User_88,
The input side of a SSR is basically an IR diode so it would operate say at 1.5V across its terminal. Therefore, one needs to use say 220 ohm series resistor to limit the current when connect it to 5V.
So let say the charging resistor is 220 ohm then. Can you estimate what value of capacitor to use in order that the capacitor voltage will reach 1.5V after 10 seconds?
Not all solid state relays are the same .....
I am suggesting the use of a DC voltage controlled SSR...
something similar to the ones shown in the link here:
**broken link removed**
It looks like the input control circuit has a little more to it than just an IR diode.
The relevant question here is at what voltage does the control voltage turn on the relay....
Assume the turn-on voltage is 3 VDC.
Assume also, that no significant current is required to turn on the SSR .... that it is essentially a voltage controlled device.
The first consideration would be to determine how much current to allow through the main resistor, R. ... A value of R=10k ohms would allow 0.5 ma to charge the Capacitor... using a 5 VDC supply. This relative magnitude prevents a large current from flowing during the 'float switch closed' condition.
From the standard formulation for charging a capacitor through a resistor:
a supply of 5VDC,
a desired period
τ of 10 seconds to reach 3 VDC,
That is, an elapsed time interval of 1 time constant.
... The reasoning here is that 3/5=0.6, which is a duration of about 1 time constant on the exponential RC charging curve.
and utilizing the 10k resistor.
Therefore,
τ=RC, or:
10 sec= 10k ohms * C farads .... solving ... C=1000
µF
That is, at t=10 sec, using the specified values of R= 10k ohms and
C= 1000
µF, the capacitor voltage should be approximately 3 VDC,
and the solid state relay should turn on.