could any one explain the problem in this circuit

[Les Jones]

With the transformer isolation method (As opposed to your neon /LDR opto isolator.) The voltage across the load resistor (Or capacitor.) due to the sum of the currents of the neons in the push buttons can be allowed to go to more than the 50 volts or so that it has to be limited to so the neon in the opto isolator does not light. So we could double the value of the load resistor (Or half the value of the load capacitor as this would double it's reactance.)

Les.

 

[akourotix]

Hi,

Can you do as in the attachment, connect a capacitor like that?

View attachment 114163

yes.

With the transformer isolation method (As opposed to your neon /LDR opto isolator.) The voltage across the load resistor (Or capacitor.) due to the sum of the currents of the neons in the push buttons can be allowed to go to more than the 50 volts or so that it has to be limited to so the neon in the opto isolator does not light. So we could double the value of the load resistor (Or half the value of the load capacitor as this would double it's reactance.)

Les.

thank you and sorry for bothering you :*

does the transformer output current{ eg..12v/0.2A)||(12/1A) } affect on the circuit ?

 

[MrAl]

yes.

thank you and sorry for bothering you :*

does the transformer output current{ eg..12v/0.2A)||(12/1A) } affect on the circuit ?

Hello,

If you get an AC rated capacitor then you can do it that way. The cap has to be AC rated and should have a nice high voltage rating.
Also a small resistor in series like 100 ohm would help to lower surge current when the switches are closed thus allowing the switches to last longer.
We could calculate a value.

 

[Les Jones]

You will only be drawing about 15 mA from the tranformer so the 0.2 amp rating will be OK. It would not matter if it was a higher current rating. Also any secondary voltage from about 5 volts to 24 volts will do. You just need to adust the potential divider ratio to the comparitor input.

Les

 

[akourotix]

Hello,

If you get an AC rated capacitor then you can do it that way. The cap has to be AC rated and should have a nice high voltage rating.
Also a small resistor in series like 100 ohm would help to lower surge current when the switches are closed thus allowing the switches to last longer.
We could calculate a value.

You will only be drawing about 15 mA from the tranformer so the 0.2 amp rating will be OK. It would not matter if it was a higher current rating. Also any secondary voltage from about 5 volts to 24 volts will do. You just need to adust the potential divider ratio to the comparitor input.

Les

thank you

can i test the circuit without using 30 buttons(if there is a way to use equivalent ckt of 30 buttons with their indication lamp) ...?!

 

[ronsimpson]

can i test the circuit without using 30 buttons

I don't know what your neon bulbs are like.
110VAC or 220vac ???
I don't know how much current flow. ???

We probably know what the voltage is across the bulb.
A standard bulb will start to turn on at 65 to 90 volts. (every bulb is different) When on its voltage is about 30% less.
A high bright bulb will fire at 95 to 135 volts and stay on at 30% less.
We don't know what kind of bulb so lets say standard bulb and it voltage will be 45 to 63v,

What resistor???
Most standard bulbs (110VAC bulbs use 100k or 220k resistors) & (220vac bulbs use 220k to 540k).
For high bright (110V use 22k, 30k. 39k. 47k) & (220V use 68k, 100k, 150k)
We don't really know that resistor. Likely 220k but who knows.

To simulate 30 bulbs in parallel:
You need to reduce the line voltage by 45 to 60 volts to simulate the voltage loss in a bulb.
Then you can use the equivalent of thirty 220k resistor in parallel. 1 bulb = 220k, 10 bulbs = 22k, 20 bulbs = 11k .......

----edited----
Much of the problem with your circuit is that 30 bulbs in parallel produces much current.

I think your circuit will not function for several reasons. The one I am think of right now is:
With all switches open, the main bulb will be on and only 2 or 3 of the thirty bulbs will be on.
Why ????
With only two bulbs, each bulb will have 1/2 the line voltage which will turn on both.
With three bulbs, the main is on and very likely the other two will get 1/2 current.
As you add more and more bulbs the voltage on the main bulb goes up and the voltage on the many bulb in parallel will drop as they share the current. The voltage on the many bulbs in parallel will fall so low that some of the bulbs will not fire. With the "fire voltage" being from 65 to 90 volts the 65V bulbs will turn on first and suck the voltage down so the 90v bulbs will not fire. Likely your bulbs will be in the 70 to 80V range but I think with 30 bulbs some/many of the bulbs will not start up.

 

[akourotix]

thank you ronsimpson

220v bulbs with 100k resistors and no other info about them


____________________________________________

Much of the problem with your circuit is that 30 bulbs in parallel produces much current.

I think your circuit will not function for several reasons. The one I am think of right now is:
With all switches open, the main bulb will be on and only 2 or 3 of the thirty bulbs will be on.
Why ????
With only two bulbs, each bulb will have 1/2 the line voltage which will turn on both.
With three bulbs, the main is on and very likely the other two will get 1/2 current.
As you add more and more bulbs the voltage on the main bulb goes up and the voltage on the many bulb in parallel will drop as they share the current. The voltage on the many bulbs in parallel will fall so low that some of the bulbs will not fire. With the "fire voltage" being from 65 to 90 volts the 65V bulbs will turn on first and suck the voltage down so the 90v bulbs will not fire. Likely your bulbs will be in the 70 to 80V range but I think with 30 bulbs some/many of the bulbs will not start up.

even shunt capacitor with main bulb used ?

now i will go to use transformer instead of (neon lamp/ldr) the same problem exist if i use transformer ?

 

[ronsimpson]

I think the problem(s) is that the main bulb is very sensitive. It will light with 0.1mA, not bright but on.
Next the remote bulbs with 100k will send 1mA at 110V and 2mA at 220V when open. (1mA each)

These switches were designed to switch a load. (old style light bulb) An old bulb will not light at 1mA. The bulbs inside the switch will not work unless there is a good load.

What to do:
>I think you want all 30 remote bulbs on and the main bulb off when all switches are open. When one or more switches are on then the main bulb is on and the remote bulbs all go off.
To do this you will need a load across the main bulb to pull the voltage down below 50 volts to keep the main lamp off and to keep the remote bulbs on.
>30 100k resistors are about the same as 3k. So adding a 3k across the main bulb will cause the voltage to be 1/2 line. So we need a lower resistor. A 1k will put 1/4 line on the main bulb. I think we need to think peak voltage not average voltage so peak at 220 is 320. 320/4 is about 80 volts. That is enough to light the main bulb. Wow, I thought 1k would work but it will light main dimly. I now think the load needs to be around 700 ohms. (math in my head, you better check it)
>From the start I think you need a load like a small old style light bulb that pulls the main voltage down to 30 volts or lower while using 60mA.

 

[Les Jones]

akourotix,
You need to be much more specific when asking questions. In post #27 you ask Ron "now i will go to use transformer instead of (neon lamp/ldr) the same problem exist if i use transformer ?" You need to explain HOW you plan to use a transformer. You should have posted a corrected version of the schematic that you posted in post #17 which showed the primary and secondary voltages and details of the potential divider that I suggested. (You could have also referred back to post #17 and described the changes that would be needed.)

Les.

 

[alhareth]

What i think if you can control the switching threshold (light intensity ) on ldr side ?!

I don't know if this can help .