Calculating power needed for laptop operating off a 12V battery

[MrAl]

I am trying to calculate the power and current needed to power my laptop off a 12V battery during astronomy sessions. I would attach a Bistek 300W power inverter to the battery to convert DC to AC and plug the laptop into the AC adapter of the power inverter.  The battery is an ExpertPower 12v 7ah Rechargeable Sealed Lead Acid Battery. the laptop is a Dell with a standard power adapter.  How would I do the calculation to determine the current draw on the battery?  I started by wondering what the current through the power inverter would be, but found my multimeter was not happy getting a resistance reading when I attached the meter probes to the leads of the cigarette lighter adapter of the power inverter (with the inverter unplugged and its switched turned on). Thanks for any help with this.

Hi,

Some good advice already in this thread i'll just add a little more...

First, nothing has been said about the life of the lead acid battery so i'll mention that first.
For these kinds of batteries you have to be careful how much you discharge it.  If you discharge it too low it will become useless quite fast.  That's the way many lead acid batteries work.  For applications like this maybe a deep discharge battery would fair better.

That said, i feel you are really following in the footsteps of the proverbial, "beating a dead horse with a stick".  That's because there will be more modern solutions now.  For one thing, there are probably lower power laptops available that will run much longer with a full charge.  That will mean longer run time right out of the box.  Second, there are mini computers out now that only use about 10 watts during the running time even with some software that uses some cpu power.  This is the basic box only (about 6 x 6 x 4 inches total size, plus wall wart).  You'd also have to get a small HDMI display panel and keyboard.  Hard disk would be from 64GB up to around 512GB depending on the model.  Total power could be under 20 watts, and some can run DIRECTLY from a 12 volt DC power supply such as a battery.

Now with that said too, back to the run time with the 12v 7 AHr battery on the laptop...
As another cute little proverb goes, "You're going to need a bigger battery".
In fact, you should probably go for a Li-ion battery not lead acid. Lead acid is too temperamental.  Li-ion is much more modern and can be recharged a LOT of times.  They come in 12v versions too that look just like a 12v lead acid battery.  To charge you get a charger that can handle those kinds of batteries.  What is also nice is you can get higher capacity with lower weight.  You can start at 20AHr and go all the way up to 100AHr or more depending on your weight limitations.  They are more expensive, but will last much, much longer than the lead acid counterpart.  You could check Amazon to get an idea how much it would cost.
Note that with a 20AHr Li-ion battery and the 12 volt mini computer solution you could get a run time of something like 10 hours, and part of the reason for that is there would be NO converter circuits between your battery and your computer.

Good luck with whatever you decide to do, and maybe you can get back here and tell us what you ended up doing and how well it is working out.

[Circuitous3]

By the way, how can you do astronomy with a laptop? Astronomers I have been around were aggressively bothered about any ambient light at all, because it would upset their night vision. Any light from a laptop screen would have made them go crazy.

I am mainly a visual observer; so far, no astrophotography for me. But recently an amateur astronomer friend gave me a digital camera (Svbony 250, an electronic eyepiece) to use to take short exposure pictures of bright objects (she's tryting to convert me over to the dark side!). The camera plugs into the laptop for capturing the images and seeing them in real time. You are quite right about astronomers shunning ambient light. I'm one of them.  However, some use imaging devices which project the images onto the laptop screen. They keep the brightness settings down and keep their distance from other observers. They can observe their images in real time and save them to the laptop. But because they are not looking through the eyepiece (their cameras are) they don't have to worry about their dark adaptation. Technology is changing everything, and astronomy is no exception.

[IanB]

I assume the lithium battery pack is similar to some of the powerbanks and would be rated for higher amp hours.

I experimented with a Ryobi tool battery, since I have some Ryobi cordless tools. A 4 Ah battery is about 75 Wh if you assume 18.5 V x 4 Ah. You can get adapters that plug on the top of the battery to use them for other purposes. For example, I looked at running a low voltage lamp in case of power outages, like this:

The cable is feeding the ~20 V DC from the battery direct to the bulb without any converters in the chain. The bulb has some internal electronics to drive the LEDs.

[james_s]

You mention trying to take a reading of resistance, that doesn't make any sense, did you mean current? I can't think of anything related to this where you'd be measuring resistance.

Keep in mind that the 7Ah rating of those SLA batteries is at a discharge rate of something like 1A, if you draw more current the Ah capacity will decrease pretty significantly. Your best bet is probably to use something like a Kill A Watt to measure what the laptop draws through the charger and then figure around 80% efficient for the inverter. It wouldn't surprise me if the battery pack in the laptop has a higher Wh capacity than a 12V 7Ah SLA battery (84Wh).

