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Why won't my TI CD4098B multivibrator give me a pulse

cliquot22

Mar 17, 2016
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I'm trying to create a ~0.25 second pulse from a leading edge trigger. I found the TI CD4098B chip (datasheet: http://www.ti.com/lit/ds/symlink/cd4098b.pdf) that shows a circuit in the datasheet that will do just what I need (leading edge non-retriggerable pulse). However, I am not able to get the chip to work.

When I trigger the +TR input with 5V (VDD) or with ground (VSS), the output Q-bar goes to 5V. It seems as long as the trigger input is not floating, the output is 5V. But more importantly, there is no output pulse. I can't get the output to give me the 0.25s pulse I need.

With the input floating, the output is around 0.6V.

Here is my circuit which, as far as I can tell, is exactly what it shows in the datasheet.
Pin 1: NC
Pin 2: NC
Pin 3: Vss (gnd)
Pin 4: Vss (gnd)
Pin 5: Vdd (5V)
Pin 6: NC
Pin 7: NC
Pin 8: GND
Pin 9: to pin 11, and output Q-bar
Pin 10: NC
Pin 11: to pin 9
Pin 12: input
Pin 13: Vdd (5V)
Pin 14: cap to pin 15 and resistor to Vdd
Pin 15: cap to pin 14
Pin 16: 5V

R=47k, C=10uF -> time ~0.25s
DSC_0355.JPG
 

Harald Kapp

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Welcome to electronicspoint.

It seems as long as the trigger input is not floating
Don't let the input float, a CMOS circuit will nottake this well. Add a pull-down resistor to ensure defined low when not triggered, connect the trigger by a pushbutton to Vcc (5V)

And show us a schematic drawing of your circuit diagram. Reading a circuit from a breadboard is tedious at best.
 

dorke

Jun 20, 2015
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Apart from not leaving inputs floating,
which you shouldn't do,as was said above by @Harald Kapp .

Why do you connect /Q to -Tr (pins 9 and 11) ?
This will cause a "locking state",
that is why your circuit doesn't work.:(

i.e once /Q will be "low" you have a "state lock".
From that point onward the +Tr has no effect.
Only a Reset will clear that lock...til the next one occurs.

Remove that connection and tie the -Tr(pin 11) to VDD,
that should fix that problem.:)

BTW,
Choosing that CD4098B is an odd thing,
unless you had it in stock;)
 
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cliquot22

Mar 17, 2016
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Attached is the schematic (before and after suggested modifications). I'll try this out later today.

The datasheet said to connect 9-11 to prevent retriggering on another input pulse. But I guess I have to send a reset to get even the first trigger. Since this is for a light sensor, I can just connect the reset to the light level.

-> Dorke: I chose the CD4098B because that was the results of my internet and Digikey searches. Is there something better for creating a pulse from a rising edge trigger?
 

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  • pulse.pdf
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Alec_t

Jul 7, 2015
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In your modified circuit the photo-sensor will have to produce about 250uA in order to bring the chip out of the reset state. Can it manage that much current?
 

cliquot22

Mar 17, 2016
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  • pulse r1.pdf
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Alec_t

Jul 7, 2015
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That looks promising. Alternatively, just increase the value of the emitter resistor from 10k to, say, 100k.
 

dorke

Jun 20, 2015
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-> Dorke: I chose the CD4098B because that was the results of my internet and Digikey searches. Is there something better for creating a pulse from a rising edge trigger?

Yes,
555 timer, CD4047, 74HC123, Schmidt trigger gates, etc.

In your "per-modified original circuit" the mono-pulse is created by a manual switch without any relation to the photo-tr.
What is it your are trying to do?
Where do the "outputs" connect-to/activate?

Is this pulse for generating a "reset signal" for a uC/uP?
There are simpler solutions for that.
 
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cliquot22

Mar 17, 2016
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Ok, I think I have it working now. Thanks for all your help.

I have replaced the switch with a connection to the light sensor. You can see the final layout.
This circuit creates a reset signal for my ESP8266 board when a light turns on and it tells the board that the light is on. I'm sure there are many easier ways to do this but this seems to work for me. I'll check out the other options when I have time.

Clipboard01.jpg
 
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