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Calculating Power Requirements and selecting suitable Power Supply for Application

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RoboWanabe

Member
Hi,

I seem to always struggle when it comes to thinking about how to power my Applications. It's easy prototyping each individual component with a bench power supply, but as soon as I need them all in one circuit everything gets harder for me. Any tuorials or reading matriel to master this dark art would be appricated :).

I need help with this specific example:

I want to use this device :https://octopart.com/wise-4012e-ae-wa-advantech-67312669
Runs at [email protected] and consumes 300mA.

I want to power this sensor separately but output attached to device above : http://www.ia.omron.com/product/item/e3z_7012f/
Runs at 12 - 24vdc and consumes 30mA - 0.36watts

I require that these devices must run for over 5 days on a battery pack.

I have found this battery : http://www.powerstream.com/PST-MP3500-I.htm
it has a value of 60wH and is about £140
or this battery : http://www.ebay.co.uk/itm/External-...attery-Charger-For-Mobile-Phone-/291613962290
It has a value of 50000maH and costs £11.49

SO! for the two battery packs about my calculations are:
for 60wH : 60wH/(1.5w+0.36w) = 60wH/1.86w = 32hours = 1 dat 3 hours
for 50000maH: 50000mAH/(330mA) = 151 hours = 6 days

Are my calculations above correct and why does the significantly cheaper option last longer?
 

AnalogKid

Well-Known Member
Most Helpful Member
Power levels first. The sensor runs on 12 V, not 5 V, so the 5 V needs to be boosted to 12 V and that boosting will be only about 75% efficient. It will take about 0.5 W at 5 V to deliver 0.36 W at 12 V, so your total load at 5 V is closer to 2 W or 400 mA.

Your power math is correct when adjusted for device efficiency.
60 W-h at 5 V equals 12 A-h. 50000 mA-h at 5 V = 50 A-h = 250 W-h. 250 / 60 = 4.167. 50 / 12 = 4.167.
60 W-h / 2 W = 30 h = 1.25 days
50 A-h / 0.4 A = 125 h = 5.20833 days
5.20833 / 1.25 = 4.167

As to why the cheaper battery lasts longer, no clear answer. You do get a lot more stuff from Powerstream. However, both are unknown devices on the innergoogle. The specs could be wrong, it could last only a few discharge cycles, etc. If it sounds too good to be true...

I'd contact both manufacturers for dimensions and detailed power capacity at my output current, and compare those numbers to other Li batteries.

ak
 

RoboWanabe

Member
Hi, thanks for the reply :)

So I'm on the right track. I think the 50000mah option is probably a cheapo and would only last a couple of cycles.

There is a 103wH version of the PST-MP3500-i, and i believe i can stack them together to double the capacity? so do they mean if I stack them on top of each other i will get 206wH capacity so they would last roughly 4 days. I think this might be the better option

thanks again
 

spec

Well-Known Member
Most Helpful Member
Hi,

I seem to always struggle when it comes to thinking about how to power my Applications. It's easy prototyping each individual component with a bench power supply, but as soon as I need them all in one circuit everything gets harder for me. Any tuorials or reading matriel to master this dark art would be appricated :).

I need help with this specific example:

I want to use this device :https://octopart.com/wise-4012e-ae-wa-advantech-67312669
Runs at [email protected] and consumes 300mA.

I want to power this sensor separately but output attached to device above : http://www.ia.omron.com/product/item/e3z_7012f/
Runs at 12 - 24vdc and consumes 30mA - 0.36watts

I require that these devices must run for over 5 days on a battery pack.

I have found this battery : http://www.powerstream.com/PST-MP3500-I.htm
it has a value of 60wH and is about £140
or this battery : http://www.ebay.co.uk/itm/External-...attery-Charger-For-Mobile-Phone-/291613962290
It has a value of 50000maH and costs £11.49

SO! for the two battery packs about my calculations are:
for 60wH : 60wH/(1.5w+0.36w) = 60wH/1.86w = 32hours = 1 dat 3 hours
for 50000maH: 50000mAH/(330mA) = 151 hours = 6 days

Are my calculations above correct and why does the significantly cheaper option last longer?
Hi Robo,

Fascinating project you are working on- I wonder what it is for.

Power requirement calculations are dead simple- the only issue is the way that manufacturers tend to present their data and the lies that second line battery suppliers tell.

