I am VERY new to electronics so most of the things I find that may be able to help me are hard for me to understand (although I am learning quickly). I made an adjustable timer using this site: http://www.kpsec.freeuk.com/projects/timer.htm

It works and does what I need...sort of. The ouput I have is not going to a bleeper it is going to another electronic device that uses a 9V charge for about one second.
I used a 9V battery to set this up and during the "countdown" I have about half that; 4.58V at the output and then it jumps to the full 9+V at the end. There are two issues here that I need help with.
The less important is: Is there a way to eliminate the 4.58V from reaching my output before it becomes 9+V? (I would like this but it is okay if I cant because the device will not operate at less than 9V)
The more important question is with the output. Currently the output has a duration similar to the input duration which I understand has to do with the capacitor discharge. I want to maintain the 10 minutes it takes to "countdown" but is there a way on the output side to limit the duration of the high (9V) signal to only a second?
Thank you to anyone who can give me an answer to this that I may actually understand.

 

What are you measuring that is 4.58V? The cathode of the red LED? It will burn without a current-limiting resistor in series. The bleeper in that circuit limits the current in the red LED.

The output of the 555 timer is its pin 3. Without a load it is at about 8V during the timing then it goes to 0.2V at the end of the timing period.
If the load draws the max allowed for the output of the 555 then its output voltage is 7V during the timing and 1.5V to 2V at the end.

What is it driving?

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Uncle $crooge

 

I actually omitted the red led from the circuit. I am measuring from the #3 pin and the connection at the #8 that would lead to the red led. This is measuring 4.58V during the "countdown" (the green led is lit during this time). Then the green led goes out and the points I am measuring go up to the full voltage in the battery which in a 9V is about 9.6V on a new one. The device is a spring powered trap that uses a 9 volt charge to release a contact points and unlatch the spring. The contact points are actually one grounded prong and one that is charged positive. 9V is enough to spark the two apart and that unlatches the spring setting the trap. A weak 9V will not power this so I don't believe the 4.58V will trip it either but I also don't want the prolonged 9V charge being send to the points in case they come in contact again which may create a hazard.

 

So you want to set the mark/space ratio so you have 10 minutes Low and 1 second High.

This may be difficult to achieve with a single 555 as it is a M/S ratio of 1/600.

However, it may be possible, so have a look at

http://ourworld.compuserve.com/homep...Bowden/555.htm

http://www.uoguelph.ca/~antoon/gadgets/555/555.html

http://www.doctronics.co.uk/555.htm#astable (follow the astable link)

http://www.williamson-labs.com/480_555.htm

http://home.cogeco.ca/~rpaisley4/LM555.html

http://www.kpsec.freeuk.com/555timer.htm

http://en.wikipedia.org/wiki/555_timer_IC

http://www.interq.or.jp/japan/se-inoue/e_ckt4.htm

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Len

 

What voltages do you measure if you measure with respect to gnd (pin 1)?

It is usual to measure voltages wrt gnd.

__________________
Len

 

I think you have a Cmos 555 like a TLC555, LMC555 or ICL7555. It is low power and its output drops to about 4.5V when it drives the green LED. An ordinary 555 has 20 times more output high current available so its output voltage high is about 7.8V with a 9V battery powering it.

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Uncle $crooge

 

Here is a suggestion for you.

Use the 555 that you have and drive the solenoid with either a transistor or MOSFET.

Attached Images

Solenoid Driver.png (3.0 KB, 15 views)

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Len