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4017 strange output level

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batman

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I have a 4017 rigged so that i can give it a clock pulse, and it moves between outputs 0 and 1, then resets, so it can only be outputting at 0 or 1 at any given time basically. Vcc is 5.2 V. Output 0 starts high, then when I clock it, it doesnt fall low and 1 doesnt go high. Instead 0 goes to 4.67, and 1 stays low. Then I clock it again, 0 goes to 5.2 and 1 stays low. Does anyone know why the outputs are not cycling correctly? This circuit worked perfectly on the proto board, and when I soldered it together, it behaves weirdly. All solder connections are correct, and all components are tested and work fine. Also, resistors and tying down the correct pins.
Thank you for any help
 

batman

New Member
The reset is pin15. It is tied low and to output 2. This way, it will be low until triggered high by the third output (output 2), and force the counter back to output 0.
 

batman

New Member
:x does anyone have any idea what im talking about? i have a cmos outputting at an indeterminate state. why is this happening :evil:
 

bogdanfirst

New Member
i don't understand why you tied low the reset?
the outpt of the 4017 is L when not H, it is not floating, so you need to use a resistor to gnd. i am not sure i understand what you actually diddid you use Q2(the third output) to reset? or you used Q1 wich is the second output?
please, a schematic will be great!!
 

batman

New Member
I tied the reset low and to Q2. This way it won't do anything until Q2 resets it. I hope this clarifies. On the proto board I had Q0 and enable tied together and to ground throught the same resistor and it worked. I did the same thing on the circuit board.
 

bogdanfirst

New Member
from what i understand you have connected your circuit as a flip flop.
but why do you did you connect Q0 and enable together and to gnd?
really, a schematic could do a lot better
 

Roff

Well-Known Member
batman said:
I tied the reset low and to Q2. This way it won't do anything until Q2 resets it. I hope this clarifies. On the proto board I had Q0 and enable tied together and to ground throught the same resistor and it worked. I did the same thing on the circuit board.
You have shorted the Q2 output to ground. Q2 cannot drive the reset if you have it shorted to ground. Ground is zero volts. Period. Shorting an output to ground could possibly cause unexplained behavior in other parts of the chip.
If you want to initialize the counter to zero count when you turn on power, post that fact here and we'll show you how to do it.
 

bogdanfirst

New Member
well, Ron H, you said it yourself in the quote. he tied them together and with a resistor to gnd. now the resistor should be a few KOhms, is it is small, it will be just as you say, a short to gnd.
 

Roff

Well-Known Member
Yeah, if he tied them to gnd through a resistor, he might be OK, as you say, if the resistor is a few Kohms. I got the impression that he had tied Q2 directly to ground.
 

batman

New Member
Nothing is shorted to ground. Theyre tied to ground through resistors in the Kohms. Yes I have it working as a flip flop.

One sate should be: Q0=5.2V, Q1=0V, the other state should be Q0=0V, Q1=5.2V.

However it behaves as follows: state one Q0=5.2V, Q1=0V, state two Q0=4.67V, Q1=0.

To my understanding, if the clock and enable pins are tied low through the same resistor, and the same high pulse is given to the clock pin and the enable pin, the counter will advance. This is how I have wired it on the protoboard when it worked.

I don't believe the issue is with the reset or enable, because the counter IS toggling between two distinct states. The problem is that one of these states is not the one I want :? :!:
 

pebe

Member
Batman, you say you are using a resistor of 'a few K' from Q2 to ground. You should need no resistor at all because CMOS devices have complentary (push-pull) outputs. If your resistor is only 1K or so, you are loading Q2 excessively.

Connect as follows. + and -ve rails to pins 16 and 8. Pin 13 (CE) to ground. Pin 14 (CLK) as input. Pin 15 (RES) to pin4 (Q2). Make all connections direct (ie not using resistors). Those are all the interconnects you require.

Take outputs from Q0 and Q1.
 

bogdanfirst

New Member
well, i see the problem now.
enable is active LOW, so when you tied it with the clock, you have the circuit disabled when you get a pulse.
 

batman

New Member
I have heard contradictory statements from different people in this forum:

1) the resistor should be a few KOhms

2) You should need no resistor at all

I wired it exactly how pebe described. I'm still getting the 4.67 state instead of 0. Any other ideas?
 

mechie

New Member
Try this

Pebe's circuit description is correct, and there is no need for any resistors.

What is the input to this circuit :?:

If you are using, say, a pushbutton and when released the input is just floating...
CMOS is so sensitive it could be picking up an input capacitively from an output, causing the 4017 to 'race', trying to count faster than is reliable, and giving unpredictable results.

Contradicting what I said above :? try a 100k resistor from 0v to the 4017 input, see if that calms it down.
NOTE:- I think 100k to 1MEG should be OK for CMOS, enough without loading the outputs too much. 10k or less is starting to get too heavy in my experience.
 
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