AK has explaining the situation but, purely out of interest, I had a go at doing a power budget too, ignoring efficiency considerations that is. What worries me is that I arrive at a much higher energy requirement to power both units for five days continuous. This is how I did it. Would someone please check my workings for errors because it is bugging me that I can't find any and the bottom line intuitively seems too high:

(1) WISE-4012ZE: 5V @ 300mA= 1.5W
(2) E3Z-R86: 12V @ 30mA= 360mW
(3) TOTAL POWER= 2.16W
(4) 5 days @ 2.16W= 5D *24H * 2.16W= 259.2 Watt Hours

If the figure above is correct, however, I suggest that your cheapest approach would be a 12V 22A/H minimum deep-discharge lead acid battery (standard auto batteries will soon fail) and a couple of dirt-cheap board-level switching regulators, one to produce 12V and one to produce 5V.

spec

PS: Batteries, as I have said over and over again on ETO, are a minefield with many rip-off products. One of the most notorious is Ultrafire. It is advisable to buy only main-line manufacture's batteries and stay well clear of the badge engineered stuff. The best manufacturer available to the general public is Panasonic (Sanyo). You will find that these batteries will cost more and will have a lesser capacity- there is a very good reason for this! You have to ask yourself why is it that the people who actually design, develop, and manufacture batteries can't meet the claims that the badge-engineering charlatans claim. :grumpy:

PPS: More and more these days you see three parameters given for a battery: terminal voltage, AH rating (stored energy), and and maximum current rating. The first two parameters are fairly straight-forward but the third is not so well known and often leads to confusion. The third parameter merely defines the maximum current that the battery can deliver without damage. It says nothing about the energy storage of the battery. In your application the current requirement is low.
 
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Les Jones

Well-Known Member
Most Helpful Member
One other thing to consider is that the ratings of batteries sold on ebay is VERY optimistic. I think these battery packs are made up of a number of 18650 lithium ion cells. As the dimensions given for the 50 AH unit is 165mm x 80mm x 22mm I think it will contain 8 of these cells. The rating of these units is given in the AH rating of the cells so by the time it is stepped up to 5 volts it will be less Batt AH * 3.3/5 * 0.9 (90% efficiency of step up converter) = 0.594 So even if the basic battery pack was 50 AH that would only be 29.7 AH. The largest capacity 18650 cells that I have seen advertised is 4.9 AH I bought some of these and none had a capacity of more than 2 AH If you Google "18650 cell capacity" you will find that many other people have also found this. I think the best genuine capacity is about 3.6 AH So if we use this figure with 8 cells we get 28.8 AH (This is at 3.3 volts) so when we correct for the output at 5 volts we get 28.2 * 0.594 = 17.1 AH (= 85.5 watt hours) I would think the one sold on ebay would probably have about half this rating.

Les.
 

spec

Well-Known Member
Most Helpful Member
One other thing to consider is that the ratings of batteries sold on ebay is VERY optimistic. I think these battery packs are made up of a number of 18650 lithium ion cells. As the dimensions given for the 50 AH unit is 165mm x 80mm x 22mm I think it will contain 8 of these cells. The rating of these units is given in the AH rating of the cells so by the time it is stepped up to 5 volts it will be less Batt AH * 3.3/5 * 0.9 (90% efficiency of step up converter) = 0.594 So even if the basic battery pack was 50 AH that would only be 29.7 AH. The largest capacity 18650 cells that I have seen advertised is 4.9 AH I bought some of these and none had a capacity of more than 2 AH If you Google "18650 cell capacity" you will find that many other people have also found this. I think the best genuine capacity is about 3.6 AH So if we use this figure with 8 cells we get 28.8 AH (This is at 3.3 volts) so when we correct for the output at 5 volts we get 28.2 * 0.594 = 17.1 AH (= 85.5 watt hours) I would think the one sold on ebay would probably have about half this rating.

Les.
Hi Les,

We have had similar experiences with LiIon 18650s. I wouldn't listen to someone wiser and got caught with the 4A to 6A versions. Unless you spend a fortune, a proper 18650 will only do around 2AH but its performance will be far superior to the claimed higher energy types. I find that a genuine 2AH battery typically has a capacity of around 2.5AH anyway.

