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