watt hours and mAh?

I was looking at an external battery, but after reading into it, I realized that there is something mathematically wrong with its designate capacity.

20000 mAh/74wh .

(watt hours = mA * voltage)
Now assuming wh means the standard watt hours:
74wh /= 20 * any voltage above 4!


can someone explain why I'm wrong, or confirm my logic?

seen here (click external laptop battery):
http://www.sales-battery.com/External_laptop_battery.html

 

I dont think there is anything wrong with the battery's capacity....
there are more like these for sale....
and what is that statement for..??

 

Its the proof of the mathematical impossibility. Basically what it shows is that its impossible for any voltage above 4 to equal 74 wh since the mA is 20 (which is basically mAh/1000).

watt hours = mA * voltage

 

Maybe the 74WH is the input power(to charge it) and the 20000mAh is how long the battery lasts? Hmm... That is kinda weird though.. Confusing advertisements. But of course they're not going to tell you that you only get like 200mAh(random number, no calculations) with the highest voltage or whatever, that would cripple sales.

 

you know what you should do, buy a 1000amp cca battery designed for a big truck and then a good 300 watt power inverter, and you can go for days.

There is definitely a mistake in that webpage. I would email them and ask them to post the real values, cause those values are clearly wrong.

K-TRON

 

The incorrect wh is not just on that website alone. As angad1608 pointed out, its on others offering the same/near identical 20000 mAh capacity.

@Syndrome. I doubt it, watt hours are used to describe capacity.

 

Wow...

Ok, a LOT needs to be cleared up here. First, some key terms:

energy - I don't have a good definition, but you all should know what it is. One unit of measurement for energy is the joule.
charge - The actual charged particles that transfer energy. A coulomb is a standard unit of charge.
voltage - potential difference, the energy of a charged particle. Measured in volts(joules per coulomb).
current - how much charge is flowing. Measured in amps(coulombs per second).
mAh - milli-amp-hours. Another unit of charge commonly used in electronics industry. This is a measure of how many charges a battery can store for later use.
power - Rate of energy transfer(joules per second). Measured in watts.

Now a few more equations:

Power = Current * Voltage (finding power for circuits)
Energy = Power * Time

There is a huge misunderstanding in this thread, and that is the difference between watts and watt-hours. Watts is a rating for power, and watt-hours is for energy. Remember I said that power is rate of energy? Rearrange this and you get Energy = Power * Time. This is where watt-hours comes from.

Ok, now to address the issue here. This battery is clearly capable of being used with multiple devices and can output different voltages. Therefore the battery must have some sort of device between the battery to increase the voltage from the battery depending on the device. My point is, the battery itself likely does not charge to one of the listed output voltages.

With a little math, a 74Whr battery with a charge capacity of 20Ah would have an output voltage of 3.7 volts. A very common voltage for your standard lithium cell. And this battery is lithium based.

Now that I do the math and see that I have gotten some similar numbers as some of you(including the OP, for one), I see you seem for the most part to know what you are doing and there were just a few typo's throughout your posts. But I'll leave all the information above in case anyone else finds it useful.

The bottom line: It is probably safe to trust that the battery is a 74Whr battery. Go by the energy rating of the battery. That's why it is there. Charge capacities(mAh or Ah) are usually just for marketing ploys and really don't matter.

 

First of all the mAh and the Wh are defined as what the laptop consumes while on battery only. Having said that it would seems that the Voltage of the battery would be 3.7V (74Wh/20,000mAh), what bothers me is what laptop is that? Since 20,000mAh sounds attractive for an external battery, I think the 74Wh is wrong.

My laptop consumes about 80Wh, and the battery capacity is 7,200mAh for 11.1V. (3.5h autonomy).

 

The advertising says External Laptop Battery, and as I said 3.7V to drive a laptop seems to me very low and I don't know any laptop using that voltage. Capacity is measured in mAh or Wh, which are normal units to measure how long the battery will last supplying energy to the device, because if you have the current then you can estimate the battery life in hours.

BTW, my battery is Lithium and the Voltage supplied to the laptop is 11.1V.

