good accuracy without requiring a crystal-controlled local oscillator. (Where the “mains” power is 50 Hz instead of 60 Hz, you would need a divide-by-50 counter - I know, I know . . .) If time permits, ask your students to think of how they could condition the 60Hz sine-wave (120 volt!) standard powerline voltage into a 60 Hz square-wave pulse suitable for input into such a frequency divider/counter circuit.
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Question 21 of 34
When counters are used as frequency dividers, they are often drawn as simple boxes with one input and one output each, like this:

Calculate the four output frequencies (fout1 through fout4) given an input frequency of 1.5 kHz:
- fout1 =
- fout2 =
- fout3 =
- fout4 =
Reveal answer- fout1 = 150 Hz
- fout2 = 25 Hz
- fout3 = 12.5 Hz
- fout4 = 2.5 Hz
Follow-up question: if the clock frequency for this divider circuit is exactly 1.5 kHz, is it possible for the divided frequencies to vary from what is predicted by the modulus values (150 Hz, 25 Hz, 12.5 Hz, and 2.5 Hz)? Explain why or why not.
Notes:The purpose of this question is to introduce students to the schematic convention of counter/dividers as simple boxes with “MOD” specified for each one, and to provide a bit of quantitative analysis (albeit very simple).





