From: gary@ke4zv.uucp (Gary Coffman)
Newsgroups: sci.electronics
Subject: Re: question about RC circuits for timing purposes
Date: 8 Apr 92 08:35:58 GMT
Reply-To: gary@ke4zv.UUCP (Gary Coffman)
Organization: Gannett Technologies Group

In article <13-PCNews-124beta@wendy.bowlgreen.oh.us> jmiller@wendy.bowlgreen.oh.us (Jim Miller) writes:
>Hello everyone,
>    I am trying to rig up a 12 volt DC circuit that will introduce a time
>delay.  In general, here's what I want to do:
>
>                 +--------+                 +--------+
>   + 12V --------| delay  |------ + 12 -----|  12VDC |-----  relay contacts
>                 | circuit|                 |  relay |
>                 |        |                 |        |-----
>   - ------------|        |------ neg ------|        |                     
>                 +--------+                 +--------+
>                 
>
>For the delay circuit, I had planned to use a a resistor and capacitor 
>in series with the positive side of the relay, and adjust the size of the
>resistor accordingly.  That is, there would not be enough voltage
>passing through the RC circuit during charging to energize the coil, but
>once the time period had passed, the relay would pick up.
>
>
>  + 12 -------/\/\/\-----||------- relay coil
>              ? ohms    ? uf

Since no DC current could pass through the capacitor, this circuit won't
do what you want. What you want is really this:

  +12------/\/\/\----x---x
             R       |   |
                    ---  \
                 C  ---  / RL
                     |   \
                     |   /
  ___________________|___|
                    
Now for the relay to pickup after being delayed by C, the coil resistance 
RL has to be much larger than the charging resistor R. Let's take the 
engineering rule of thumb of 10x. So if we have a very sensitive relay 
with a coil resistance of 5000 ohms, then R can be no larger than 500 ohms. 
Since one time constant charges the capacitor to 63% of the supply voltage,
two TC will bring the voltage up to the pullin voltage of the relay.
For a one second delay that works out to a C of about 1000 uf. When
power is removed, RL discharges C to 1/3 it's voltage in one TC or
about 5 seconds. Now to be accurate, you need the actual pullin and
dropout voltages of the relay, and you should allow for the 1/10 discharge
that is going on through RL while the capacitor is charging through R,
but you should have the idea by now.
              
>This circuit does not have to be very accurate at all; I just need to
>introduce a second or so delay between the input to the circuit and
>the time the relay picks up. Also, the circuit will not need to be operated
>very often, so it wouldn't need to recover quickly when the +12 is taken
>away.
>
>If this is indeed a proper solution to the problem, how should I
>choose the values of R and C?  Will I have a problem with the capacitor
>holding the relay up long after the +12 is taken away? 

Well, if you can tolerate a 5 second delay, you should be OK. You can get
a fast release by using an extra set of contacts on the relay to remove
the capacitor from the circuit and connect it to a discharge resistor after
the relay pulls in. If the relay you need to drive is not a sensitive type, 
you'll have to use a *very* large capacitor. In that case, you'd want to go 
to an electronic delay driver using a FET switch to drive the coil.

Gary