 

Your laptop lithium battery has more than one cell. Probably 6. If you wire three 3.7volt lithium's in series, you get 11.1volt output voltage. If you take another three and wire them in parallel with the first three, you still get an 11.1volt battery with double the capacity.

That battery plugs into the normal power input on your laptop. It does not connect to where the normal laptop battery does. The input voltage for different laptops varies, and while your laptop is designed to be powered from an 11.1volt battery via the battery bay, it is probably designed to be powered by a different voltage via your AC adapter(where this battery plugs into your laptop).

mAh is a charge capacity rating, and Whr is an energy storage rating. They are quite different.

You can do that. But if you know the energy capacity of the battery and you know the power draw of your device, you can also estimate battery life. Also, when a battery becomes part of a complex system like this one, it is much easier to deal with energy.

You are missing something very important- the device the OP linked to is more than just a battery. It is a battery with some far from simple electronics between the battery and whatever you choose to power with it. The electronics not only monitor the battery and display information via the screen, but it changes the electricty from the battery so that it is appropriate for and will not damage the laptop you plug it into. It changes the voltage from whatever the battery output voltage is(probably 3.7volts) to what it thinks the laptop is supposed to get. It order to do this, lots of low energy electrons(high current low voltage) come from the battery and go to the voltage adapter, then very few high energy(low current high voltage) electrons leave to go to the laptop.

This is why energy capacity(watt-hour) is more useful/easier to use than charge capacity(mAh). If you connect a laptop that takes 2000 milli-amps(pretty standard, at 15 volts that would be 30 watts) to this battery, it would seem that the laptop would get 20,000mAh/2,000mA = 10 hours battery life. This is not the case. It is actually about two and a half.

For the device to put out 2,000 milliamps at 15 volts, it would take 8108 milliamps at 3.7volts from the battery. This gives 20,000mAh/8101mA = 2.46 hours. Don't believe me? I'll do it the simple way. 74Whr battery / 30watts = 2.46 hours. See why Whr is good and mAh is just a marketing ploy?

 

I actually have 9 cells.

I was actually wondering that. In that case the situation is worst, because normally the voltage at the laptop power input is higher than the voltage from the battery to the laptop, in my case the power adapter supplies 19V and about 4.74A. So, 3.7V will definitely adapt very badly to the DC/DC laptop internal converter.

Can you relate the mAh to the Wh?

Is not just a question of compensating with current, the battery must supply the required voltage to the DC/DC internal bay, it cannot increase the current to keep the same power, is an interface requirement.

That was exactly my point, what is the advantage of having an external battery that supplies less autonomy than the internal one? I think 10h is an excellent proposition, that's why I think the 74Wh is wrong.

Ummmmm, here you are going a little bit too far, battery autonomy is always associated with mAh, Wh is rarely used. If I have a system requiring a certain power, VxI, and I know the Voltage requirement, then I can know how much current the system draws, and if I want the system be 8h autonomous, then I would do Ix8h, this is universal.

 

As OP suspected the math is wrong. I spent an hour on it a while ago, like hours. First before we criticize the add. To OP w=mA*v is wrong! Does not help to clear things up! watt hours? No sh#t! Call them watts as it is given. A watt hour I believe is a watt by definition. Volt has no time constraint.

I have already complained w=mA*v? Wrong w=A*V that is it! I do see OP making the conversions so just miswrote the formula but still frustrating.

OK Amps are done per/s so to convert A to milli Ahm one might times by A times 1,000 to get mA's? But then to get mA to mAhm? Well since A are based on seconds? I figure you must times (X) by 60? To get minutes. Then to get hr's times (X) 60 again to get hours? Then you have your numbers and they still don't work? In simple equations?

Here is a definition?

The ampere, in practice often shortened to amp, (symbol: A) is a unit of electric current, or amount of electric charge per second. The ampere is an SI base unit, and is named after André-Marie Ampère, one of the main discoverers of electromagnetism.

That is the trick, take a flow rate and make it a quantity? It involves a much more complicated equation of which we lack the variables? Accident? I doubt.

Some one make it work or shut up, please!

Three wires? voltage? You Kids!

