could any one explain the problem in this circuit

[akourotix]

hi,

i need a neon lamp to be on when any of parallel buttons pressed .
i test circuit 1 it works as i need.

but if i used buttons with their indication lamps as in circuit 2 the neon lamp always on even any buttons not pressed.

could anyone explain the problem ?and how can i solve it ?

 

[gophert]

The current from the three indicators have enough current in total to lite the single lamp.
The indicators shouldn't be in series with the lamp, they should be in parallel as follows...

(Remove connections covered in red, add those in black.)

 

[ronsimpson]

With all switches open you will have current in R2 and bulb. and (1/3 current in R3) + (1/3 current in R4) + (1/3 current in R5 and bulb)

Almost true: In circuit 2 you have R2 bulb, R3 bulb. This puts 1/2 line voltage on each bulb.

Is true: Circuit is R2---bulb---(R3 || R4 || R5) (R3 and bulb) is parallel with (R4 and bulb) is parallel with (R5 and bulb)

 

[akourotix]

thank you ..

the problem is i cant change the indication lamp connection as it is included inside the button and only two wire available at the point where the button placed and there more than 15 buttons (the distance from neon lamp to buttons (5m-30m)) which make adding the third wire difficult to me ..

so can i solve the problem without changing the indication lamp connection ?

 

[Les Jones]

If you put a load of a suitable value in parallel with the common neon then the voltage due to the small current leakage from the neon/resistor in the push will not be enough to give a voltage across it will not be enough to light the common neon. What is the value of the resistor in each push button and what is the number of push buttons ? (More than 15 could be 1000 !) This sounds like a number of buttons next to doors in a building that switches on lighting in corridors. If I am correct then there must be something else connected to the common neon to trigger a timer. Can you give some more details The load could be a resistor but that would generate some heat. A suitable capacitor could be used and that would not generate any heat.

Les.

 

[alec_t]

the small current leakage from the neon/resistor in the push

Working on the assumption that it is about 1mA per push button (neon indicators for UK mains typically have ~220k in series with the bulb), the total current would be at least 15mA.

 

[akourotix]

What is the value of the resistor in each push button and what is the number of push buttons ? (More than 15 could be 1000 !)

.

i cant see the resistor i only see the terminal of the buttons and i measure the voltage across one button when it not pressed and it was 160V.

i count the buttons it was 30 .

This sounds like a number of buttons next to doors in a building that switches on lighting in corridors.
.

yes,i want to get a signal to mcu using ldr to turn lamps on for a certain time

 

[Les Jones]

If we assume that the resistor in each button is 220K (Which I think is a reasonable figure.) then 30 of them in parallel is 220k/30 = 7.3K The peak voltage of 220 V RMS is 220 x 1.414 = 311 volts If we assume the voltage across the neons in the push buttons is 60 volts then would have 251 volts (peak) minus the maximum voltage we can allow across the common neon without it striking, If we assume this is 50 volts then we have 201 volts across the resistors in parallel. 201/7300 = 28 mA (Peak) so we need a load resistor of 50/0.028 = 1.8K. When one of the buttons is pressed there will be 220 volts RMS across this resistor so it will dissipate 220^2/1800 = 27 watts. I think using a relay or small transformer would be a better way to provide isolation than the neon/LDR method.

Les.

 

[akourotix]

If we assume that the resistor in each button is 220K (Which I think is a reasonable figure.) then 30 of them in parallel is 220k/30 = 7.3K The peak voltage of 220 V RMS is 220 x 1.414 = 311 volts If we assume the voltage across the neons in the push buttons is 60 volts then would have 251 volts (peak) minus the maximum voltage we can allow across the common neon without it striking, If we assume this is 50 volts then we have 201 volts across the resistors in parallel. 201/7300 = 28 mA (Peak) so we need a load resistor of 50/0.028 = 1.8K. When one of the buttons is pressed there will be 220 volts RMS across this resistor so it will dissipate 220^2/1800 = 27 watts. I think using a relay or small transformer would be a better way to provide isolation than the neon/LDR method.

Les.

what is the type of the resistor
i appreciate your effort

i used 220v coil relay method and optocoupler method i have the same problem .
the same solution for neon\ldr method can apply on relay or optocoupler method ?

how can i use transformer ?the same problem exist if i use it ?

thank you .

