Design current?

[Newtothis]

Hey new to the forum

And learning to be an electrician.

I have a few questions about voltage drop.

I understand the formula and all that

for example if I had 25 led lights at 24w each spaced evenly over a distance of 65m fed off a type c 10 amp mcb

would the design current (lb) used in the formula be 10 amps or would it be the 24watts x 12 = 288watts and in amps that is 1.25A

so would my formula us the breaker size of 10 amps or the load of the lighting circuit for voltage drop?

so if it was in 1.5mm it would be
mV/A/m x IB x L /1000
29 X 1.25 x 65 /1000 = 2.3 volts

or would it be
29 x 10 x 65/1000 =18.8 volts

 

[Lucien Nunes]

Hello and welcome, do put up an intro post when you have a chance. You might like to request trainee access from the mods too.

The design current is whatever you, as the designer, want it to be. If your lights consume 2A and you don't envisage any other lights ever being used on the circuit, you could choose your cable for VD based on 2A regardless of whether the MCB is 6A or 10A etc. For a regular indoor lighting circuit, you would be cursed if someone came along to add another light and found that the cable was already on the limit for VD and that a new cable would need to be run. OTOH if the circuit was 300 metres long and the load unlikely ever to be increased, you might reasonably stick to the 2A to avoid unnecessary expense on the cable. For a circuit where you have no control over the amount of load up to the In rating of the OCPD e.g. a general purpose socket-outlet circuit, you should allow for the full circuit rating in your VD calcs.

Minor point, perhaps this is what you have been advised but I would show the working for the drop differently. You have calculated a current of 1.25A to use in the VD equation based on half the real number of lights, because they are distributed along the length of the cable. I would do it the other way, and show the real total current (25 x 24/230) but apply it to half the length of the cable. Just seems more logical.

 

[pc1966]

for example if I had 25 led lights at 24w each spaced evenly over a distance of 65m fed off a type c 10 amp mcb

would the design current (lb) used in the formula be 10 amps or would it be the 24watts x 12 = 288watts and in amps that is 1.25A

Firstly I would be very surprised in almost all cases of either 10A or a C-curve MCB being justified.

Yes, LED lights have a significant switch-on surge and worse with simultaneous switched sets (large office, lighting for shop display, etc) but you are still only talking 288W worth, not kW region.

For now the IET procedure still assumes 100W per fitting, even though 100W lamps are pretty much history, so when designing a generic circuit with user-replaceable lamps that is the assumption. It may be obsolete, but it is still used for diversity planning, etc.

But for specific systems you would use the expected current (i.e. your selection of lights, not the remote possibility that someone replaces every lamp with a 100W bulb they had hidden in the basement for the last decade or so).
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Ah, just seen that Lucien Nunes has replied with useful info as well!

 

[Newtothis]

Hello and welcome, do put up an intro post when you have a chance. You might like to request trainee access from the mods too.

The design current is whatever you, as the designer, want it to be. If your lights consume 2A and you don't envisage any other lights ever being used on the circuit, you could choose your cable for VD based on 2A regardless of whether the MCB is 6A or 10A etc. For a regular indoor lighting circuit, you would be cursed if someone came along to add another light and found that the cable was already on the limit for VD and that a new cable would need to be run. OTOH if the circuit was 300 metres long and the load unlikely ever to be increased, you might reasonably stick to the 2A to avoid unnecessary expense on the cable. For a circuit where you have no control over the amount of load up to the In rating of the OCPD e.g. a general purpose socket-outlet circuit, you should allow for the full circuit rating in your VD calcs.

Minor point, perhaps this is what you have been advised but I would show the working for the drop differently. You have calculated a current of 1.25A to use in the VD equation based on half the real number of lights, because they are distributed along the length of the cable. I would do it the other way, and show the real total current (25 x 24/230) but apply it to half the length of the cable. Just seems more logical.

I think i meant to use 24 instead of 12 xD
Ah cool so my circuit would comply with VD with a drop of 2.3volts?

im just confused because in the onsite guide it says you can have a 90 meter circuit on a 6 amp breaker and a 35 meter circuit on a 10 amp breaker but when you work out the VD they don't comply

 

[Newtothis]

Firstly I would be very surprised in almost all cases of either 10A or a C-curve MCB being justified.

Yes, LED lights have a significant switch-on surge and worse with simultaneous switched sets (large office, lighting for shop display, etc) but you are still only talking 288W worth, not kW region.

For now the IET procedure still assumes 100W per fitting, even though 100W lamps are pretty much history, so when designing a generic circuit with user-replaceable lamps that is the assumption. It may be obsolete, but it is still used for diversity planning, etc.