 

Ohh, now I get it, the family Watts is the marriage between Mr. André-Marie Ampère and Mrs. Volts?

Thank you pp, now I can go to bed feeling more intelligent than yesterday

 

Then you've got 3 sets wired in parallel of 3 cells wired in series.

What exactly do you want to know? Give me a question and I'll answer it.

 

I hate to double post, but... you seem have no idea what you're talking about. You're just playing with numbers, prefixes and definitions.

First, there are a number of typos in the first post. Just look at the numbers he used if you can't figure out what he meant to write.

There is a big difference between watts and watt hours. You also seem to misunderstand a number of other things. Read through my first post if you want to learn more of the basics.

All the necessary equations are here.

I already "made it work", so don't yell at me for "just complaining". I'm sorry I had only negative things to contribute in this post, but I already explained everything and have nothing more to say.

 

WOW JPZ? I am confused?

 

Confusion of watts and watt-hours

Power and energy are frequently confused in the general media. Power is the rate at which energy is used. A watt is one joule of energy per second. For example, if a 100 watt light bulb is turned on for one hour, the energy used is 100 watt-hours or 0.1 kilowatt-hour, or 360,000 joules. This same quantity of energy would light a 40 watt bulb for 2.5 hours.

 

flip! Answer the question. We know the theory! Give us the answer!

 

What answer do you want? The first post in this thread is correct, it is a 74Whr battery, and it should last from 2.5hours to 5 hours depending on how much power the laptop draws(I used 30W to 15W for estimates). You can expect this battery to last as long as the standard battery that came with your laptop.

 

I agree with JPZ: Watts = Volts x Amps (for DC power) and watt-hours is the only useful indicator of power capacity and can be made up by different combinations of volts and amp-hours. The Amp-hour capacity of the battery in the original post is if all the cells are in series so it 20 Amp hours x 3.7 volts.

If my computer uses 12W power and the battery capacity is 48 Amp-hours then it can (in theory) run for 4 hours.

John

 

Even if you understand? I have serious doubts? You lack all ability to express! You should consider that? You are not dealing with dumb people here. And my impression until you answer in a concise way? Do not baffle us with your brilliance! do the math which up to now you have not! 74 vs 20000 do it! I do not think you even get the way they twist a term like amps which is per/s and convert to hours! Explain or shut up!

 

OK. Let's try and put it more simply:

It is common for battery capacities to be expressed in amp-hours. For example, an automotive battery may be 60 amp-hours which means it can (in theory) deliver 60 Amps for one hour or 6 amps for 10 hours or 600 amps for 0.1 hours. Since most automotive batteries use a standard 12V, so no one bothers to give that info. However, if you include the voltage then the battery capacity can be aso expressed as 720 watt-hours (where, as already said, watts = volts x amps).

The battery in the original post has a rating of 20000mAh = 20 amp-hours. It also has a rating of 74 watt-hours. Nowhere is the voltage stated.

However, turn around the formula watts = volts x amps and you can get volts = watts / amps. Therefore the voltage in that battery is 3.7V.

As already noted, 3.7V is the normal voltage for a single Lithium-Ion cell.

Still not clear?

John

 

Not only did I already explain everything you have asked, but you tell me that I cannot effectively "express" myself when I can barely understand what you write. Maybe if you wrote complete sentences I would understand what you're trying to ask; maybe if you read my posts you would understand the basics of electricity, power, and energy.

First of all, you can't "convert amps to hours". Current(amps) is the number of charged particles that flow through the circuit per second. You can't convert particles to time.

P = IV (power = current * voltage)
P=W/T(power = work[energy] / time)
By rearranging P=W/T, you get:
W=P*T(work[energy] = power * time)

If you substitute P=IV for P in W=P*T you get:
W=IV*T or for clarity, W=IT*V(notice unit analysis- amps*hours*volts)

This is how to get Whr rating for a battery. Whr= amps*time(mAh) * volts

You want me to calculate Whr from mAh(since you so clearly disliked it when I calculated V from mAh and Whr like the OP asked)?

