Safely getting from ~12V to 9.5V -- high capacity?

[Nazo]

Ok, so I just bought one of these: WeatherX RPS8807DT 200-Watt Portable Power System It was my hope to be able to use this thing to run my EeePC for extended periods of time off of this and this was one of the major reasons I bought it (though I would add, obviously not the only one as it has a lot of uses obviously.) But, most of the battery power sources of the nature I was looking for don't go up to a high wattage, do but have poor capacity, or cost a LOT. This one seemed a great balance with incredible running time for its price mainly just at the cost of being a lot heavier due to using a lead-acid battery instead of nickel-metal hydride or nickel-cadmium (or lithium-ion, though I don't think I've seen any of these in a power device of this nature yet.) So basically this thing is using the same type as a car battery, just on a much smaller scale. There's only one thing bugging me. To generate AC it runs a motor rather than the newer methods of using voltage conversion and generating a sine wave. This looses efficiency obviously by quite a notable margine (according to the product page, 60 watts will only get about 2 hours of power via AC and it will get about 3 hours DC.) Also, I was a little concerned at first because when lower powered devices are plugged in (for example, a LED nightlight) the produced voltage can go up to as much as 144V (but when I plugged in an AC powered fan that runs at some 350mA it dropped to 110V, so I feel less concerned about that now.) This could probably be risky with lower powered devices such as some chargers, right?

Anyway, for the moment my biggest concern is my EeePC. It can use up to some 22 watts or so (though a fair bit less when just charging while off -- I haven't gotten around to testing how much it really uses yet, but the plug is rated at 22 watts and I can run the EeePC off of my 16 watt maximum sustained supply if I pull out the battery or charge it if I leave the system off, so I think it comes close to that 22 watts when fully used, or at least I know it's more than 16) so I guess it would be ok in the voltage department as that should drop it really close to the rated 120V -- well within safety ranges. Of course, this still puts it at a high enough power usage that the efficiency difference has enough of a significant effect to be worth trying for something more efficient. Instead of going from DC to AC to DC again, perhaps I should just go from DC at approximately 12 volts to DC at the 9.5V the EeePC is rated at. Unlike in a car electrical system, the DC does seem to be somewhat reliable with the maximum being close to or very little over 12V, but at the same time I think this may mean that if it drops low enough it could maybe go belowe 9.5V?

Basically, my biggest concern here is that this thing has a LOT of capacity behind it. Not as many amps as a car battery, but definitely enough to do some serious damage if things went wrong. Also, I wonder if a normal voltage converter could handle that much power? Most importantly though, I must admit that I've never put together any sort of voltage converter before. I'm not really sure what the little voltage converters themselves are called to look them up on RadioShock/Mouser and I'm not really sure exactly how to use them as they usually have three pins rather than just an in and an out. If anyone could at least help point me in the right direction, I'd be grateful.

Also, just out of curiosity, what do you people think about plugging in a PC? Obviously it wouldn't be very often, nor for very long, but, it says that this thing can handle up to 200 watts. I'd have to get my video card in low power mode or something probably, but I could probably run even this PC on that as its normal voltage when not being pushed is in that vicinity and I have some older hardware that I'm almost certain would use less. The main thing is that the voltage would probably drop pretty low with that much load. Could the PSU handle that much of a drop?

 

[Thor]

i was trying to come up with a good way to make sure that the devices will play well together, why they didnt just make the eee 12v to begin with i dont know.. then you could just plug it right into a lighter outlet and be done with it..

that said, the only other thing i can think of would be to build your own batery box, as the batterys are a known voltage, if you want longer life you run then in series+parellel (get your voltage right, then stack the batterys to achieve a higher amprage, ,

the down side to this is that you need to get to 9.5 v which is not doable, 9v is, and 10.5 is.. but not 9.5, there is a chance that the eee may not care about 9v vs 9.5.. it may just use more amps to make up the difference depending on how it deals with regualtion.. most laptops wont care about a .5v difference. so i doubt it will..

useing 6 sanyo HR-DU NiMH 10,000MaH batterys will get you about 5.5 hours of run time for about 115$ .. but if you run 2 or more pack in parrellel (12,18,24 cells) the time will grow accordingly (24 cells would be about 22 hours of runtime. but these are slightly larger (i beleave) than d cells batterys, so they are big... and a 24cell box will be considerably large, slightly smaller than a car battery, but alot lighter. so.. 450$ worth of batterys to power a 400$ laptop.. sounds about right to me.