The same principle applies to auto batteries. I have bought cheap badge-engineered versions that only lasted a couple of years and never performed that well anyway. My current battery is a main-line product that was twice the cost of the cheapest type, but even after 14 years it still whips the motor over smartly on the coldest days. Mind you, Mobile One engine oil helps a lot in that respect- I love it and Comma Petrol Magic (injector cleaner) and try too have a glass or two of each every day :D

spec
 
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spec

Well-Known Member
Most Helpful Member
Hello again Robo,

One more thing about batteries is the capacity variation with temperature, especially as I imagine that you will be leaving this equipment in a remote location with no heating. It can get quite nippy in Coventy (UK?). Batteries loose capacity as their internal temperature falls. Some batteries only have half their capacity at 5 deg C compared to 25 degC.

Also you should check the minimum temperature that the battery will stand. The problem is the electrolyte crystallizing or freezing which will write the battery off.

spec
 
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AnalogKid

Well-Known Member
Most Helpful Member
(1) WISE-4012ZE: 5V @ 300mA= 1.5W
(2) E3Z-R86: 12V @ 30mA= 360mW
(3) TOTAL POWER= 2.16W

If the figure above is correct,
It is not. 1.5 + .36 = 1.86, not 2.16

Also, what is the reason for ignoring power losses due to power conversion efficiency once the OP says he wants to buy real world batteries and operate a real world device?

ak
 

RoboWanabe

Member
Hi Spec.

I'm afraid its top secret, but I think you can deduce what I'm trying to do your self :)

Cheers guys. I think i will do abit more digging around, at least i know i'm roughly calculating it correct.

But! Analogue Kid. About his calculation : 60 W-h at 5 V equals 12 A-h. 50000 mA-h at 5 V = 50 A-h = 250 W-h. 250 / 60 = 4.167. 50 / 12 = 4.167.

Is that not just the increase in capacity of the 50AH battery to 60Wh? I'm not sure why you use the term efficiency? efficeny in this case would be Power out / Power In?

Thanks Again
 

AnalogKid

Well-Known Member
Most Helpful Member
The sensor runs on 12 V, not 5 V, so the 5 V needs to be boosted to 12 V and that boosting will be only about 75% efficient. It will take about 0.5 W at 5 V to deliver 0.36 W at 12 V, so your total load at 5 V is closer to 2 W or 400 mA.
No power conversion circuit is 100% efficient. A typical number for a small boost converter is 75% efficient. That is, for every 10 watts that go into the converter from the power source, only 7.5 watts are available at the output. In your case, delivering 0.36 W to the sensor will require o.5 W from the battery.

ak
 

spec

Well-Known Member
Most Helpful Member
It is not. 1.5 + .36 = 1.86, not 2.16

Also, what is the reason for ignoring power losses due to power conversion efficiency once the OP says he wants to buy real world batteries and operate a real world device?

ak
I had 1.8W total initially but in the many recalcs, because I didn't believe the size of the bottom line, I did make and error and got 2.16A:banghead:

Power efficiency is not mentioned because I did not want to cloud the issue. I stated that efficiency was not included, so no need for the lecture. You did not mention temperature effects which has a major effect, but I didn't leap on your post in an accusing fashion, something I could do to many circuits on ETO if I was so inclined.

Here are the revised calcs:

(1) WISE-4012ZE: 5V @ 300mA= 1.5W
(2) E3Z-R86: 12V @ 30mA= 360mW
(3) TOTAL POWER= 1.8W
(4) 5 days @ 1.8W= 5D *24H * 1.8W= 218 Watt Hours

The salient point is that, even with this corrected lower value power, the five-day energy requirement is still huge and would be a big challenge to supply with LiIon batteries, and this is the main point.

As a result, I still advise that a deep discharge lead acid battery with two switch mode PSUs, one for 5V and one for 12V would seem to be the best way to go.

So, to include 75% efficiency for the regulators the total energy requirement would be 1/0.75 * 218 Watt Hours = 290.67 WH

Thus the minimum battery capacity would be 290.67 WH/12V= 24.22 AH

While I am on air and being particular, the temperature effects mentioned in a previous post should be taken into account, not only in A/H loss but also terminal voltage drop, in order to arrive at a real world design specification for the battery. It would be out of place to provide these calcs at this initial stage because they are fairly detailed, and would cloud the issue.

Other advantages of a lead acid battery, in addition to potentially lower cost, is the simple charging requirements and general robustness of lead acid batteries.