Well there's a huge problem here. What do I use for voltage? Look at the battery page which the OP linked us to. Do I use 19 volts? Do I use 16 volts? Do I used 12 volts? What about 5 volts? Here, I'll do all of them.

Whr = ITV(mAh * V)(in order to get Whr as final units, I need to convert to amp-hours(Ah) from mAh so I will use 20Ah instead of 20,000mAh for the battery)
Whr = 20Ah * 19 volts = 380Whr
20Ah * 16 volts = 320Whr
20Ah * 12 volts = 240Whr
20Ah * 5 volts = 100Whr

Does this mean the Whr rating depends on what laptop you plug into the battery and what voltage the laptop needs? Hell no. The Whr rating of the battery is constant. Can you see now why the 20,000mAh rating is useless? This is a 74Whr battery. You want to know how long it will last? Use Power = Work / Time. Rearrange to get Time = Work / Power. Battery life = Work(Whr) / power(watts consumed by your laptop).

Nothing new in this post. Nothing "brilliant" about this post, or any other one in this thread.

 

So, your conclusion is 20,000mAh is useless, and 74Wh is constant and the key number.

Well, let's do this very painless, and tell me where I'm wrong, just say it beside my point, and state why I'm wrong.

1. VxIxt = cte. This is what you said, right?

2. V is cte, we know this because is an interface requirement, is this still fine?

3. If V is cte, then forcefully Ixt must be also constant, is this fine?

4. Then I and t will change in opposite directions, if a system need more current, then t will go down, and if the system needs less current, then t will go up, am I still in line here?

5. Then if Ixt is cte, then the mAh is constant and therefore representative of the capacity of the system.

Now let's play a little, you are saying that 74Wh is the right number, so let's see if this battery will work with my laptop. My laptop consumes 22.83W as per the manufacturer.

6. Then 74/22.83 = 3.24h, is this fine?

7. And V = 11.1V (manufacturer spec), then I = 22.83/11.1 = 2.057A, is this right?

8. Finally, 2.057x3.24 = 6,664mAh.

Then, why is the spec saying 20,000mAh?????????????

 

I'm sorry, could you explain what CTE is/stands for? IVT=Energy(from your first point). But you then(in no. 2) say V = CTE, so either V is something other than voltage, or perhaps CTE means something else. The most logical answer to me is that CTE means that something is constant.

 

Yes, that's correct, cte means constant, sorry for not being clear.

And shorter your answers the best, I fall asleep when the responses are too long

 

Ok, in that case:

This is not constant. It depends on the device plugged into the battery device which the OP linked us to. Check out the page; it is rated for 19V, 16V, 12V, and 5V.

Yes, assuming that the battery has been accurately rated for 74Whr.

This means that the battery is equivelant to an 11.1V 6664mAh battery. The actual battery cells are not 11.1V; they are much lower(likely 3.7volts).

You seem to be missing the point that this is not a simple battery connected to a simple electronic device. The device which the OP linked us to is more than just a single battery cell- it contains a voltage regulator. There are two separate circuits- the battery cells connected to the voltage regulator, and the votlage regulator connected to the laptop.

EDIT: To make my final point clearer, since voltage input to the regulator does not equal the voltage output of the regulator(that's the whole point of the regulator), current into the regulator cannot be equal to current out of the regulator, otherwise there would be no conservation of energy(e.g. you could put 1 amp at 5volts in(5 watts) and get 1 amp at 10 volts out(10 watts)).

 

Are you saying that this device will work as an autonomous charger?

 

Define autonomous charger.

 

Imagine a normal power adapter that is connected to the AC to 1. Power your PC and 2. charge your pc battery. But now you are on the road and don't have that adapter.

Now, you get this device, you charge it, and then is charged. Then you are working with your laptop and run out of power, then you connect this device exactly where the power adapter goes (power in) and it will provide power to the laptop and at the same time charge the battery, all this without being connected to the AC, is this the use of this device?

 

You could use the device like that, or you could plug the device in to begin with, your laptop will run for a few hours as if it were plugged into AC, then when the external unit dies, you can unplug it and run your laptop from the battery like normal.

I don't see how this applies to the life of the external battery unit from the first post.