i suggest trying to see if they have a car adaptor out there and just use that instead

/edited math was off
/edit again 1 6cell pack, after boxing will weigh in at about 2.5 (2.28 not includeing the box and wireing) pounds... so 24 cells would be slightly less than 10 pounds

 

[Michael Daly]

Why not just use a UPS for the Eee PC? It might cost a bit more depending on the model, but it would be much more reliable and you know it's specifically designed for that use.

Otherwise, you could get a generic 12V sealed lead acid battery (not a car battery) and connect a 200W inverter to it. Just choose an amp-hour rating that gives you the run time you want. You'll need a charger as well.

That gadget is not really worth it for powering a PC, since you're paying for a lot of other stuff.

 

[Mohonri]

You can step 12V down to 9.5V a couple of ways. The first way is to use a DC-to-DC converter, which is a bit complicated to build, and I'm not sure if there are commercially available ones that will give you 9V. The second way is to use a series of five diodes to drop the voltage down. Assuming 22W @9.5V, that's about 2.3A. If each diode drops about 0.6V, that's close to 1.5W that each diode will have to dissipate. You'd have to make sure that you choose diodes that can handle that kind of power dissipation.

One more option (similar to the second one mentioned above) is to use a linear regulator, which would function the same as the series of diodes, but do it all in one package. Since it would be dropping [email protected], you'd need one that can dissipate at least 6W, so that means you'll need a heatsink. It's worth noting that with both the diode and linear regulator concept, you'll be wasting about 20% of your power as heat. The DC-DC converter will be much more (>95%) efficient. However, the regulator is very simple--the circuit will consist of as few as two parts--the regulator and a resistor to set the voltage.

 

[Nazo]

You can step 12V down to 9.5V a couple of ways. The first way is to use a DC-to-DC converter, which is a bit complicated to build, and I'm not sure if there are commercially available ones that will give you 9V.

I have no idea how one would build one. I looked these up and they are definitely pretty expensive though. I actually found one that could do 9.5V, but I'm pretty sure that they don't sell one seperately. (Mind you, they were selling them at a price that would work out to something around $2 if they did, but I'm sure that without that bulk it would cost a LOT more and a typical price for one seperate among the others there was more around $30...)

The second way is to use a series of five diodes to drop the voltage down. Assuming 22W @9.5V, that's about 2.3A. If each diode drops about 0.6V, that's close to 1.5W that each diode will have to dissipate. You'd have to make sure that you choose diodes that can handle that kind of power dissipation.

That sounds like a lot of power dissipation there and a lot of waste... I'm not sure how to do it with diodes, but wouldn't keeping them from tearing up be a bit of a challenge? It does strike me that if there is no way to get it right to 9.5V, maybe it could be gotten close and then a diode go the rest of the way maybe? Well, it probably doesn't have to be exactly 9.5V. I have no idea of tolerance levels, but if I get it within, say, 5%, surely it can handle that much? After all, its AC-to-DC plug is a simple very small little thing designed for travel. Not at all the big power supplies you get on most laptops these days. It probably has to tolerate a bit of inconsistency. I'd rather assume a relatively low percentage though since I'm talking about running this thing for relatively long periods of time.

One more option (similar to the second one mentioned above) is to use a linear regulator, which would function the same as the series of diodes, but do it all in one package. Since it would be dropping [email protected], you'd need one that can dissipate at least 6W, so that means you'll need a heatsink. It's worth noting that with both the diode and linear regulator concept, you'll be wasting about 20% of your power as heat. The DC-DC converter will be much more (>95%) efficient. However, the regulator is very simple--the circuit will consist of as few as two parts--the regulator and a resistor to set the voltage.