[Siwastaja]

You can reduce the wasted power by replacing the AC adapter - inverter combination by a 12V input laptop adapter, those things exist. I also suggest you measure the actual consumption. Maybe if your CPU load is not that high and you are able to dim down the screen, it might not be nearly that bad.

Are you suggesting a 12 V input laptop adapter like this?  https://www.ebay.com/itm/Nippon-America-DVCS-3500-12VDC-to-15-24VDC-3500mA-Universal-Laptop-Power-Adapter-/362681873610

Yes, something like that; I don't know about the quality of that particular brand. It's a simple boost converter and even if it's crap, it's still at very least 80% efficient, possibly more. Way better than the combination of inverter + ac power supply.

[Circuitous3]

Hi,

Some good advice already in this thread i'll just add a little more...

That said, i feel you are really following in the footsteps of the proverbial, "beating a dead horse with a stick".  That's because there will be more modern solutions now.  For one thing, there are probably lower power laptops available that will run much longer with a full charge.  That will mean longer run time right out of the box.  Second, there are mini computers out now that only use about 10 watts during the running time even with some software that uses some cpu power.  This is the basic box only (about 6 x 6 x 4 inches total size, plus wall wart).  You'd also have to get a small HDMI display panel and keyboard.  Hard disk would be from 64GB up to around 512GB depending on the model.  Total power could be under 20 watts, and some can run DIRECTLY from a 12 volt DC power supply such as a battery.

In fact, you should probably go for a Li-ion battery not lead acid. Lead acid is too temperamental.  Li-ion is much more modern and can be recharged a LOT of times.  They come in 12v versions too that look just like a 12v lead acid battery.  To charge you get a charger that can handle those kinds of batteries.  What is also nice is you can get higher capacity with lower weight.  You can start at 20AHr and go all the way up to 100AHr or more depending on your weight limitations.  They are more expensive, but will last much, much longer than the lead acid counterpart.  You could check Amazon to get an idea how much it would cost.
Note that with a 20AHr Li-ion battery and the 12 volt mini computer solution you could get a run time of something like 10 hours, and part of the reason for that is there would be NO converter circuits between your battery and your computer.

Good luck with whatever you decide to do, and maybe you can get back here and tell us what you ended up doing and how well it is working out.

Yes, thanks for the good advice and information. I'm actually due for a laptop upgrade anyway. And the lithium-ion battery sounds like a good way to go. I only started thinking about how to power a laptop in the field because of the digital camera I was given for the telescope. All the good information in the thread will help me figure out a good way to go.  I may just go with a temporary solution for now, to see how much use of the camera I will get.  But I've also been wanting to learn more about electronics and studying it recently. I'm finally making a dew heater system for my telescopes, something I've been thinking about for a number of years. The heater controller has 4 pulse width modulators controlling the voltage to each of 4 output channels. The heater elements plug into each channel and warm different accessories on the telescope to prevent dew formation (on the eyepiece or finder for instance). The current for each channel is only about 0.2 amp, so that 7 Ah SLA battery should power the elements for many hours.

[Circuitous3]

Very nice Ian! You'll be all set for the next power outage.

[Circuitous3]

You mention trying to take a reading of resistance, that doesn't make any sense, did you mean current? I can't think of anything related to this where you'd be measuring resistance.

Well, I was thinking that to get a sense of the power requirement of the power inverter-laptop system I should try to get a measure of resistance. Then I could easily calculate current give a 12 V supply. Then I wondered what the resistance of the power inverter would be, so tried to get a measure of that. But, I realize I would need to know the resistance of the inverter-laptop combination.  I suppose I could connect the inverter-laptop to the 12 V battery and then measured the current across the battery terminals, but I wasn't set up to make those connections, so was only thinking in the abstract. Was I thinking about this the wrong way? Having followed the thread, I realize now there are easier ways of meaasuring the power, such as with the Kill A Watt (which I would need to get).

[wasedadoc]

You mention trying to take a reading of resistance, that doesn't make any sense, did you mean current? I can't think of anything related to this where you'd be measuring resistance.

Well, I was thinking that to get a sense of the power requirement of the power inverter-laptop system I should try to get a measure of resistance. Then I could easily calculate current give a 12 V supply. Then I wondered what the resistance of the power inverter would be, so tried to get a measure of that. But, I realize I would need to know the resistance of the inverter-laptop combination.  I suppose I could connect the inverter-laptop to the 12 V battery and then measured the current across the battery terminals, but I wasn't set up to make those connections, so was only thinking in the abstract. Was I thinking about this the wrong way? Having followed the thread, I realize now there are easier ways of meaasuring the power, such as with the Kill A Watt (which I would need to get).

Measuring the resistance of an inverter input using only an ordinary multimeter does not give any meaningful information about how much power it takes when in normal operation.