 

The way they market it is as an "external battery". So, if my battery is dead, and that I connect this device in the power in, then just because the internal circuit of the laptop, it will power the laptop and charge the battery, unless I remove the battery, which I don't suggest to do.

I think the 20,000mAh has to do with the overall current, in another words, the device is acting as a charger, providing current to the laptop but it also providing current to charge the battery.

Then I agree that the mAh is bigger. The normal mAh we are used to work with is the one coming from the battery into the laptop, no charger is connected. But when the charger is connected then the current almost triple, because is charging the battery and at the same time powering the laptop.

Anyway, nice talking to you mate, have a good night.

 

This device plugs into your laptop where the AC adapter does; it's basically as if you had access to AC for a few hours- you can use it however you want, to power the laptop, charge the battery, or do both at the same time(although that depends, the device must have a current/power output limit). This is really unrelated to the topic of this thread.

No, that is definitely wrong.

If the battery in the laptop is already charged, then the external battery will not have to charp the laptop battery. There are a lot of things wrong with what you said.

I don't know about you, but I'm going to be up for another 6 hours or so.

I still think you're missing something REALLY big here. I'll quote it from my previous post. If you're starting to fall asleep, now is the time to wake up:

Since the actual 3.7volt cells are so low in voltage compared to an output voltage of say, 12 volts, the cells have to put out more current at low voltage to get less current at a high voltage. This explains the need for such a high mAh battery in this device, and why such a high mAh cell doesn't last very long.

For example:

Say you've got your laptop(for simplicity I'll use 12 volts as the DC input from the AC adapter for your laptop and 24 watts as the power draw of your laptop; the numbers you gave me are inaccurate anyway).

24 watts / 12 volts = 2 amps(find the current requirement for your theoretical laptop).

So your laptop is drawing 2 amps at 12 volts from the battery, for a total of 24 watts.

Assuming a 100% efficient voltage regulator, the batteries must put 24 watts into the regulator to get 24 watts out to power your laptop. With a battery voltage of 3.7, this gives:

24watts / 3.7volts = ~6.5 amps

Even though your laptop is getting 2 amps, the battery cells have to put out 6.5 amps!

Remeber we agreed that battery life will be ~3.25 hours? To find the mAh of the actual battery itself, we get:

6500milliamps * ~3.25hours = 21125 mAh for the battery. However, 3.25 hours is incorrect(that was for your 22.xx watt laptop, I used 24 hours for this example. If I calculated the hours exactly, this number would have come out to be 20,000mAh).

So this is a 3.7 volt 20,000 mAh battery. Or an 11.1 volt 6664mAh battery. or a 7.4volt 10,000mAh battery... etc.

 

Not exactly, you use it like a secondary battery, you charge this and your laptop battery, then you let the external battery drain first and you finish the job on your laptop battery.


On another note, how on earth did Powerpack got banned???

 

That would be the common way to use a device like this, as I already explained. However, you make it sound like you have to use it like this, which is certainly not the case.

 

He was banned for consecutive offensive posts he made on other threads. Hes only banned for a few days. It should be a time for him to reflect at the posts he makes.

But anyway, i finally caught up at what you guys were talking about.

 

Powerpack wasn't one day good and another day bad, he was just a human, he would peacefully contribute and also become impatient when people would not answer directly a question, then he would kind of shake the cage a little. He also had a "hot pepper" humor and I think this is what really got him in trouble.

I would love to think that is gone just for a few days, but he was "permanently" banned.

 

First of all, is not good to have two battery connected to a laptop at the same time, so if you connect this device, then you would have to disconnect the main battery, which I don't recommend because laptops don't work well in the absence of the main battery and just being powered by the power adapter or this universal device.

If you connect this device where the power adapter goes, and the battery is also connected, then you will have two sources of power, and you risk to have currents going among both batteries.

My last concern is that the laptop power in connector is not an input for a battery, it doesn't have all the required pins interface, please see the attached drawing.

A last concern, an universal power supply like this device, doesn't have good voltage regulator circuit and you could fry your laptop.

 

Why is it inaccurate?

