0
\$\begingroup\$

I need to calculate the amount of money I pay for charging my phone with this charger

Samsung phone charger

I know that I need to calculate the watts then kWh, so I multiplied 5 V by 1.55 A to get 7.75 watts and then use that to calculate kWh. I'm so confused though because whenever I asked anyone in my school they would say, "you don't pay for the output, you pay for the input, so what you did is wrong".

Are they right? Why or why not? What is the correct method?

Note: I'm not looking for a very accurate number because I don't wanna use a Kill-a-Watt. In addition, I watched a video of someone using it on their charger and it gave a number around 4 or 5, that makes me think my teachers are wrong.

Phone battery I'm adding info about the battery since some of you asked for it. It's a Samsung secondary Li-on battery. Nominal voltage 3.85 V / 11.55 Wh. Charge voltage 4.4 V / 3000 mAh.

\$\endgroup\$
  • \$\begingroup\$ Do electric companies charge less at night? \$\endgroup\$ – user16307 Nov 16 '17 at 18:25
  • \$\begingroup\$ @user16307 Depends on the electric company. Irregardless, this load is not significant. \$\endgroup\$ – StainlessSteelRat Nov 16 '17 at 18:27
  • 1
    \$\begingroup\$ "you don't pay for the output, you pay for the input" Yes, they are correct because the energy input (the energy supplied by the company and hence charged for) includes the wasted heat energy (due to inefficiencies in your charger) as well as the output energy. \$\endgroup\$ – JIm Dearden Nov 16 '17 at 18:59
  • 1
    \$\begingroup\$ Refer to this stack exchange question for your references. \$\endgroup\$ – Spehro Pefhany Nov 16 '17 at 21:45
  • 1
    \$\begingroup\$ @AliChen added battery info in main post. \$\endgroup\$ – 1da Nov 16 '17 at 22:59
0
\$\begingroup\$

Those numbers tell you the current that the supply can provide, not how much your phone needs. Your phone will only draw as much as it's designed for, and so you don't know how much power is coming from the mains. So, they are correct. You can't get anything meaningful from the adaptor text itself.

There are two ways to get this figure. You can measure the current going into the adaptor itself, or you can measure the current going to your phone. The latter means you also need to know the efficiency of the adaptor as there is some loss in the adaptor when it converts mains voltage to 5 V. You could potentially estimate it from the mains current input (0.3 A), but that doesn't say what voltage or frequency it's at.

Adding calculations for the second method Ok, so your phone is probably charged over USB which used to be limited to 500 mA. Also, that 1.55 A is the maximum the supply can provide, you don't want to have that all the time. So, let's assume (crudely) that \$ I = 500\ \mathrm{mA} \$. Power to your phone is \$P_{out} = IV = 0.5 \times 5 = 2.5\ W\$. Then, let's assume the adaptor is 80% efficient, so \$P_{in} = \frac{P_{out}}{0.80} = 3\ W\$. The multiply by the time to completely charge, and you're there.

\$\endgroup\$
  • \$\begingroup\$ Let's talk about the latter, measure the current going to my phone, well can't I just say it's 1.55A (the maximum)? I know it's inaccurate right. Now what's the next step after measuring the current? Sorry, I don't get it. \$\endgroup\$ – 1da Nov 16 '17 at 18:17
  • \$\begingroup\$ Ok, so your phone is probably charged over USB which used to be limited to 500 mA. Also, that 1.55 A is the maximum the supply can provide, you don't want to have that all the time. So, let's assume (crudely) that \$ I = 500\ \mathrm{mA} \$. Power to your phone is \$P_{out} = IV = 0.5 \times 5 = 2.5\ W\$. Then, let's assume the adaptor is 80% efficient, so \$P_{in} = \frac{P_{out}}{0.80} = 3\ W\$. The multiply by the time to completely charge, and you're there. \$\endgroup\$ – awjlogan Nov 16 '17 at 18:22
  • \$\begingroup\$ Sure, if you don't care about accuracy you can pick whatever number you want. If you do care about accuracy you not only have to measure the current, but you have to do it over a full phone charge cycle and based on your use case estimate the cost to charge the phone that way. \$\endgroup\$ – John D Nov 16 '17 at 18:24
  • \$\begingroup\$ @JohnD Absolutely right. I suspect this question is more about getting a feel for the size of figures, which is a hugely important skill for school kids. If you justify your assumptions then it's fine. \$\endgroup\$ – awjlogan Nov 16 '17 at 18:25
  • 1
    \$\begingroup\$ @1da, it is likely that you will never recoup your investment into solar charger, sorry :-( \$\endgroup\$ – Ale..chenski Nov 16 '17 at 21:46
5
\$\begingroup\$

They are correct that you pay for the input. Now the question is, "How do I calculate the input when I only know the output?"

