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This question already has an answer here:

I have been reading your articles on power supply in readiness for starting to build a home automation system using Loxone equipment.

I feel confident in the project but the power supply situation leaves me a bit perplexed.

I am intending to use 8 TDK Lambda (DRF240-24-1) power packs to drive the 24V DC LED lighting in my new house.

Each power pack has a rated input of 100 – 240 VAC 2.7A. The output is 24 – 28 V DC 10 – 8.6A.

So that is a power requirement of 648W for an output of 240W.

My first question is where is the missing 408W? I realise that the pack itself will require some energy to work, but I would have thought that would be around 20% of the input.

Secondly; if the input is rated at 2.7A x 8 power packs then I will need an RCB in the consumer unit rated at over 21.6A, - or more likely several lower rated RCBs to split the load into zones.

Am I missing something here?

From the 8 Power packs there is a total power available at 24V DC x 80 A = almost 2,000W whereas to provide this I have to supply 21.6A x 240VAC = over 5,000W. Am I going wrong somewhere?

Any observations would be appreciated especially if I am completely failing to understand this issue.

Thanks in anticipation.

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marked as duplicate by winny, laptop2d, Finbarr, Bimpelrekkie, Sparky256 Jun 8 '18 at 2:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ According to the datasheet, it's max power is 240 W. \$\endgroup\$ – Long Pham Jun 3 '18 at 13:45
  • \$\begingroup\$ Is "The output" your load ? You barely need about one PSU. \$\endgroup\$ – Long Pham Jun 3 '18 at 13:48
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    \$\begingroup\$ You should consider 48Vdc for home grid not 24V due to AWG16 V drop at shared house lighting currents unless each lamp has its own DCDC converter . This has been my design for 10 yrs and believe practical for future. \$\endgroup\$ – Sunnyskyguy EE75 Jun 3 '18 at 14:01
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    \$\begingroup\$ I reckon the 2.7A is at the lowest voltage input, and isn't all watts, but VA. \$\endgroup\$ – Neil_UK Jun 3 '18 at 14:11
  • \$\begingroup\$ James, in rated input of 100 – 240 VAC 2.7A, 2.7A is the maximum the supply will draw under worst-case conditions. Worst-case being lowest input voltage (100v), and maximum load. Double the input voltage, and the current will be nearly half. Specs are also generally over-rated; get a kill-a-watt meter, plug one of your supplies into it, and get some real values. \$\endgroup\$ – rdtsc Jun 7 '18 at 12:21
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You've misread the intent of your input specs. 2.7 A is the maximum current, which occurs when the input is 100 volts. It's useful to specify this maximum current, since that determines the size of the wire required for safety.

However, assuming 270 watt in, at 240 volts the current will only be 270/240, or 1.1 amps.

TL;DR - 2.7 amps doesn't occur at higher input voltages, so the input power does not increase then, either.

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100 – 240 VAC 2.7A. The output is 24 – 28 V DC 10 – 8.6A.

The input current will be maximum at the minimum input voltage. So \$ P = VI = 100 \times 2.7 = 270\ \text W \$.

Output is \$ 24 \times 10 = 240 \ \text W \$ or \$ 28 \times 8.6 = 240.8 \ \text W \$.

Efficiency is \$ \frac {P_{OUT}}{P_{IN}} = \frac {240}{270} = 89% \$ which is OK.

From this information we can work backwards and see what the input current will be at 240 V: \$ I = \frac {P}{V} = \frac {270}{240} = 1.1\ \text A \$.

My first question is where is the missing 408W?

It's still back at the power station. Everything is OK.

If the input is rated at 2.7 A x 8 power packs then I will need an RCB in the consumer unit rated at over 21.6 A.

You have omitted location and supply voltage details from both your question and your user profile. If you are in 110 V land your max current will be probably 2.5 A per unit. In 230 V land it will be about 1.2 A per unit.

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Another way to approach the operating input current is to start with the load. The input power equals the output power divided by the efficiency of the supply. Assuming you are not running each supply at full power all the time, the input current will increase and decrease with the load current.

Note that the typical efficiency on the datasheet (94%) is at full load (even though not stated in the notes). At lower output currents the supply's efficiency decreases. Don't be surprised if it measures something like 75%-80% at half power.

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Thank you to all – I knew I was missing something obvious. You can only use as much energy as the ultimate load requires! This was illustrated by Olin Lathrop using apples in another thread.

I should have said that I am in the UK where the supply voltage is 240V.

Just to put this to bed; I should work back from the final load i.e. assume that each power pack is only loaded to say 90% of its maximum output of 10A that is 9A and at 24V that is 216W. Then assume that the efficiency of the pack is around 80% so 216W output requires 270W input and at 240V that is 1.125A. With 8 units that is now only 9A in total

So by carefully selecting zones for the lighting I can easily supply the system through two 5A RCBs. This also assumes that all of the 24V lighting will be on at the same time which is unlikely. Additionally I am sharing the load between the power packs in such a way that none of them will come close to the maximum output load.

I hope I have now got it.

Thanks for putting me on the right track.

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