But for specific systems you would use the expected current (i.e. your selection of lights, not the remote possibility that someone replaces every lamp with a 100W bulb they had hidden in the basement for the last decade or so).
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Ah, just seen that Lucien Nunes has replied with useful info as well!

the expected current being the load you will install onto the circuit and not the rating of the mcb
So a better breaker even with a bank of 12 lights would be 6 amp type B?

Also thanks for both your replies

 

[telectrix]

it's more complex that a simple calc. say you have 10 lights. you calculate the VD to the 1st light, using the flc.and the length to the 1st light, then from 1st to 2nd using 90% FLC. then from 2nd to 3rd using 80% FLC.............. from 9th to 10th, current is ony 10%. i.e. the current draw of 1 light.

 

[pc1966]

As telectrix says, it is a bit more complicated in both ways.

Usually for light circuits the assumption is there are a lot of lights evenly spread out over a building, etc, so the simplest assumption is the drop is half of that if all lights were at the far end (as also said by Lucien Nunes). But if you have a long feed and a bunch of lights at the far end of it you might need to compute the VD in two steps, one for feeder, other for "half length" on the distribution bit (assuming it is also long-ish).

For MCB sizing it is more difficult as the inrush current on LED drivers is often very high (tens of amps) but only for fraction of a millisecond as a few uF of capacitance is charged up, and the normal response time of the magnetic trip of a MCB is in the couple of millisecond region.

So while a 6A B-curve MCB would definitely fire on a 10ms half-cycle overload of 5*6A = 30A, it might take 300A to fire it in 0.1ms. Not many manufacturers give with the LED inrush characteristics, nor the MCB response to very short overloads.

Some examples are given here:

Learning Center - Application note - How to choose the right circuit breaker | eldoLED

When switching on an LED lighting system and nothing happens, the inrush currents of the LED driver triggered the circuit breaker. Read more.

www.eldoled.com

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Based on the eldoled example, your 288W total LED example is marginal for a 6A B-curve MCB if simultaneously switched, so it may be a case for a C-curve here.

Of course if you are computing for a house with 25 LED lamps that are largely independently switched you won't have the massive surge of all charging simultaneously and a 6A B MCB/RCBO would be fine.

 

[Newtothis]

it's more complex that a simple calc. say you have 10 lights. you calculate the VD to the 1st light, using the flc.and the length to the 1st light, then from 1st to 2nd using 90% FLC. then from 2nd to 3rd using 80% FLC.............. from 9th to 10th, current is ony 10%. i.e. the current draw of 1 light.

So if i had 12 led lights on 1.5mm twin and earth one circuit of 24w each and the circuit was say 90 meters and each light was evenly spaced at 7.5 meters

so the VD would be something like this at each point of the circuit?

1 24w/230 = 0.1a (29 x 0.1x7.5)/1000 = 0.02v
2 48w/230 = 0.2a (29 x 0.2x7.5)/1000 = 0.04v
3 72w/230 = 0.3a (29 x 0.3x7.5)/1000 = 0.068v
4 96w/230 = 0.41a (29 x 0.41x7.5)/1000 = 0.09v
5 120w/230 = 0.52a (29 x 0.52x7.5)/1000 = 0.11v
6 144w/230 = 0.6a (29 x 0.6x7.5)/1000 = 0.13v
7 168w/230 = 0.73a (29 x 0.73x7.5)/1000 = 0.15v
8 192w/230 = 0.83a (29 x 0.83x7.5)/1000 = 0.18v
9 216w/230 = 0.93a (29 x 0.93x7.5)/1000 = 0.2v
10 240w/230 = 1.1a (29 x 1.1x7.5)/1000 = 0.22v
11 264w/230 = 1.14a (29 x 1.14x7.5)/1000 = 0.24v
12 288w/230 = 1.2a (29 x 1.2x7.5)/1000 = 0.27v

giving me a VD of the sum of the volts : 1.71 volts?
or would it just be 0.27v

 

[telectrix]

your first light feed will draw the full current of all lights. the second feed will draw less (11/12 of total.) then 3rd will be 10/12 of total. etc., etc. so when you get to the last light, that will be your total VD.

 

[Newtothis]

so was what I did right just 12 needs to be 1?
Source URL: Design current? - https://www.electriciansforums.net/threads/design-current.188475/

 

[Newtothis]

your first light feed will draw the full current of all lights. the second feed will draw less (11/12 of total.) then 3rd will be 10/12 of total. etc., etc. so when you get to the last light, that will be your total VD.

so was what I did right just 12 needs to be 1?

