Re: Light sensitive relay switch alternative?


Andy Turner wrote:
On Sun, 24 Jul 2005 15:42:52 GMT, Kristian Ukkonen
Andy Turner wrote:
Well sorta.. the problem is with the switch implementing a "delay
circuit to avoid cycling", which I guess I should have spotted when
ordering it.. So this means that it's simply not fast enough for the
application. I presume the delay is implemented in the chip so there'd
be nothing I could do about removing it? Or does the capacitor do the
delay (smaller capacitor, smaller delay?). 

Yes, make the capacitor at + input of opamp smaller
for a smaller delay.

Cheers! I've noticed that it takes longer to switch one way than back
again. Also, I notice that there's two capacitors, one 100uf and one
470uf. Is this implementing the two different delays? What would
happen if I took the capacitors off and just wired direct? Would that
give me no delay or would it stop working or blow something?


Do notice: "CAPACITOR AT + INPUT OF OPAMP"

The other capacitors, according to circuit diagram at
the webpage selling the circuit, are just for powersupply.
http://www.maplin.co.uk/Media/product_pdfs/qp97.pdf

I apologise for not knowing too much about what must be GCSE level
electronics. I do very much regret opting for a different subject at
school..!


Schools are just for realizing that by reading books
you can find out ANY information. For learning electronics,
find a book on electronics, read it and start doing stuff.
You can test circuits with free switcher cad 3 download:
http://ltspice.linear.com/software/swcadiii.exe
Most simple electronics, like this circuit, are very easy
to design - it is just resistive voltage dividers and basic
stuff like that.

One resistive divider, 100k potentiometer, gives the
comparator switch level to - input.

Another resistive divider 150k, VDR, 15k gives the
other voltage. There is, in addition, a "delay" circuit
by the 17k and 180k resistors, diode and 100uF capasitor.

When LDR has high resistance (dark), voltage is high and the
100uF capacitor charges slowly to this high voltage (and some
current goes through 100k but to simplify forget this now).
When LDR resistance drops (light), diode prevents the lower
voltage from charging the capasitor, and capasitor discharges
through the 100k resistor until eventually voltage is
lower than the voltage set by the 100k potentiometer, and
the state of the relay changes.

Now, when LDR again has high resistance (dark), the 100uF
cap is starting to charge and voltage eventually becomes
higher than the voltage from 100k pot, and the state of
relay changes again.

The opamp (the triangle part with inputs marked "-" and
"+" is a comparator - it compares whether - or + input
has a higher voltage, and changes output accordingly.

LDR = light dependant resistor

I hope this helps you understand how the circuit works.
You can enter the circuit to switcher cad 3, and replace
LDR with a switch.. See how it works!

To answer your original question: Only the 100uF capacitor
matters for the delay. Take it away and you have no delay.
I'd suggest setting a little delay. Simulate the circuit
with Switcher cad 3, and find out correct size for cap!!

.

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