We need one more piece of information: the efficiency of the PSU. For this type of device it should be about 85%.

$$ P_{OUT} = P_{IN}Eff $$

where Eff is efficiency.

You can rewrite this as $$ P_{IN} = \frac {P_{OUT}} {Eff} $$

From this you can calculate your watts and multiply this by your hours of use per year. Note that your result will be on the high side because the charger will taper off the current as the battery becomes charged and eventually drop to a trickle.

\$\endgroup\$
2
\$\begingroup\$

Yes they are correct, you need to consider the efficiency of the supply. Also the nameplate will not necessarily reflect what is happening under the hood, as it were. The phone will draw a great deal of current while charging and the adapter will waste a certain amount getting that power from the mains plug. When the charging is finished the current will drop, but not to zero since most phones will continue to operate from the adapter while plugged in, leaving the battery at 100%. Finally, when you unplug the phone but leave the adapter in the wall, the power supply will waste power just sitting there (efficiency drops to zero since there is no useful output current).

The minimum (active) efficiency of that Level V supply is 0.0750 * ln(5V*1.55A) + 0.561 = 71.4%. Assuming your supply takes full current for 2.5 hours to charge it will use no more than 5 * 1.55/0.714 = 10.8W for 2.5 hours or 0.027kWh.

After the battery is charged it will drop considerably. With the phone unplugged the draw will be no more than 0.3W, so if we assume the phone is drawing no more than 0.5W with efficiency in the 70% range, we get 1W consumption while the phone is just sitting there after having charged. If you leave it plugged in 24/7 the total consumption is less than 0.05kWh/day. If your electricity costs you $.15/kWh it will cost around 50 cents a year to keep your phone charged.

In fact the charger is bound to be a bit better than the minimum required, so probably somewhat less. Also it will never discharge if you leave it plugged in all the time.

Note that the charger can use a significant part of the total consumption just plugged in and idling with no phone connected. This "vampire power" adds up with millions of adapters so connected, and governments have been tightening the regulations on quiescent consumption.

\$\endgroup\$
  • \$\begingroup\$ +1 for vampire power. Funny I was just commenting on that when you answered. \$\endgroup\$ – Trevor_G Nov 16 '17 at 18:59
  • \$\begingroup\$ But the input energy to provide the output required from the electricity power station is also worth considering due to all the losses between the source energy and point of use : generation losses, transmission losses, digging coal out of the ground etc etc Which may be a good comparison with your solar panel... @1da \$\endgroup\$ – Solar Mike Nov 16 '17 at 20:04
2
\$\begingroup\$

This is like nailing jello to a wall.

First, the cell phone itself will not have a constant draw down the USB side. Obviously it will throttle back to almost nil when fully charged, but it will also vary during charging based on the battery charging curve and other phone load (e.g. Plugged-in usage).

Second, the charger is not 100% efficient, and even that efficiency is not linear.

The 0.3A figure is the nameplate rating and is useless. Its purpose is to tell electricians how much power to provision for the device; i.e. If she is wiring a cellphone factory or cellphone store, can they wire all the kiosks with one 20A circuit or will they need 3. This figure accounts for worst case use x power factor** x whatever fudge factors UL wants to see in order to approve the item. On your charger the difference between rated input and output is 28 watts, that can't be real, something that small would get too hot!

Back to the Jello-nailing, I would guesstimate the rating of the phone's charger (typically 5w) and then double that to account for inefficiency (10W) , for the runtime of the recharge (e.g. 2 hour). Watts x hours = watt-hours. Divide by 1000 for kilowatt hours.

Americans pay typically 10 to 20 cents per kilowatt-hour of power.

If you pay 12 cents a kwh for power, 1 watt continuous 24x7 costs $1/year.