Hmm. This is what I was asking about. However, these waste the energy as heat? You say about 20%, so this sounds like you're saying that it's actually converting each volt not to be used directly to heat. That seems like it could be a lot. Are you sure though? I have a device that sits in my car's drink holder that provides power to USB devices and this thing is about as simple as it gets. It uses a voltage regulator (I see I had the right term for what I was asking about earlier) to go from 12+V to 5V. If as much as you say were converted directly to heat, this thing should be pretty rough on it... Ok, USB is a lot less power obviously (0.5A is what the spec requires it to provide) but that still works out to dropping 7V @ 0.5A so around 3.5 watts by your model, but to have no heatsink or, in fact, any form of airflow or anything I wonder if it's really converting every last bit of that directly to heat? Considering that I left this thing plugged in all the time, wouldn't it be practically melting through that thin plastic casing?

Well, anyway, I have a lot of left over heatsinks and thermal compounds (and a dremel tool should one otherwise perfect turn out to be too big.) I can definitely do that much. I guess the real question is, where would I find such a regulator and how would I build a circuit that uses it? The USB power device seems pretty simple, but I'm not quite clear on what it's doing. I see a 220uF capacitor across two of the pins for some reason. Assuming I was to go with such a method and could find a correct regulator somewhere, what would I need on mine?

Of course, I guess it doesn't matter if I can't find one... Mouser seems to have failed. I see only converters there...

i was trying to come up with a good way to make sure that the devices will play well together, why they didnt just make the eee 12v to begin with i dont know.. then you could just plug it right into a lighter outlet and be done with it..

Not in a car... Car electrical systems are messy things. A car's 12V can easily be something like 14V in real life in fact. Those things that do plug right in have voltage regulation built inside.

that said, the only other thing i can think of would be to build your own batery box, as the batterys are a known voltage, if you want longer life you run then in series+parellel (get your voltage right, then stack the batterys to achieve a higher amprage, ,

That would take a LOT of batteries to get close to this. Plus I'd have to use rechargables or it wouldn't be anywhere nearly as elegant of a solution. I suppose I could find some Cs or Ds instead of AAs or whatever, but those are even harder to find a rechargable of due to their sheer lack of use in today's hardware. (Not to mention the cost.) Anyway, I got this thing because it's a very elegant solution to a lot of problems of which this is only one -- more importantly, I have this thing. Also, my bank account isn't exactly overflowing with money.

useing 6 sanyo HR-DU NiMH 10,000MaH batterys will get you about 5.5 hours of run time for about 115$

And this thing will get me about the same number hours for half the cost even if I just use the built in inverter. Remember, it claims 60 watts of power will get some 2 hours of power that way. As this will be 22 watts maximum (and likely not that) I'm looking at some 2.72 longer than that assuming the laptop always runs at the maximum (and it does not) which means about 5.45 hours of power at nearly half the cost (and, btw, I don't get shipping back even if I go the route you and Michael Daly would have me go of running to return this thing simply because it's not 100% perfect to buy alternatives, plus I have to actually ship it in, costing me even more money that I don't have.) Heck, I could get an inverter for the 12V auto plug on this thing even if I were going to go that route and it still cost me less. Yes the battery is less efficient, but it's a lot less inefficient at cost. I can live with the weight. Newegg claimed it was about 15 lbs of weight, but, honestly, when I lift it it feels like 10.

i suggest trying to see if they have a car adaptor out there and just use that instead

They do. But the only one I ever saw was this: Incipio Technologies TK-202 - $39.99 (with the lowest by a third party being something like $35.) Speaking of third parties, this is as third party as you get, so I can't be sure just how well they followed specs (though at that price I would suspect that they did. Still, you never know. A lot of companies assume internal voltage regulation will work a lot better than it actually does in real life under the assumption that you'll never plug things in for long through their method. I've had one power adapter burn out due to this once and I'm just lucky that it merely tore up on the inside and simply stopped producing power without doing a lot more damage than that in the process.) If I can build something, about the highest costs I can possibly envision would be some $10 -- counting an enclosure...

Why not just use a UPS for the Eee PC? It might cost a bit more depending on the model, but it would be much more reliable and you know it's specifically designed for that use.