Another thought is that solar panel charging may be possible, depending on the deployment of the system. Once again, this would be much simpler for a single lead acid battery than huge banks of LiIon cells.
 

spec

Well-Known Member
Most Helpful Member
Hi Robo,

Hi Spec.
I'm afraid its top secret, but I think you can deduce what I'm trying to do your self :)
Got it :cool:

But! Analogue Kid. About his calculation : 60 W-h at 5 V equals 12 A-h. 50000 mA-h at 5 V = 50 A-h = 250 W-h. 250 / 60 = 4.167. 50 / 12 = 4.167.
This is a mixture of beans, bananas, peas, and plums. You need to clearly divide power, current volts and energy. My energy budget shows a simple way to do this. The bottom line is that the batteries you are considering would be a very expensive and complicated approach. You would probably need 5 of the 60WH batteries to meet your 5 day duration, and that would be if the battery suppliers figures were real world, which seems unlikely.

I'm not sure why you use the term efficiency? Efficiency in this case would be Power out / Power In?
Just to reiterate what AG and Les have said:
No process is 100% efficient in the real world. If you take your car for example, if it is a diesel it will have an efficiency of around 25%. This means that only 25% of the energy in the fuel is converted into useful power at the engine flywheel. The other 75% of the energy is wasted in heat.

By the same token, a switch mode power supply (SMPS) has a typical efficiency between 70% and 90%, depending on many factors. We have assumed 75% efficiency for the inverters in your system so consider the 5V inverter in the system that I outlined. The required output voltage is 5V and the required output current is 300mA, which gives an output power of 5* 300mA= 1.5W. But the inverter efficiency is only 0.75%. This means that the input power that the inverter will require is 1.5W/0.75= 2W. ie 500mW will be lost.

In most cases, energy is lost in conversions as heat, and that is the case with SMPSs, so in this example the SMPS will be dissipating 500mW as heat.

That's all there is to it, but it is important to make sure you do not mix parameters, and the old adage in design always applies: Keep It Simple. :)
 
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spec

Well-Known Member
Most Helpful Member
Would the temperature effects reduce the available power by 25%?

ak
Hi AK,

With some battery chemistries the loss of energy of a battery can be radical going from 25 deg C to 5 deg C, but 50% loss is common. So a 3.7V 1AH battery may only provide 3.4V at 500mAH at 5 deg C.

In Robo's application it all depends on the temperature of the environment. In the winter months, in mid area of the UK, the temperature can average around 5 deg C so loss of energy stored would probably be considerable. I will have to have a look at the temp/energy figures for LiIon and lead acid because I have forgotten the details.

To answer your question about power drop at low temperatures; yes, the available power also drops a lot because the terminal voltage drops and also the maximum current available drops. But in this application this will not be significant because of the low power requirement. Incidentally the energy capacity is, to a rough order, inversely proportional to power used. Normally a battery is specified at a current drain of AH/10, so in the case of a 1 AH battery the AH rating would be at a current of 100mA.

The reason for this temperature dependence is that, in very general terms, a battery relies on a chemical reaction, which increase in vigor as the temperature increases, up to a point that is.

Cheers

spec

Link to a site describing some of the complications involved with rechargeable batteries: http://www.mpoweruk.com/performance.htm
 
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spec

Well-Known Member
Most Helpful Member
Robo,

Here is a notional example of deep discharge battery that should give your application around 10 days duration with the temperature range expected in your part of the UK. It only costs £55 UK, but it is a bit big and heavy, although I haven't compared it with the LiIon approach:

https://advancedbatterysupplies.co.uk/product/abs-lp60/

(please note that I haven't analyzed this battery in detail. It is shown for information only)

spec
 

RoboWanabe

Member
Thanks for that Spec :)

But i have to make this design as un-obtrusive as possible, A battery like the one above will not go down well in a factory :p ( I suppose i should of told you that)

LiIon are usally compact and take up less room, the lead acid option above looks like it has the right specs but it is too big, are there any smaller options?

Please bare in mind guys that the PST-M3500 Has a 5v volt USB output for the Wise unit and also another DC output which can be set to a voltage between 5v and 19v and supply up to 4 Amps, so no extra circuitry needed?

The end design needs to be able to Run the WISE unit, an Opto Reflective sensor for counting items and also maybe up to 3 light intensity sensors.

