Richard said:
Hi,
I need a circuit that activates on a 60 second delay, the input must be
9v battery and the output device requireds 9v. Any one have a diagram for
something simple like this.
Is it posible to do this simply with a capacitor and some resistors?
Thanks ever so much for your time.
You can do this with an RC delay. One way would be to use a switch
transistor, as follows:
(look at this with a fixed-width font, or it won't make sense)
On-Off
/
/ s d
.--o o----o---------+^+-------------.
| | ||| |
| | === |
| | |g |
| | | |
| | | |
| | | |
| --- | |+
| --- | .------------------.
| 9V | | | |
--- | | | |
- | | | |
| o---------' | Your |
| | | Circuit |
| | | |
| .-. | |
| | | | |
| | | | |
| '-' | |
| | '------------------'
| | |-
| | |
'-------------o-------------------------'
(created by AACircuit v1.28 beta 10/06/04
www.tech-chat.de)
The idea is that when the switch is closed, the capacitor will charge up
to 9V, and slowly decay down to 0. The P-MOFSET will turn on when it's
gate at some point as the voltage decays. The voltage at which it turns
on will be determined by the cap and resistor values.
The problem with this circuit, through, is that the transistor will turn
on slowly. Because of the fact that you want a long delay, the
transistor will stay in it's 'linear region' for a long time, causing
the voltage across your circuit to come up quite slowly.
A better way is to use an opamp (or comparator) to sense the delay and
turn on a drive transistor with. Here is a rough sketch
On-Off
/ P-MOSFET
/
.-o o---o-------o------------o---------+^+-----------.
| | | | ||| |
| | |+ | === |
| | --- 47uF | | |
| | --- | | |
| | | | | |+
| | | | | .-----o-----.
| | | | | | |
| | | | | | |
| | 1MEG | | | | Your |
| 9V .-. | |\| | | CIrcuit |
--- | |<----------------|-\ | | |
- | | | ___ | >--------o | |
| '-' o-|___|-o -|+/ | | |
| | | 1MEG | |/| ___ | | |
| | | '--------|___|-' | |
| | | | 10MEG '-----o-----'
| | .-. | |-
| | | | 1.8MEG | |
| | | | | |
| | '-' | |
| | | | |
| | | | |
'---------o-------o------------o-----------------------'
(created by AACircuit v1.28 beta 10/06/04
www.tech-chat.de)
When you throw the ON switch, the 47uF cap will be dragged up to 9V, and
will gradually decay down to 0V. The 1MEG potentiometer is used to
adjust when the P-MOSFET comes on. Depending on the leakage of the 47uF
cap and the mosfet, it's possible that the thing will decay more quickly
than expected, so the potentiometer can be used to adjust the time
delay. You can get LM324 chips from Radio shack (it's an opamp). There
are 4 opamps on it, so if you use that chip, make sure you disable the
other opamps by tying the - leads to 9V, and the + leads to 0.
Otherwise, it might cause some kind of oscillations. If you can get a
comparator instead, you may need what's called a 'pull up' resistor on
the output. Look at the package. If so, a 1k resistor from the output to
9V (after the swtich, of course) will do the job.
The P-MOFSET should have an off voltage of something like 3V, since the
LM324 will only get up to about 7V, and it should be completely off. The
ON voltage of the P-MOFSET must be less than 9V. My guess is that all of
the P-MOFSETs you can get at radioshack will fall into this category.
The voltage at the - input of the opamp will be near
V = 9 * exp(-t/RC).
Where t is the time after the switch is closed.
You can probably get the parts for a few dollars.
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.