** Power factor is the ratio between the power you actually use (cherrypicking parts of the AC sinewave), and what the transformer/wiring must carry to deliver the entire AC sinewave. For instance if you use a half wave rectifier, your PF is 50%.

\$\endgroup\$
0
\$\begingroup\$

Assumptions: 120V, 2 h to charge, 12¢/kWh

P = VI = 120V × 0.3A = 36W

Energy = P t = 0.036kW × 2h = 0.072 kWh

Cost = Energy × Cost/Energy = 0.072 kWh × $0.12/kWh = $0.00864

Or 0.864 of a cent. You should be able to follow this.

\$\endgroup\$
  • \$\begingroup\$ WOW.. that would be a horrendously inefficient wall wart.. but could well be. \$\endgroup\$ – Trevor_G Nov 16 '17 at 18:27
  • \$\begingroup\$ hehehe. Let's call it a worst case calculation. \$\endgroup\$ – StainlessSteelRat Nov 16 '17 at 18:28
  • \$\begingroup\$ This is exactly what my teacher told 😔 So, this is the worst scenario I think. \$\endgroup\$ – 1da Nov 16 '17 at 18:42
  • \$\begingroup\$ Energy is cheap. The actual consumption depends on lots of factors, which others have highlighted. Now 36W in and 7.75W out, does not make a lot of sense, but without getting into the guts of the charger, you have to go with nameplate ratings. A range of voltages are given, but 30W at 100V still seems too large. \$\endgroup\$ – StainlessSteelRat Nov 16 '17 at 20:57
  • \$\begingroup\$ @1da, your teacher is about 5X off. He/she needs to get acquainted with modern requirements and standards for energy efficiency of consumer devices. \$\endgroup\$ – Ale..chenski Nov 16 '17 at 20:57
-1
\$\begingroup\$

There is another, potentially more accurate way to estimate the cost of one phone charge. The idea is based on capacity of phone battery. Say, you have an old Samsung S4 phone. It has an internal battery with 9.88 W*h capacity, say 10Wh.

Since your power meter is always on external AC side, you need to consider efficiency of the entire power supply chain. You have two converters, one is the wall adapter. It is a modern adapter, and therefore should be no less than 80% efficient (you need to check with standards). The second converter is inside the phone charger circuitry, so assume the same 80%. The battery also has some losses during charging process. I would say 95% efficiency.

As result, the INPUT amount of energy must be 10/0.95/0.8/0.8 = 16 W*h for a full charge. After that the charger will consume negligible power.

So, assuming US electricity rate of 10c per kWh, the cost of one-time charge is about 16/1000 * 10 = 0.16 cents.

\$\endgroup\$
  • \$\begingroup\$ I am wondering what would be the reason to downvote my answer? Anyone, come forward please. Not that I care much, just curious... \$\endgroup\$ – Ale..chenski Nov 16 '17 at 23:08
  • \$\begingroup\$ Wow, no idea why the -1's. Have a +1 to help, it's not a bad answer. \$\endgroup\$ – Bryan Boettcher Dec 1 '17 at 14:44
-1
\$\begingroup\$

Given the statement by Transistor this post has been heavily edited for his and others elucidation.

Charger Samsung Galaxy Alpha Network, Note Edge, (EP-TA50EWE). White technical data: Input: 100-240V @ 0.3 A Output: 5V @ 1550mA Standard USB jack

1W = 1 j 7.5W = 7.5j

3600 seconds in 1 hour

so 7.5j x 3600 = 27kj/h

Power(Watts) = Energy(Joules) / time(in seconds)

An energy expenditure of 1 Wh represents 3600 joules.

So we now have 27kj per hour divided by 3600seconds = 7.5Wh or 0.0075kWh

W and Wh are interchangeable.

Regardless of the comments below, there is a bit of nit picking going on here with regards to kW and KW, I am sure people understand what is being meant and that temperature is not being talked about.

That over with, just to explain a few points with my earlier post for those who saw it.