I am currently liaising with SICK sensors to find the right type, low power etc. also for the WISE units Digital inputs i need the sensor to have a dry contact ? i.e. short D0 to GND. Please would somone be able to confirm that.

In the spec it says Dry Contact 0:eek:pen 1:close to GND. But i've measured the voltage between D0 and GND and there is a 5 volt potential. So i think the Sensor needs to be the dry contact ?

Robo
 

spec

Well-Known Member
Most Helpful Member
Thanks for that Spec :)

But i have to make this design as un-obtrusive as possible, A battery like the one above will not go down well in a factory :p ( I suppose i should of told you that)

LiIon are usally compact and take up less room, the lead acid option above looks like it has the right specs but it is too big, are there any smaller options?

Please bare in mind guys that the PST-M3500 Has a 5v volt USB output for the Wise unit and also another DC output which can be set to a voltage between 5v and 19v and supply up to 4 Amps, so no extra circuitry needed?

The end design needs to be able to Run the WISE unit, an Opto Reflective sensor for counting items and also maybe up to 3 light intensity sensors.

I am currently liaising with SICK sensors to find the right type, low power etc. also for the WISE units Digital inputs i need the sensor to have a dry contact ? i.e. short D0 to GND. Please would somone be able to confirm that.

In the spec it says Dry Contact 0:eek:pen 1:close to GND. But i've measured the voltage between D0 and GND and there is a 5 volt potential. So i think the Sensor needs to be the dry contact ?

Robo
Hi Robo,

No worries.

Yeah, I thought the lead acid deep discharge battery might be on the big side, although you may be able to get a deep discharge lead acid battery of half the volume that would give you around 5 days duration worst case.

As you say LiIon is most power dense and light of the commonly available battery chemistries so, in view of you size limitations, that would be the avenue to explore, but off the top of my head, I think it would be expensive.

Have you thought about continuously turning the system on and off to save power. Of course it depends on the application and how well the system components power down and power up- just a thought.

As an exercise, and for illustration purposed only, here are the basic parameters for an 18650 LiIon approach:
You need quite a bit of energy, even without any new requirements, so taking my last bottom-line five-day total energy figure of 290.67 WH:
(1) Allow 25% say for battery temperature and other effects brings 1.25 * 290.67= 363.33 WH.
(2) Take the energy of an industrial grade 18650 rechargeable battery as 3V * 2.5AH= 7.7W H.
(3) This means that you would need 363.33WH/7.7WH = 47 individual 18650 LiIon batteries.
(4) The total weight would thus be 47 * 46.5g = 2.34Kg
(2) The net volume would be 4.97 cc* 47= ???cc
(3) At £7 UK per 18650, the cost would be 47 * 7 = £329 UK.

spec
( I am in a rush so haven't had time to checked this post)

Data Sheets

(1) Panasonic NCR18650
http://industrial.panasonic.com/cdbs/www-data/pdf2/ACA4000/ACA4000CE240.pdf
 
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RoboWanabe

Member
Hi Spec,

I have thought about that and I'm currently looking for a different Wise unit to use which checks the state of the inputs less frequently thus hopefully reducing power.

I Have found this alternative : http://downloadt.advantech.com/ProductFile/PIS/WISE-4060[2F]LAN/Product%20-%20Datasheet/DS_WISE-4000_LAN_EN20150902152116.pdf

If in the data sheet it says :
Power Input 10 ~ 30 VDC
Power Consumption WISE-4010/LAN: 1.2 W @ 24 VDC WISE-4050/LAN:

If i choose to Run at the lower Voltage end, 10V, will my power calculation be :

0.05 * 10V = 0.5w ? i.e. is the current consumption the same over all voltage ranges?

( I have a feeling it might be a stupid question so sorry in advance)

Thank you
 

Les Jones

Well-Known Member
Most Helpful Member
Hi Robo,
What you say is correct if the unit uses a linear regulator to drop any input voltage between 10 and 30 volts down to say 9 volts. It could use a switch mode power supply which would always take about 1.2 watts. It would do this by consuming more current at a lower voltage input. You would need to contact the manufacturer to see if it would take reduced power on 10 volts. It's not a stupid question.

Les.
 

RoboWanabe

Member
Thanks Les :)

I just sent them an email asking. I can't see what the advantage would be by using a Switch mode regulator to maintain the Power consumption, especially when everything would still work by consuming less power?

Thanks
 
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