I have already gave you the real numbers, my laptop is an IFL90, it has a 79.92Wh/7200mAh sony battery, and the battery voltage to the DC/DC converter block is 11.1V. And the battery last in average about 3.5h. Do the maths yourself.

How can you drive my laptop with 3.5V? My laptop needs 19V at the power in connector (from the power adapter), or 11.1V at the battery - DC/DC interface. Is not just current, the powering device MUST meet the voltage requirement, otherwise it won't work.

 

You just don't understand that there is a voltage adapter built into the battery the OP linked to us. The cell(s) inside the device output 3.7volts to the adapter, and the adapter outputs(hopefully) the 19volts your laptop needs.

Because this device plugs into the DC jack on your laptop, not by your laptop's battery interface, and it requires 19volts(not 11.1volts). Also, the voltage of your battery pack is not constant, it starts probably around 12volts and works its way below 11 volts, possibly down to 10volts depending on the cutoff voltage set for your particular battery.

I didn't have the voltage requirement for the DC in on your laptop(19 volts). Also, your supposed power requirement of 22.xx watts is probably wrong, especially if the manufacturer told you that as I remember you claiming.

There's nothing wrong with having two batteries connected to your laptop. Your laptop battery is after all 6 little batteries(or 9, I don't remember if you have a 6 cell or a 9 cell).

Who told you that you can't power a laptop without the battery just from the DC in? There's nothing wrong with doing so except that you won't have a battery backup in case there is a power failure or the external battery dies.

This is not a problem. There are very complex power circuits inside laptops. Many laptops even have dedicated circuit boards for power regulation. Your laptop was actually designed to have two power sources connected, and there is no risk of current going from one to the other(except for charging the internal battery from the external power source if the internal battery is not full, which laptops are designed to do).

Right. That doesn't mean you can't use it. It just means that all you can do is power your laptop- you can't monitor the battery and your laptop will have no way of knowing when the external battery is going to die. This is why that external battery has an LCD on it for monitoring.

It may or may not- this one looks rather cheap and probably doesn't have a very good power circuit. However, your laptop has an excelled power circuit and I highly doubt you could fry your laptop with this device.

I didn't say from day to day... I said from time to time. In some of his posts he seems like a polite, perfectly reasonable person. Yet in other posts, he makes no sense, can barely even speak English, and did nothing but write abusive things. To me, it seemed like there was another person using his account at times.

 

I don't where are you getting that I don't understand, I have already said that my laptop power in needs 19V and that the OP device supplies that. Is you who is mixing things up talking about the 3.7V. If you are working this device as an external battery, how are you planning to get the 3.7V up to 19V, using a magic op-amp, who will feed the om-amp, the fluorescent lights?

Please read my entire post, I have already made this point very clear, no need for you to claim that I have said the contrary.

Oh, I'm sorry, next time when I'm flying and that the pilot says that the plane speed is 450mi/h I will challenge that because there is not such a thing constant speed, because you have so many factors accelerating or decelerating the plane.

Interesting, I have given you the 79.92Wh/7200mAh battery capacity, and told you that the battery last 3.5h and you cannot calculate the power consumption? Take an electricity class dude!

The manufacture doesn't recommend it, Compal, call them and tell them they are wrong so you give me a break.

Thank you, one credit for me.

When you spend over $2k for a laptop you don't stay in front of such device and, hmmmmmm, may be I fry my laptop, may be not, may be yes, may be not............... You want to make sure, those universal power supplies are known for having a bad voltage ripple and can damage sensitive circuits of a laptop for example.

 

You still don't get it. I guess I'll have to draw a picture. The cells in the external device output 3.7 volts to the regulator inside the external device. The regulator in external device brings 3.7volts up to what your laptop needs(in your case 19 volts). The external device outputs what your laptop needs(in your case 19 volts).

At the time when I posted(two of my posts back, when I said your numebers were inaccurate) I didn't know your laptop required 19volts at the DC in. You hadn't told me yet, but I knew it wasn't 11.1volts.

Hey, you asked why I said it was inaccurate. There's a reason I didn't say why in that post. Anyway, my was that 11.1volts for DC in on your laptop is inaccurate, and because of that, your estimated current was off. MY second main point was that the power draw of your laptop depends on what you are doing, and it probably won't be the 22.xx watts your manufacturer tells you it is.