PSU's designed for the Eu and USA have strict tolerances they must meet, however that does not mean that every PSU made outside of those areas will comply with those regulations. Some low current PSU's are half wave rectified and rely on a tuned circuit at mains frequency to provide 5V @ XmA for use in low power circuits. Please see http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf

However I ripped apart a Samsung charger yesterday to find a switch regulator circuit. The efficiency of these are stated to be between 75% - 90% + any losses within the circuit. Again these losses are so small that its really not worth mentionig, unless you want precise, which is just not worth it in every day life or it is a cumulative circuit where the sum of the losses are important. I tend to look at switch regulator supplies as lower current devices, just to differentiate between switched mode regulator and switched mode power supply(multiple Voltages and higher currents), although strictly speaking you will find a switched mode regulator at the heart of a switched mode PSU.This stems from a work nomenclature where I want a SMPS does not mean I want a switched regulator.

There are a number of factors here with regards to charging and efficiencies. the latter you only need a rough figure for and best to take worse case scenario( and easiest calculation for) of 75% to calculate power loss.

So if we wanted we could just say that the actual W used would be 25% more, than that quoted for the output power, which would be approx 9.375W or if you want to go through the rigmarole of calculating Wh to find it means the same thing go ahead.

I could go on about charging batteries and characteristic curves for each of the devices used, however all you wanted was a rough figure and a way to get to it :)

So saying that I have seen others state things that need modifying a little. When a flat battery is put on charge it will, given the Voltage stated, want to suck current at its maximum of 3A, however as the PSU cannot provide this it will hit the limits of the supply current and the % efficiency may drop as low as 50%/ As current rises above the stated value, the efficiency of a switched regulator goes down Many Switch mode PSU's however have fold back current limiting to stop this, unlike these little switched regulator supplies. What will happen here, if left unfettered, is that until the battery has enough power to balance out at the figures given, charge rate will be slow and power consumed will be high. As the Battery power levels increase, the stated figures will apply and the efficiency will increase, that is until a low charge current is required and again efficiency will decrease. Most cell phone batteries do have battery management so can turn off and limit the current supplied for each of these conditions. Whether they do or not is solely down to the manufacturer/design team. Fast chargers work on the principle that the temperature of a battery is used to regulate charge applied, ergo hitting a battery with a large current will charge it fast but temperature rises at the same time. At a given temperature current is turned off, and the battery is left to cool down below a set limit, before charge current is again applied. Doing it this way should show you that calculating the power used is not simple, however fast charging is never very good when used regularly, as the battery material will degenerate faster. Maybe the next generation of fast charge batteries will be better, 12 minutes for a 4000mAh battery is touted.

\$\endgroup\$
  • 1
    \$\begingroup\$ (1) watts are expressed in watts (W) not watts per hour. (Capital 'W' when abbreviated. Lowercase when spelled out as per SI convention.) (2) "*To use 1KW ...". Two errors: 'KW' = kelvin-watt. You mean kW. Second, kW is a power measurement. Energy measurement is in kWh. (3) The power supply does not work by "leakage current through a capacitor". It's a switched mode PSU. Efficiency likely > 80%. A major edit is required to correct the errors in this answer. \$\endgroup\$ – Transistor Nov 29 '17 at 16:23
  • 1
    \$\begingroup\$ Thanks for the update and elucidation. Your understanding is not correct, however. (1) "1 W = 1 j (sic)". This is wrong. 1 W = 1 J/s is correct. Power is measured in watts; energy in joules. They are not the same. (2) "W and Wh are interchangeable." No they are not. The watt (W) is a measurement of power. The watt-hour (Wh) is the integral of power with respect to time and it is a measure of energy. \$\endgroup\$ – Transistor Dec 1 '17 at 18:50
  • \$\begingroup\$ "7.5j x 3600 = 27kj/h". This is also not correct. 7.5 J/s × 3600 s = 27 kJ = 7.5 Wh is correct, and makes sense because running 7.5 W for an hour will consume 7.5 Wh. \$\endgroup\$ – Transistor Dec 2 '17 at 0:30
  • \$\begingroup\$ Oh dear, we do like to be pedantic about our little quibbles as regards symbols, so I neglected to put a few symbols in the correct way for you, it still does not get around the fact that your first comment was nonsense. The 7.5W stated still works out at 7.5Wh and PSU's are made in the way I said in my first iteration. \$\endgroup\$ – LateDev Dec 2 '17 at 11:57
  • \$\begingroup\$ Thanks for the feedback. Can you clarify what you think was wrong with my first comment? (I numbered the points.) 7.5 W only works out at 7.5 Wh if the unit is run for an hour. Units matter. Capitals matter. \$\endgroup\$ – Transistor Dec 2 '17 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.