4 bit synchronous binary counter

Thread Starter

raphaelriv29

Joined Nov 4, 2013
10
Hello all

I'm trying to learn how counters work so I picked this 74HCT4520 synchronous binary counter up and I did the following

pin 1 - clock input
pin 2 - set high
pin 3 - LED1
pin 4 - LED2
pin 5 - LED3
pin 6 - LED4
pin 7 - set LOW
pin 8 - GND
pin 16 - 5V

so viewing the state of the LEDs I thought upon resetting the counter once it would start counting from state 0000 ---> 1111 (2^4 = 16 values 0-15)
but the counter its random, I don't see a nice count such as:

0000
0001
0010
.
.
.
.
.
1111

instead I see the values all over the place, worth saying that there are 16 values after it starts to count again so at least that seems to make sense. What concept am I missing? :confused:

also what is the point of pin 2 and 10, looking at the figure 6, the timing diagram shows that when the regular clock goes low if a clock is fed into this pin than the count continues? :confused:

Thanks

Newbie
 

tshuck

Joined Oct 18, 2012
3,534
You aren't trying to use a pushbutton/switch as a clock, are you?

If so, look up switch bounce and understand why you can't use a pushbutton as a clock (unless you denounce it first).

If not, what is the frequency of your clock?

Yes, it would appear that pins 2 and 10 allow the counter to count when the clock is low.
 
Last edited:

MrChips

Joined Oct 2, 2009
24,593
If you wish to avoid having to debounce a pushbutton simply use a 555 timer circuit set for a low frequency. Even a simple 74LS14 or 74HCT14 inverter oscillator would do the trick.
 

Thread Starter

raphaelriv29

Joined Nov 4, 2013
10
hey guys,

I have another question, so I was able to feed a proper clock signal that gives me a nice count. The following output was obtained from the first binary counter of the HC4520:

Output 1: 3.2 KHz (frequency) 128.9 μs(width)
Output 2: 1.6 KHz (frequency) 250.9 μs(width)
Output 3: 802 Hz (frequency) 580.0 μs(width)
Output 4: 400 Hz (frequency) 1107.7 μs(width)

so all of those make sense but shouldn't the width of Output 1 be approximately twice that of the Clock input?

My second dilemma is, when I feed "Output 4" to pin 10 of the second counter and grounded the CLOCK input pin 9. This gave me the following outputs:

Output 5: 200 Hz (frequency) 2496.1 μs(width)
Output 6: 200 Hz (frequency) 14.81 μs (width)

What I noticed is that output 6 ends up having the same width as the original input, but with a frequency of 200 Hz (which makes sense) but how is the width working here?

thanks
 

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MrChips

Joined Oct 2, 2009
24,593
Depends on what you mean by width.

You should be measuring the period of the waveform, not the width of the pulse.

The width of the input clock pulse has no effect on what happens at the outputs. The outputs should have 50% duty cycle, i.e. the width is half the period.
 

Thread Starter

raphaelriv29

Joined Nov 4, 2013
10
that's correct, all outputs have 50% duty cycle (by width I meant the pulse width) but I only started fixating on this because I noticed a difference and I wasn't sure why it was happening.

I'm going to play with this some more but if I feed a non-50% duty cycle clock vs a 50% duty cycle clock signal, it makes no difference?
 
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