I don't remember you telling me that your battery currently lasts 3.5hours in practice. I could easily do the math, but there's no point. This is irrelevant to the topic.

Please, don't start going the way of PowerPack. I could easily have told you that you were blatantly wrong and have no clue what you're talking about a dozen posts back and left it there. Instead, I'm trying to help you understand this and to teach you something.

If don't know the circumstances, but Compal probably recommended that you leave the battery in because there is no point in taking it out, and that it is a good idea to keep the battery in as a UPS. Did Compal specifically say that it would damage your laptop?

Here on NBR we have entire threads full of people asking about using their laptops on AC only and taking the battery out. Everyone seems to think that using a laptop with the battery in and plugged into AC is horrible for the battery and will destroy it.

Personally, I always keep my battery in. You're one of the few if not only other person on this forum who thinks it is a good idea to leave the battery in, so don't think that I'm trying to argue that leaving the battery in is bad. I'm just don't think that leaving the battery out will physically harm a laptop. If you want to continue this subject about batteries in/out, it should be taken to a different thread. I'm done talking about it here.

What's this about credits? You get a credit for saying that a DC jack is DC in and nothing more? That doesn't even pertain to the subject at hand...

This is also going off topic. I'll address it one last time, but that's it for here in this thread. If you paid so much money for your laptop and you truly care about it, why would you even consider buying some really cheap third party device from a no-name company, especially to power your expensive laptop? Objectively, it may or may not fry your laptop. It probably won't. Subjectively, there's no good reason to test the quality of some cheap 3rd party device or the quality of your laptop.

I'll get to drawing that picture for you now.

 

Sorry to double post, but forgot about this and just couldn't let it go-

Op-amps... first of all, op-amps are for amplifying signals, not for powering devices. Op-amps are for lower power amplications and have ridiculously high impedances. Yeah, you could use them as part of a power circuit, but you would never use an op-amp to directly power a circuit. Also, the output of an op-amp can't exceed it's supply voltage- I figure this is probably why you said I would need a "magical" op-amp(otherwise known as an ideal op-amp) but this alone would make it pointless to try to amplify 3.7volts with an op-amp because you wouldn't be able to get more than 3.7volts from the op-amp.

 

Count me in. I also believe that it is good practice to keep the battery in place, even when running on mains. The charging circuit should only charge the battery when necessary.

John

 

Great! That makes three of us and counting...

On a side note, I really don't feel like drawing any pictures right now and I need to get back to some physics work.

 

Ok, so how do you bring the 3.7V up to 19V when you are not connected to the AC?

 

If you agree with me, why are you arguing with me? Re-read your sentence before posting, if you want to contradict me, then be consistent at least.

John, I'm the first to say that the battery should be kept in the laptop, JPZ just contradicted me for the sake of it.

Who is the third??????

And if you cannot calculate power I doubt you can draw as well.

 

This is an unclear question, but I think I know what you're trying to ask. I assume you're talking about the cells in the external device and not the 3.7volt cells in your internal laptop battery because you talk about increasing electrical potential from 3.7volts to 19 volts(presumably for DC in). You say not connected to AC, which I also assume means that the external device is plugged into the laptop rather than the DC out from the AC adapter.

Assuming the above, I could rephrase your question to:

How does the voltage regulator inside the external battery device output 19 volts when it only gets 3.7volts from the cells inside?

That is a very good question. Only the manufacturer can answer that question. I don't have any information about how they designed the electronics for this device. For this reason, we have to treat the regulator as a mysterious "black box".

 

I think you need a coffee.

How do you bring the 3.7V from the OP device up to 19V? Is this device creating energy?

 

Read what I wrote. I read it multiple times before posting. I specifically wrote:

I also wrote:

That was my argument.

I am the third. I specifically said so.

If you continue to make comments like this I'm not going to help you. Have you heard the phrase "biting the hand that feeds you"? I have more important work to do. If you won't even take the time to read through my posts and continue to act as if I don't know anything, you will be none the better for it.