 

[Les Jones]

I agree that you would need a load of some sort unless the relay was a large contactor that required quite a large current to pull it in. Using a transformer you could probably use a higher value load resistor (Which would make the neons in the push buttons dimmer.) but you could rectify the secondary and have an adjustable threashold detector. If you used a capacitor as a load you would need about 2 uF (It would need to be X or Y rated for this sort of application.

Les.

 

[akourotix]

I agree that you would need a load of some sort unless the relay was a large contactor that required quite a large current to pull it in. Using a transformer you could probably use a higher value load resistor (Which would make the neons in the push buttons dimmer.) but you could rectify the secondary and have an adjustable threashold detector. If you used a capacitor as a load you would need about 2 uF (It would need to be X or Y rated for this sort of application.

Les.

the value of capacitor depends on the number of buttons ? using capacitor suitable for all method ?
and 400v capacitor enough ?

regards

 

[Les Jones]

The capacitors must be at least 250 volts AC rating and X or Y class (This specifies how they behave when they fail.) Contact supressors**broken link removed**could be used. As these are 0.47 uF you would need four of them in parallel to give a reactance of about 1.8 K. The fact that they have a built in series resistor is an advantage as it would reduce the large current pulce if the button is pressed near the crest of the waveform.

Les.

 

[ronsimpson]

I don't understand what you are trying to do. Here I added a small light bulb. Now the "neon lamp R2" will be off when all switches are off. AND All switch lights will be on.

 

[akourotix]

The capacitors must be at least 250 volts AC rating and X or Y class (This specifies how they behave when they fail.) Contact supressors**broken link removed**could be used. As these are 0.47 uF you would need four of them in parallel to give a reactance of about 1.8 K. The fact that they have a built in series resistor is an advantage as it would reduce the large current pulce if the button is pressed near the crest of the waveform.
Les.

I don't understand what you are trying to do. Here I added a small light bulb. Now the "neon lamp R2" will be off when all switches are off. AND All switch lights will be on.
View attachment 114026

thank you

what is the best ? using resistor or capacitor or small light bulb ?

regards

 

[akourotix]

hi ,

using protective elements is a must with this kind of circuit ??!

 

[Les Jones]

The OP has taken the thread to the conversations section. This breaks the continuity of the thread.
This is his last question in the conversation section

"so if i use relay instead (neon-ldr) i should open the circuit of the buttons to overcome heat problem ..?
is there any other method better than use (neon-ldr) and relay methods ?
what about use transformer ?

regards"

The idea of using the timer output to remove the power to the push buttons will work with any of the suggested loads. (Resistor, lamp, capacitor, or contactor that takes a large enough current so it does not pull in due to the current through the neons in the push buttons.) (I only thought of that idea yesterday to get round the heat produced with a resistive load if a push button was pressed for a long time.) I would personaly use the transformer method for isolation with an X or Y rated capacitor for the load. In practice I would use four 0.47 uF "contact suppressor capacitors" connected in parallel. The transformer method makes it easier to set the switching threashold from the DC voltage (After rectification) from the low voltage secondary. (12 or 24 volts) A comparitor circuit or a small relay with a zener diode in series with it's coil could be used.

Les.

 

[akourotix]

you mean like this ?

what is the purpose of comparator ckt?

 

[Les Jones]

Almost right. You would only want the trigger signal when the voltage across the primary was close to 220 volts. So if the transformer had a 12 volt secondary the voltage across C2 would be about 15 volts when the primary voltage was 220. (12 x 1.414 - (2 x 0.7)) You would need a potential divider to drop the 15 volts down to 4 or 5 volts otherwise you would not be able to set it trigger above about 75 volts on the primary. You could probably get away with using a 1 uF capacitor (Or a 3.6 K resistor which would reduce the power disipation to 13 watts when a button was pressed.) This would slightly reduce the brightness of the neons in the push buttons.

Les.

 

[akourotix]

You could probably get away with using a 1 uF capacitor (Or a 3.6 K resistor which would reduce the power disipation to 13 watts when a button was pressed.) This would slightly reduce the brightness of the neons in the push buttons.

Les.

You mean use another capacitor or reduce C to 1uf ?

Ak.

 

[MrAl]

hi,

i need a neon lamp to be on when any of parallel buttons pressed .
i test circuit 1 it works as i need.

but if i used buttons with their indication lamps as in circuit 2 the neon lamp always on even any buttons not pressed.

View attachment 114008

could anyone explain the problem ?and how can i solve it ?

Hi,

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