Well, firstly, it will cost a lot more. I actually checked UPS systems first and the cheapest that didn't look like it would tear up easily was more expensive (my parents have been using various UPS systems for years now and it seems most of them tear up surprisingly quickly.) The thing about having so little money is you quickly learn that it's not ok to try to skimp so much that you buy something which will cost you more in the long run, so I'd be looking at notably more expensive UPS systems to get something that will hold up for as long as this thing should. Secondly, this thing has a lot of uses ranging from simple things like an emergency light I can use in the house for when power is lost to more useful things like NOAA weather radio (and a line-input on some surprisingly capable if a bit too bassey speakers) and even a removeable rechargable flashlight that charges through it and is almost frighteningly bright (I have another LED flashlight that uses more LEDs, but clearly the LEDs in that flashlight are much higher powered or something because looking directly into this thing is literally painful.) Why give all this up just for a minor convenience of a UPS even if the UPS didn't cost significantly more? After all, those still use the inefficiency of an inverter, even if they use a better inverter (but I could buy a seperate inverter if I really wanted and I can get a better one probably.) What I'm trying to do here is skip the inverter and go straight to the battery for maximum efficiency since I see no reason to go from DC to AC back to DC -- the AC is totally unnecessary...

Otherwise, you could get a generic 12V sealed lead acid battery (not a car battery) and connect a 200W inverter to it. Just choose an amp-hour rating that gives you the run time you want. You'll need a charger as well.

When you add all that together, it's going to cost around the same amount or more. Especially since it's pretty hard to actually find such a battery outside of the auto store... (But don't underestimate the cost of the chargers that can be counted on.) Not to mention the advantage here of portability. This thing is easily carried around and can be powered by one small 12V plug in a wall or a small car plug (it doesn't really bother with any external voltage convertion -- apparently it can handle up to 15V according to the manual.)

That gadget is not really worth it for powering a PC, since you're paying for a lot of other stuff.

Stuff I'm using... Also, it's not a PC. That was just a side thought as this thing offers a lot of wattage, so it might be useful for getting data off of a harddrive in a hurry maybe. Not something I'd be using very often obviously.

 

[Mohonri]

There are several places online where you can order parts. mouser, newark, and digikey are fairly popular around here. All three have pretty impressive search functionality. The LM350T ($0.86) (datasheet) might be what you're looking for--it can supply the voltage and current you need. The datasheet even includes a very helpful "typical application" on page 5. It requires two resistors and (optionally) 2 capacitors. The caps are basically there just to smooth the output. In this application, where there's an additional stage of voltage regulation inside the laptop, I wouldn't be too concerned if you don't put those caps in. (if you have them lying around, say from a busted-and-gutted PSU.....not that I've ever done that...*ahem*...go ahead and include them)

The little car USB adapter is only dropping 3.5W as you say. It's not an enormous amount, and neither is 6W. Getting rid of the heat (and yes, all 6W will be dissipated as heat) isn't a big issue at all--a very small heatsink, such as a chunk out of a PSU heatsink, will be plenty. The only issue is that you'll get 10-20% less run-time from an inefficient (though simple) linear regulator than you would if you had a more efficient DC-DC converter.

EDIT: EDIT: please disregard previous edit about an even cheaper regulator--I thought it was 5A, it was really 1A

 

[Nazo]

Well, I was thinking that such a thing could be useful in a car as well. It would be kind of nice to be able to power my EeePC in my car I think. Assuming the output was clean enough this would presumably be useful for that as well. And I agree about the capacitor. There's no reason not to put one -- in fact, I think it would be stupid not to when it's so easy and has some potential benefits. I'd rather preserve the circuitry in the EeePC more than that outside of it. I can easily replace outside parts. Inside parts, not so easy. (Short term there may be no difference, but I'm kind of hoping this EeePC will last a lot longer than laptop manufacturers assume people mean to use laptops these days. This thing was cheap and it's extremely small and convenient so I can easily see myself using it some five years from now for diagnostics and such...) The question here is what capacitor should I use?

Assuming I was to go with that, how exactly do I control the output voltage? I presume it's through the use of the resistors, but must admit that I'm not sure what sort I'd need. I think I'd like to have one of these anyway so I could use it in the car from time to time if it can be done well enough to be safe. It wouldn't rely on the input to be the same would it? I think I can't count on the input not to change (in fact, there's probably a risk that if the battery goes low enough it could fall below an operable voltage for the EeePC, but at least I could get some time out of it presumably more efficiently than with the AC-to-DC adapter.)

