Discuss What power rating would this garage door be? in the Electrical Forum area at ElectriciansForums.net

happyhippydad

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I have a roller garage door that my friend has fitted which i've noticed is connected to a socket to give it power, but then has a flood light attached to the control panel which is also connected to the lighting circuit. In other words you need to isolate sockets and lights to make sure the control panel is properly isolated.

I would like to put this garage door on the lighting circuit. I have rang the garage door company and they said that the power was 8W so clearly they have no idea. It is a galvanised door approx 3m x 3m. Direct drive.

Does anyone have a rough idea of what power these types of motors would be?

Cheers in advance :)
 

telectrix

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look on the motor. it should have a rating plate. it will be a FHP motor so will be under 5A, should be OK on the lighting circuit unless the start surge requres a type C breaker.
 

Pete999

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Fit a FCU to the socket and feed the lo from that.
 

westward10

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If the lighting circuit is wired using 1.0 cables then it is inadequate for a power circuit. See Table 52.3
 

Sparkyboy

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You should fit a FCU fed from the socket fitted with a 5A fuse, I recently wired a garage with two roller shutter doors.
 

Spoon

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I'm with @Pete999 post #3.
Out of curiosity, what size fuse is in the plug?
You would get one of these that could tell you how much it is drawing?
 

Lucien Nunes

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Electrical behaviour and starting load will depend on whether it's an AC or DC motor as both are used, screw or chain drive, and on how well the door is counterweighted. I would expect it to be in the range of 250-500W when running, but would allow 1kW / 5A for starting. Why do you want to run it from the lighting circuit?
 
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happyhippydad

happyhippydad

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Electrical behaviour and starting load will depend on whether it's an AC or DC motor as both are used, screw or chain drive, and on how well the door is counterweighted. I would expect it to be in the range of 250-500W when running, but would allow 1kW / 5A for starting. Why do you want to run it from the lighting circuit?
At present the control panel is fed from a socket. In the control panel there are terminals for a light to be wired into it, which it has. This light is also connected to the lighting circuit (so even if the sockets were off the S/L in the control panel could become live if the light was switched on). It would be a right hassle to take the light out of the lighting circuit, so I thought to put the garage door on the lighting circuit so it can be isolated correctly.
 

Lucien Nunes

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Are you saying it is energising the light (that is part of the lighting circuit) from the line obtained via the plug, or just that it closes an electrically separate contact in parallel with the light switch to operate the light? The former case is wrong and needs to be fixed as the two circuits are not electrically separate - in theory the plug pins could become live too, from the lighting circuit. If the cable run to the light or switch can have PL and SL connected, then this problem can be solved with a small relay in the panel.

The latter case is electrically fine, merely needs a label on the front identifying the two sources of supply. You would need to isolate the lighting circuit anyway so pulling out the plug is hardly an extra complication. I know it's not so common in domestic situations but in industrial plant it's quite normal to have multiple circuits within a panel.
 
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happyhippydad

happyhippydad

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Are you saying it is energising the light (that is part of the lighting circuit) from the line obtained via the plug, or just that it closes an electrically separate contact in parallel with the light switch to operate the light? The former case is wrong and needs to be fixed as the two circuits are not electrically separate - in theory the plug pins could become live too, from the lighting circuit. If the cable run to the light or switch can have PL and SL connected, then this problem can be solved with a small relay in the panel.

The latter case is electrically fine, merely needs a label on the front identifying the two sources of supply. You would need to isolate the lighting circuit anyway so pulling out the plug is hardly an extra complication. I know it's not so common in domestic situations but in industrial plant it's quite normal to have multiple circuits within a panel.
This is what I have Lucien...
garage door.jpeg
 

ipf

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From what I can make out, you have two supplies at the dp switch. Needs a couple of relays controlling the light, one for manual and one for the sensor, I'd say.
 

Lucien Nunes

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That looks very naughty! It's tying the two circuits together and could make the plug live. Where you have a DP switch, replace with one SPST relay with 230V AC coil and 5A rated contacts. Connect the coil to the 'lamp output' from the door opener, and the contacts between the two SL's (lamp direct and via sensor) No wiring alterations needed. Observe electrical separation between the coil circuit and the contacts.

The door opener will however only energise the light when the PIR sensor is also enabled at the 2g switch. If you want the opener to have independent control of the light, you will need to get a PL to the relay from the 2g switch, I don't know if that's adjacent or distant. Optionally bypass the DP switch and put the relay adjacent to the 2g switch.
 
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Lucien Nunes

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For a light?
 

Wilko

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Hi - sorry to say, but I'd junk all that and attach a small LED fitting directly to the light terminals. That way the 2 circuits remain separate and the new LED light comes on for a timed period as the door opens or closes (guessing at what the board logic is...) :) .
 

ipf

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Just thinking. Why not just remove the light circuit supply and pick up from the dp switch? 3 wires to the override and sensor if no neutral required.
 
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happyhippydad

happyhippydad

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That looks very naughty! It's tying the two circuits together and could make the plug live. Where you have a DP switch, replace with one SPST relay with 230V AC coil and 5A rated contacts. Connect the coil to the 'lamp output' from the door opener, and the contacts between the two SL's (lamp direct and via sensor) No wiring alterations needed. Observe electrical separation between the coil circuit and the contacts.

The door opener will however only energise the light when the PIR sensor is also enabled at the 2g switch. If you want the opener to have independent control of the light, you will need to get a PL to the relay from the 2g switch, I don't know if that's adjacent or distant. Optionally bypass the DP switch and put the relay adjacent to the 2g switch.
Could it definitely?
The only time the light comes on from the control panel is when the garage door is being opened. So at this point the PL in the control panel makes continuity with the SL of the light in the panel, like a switch. However if the plug was removed then the door could not be opened, therefor the control panel would not make the connection between Pl and SL. If the light was then switched on manually would its pathway not just end at the SL in the panel?

Although if the door was already opened and the plug was then removed would there still be contact between PL and SL in the panel?
 
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happyhippydad

happyhippydad

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Just thinking. Why not just remove the light circuit supply and pick up from the dp switch? 3 wires to the override and sensor if no neutral required.
But there is only a light circuit supply to the DP switch? The only connection the DP switch has with the panel is the SL and N for the light output.
 

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