[james_s]

Well, I was thinking that to get a sense of the power requirement of the power inverter-laptop system I should try to get a measure of resistance. Then I could easily calculate current give a 12 V supply. Then I wondered what the resistance of the power inverter would be, so tried to get a measure of that. But, I realize I would need to know the resistance of the inverter-laptop combination.  I suppose I could connect the inverter-laptop to the 12 V battery and then measured the current across the battery terminals, but I wasn't set up to make those connections, so was only thinking in the abstract. Was I thinking about this the wrong way? Having followed the thread, I realize now there are easier ways of meaasuring the power, such as with the Kill A Watt (which I would need to get).

Measuring resistance only works for a purely resistive load such as a resistor or heating element. The inverter is an active circuit, the power it draws will have no relation to whatever resistance you measure across the input circuit. It doesn't even make sense to talk about the "resistance" of an active circuit. You have to measure the current it draws while it is operating, which will vary depending on the load you apply to the output of the inverter. The power the laptop draws will vary widely too depending on state of charge of the internal battery and how hard the CPU is working.

[Circuitous3]

Measuring resistance only works for a purely resistive load such as a resistor or heating element. The inverter is an active circuit, the power it draws will have no relation to whatever resistance you measure across the input circuit. It doesn't even make sense to talk about the "resistance" of an active circuit. You have to measure the current it draws while it is operating, which will vary depending on the load you apply to the output of the inverter. The power the laptop draws will vary widely too depending on state of charge of the internal battery and how hard the CPU is working.

I'm confused about why the power drawn by an active circuit has no relation to the resistance of that circuit.  Just as I can measure the current through a circuit (say the inverter-laptop system) to get a measure of the power it uses, can't I also get the resistance of that circuit (since we know the voltage, say 12 V)? And can't I also measure the resistance of that circuit to get the current that it should draw?

[james_s]

I'm confused about why the power drawn by an active circuit has no relation to the resistance of that circuit.  Just as I can measure the current through a circuit (say the inverter-laptop system) to get a measure of the power it uses, can't I also get the resistance of that circuit (since we know the voltage, say 12 V)? And can't I also measure the resistance of that circuit to get the current that it should draw?

Because when you measure resistance, that is done by passing a very small current through the circuit from a low voltage. When you do this with an active circuit, the circuit will not be operating, when it is not operating what are you going to measure? Resistance is for measuring resistors, a switching power supply is not a resistor, it's as simple as that. If you measure the power draw you can work backwards and come up with an equivalent resistance that will draw the same amount of current but it still doesn't make sense to talk about the resistance of an active circuit. Something else you could read up on is impedance, with DC impedance and resistance are the same thing, but with AC they are wildly different, the concept is related.

For a very simple example take the humble incandescent light bulb. Technically it's a purely resistive load, however the properties of a tungsten filament are such that the resistance when cold is dramatically lower than at operating temperature. If you measure the resistance of a bulb you will be measuring it under conditions far outside that in which it operates so the resistance value you measure will be perhaps 10% of what it will be at operating temperature and your calculated power will be off by a factor of 10. To take a meaningful measurement of resistance, you have to do so under operating conditions.

[BeBuLamar]

I'm confused about why the power drawn by an active circuit has no relation to the resistance of that circuit.  Just as I can measure the current through a circuit (say the inverter-laptop system) to get a measure of the power it uses, can't I also get the resistance of that circuit (since we know the voltage, say 12 V)? And can't I also measure the resistance of that circuit to get the current that it should draw?

Because when you measure resistance, that is done by passing a very small current through the circuit from a low voltage. When you do this with an active circuit, the circuit will not be operating, when it is not operating what are you going to measure? Resistance is for measuring resistors, a switching power supply is not a resistor, it's as simple as that. If you measure the power draw you can work backwards and come up with an equivalent resistance that will draw the same amount of current but it still doesn't make sense to talk about the resistance of an active circuit. Something else you could read up on is impedance, with DC impedance and resistance are the same thing, but with AC they are wildly different, the concept is related.

For a very simple example take the humble incandescent light bulb. Technically it's a purely resistive load, however the properties of a tungsten filament are such that the resistance when cold is dramatically lower than at operating temperature. If you measure the resistance of a bulb you will be measuring it under conditions far outside that in which it operates so the resistance value you measure will be perhaps 10% of what it will be at operating temperature and your calculated power will be off by a factor of 10. To take a meaningful measurement of resistance, you have to do so under operating conditions.

I addition to what you said. If you measure the voltage supplying to an inverter and the current that it's drawing you will have a wildely varying current draw depending on the load connected to the inverter. So there is no way you can figure out how much current the inverter will draw by trying to measure its resistance.