Also, do you think a car inverter might be better than this overall and better than the motor method this thing uses? Actually, I do already have one, but it was relatively cheap. I think it could manage up to something like 40 watts. The problem is, when I found these on sale for pretty cheap in a store and got one, I told someone else and they got one too. They reported a short while later that it tore up when they used it and I think they even said it messed up the device they used it on. The only thing is, this wasn't a normal device... It was an ozone/ion generator. Basically it ran up a huge voltage with practically no real capacity and had that voltage arc from one point to another so it would generate ozone essentially the same way lightning does on a considerably smaller scale (and btw, these really do work -- or at least the ones they were using do.) Perhaps this caused problems just due to the way it works? I noticed when I used a much larger scale version of the same basic thing that it would get a lot of feedback where you could get sparks from other places if you weren't careful. I don't think it could have gone through the adapter the thing was using, but I'm not really sure if that couldn't have been related to it tearing up. After this I was afraid to really use it for anything important and never had any reason to use it for something unimportant...

Also, isn't it just automatically more inefficient going to AC and back again (for that matter, isn't AC supposed to be pretty inefficient on its own?) I can't help but to think that even with that 20% inefficiency, this method is probably more inefficient still. I noticed that it says 3 hours for 60 watts DC and 2 hours for 60 watts AC, so that means the motor method is loosing probably some 33% efficiency assuming nothing outside of that process has any inefficiency whatsoever...

If a third party inverter is more efficient, I can just use that most of the time. I do like the idea of having a cigarette lighter plug that can power my EeePC though and definitely want one if I can do that. What do you think?

EDIT: Doesn't the current rating of the cheaper regulator specify 1 amp? Oh, btw. If it helps any, the EeePC specifically states +9.5V @ 2.315A for (obviously rounded) 22 watts on the power statement on the bottom (just to give you a more exact idea of what it's supposed to use at a maximum.)

 

[Michael Daly]

Eee PC car adapter. Ditto. Result of a quick google search.

 

[Mohonri]

If you look on the "typical application" page of that datasheet, it explains how to choose the resistors in order to set the output voltage. Any resistors will do in this case--they're not sinking any significant amount of current, so you can use even 1/8W resistors. As far as capacitors go, anything bigger than about 1uF and rated at above, say, 16V should be fine.

Such a circuit doesn't really care what the input voltage is, provided it's above (Vout + Vdropout), so above 10V should be fine. It'll just drop the excess voltage on the input and provide the chosen output voltage. Since your input voltage is liable to go up to about 14.4V in the car, that'll mean you could potentially be sinking (14.4-9.5)*2.3 = almost 12W. It's still not a huge amount, and it's a worst-case scenario, but it's worth considering. If the voltage goes below (Vout + Vdropout), the regulator will simply stop providing power on the output.

I'm not sure what you're talking about when you mention motors--there's nothing involving motors here. You *could* use an inverter, but as you suspect, it's not particularly efficient. I don't know how the efficiency compares to the 75-80% we'd get with the linear regulator. If you're powering it with the car, efficiency doesn't really matter anyway.

You're right--the cheaper regulator I linked is only rated for 1A. Don't know how it showed up in the search results when I specified 3-5A capacity....

What Michael Daly suggested is good advice. If you crack open one of those, you'll find either a series of diodes or a linear regulator like the one we're discussing. The low parts count lets them sell it for that cheap, and I'd recommend you buy that instead of building your own.

 

[Nazo]

If you look on the "typical application" page of that datasheet, it explains how to choose the resistors in order to set the output voltage.

Well, I got that much, but I don't quite know how to actually calculate it. It says:
VO = 1.25V(1+R2/R1) + IADJ R2
With this diagram:

Below R1 it says 120, so R1 should 120 ohms? Also, what is IADJ? The way they put it on there makes me think it could actually mean that it's something seperate. Otherwise it shouldn't be part of the formula like that should it?

Sorry, I just don't do a lot of math with electronics exactly.

I'm not sure what you're talking about when you mention motors--there's nothing involving motors here.

I was asking if one of those inverters you can normally buy for this might be more efficient than what this thing uses. It runs a motor to make the DC-to-AC conversion apparently as opposed to the simulated sinewave most use these days.

You *could* use an inverter, but as you suspect, it's not particularly efficient. I don't know how the efficiency compares to the 75-80% we'd get with the linear regulator.