12 24w/230 = 0.1a (29 x 0.1x7.5)/1000 = 0.02v
11 48w/230 = 0.2a (29 x 0.2x7.5)/1000 = 0.04v
10 72w/230 = 0.3a (29 x 0.3x7.5)/1000 = 0.068v
9 96w/230 = 0.41a (29 x 0.41x7.5)/1000 = 0.09v
8 120w/230 = 0.52a (29 x 0.52x7.5)/1000 = 0.11v
7 144w/230 = 0.6a (29 x 0.6x7.5)/1000 = 0.13v
6 168w/230 = 0.73a (29 x 0.73x7.5)/1000 = 0.15v
5 192w/230 = 0.83a (29 x 0.83x7.5)/1000 = 0.18v
4 216w/230 = 0.93a (29 x 0.93x7.5)/1000 = 0.2v
3 240w/230 = 1.1a (29 x 1.1x7.5)/1000 = 0.22v
2 264w/230 = 1.14a (29 x 1.14x7.5)/1000 = 0.24v
1 288w/230 = 1.2a (29 x 1.2x7.5)/1000 = 0.27v

so the total would be the sum of all of the above voltages?

 

[pc1966]

so the total would be the sum of all of the above voltages?

Yes. It should be close to total current and half the cable length of drop, but depending on how your rounded the numbers it might to add up that way.

Normally you only do the full calculation for VD if it is not a uniform distribution of loads, or they are of quite different values. And even then only if you have reason to think it is going to exceed the 3% limit. The OSG table of standard circuits / lengths saves time!

 

[Lucien Nunes]

As mentioned above, there is sometimes a longish 'home run' between the DB and the first point, which obviously carries the full current and incurs the highest drop per metre of any section. Therefore I would always take the effective length of the circuit as the length of the home run, plus half the remainder, then calculate the drop using the total load.

In your example that's:
7.5m [home run] + 82.5/2 [half the remainder] = 48.75m effective circuit length for VD purposes.
Total current = 12 x 24/230 = 1.252A
Drop for 1.5mm² = 29mV/A/m
Drop at furthest point = 1.252 x 29 x 48.75 / 1000 = 1.77V

Your calculation adding the drops length by length will give the same result if you redo it without the rounding errors!

Another note: don't forget that not all electronic loads have unity power factor, so the current might be higher than P/V.

 

[Newtothis]

As mentioned above, there is sometimes a longish 'home run' between the DB and the first point, which obviously carries the full current and incurs the highest drop per metre of any section. Therefore I would always take the effective length of the circuit as the length of the home run, plus half the remainder, then calculate the drop using the total load.

In your example that's:
7.5m [home run] + 82.5/2 [half the remainder] = 48.75m effective circuit length for VD purposes.
Total current = 12 x 24/230 = 1.252A
Drop for 1.5mm² = 29mV/A/m
Drop at furthest point = 1.252 x 29 x 48.75 / 1000 = 1.77V

Your calculation adding the drops length by length will give the same result if you redo it without the rounding errors!

Another note: don't forget that not all electronic loads have unity power factor, so the current might be higher than P/V.

i rounded because i was lazy 8)

Ah thank you I get it now
I also realised if you are using LED lights its going to be pretty hard to exceed VD if the loads are distributed .

for electronic loads how would you be able to find out the unity power factor?
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Yes. It should be close to total current and half the cable length of drop, but depending on how your rounded the numbers it might to add up that way.

Normally you only do the full calculation for VD if it is not a uniform distribution of loads, or they are of quite different values. And even then only if you have reason to think it is going to exceed the 3% limit. The OSG table of standard circuits / lengths saves time!

I looked at the on site guide
and its shows 10 amps on 1.5mm as being able to be 52 meter....

but if you used the whole 10 amps
( so 52/2 for the effective length for VD is 26 meters )
Total current = 10A
Drop for 1.5mm² = 29mV/A/m Drop at furthest point = 10 x 29 x 26 / 1000 = 7.54 V

 

[pc1966]

I looked at the on site guide
and its shows 10 amps on 1.5mm as being able to be 52 meter....

but if you used the whole 10 amps

Yes, these are for general circuits that can be loaded up to the MCB rating.

More importantly is to check the limit on typical Zs for fault clearing. That can be less than the VD limit due to T&E smaller CPC and/or the assumed Ze being significant (generally for high current MCB using C/D curves).
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You can rely on a RCBO tripping at high Zs, but it is not something I ever feel terribly happy with unless there is not a sane alternative.