Stupid battery-operated projector
October 8, 2018 5:31 PM Subscribe
I've got this amazingly tiny projector (axaa p2b) but it has a terrible design flaw: It only runs from the battery. I don't care about the battery feature, and i'm fine with always having it plugged in. So, am I crazy to just rip out the battery and hook it up to an adaptor?
Now, it has an AC adaptor, but that only charges the battery - and doesn't seem to charge it while running. So after the battery runs out (which might or might not be at the end of your movie, it just dies. Brutal.
The batter is a 7.8 V lithium-ion battery. It has 2 black leads, and 2 red leads. (Both are 7.8V) The packaging states that the projector runs at 11 Watts. I can't find any markings on the battery (it's just a silvery rectangle) - so I'm calculating the battery using ohm's law at: P = VI: 11 = 7.8I I = 1.41A
So, do you think I can get 1500mA adaptor that can do 7.8V and connect the + to the red and the negative to the black wires? (Could there be some significance with having two red and two black wires?)
Any encouragement if this sounds like a sane plan would be appreciated. (I keep thinking: It's so simple, why the hell didn't the manufacturer do this? What am I missing??)
Now, it has an AC adaptor, but that only charges the battery - and doesn't seem to charge it while running. So after the battery runs out (which might or might not be at the end of your movie, it just dies. Brutal.
The batter is a 7.8 V lithium-ion battery. It has 2 black leads, and 2 red leads. (Both are 7.8V) The packaging states that the projector runs at 11 Watts. I can't find any markings on the battery (it's just a silvery rectangle) - so I'm calculating the battery using ohm's law at: P = VI: 11 = 7.8I I = 1.41A
So, do you think I can get 1500mA adaptor that can do 7.8V and connect the + to the red and the negative to the black wires? (Could there be some significance with having two red and two black wires?)
Any encouragement if this sounds like a sane plan would be appreciated. (I keep thinking: It's so simple, why the hell didn't the manufacturer do this? What am I missing??)
Best answer: Something like an LM317 might work
Something like this adjustable switching regulator module might work even better.
I keep thinking: It's so simple, why the hell didn't the manufacturer do this? What am I missing??
You might be missing the fact that the manufacturer offers this power supply accessory that it specifically states will provide continuous power to run the projector indefinitely.
posted by flabdablet at 12:07 AM on October 9, 2018 [8 favorites]
Something like this adjustable switching regulator module might work even better.
I keep thinking: It's so simple, why the hell didn't the manufacturer do this? What am I missing??
You might be missing the fact that the manufacturer offers this power supply accessory that it specifically states will provide continuous power to run the projector indefinitely.
posted by flabdablet at 12:07 AM on October 9, 2018 [8 favorites]
Dumb question, but have you tried just pulling the battery, plugging in the charger and operating the device? I know I've had laptops with bad batteries that that was the only way to force it to draw exclusively on the charger instead of trying to draw from the battery and let the battery draw from the charger.
posted by solotoro at 7:13 AM on October 9, 2018
posted by solotoro at 7:13 AM on October 9, 2018
Looks like they sell a higher powered wall charger to power the device continuously
http://www.aaxatech.com/store/products/HDPico_power_adapter.html
posted by jmsta at 10:27 AM on October 9, 2018
http://www.aaxatech.com/store/products/HDPico_power_adapter.html
posted by jmsta at 10:27 AM on October 9, 2018
Their "special charging adapter" looks to just be a MicroUSB wall wart with a reasonably-high current rating (3A).
Is it possible you just haven't tested it using a AC adapter that can source enough current? On a low-current adapter, you'd get exactly the behavior you describe—it would draw down the battery and when the battery voltage fell below a critical threshold, it would turn off. It would take longer to run down than running on the battery alone, of course—a test you could perform to confirm this behavior.
Anyway, get a 3A or higher MicroUSB and give it a shot. Can't hurt, might work.
posted by Kadin2048 at 10:08 PM on October 9, 2018
Is it possible you just haven't tested it using a AC adapter that can source enough current? On a low-current adapter, you'd get exactly the behavior you describe—it would draw down the battery and when the battery voltage fell below a critical threshold, it would turn off. It would take longer to run down than running on the battery alone, of course—a test you could perform to confirm this behavior.
Anyway, get a 3A or higher MicroUSB and give it a shot. Can't hurt, might work.
posted by Kadin2048 at 10:08 PM on October 9, 2018
Be a little careful with using just any old adapter. The 5V 3A adapter that others have linked to above is for the HD Pico projector; the one offered for your P2-B projector is specced at 12V 1.5A which is in no way the same.
Even though both pages use the same image to illustrate the power supply, I strongly doubt that any manufacturer would actually design a standard micro USB plug onto a 12V power supply.