Well, the internal inverter probably looses some 33% or so according to the manual's time chart (things running off of AC get 1/3 less time than the same amount of power running off of DC,) versus this being 20% (and this assumes the AC-to-DC adapter itself doesn't loose any.) I have no idea about other inverters.

What Michael Daly suggested is good advice. If you crack open one of those, you'll find either a series of diodes or a linear regulator like the one we're discussing. The low parts count lets them sell it for that cheap, and I'd recommend you buy that instead of building your own.

Well, firstly I could build this for cheaper. As I said, I already have most of the parts. It's only the regulator that I would need to get. Secondly, the first one in the link is $18 BEFORE shipping... The second one is almost somewhat reasonable, but I'm pretty hesitant about buying from these people after reading some of what people are saying about them (I didn't bother to look up the first site.) Even assuming I had no problems getting the item from them, they don't really stand behind the quality of their merchandise. It could tear up my EeePC and they wouldn't really care. From what I'm seeing there, it seems a lot of their items are cheap because of low quality. Besides, this is a relatively simple device. My only confusion here is in getting the right voltage. Once I do that, I shouldn't have any problem building one. All for a lot cheaper than that. We're talking about just a few dollars assuming I add a pot (I almost certainly have capacitors though I'm not sure if I have one for 0.1uF.) I did do a search beforehand btw. The problem is, there are only so many sites you can reliably buy from.

 

[Mohonri]

Yes, R1 is 120 Ohms. As stated in the datasheet, the IadjR2 term will be very small compared to the 1.25*(1+R2/R1) term, so you can ignore it. Thus the equation becomes Vo = 1.25 * (1+R2/R1). You already know Vo and R1, so you can solve for R2 to find the right value.

I'm not sure you'd want to use a motor-based inverter. Motors introduce additional inefficiencies, EMF, maintenance issues, and noise.

Your choice of capacitor isn't really all that important, as long as it's rated for 16V or more. So if you have a spare 2200uF cap sitting around, you could use that instead of 1uF. It's not a critical application here.

 

[Nazo]

Yes, R1 is 120 Ohms. As stated in the datasheet, the IadjR2 term will be very small compared to the 1.25*(1+R2/R1) term, so you can ignore it. Thus the equation becomes Vo = 1.25 * (1+R2/R1). You already know Vo and R1, so you can solve for R2 to find the right value.

Ok. Is that more internal or something? Well, if I can ignore that last part it does make finding R2 a lot easier since that leaves it down to just one variable. Which is good because I forgot how to solve for two variables, lol. Thanks for clearing that up.

I'm not sure you'd want to use a motor-based inverter. Motors introduce additional inefficiencies, EMF, maintenance issues, and noise.

I didn't think of it that way. Good point. So it's probably best to use another inverter when I can anyway. I guess the only question here is, how safe are cheaper inverters? I'm still feeling rather less than enthusiastic about the one I currently have.

Your choice of capacitor isn't really all that important, as long as it's rated for 16V or more. So if you have a spare 2200uF cap sitting around, you could use that instead of 1uF. It's not a critical application here.

I understand that, but that's for smoothing the output, right? It also shows another capacitor, C1 going on the input, which it says is 0.1uF. While even they agree that the output capacitor's (Co) value isn't as important, it seems like they never said this about C1. Well, even if I have to buy a capacitor just for the input we're still talking about pocket change here. I wonder if I can get away with other values on it though?

 

[Mohonri]

The capacitor on the input is there to reduce the level of noise coming into the regulator. Since this is attached to a fairly complex electrical system, it won't do a great deal of good on its own. You might consider putting a (fairly hefty, remember it'll be dissipating some power) diode on the incoming line, upstream of the input capacitor. That way, the input capacitor will do a lot more good, and you'll have the added benefit of dropping less voltage across your regulator.

 

[Nazo]

Seems to me like the resulting circuit from what we're talking about would likely be of a higher quality than anything I could buy as it's going a bit above and beyond the call of duty. I'll bet even the more expensive ones don't do stuff like that. It's easy and it's cheap, but it isn't 100% required from the point of view of a penny pinching company. Well, I think maybe I have the gist of it all now. I'd like to thank you for taking all of this time to help me out like this too. I really appreciate it.