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pull up/ pull down resister with switch (only need a pulse add a capaciter?)

A

Andrew Crook

Jan 1, 1970
0
1. I have used a voltage divider to add push buttons to a logic circuit.
2. I have also used capacitors with a voltage divider to make a auto reset
on power up on simple logic circuits.

However I have a device that acts as a switch, I need into act as a push
button i.e. a single pulse (high or low) when turned on (then opposite after
N secondes). Can I combine 1 and 2 in some way, have tried but don't get the
correct results.

regards

Andrew
 
R

R. Steve Walz

Jan 1, 1970
0
Andrew said:
1. I have used a voltage divider to add push buttons to a logic circuit.
2. I have also used capacitors with a voltage divider to make a auto reset
on power up on simple logic circuits.

However I have a device that acts as a switch, I need into act as a push
button i.e. a single pulse (high or low) when turned on (then opposite after
N secondes). Can I combine 1 and 2 in some way, have tried but don't get the
correct results.

regards

Andrew
------------
Experiment with this:

A PB with a resistor pull-up, or pull-down, the node of the two goes
to a series cap, the other side of the cap goes to another pull-up or
pull-down. Observe the effect on the other side of the cap with respect
to ground when you push the PB. You'll figure it out.
-Steve
 
R

Robert Monsen

Jan 1, 1970
0
Gasp! I didn't realize he'd cross posted...

Apologies for the attachment~!
 
A

Andrew Crook

Jan 1, 1970
0
Thanks for your reply,

what I need to happen is when the button is depressed the output to go low
then rise over nsec (0->1) and when pressed just be (1). This is used to
trigger a latch circuit that I have made.

What I have at the moment does the opposite! :-(.

I am a little worried when the button is press because I get a negative
reading on my meter. Should this be okay with a electrolytic capacitor?



________________________ +9V

|

|

R1 10K

|______+(C1 10uf)_________ Output

| |

PB |

| R2 1M

| |

_____________________ 0V



Regards



Andrew
 
R

Robert M. Urban II

Jan 1, 1970
0
Andrew,

You might try a r/c series circuit to ground. Bur have a normally open
pushbutton switch in parallel with the cap. The rise time will be dependant
on the values of r and c. What happens in this circuit is that the cap will
charge up on power up. The junction of r and c is the junction that will
connect to your latch. When you depress the pushbutton you will discharge
the cap, and thus it will rise to a full charge according to the vaules of r
and c.

I hope that this helps.

Rob
 
A

Andrew Crook

Jan 1, 1970
0
Thanks!

Used my reset setup (capacitor in voltage divider circuit with a chosen
resistor value).
put a PB in parallel with capacitor as suggested and when the button is
pressed output is 0 and when the button is depressed it goes from 0->1, then
stays at 1.

This is almost correct, however, when the button is press I need the circuit
output to be 1. Then when
depressed drop to 0 then rise back to 1 over a period of n.

almost there

many thanks for you help

Andi
 
R

Robert M. Urban II

Jan 1, 1970
0
Andrew,

You might try putting the cap in series with the voltage divider. Set the
divider so that the output is = 1 and keep the switch in parallel with the
cap.

Hope this helps.

Rob
 
A

Andrew Crook

Jan 1, 1970
0
Thanks, That what i thought i had done see my diagram. when the button is
press I need the circuit
output to be 1. Then when depressed drop to 0 then rise back to 1 over a
period of n.


________________________ +9V
|
|
R1
|
|
|_________ output
|
|______
| |
C1 PB
| |
|______|
|
|
|_________________________ 0V

Andi
 
R

Robert M. Urban II

Jan 1, 1970
0
Andi,
Not quite...
Open the attachment of my last post. There is a 2 pole pushbutton. One
delivers the constant output while the other discharges the cap. When
depressed, the output only "sees" the rise of n seconds.

Regards,
Rob
 
A

Andrew Crook

Jan 1, 1970
0
ahh thanks.... never saw or got the attachment, would you be so kind to send
it please.

many thanks

Andrew
 
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