# Need help with relay calculation

Discussion in 'General Electronics Discussion' started by Lei Reyes, Aug 13, 2014.

1. ### Lei Reyes

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Jul 2, 2014
I have the setup below, and I want to theoretically compute (without the use of a multimeter) the total current draw and wattage (power consumption) of the system as a whole. I'm really confused with the datasheet of the relay so i can't really theoretically measure what i'm looking for ...

i want to know the current draw and wattage whenever the switch is triggered on (the relay being energized)
tnx for the help

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2. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I've fixed the ALL CAPS in your post. This is bad netiquette. It's as if you're SHOUTING at us.

You've practically answered your own question. The relay coil has a resistance of 160 ohms. Using Ohm's Law you can confirm that at 12V, it will draw 75 mA:

I = V / R
= 12 / 160
= 0.075A
= 75 mA.

This current is drawn from the 12V supply, through the switch.

Power can be calculated as P = V × I
= 12 * 0.075
= 0.9 watts

3. ### Lei Reyes

60
2
Jul 2, 2014
Oh im very sorry with that sir, really sorry sir .
ok sir and how about the led lamps connected to the relay sir ??

4. ### KrisBlueNZSadly passed away in 2015

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The other power used by the circuit is drawn from the 230 VAC mains supply. The way you have wired it, the bottom lamp is powered all the time, and the top lamp is powered when the relay closes. They are not connected in parallel as stated on the diagram.

Each lamp has a box that represents a power supply, I guess, that is marked "28V 7W". I guess 28V is the output voltage to the LED lamps.

The 7 watt rating for the power supplies could be the output power or the input power. If 7W is the input power, then there will be 14W load on the 230V supply when the relay contact is closed.

If it's the output power, then the input power will be higher than 7W because of losses (inefficiency) in the power supply. Assuming it's a switching supply, you could estimate the efficiency at around 75%. In that case, the input power will be 7W / 0.75 which is 9.3W each, and 18.6W for both when the relay is closed.

If you tell us the reason why you want to know this, we may be able to give you some more useful information.

5. ### Lei Reyes

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Jul 2, 2014
oh yes sir, i think the diagram below makes sense now . so sir u are saying that:
knowing that the lamps are connected in parallel:
if the 7w is the input power: there will be 14W load on the 230V supply and current draw of the lights will be 14W/230V or 14W/28V ???
if the 7w is the output power: there will be 18.6W on the 230V supply and current draw of the lights will be 18.6W/230V or 18.6W/28V ???

i want to know how to actually compute sir the loads (current and power) just for my paper works in school, some documentations about the load rating of my home work ...

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6. ### Lei Reyes

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Jul 2, 2014
and also sir, do we need to consider also the Contact Rating of the relay if we are computing the power and current loading ?

7. ### KrisBlueNZSadly passed away in 2015

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You're talking about the power going into the power supplies, so that would be 14W at 230V AC. But with most power supplies, the input current waveform is not the same shape as the voltage waveform. The voltage waveform is a sine wave, but the current waveform is not. Here is a good illustration from https://www.yokogawa.com/ymi/tutorial/tm-tutorial_wt_03.htm:

So describing the input current for a power supply that is powered from AC is not as simple as dividing the input power by the input voltage.

If 7W is the output power, i.e. the power dissipated by each lamp, and there is 28V across each lamp, the current in each lamp will be 7W / 28V = 0.25A. If 7W is the output power from each power supply, the 9.3W and 18.6W figures relate to the input side of the power supply, where the power is higher, because of power loss in the power supply.
The contact rating won't affect the amount of power used. You just need to make sure that the contact rating is high enough for the loads you're using.

BTW there's no need to call us "Sir"!

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
:-o Thanks Kris!

I think you've described it pretty well.

All that Lei needs to realise is that the power required is the sum of the power dissipated in every load. In his case, we assume the loads are the relay, and each of the two LED globes.

For simplicity we assume that the 12V 1A power supply is 100% efficient, and that there are no other losses in the circuit. We also assume that the power consumed is what is written on the label.

All of these power ratings will be steady state, and/or averaged over time, so we can't say instantaneously what the exact power demanded by the circuit will be. But over a reasonable time (and this may be 1 cycle of the mains) it will average out to some value.

Kris gave a perfect example above where the power is zero for most of the cycle and then jumps up to some higher amount (which itself varies) for a small part of the cycle. Pretend that this is the V/I graph for one of your 7W lamps. For that small time in the middle the lamp may be consuming 21W, but it does so only a third of the time, so the average power is 7W. There are other odd circuits which store power in one part of the cycle and give it back in other parts. For those the power can actually be negative during part of the cycle.

Here Kris has made some assumptions about some discrepancy in measurement being caused by losses in power supplies. This relates directly to my point above where I assumed there was no loss in power supplies. In real life this isn't true.

Also note that Kris is giving the average current. If the device draws power in spikes, the maximum current might be higher (and even if it draws it "evenly", it may vary through the AC cycle). In most cases we can ignore this, but it may be worth bearing in mind that for this device the maximum current during some period might be 50% higher than this, or several times higher, depending on the nature of the load. This is not something that can easily be calculated from the sort of block diagram you're using.

In most cases for things other than loads (i.e for cable, power supplies, switches, etc.) any mention of current is typically an upper limit that you should not exceed, not what will or must be used.

I'll second that. We both have names and neither of us (to my knowledge) wear shiny armor whilst riding out on horseback to kill people.

9. ### KrisBlueNZSadly passed away in 2015

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You shouldn't be surprised!
Yes, that's the basic point here. Calculate the power used by each load, and add them all together.
I shouldn't have assumed that. I really don't know what Lei wants this information for. If he wants to know the total power consumed from the mains, then he should factor in the losses in the 12V supply.
Yes, and that would normally be the input power. But with an output voltage of 28V, if the 7W is the output power, not the input power, that's an output current of 250 mA. Coincidence?
Yes, nicely put.
I was assuming that the lamps are powered with DC from switching supplies, so the funny current waveform would apply at the mains input to the supplies.

I was going to ask you to explain, with this funny current waveform, how this relates to VA vs. watts, and how you would specify the power consumption of a power supply like that, in terms of what would be measured by your household electricity meter.
Right. If the load is non-linear, like the input to a switching supply without power factor correction, that's going to happen.
Right. You would need to know some detail about all three power supplies.
Or in the case of a power supply, it would be an upper limit that the power supply would not exceed, except during power-up. It wouldn't be the RMS current, or the mean absolute current.
And neither of us have been tapped on the shoulders by a sword held by a monarch. "Sir" is a mark of deference and respect, and a word I occasionally use when talking to old people, but being called Sir makes me uncomfortable. When salespeople in shops call me Sir, I call them Sir or Ma'am in return!

10. ### Lei Reyes

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Jul 2, 2014
ok and for the relay .
if i'm using the relay in the diagram which is a 3PDT relay,

does this mean that i can use 5A for each of the 3 pole ?
OR
i can only use a total of 5A for the 3 poles together ? sum of 15a for the 3 poles

11. ### KrisBlueNZSadly passed away in 2015

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If it's a 5A relay, that rating is for each pole. So a three-pole 5A relay can switch up to 15A. But I don't think you can connect two or three poles in parallel and switch more than 5A with them, because the poles won't close and open at exactly the same time. You could switch three separate circuits of 5A each though.

Anyone else got some better advice here?

(*steve*) and Lei Reyes like this.