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HEF4059 divide-by-n counter - doesn't divide

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captainhannes

New Member
Hi all!

I currently have a project with a HEF4059 divide-by-n counter.

Datasheet:
http://www.electro-tech-online.com/custompdfs/2003/08/HEF4059B_CNV_3.pdf

I'm using it in mode 8, which allows a division of max. 21327.
My division # is 21043.

Here is my calculation:
21043 / 8 = 2630 remainder 3
=======================
n = 8 ( 1000 x 1 + 100 x 15 + 10 x 13 + 1 x 0 + 3)
===================================
Code:
____3___ 1  ____0______ ______13______ ______15_______
J1 J2 J3 J4 J5 J6 J7 J8 J9 J10 J11 J12 J13 J14 J15 J16
1  1  0  1  0  0  0  0  1  0    1  1    1  1    1  1
Mode 8:
LE = L ; Ka = L ; Kb = L ; Kc = H

I'm also using an reset curcuit to start the divider in master-preset:
Code:
____
    |
    ---
    ---  1uF
    |_______ to a HEF4025 NOR gate wired as inverter ___ Kc 4059
    |
    R  1k
    R
  _|_
The problem is that the output of the 4059 is always low.
I checked:
* supply voltage - OK
* pin voltages high/low - OK
I don't have an HF freq. counter, but the clock input of the 4059 shows a voltage of half of the supply voltage, so I assume there is a clock signal.
My multi-meter says the frequency is to high to be measured. If the input would be DC, it would say 0Hz.
BTW, the clock is 1.2288 MHz

Is there anything wrong with my calculation and/or the wireing of the 4059 ?

Thank you all very much for any tips!! Very much appreciated!!

Best regards from Austria,
Hannes.
 

rizzo

New Member
4059 circuit

Hi Hannes!

I studied the 4059 data sheet and your schematic diagram, and I found nothing bad. I think it should work.
(The 6K2 resistor in the crystal oscillator seems to me a somewhat high value; but, if you found that the clock, after the divide-by-two flip flop, has a mean value of half the supply voltage, the oscillator is working).

So, my best guess is that the pulses are really going out of pin 23, but you cannot see them. They are short pulses, one clock cycle long each, i.e. less than 1 µs, repeating about every 17000 µs. So, you cannot see their effect on the mean value, and (probably) even the multimeter, as a frequency meter, cannot see them because they are too short.

My advice is: make another divide-by-two circuit, just like the one you used after the crystal oscillator, and feed the Q (pin 23) output to his input (clock).
Look at the output with the multimeter. If it is half of the supply voltage, the pulses are there. If not, my second best guess is that there is a short between pin 23 and Vss (ground).

Best wishes

Ezio Rizzo
 

rizzo

New Member
4059 circuit

I posted a tentative answer to your question in the forum.

If it is not useful to you, or you want to contact me, please email eziorizzo AT libero.it <Edited to avoid spam. Too late.>
as I do not follow regularly the forum.

Ezio Rizzo
 

captainhannes

New Member
Ciao Ezio!

Thank you very much! That was exactly the reason, why I couldn't see the output of the 4059. The pulses were too short to be recognized by the VM.

Mille grazie :D
Hannes.
 

ralf_altman

New Member
Hi Annes & Ezio,

If it's possible, could you please send me a copy of the schematic. I would like to have it to use as a reference for my design with HEF4059. I am havinf trouble with it.

Thanks,
Ralf
 
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