I have an Acer Iconia tablet that also needs a 12V 1.5A charger. The socket on the tablet that the charger plugs into is capable of accepting either a standard micro USB plug or the special charger plug, which looks very much like micro USB at first glance but has an additional bump on one face to block it from plugging into a standard micro USB socket, as well as two extra internal contacts for the 12V charge supply.
If you plug a standard micro USB plug into that socket, the tablet won't charge. It's entirely possible that your P2-B projector has a similar arrangement.
posted by flabdablet at 10:43 PM on October 9, 2018
Even though both pages use the same image to illustrate the power supply, I strongly doubt that any manufacturer would actually design a standard micro USB plug onto a 12V power supply.
I have an Acer Iconia tablet that also needs a 12V 1.5A charger. The socket on the tablet that the charger plugs into is capable of accepting either a standard micro USB plug or the special charger plug, which looks very much like micro USB at first glance but has an additional bump on one face to block it from plugging into a standard micro USB socket, as well as two extra internal contacts for the 12V charge supply.
If you plug a standard micro USB plug into that socket, the tablet won't charge. It's entirely possible that your P2-B projector has a similar arrangement.
posted by flabdablet at 10:43 PM on October 9, 2018
Could it overheat if you run it for longer than the 90 minute time it's designed for? Maybe there's a way to heat-sink it or add more fans, just to be sure?
posted by TreeHugger at 9:55 PM on October 10, 2018
posted by TreeHugger at 9:55 PM on October 10, 2018
Given the existence of an accessory specifically advertised as follows,
posted by flabdablet at 11:44 PM on October 10, 2018
Provides continuous power to run the projector indefinitely. Ideal for users who need to use the projector for long durationsit seems to me that overheating is extremely unlikely to be an issue. Also, I would be astonished to discover that any piece of equipment as physically small as that projector would take anything like 90 minutes to reach overheating temperature, if that is indeed what it was ever going to do.
posted by flabdablet at 11:44 PM on October 10, 2018
Response by poster: Sorry for the delay
I totally forgot that there was a special adapter for continuous operation. I'm not in the US, so it was expensive/difficult and ruled it out as a option. I tried just plugging in a 3A USB external battery to the projector, but it didn't seem to do anything, so like flabdablet suggested, it looks like they are doing something funky and sending 12V with their special "USB" adaptor.
I disassembled the battery a bit - both the black and red leads are soldered together closer to the battery. So that simplifies things. I read the voltage from the solder points on the battery and it's now 8.4V so I guess my next step is to figure out how to wire up an LM317 to reduce the voltage. (Or just take a chance that 9V in won't blow anything up.)
posted by kamelhoecker at 6:24 PM on October 16, 2018
I totally forgot that there was a special adapter for continuous operation. I'm not in the US, so it was expensive/difficult and ruled it out as a option. I tried just plugging in a 3A USB external battery to the projector, but it didn't seem to do anything, so like flabdablet suggested, it looks like they are doing something funky and sending 12V with their special "USB" adaptor.
I disassembled the battery a bit - both the black and red leads are soldered together closer to the battery. So that simplifies things. I read the voltage from the solder points on the battery and it's now 8.4V so I guess my next step is to figure out how to wire up an LM317 to reduce the voltage. (Or just take a chance that 9V in won't blow anything up.)
posted by kamelhoecker at 6:24 PM on October 16, 2018
I guess my next step is to figure out how to wire up an LM317
If you're going to go the battery replacement route, I strongly recommend using one of the 95 cent adjustable switching converter modules I linked above instead of that LM317. Not only are they ridiculously cheap for what they are, but they'll run cooler and waste less power than any design based on a linear regulator, plus you don't have to fabricate a circuit board or do anything else resembling actual design, just box it up.
1.5A is also right on the maximum specified output current for the LM317 while the switcher is actually rated at 2A, or 3A if you stick on a heatsink, giving you a comfortable amount of headroom.
posted by flabdablet at 3:37 AM on October 17, 2018 [1 favorite]
If you're going to go the battery replacement route, I strongly recommend using one of the 95 cent adjustable switching converter modules I linked above instead of that LM317. Not only are they ridiculously cheap for what they are, but they'll run cooler and waste less power than any design based on a linear regulator, plus you don't have to fabricate a circuit board or do anything else resembling actual design, just box it up.
1.5A is also right on the maximum specified output current for the LM317 while the switcher is actually rated at 2A, or 3A if you stick on a heatsink, giving you a comfortable amount of headroom.
posted by flabdablet at 3:37 AM on October 17, 2018 [1 favorite]
Ooh, I have an easy idea. Take your 9v supply, stick a single standard rectifier diode rated at >1.5 A in series (diode will have an approx .65v drop) and you'll have 8.35v out ...which would almost definitely be close enough. ASCII schematic diagram
+9v ----->]----- +8.35v out . It wouldn't be a bad idea to also tack a 100uf/16V electrolytic capacitor (pay attention to polarity!) between the 8.35V output and ground.
Agree with flabdablet that a switching converter module is a much more elegant way to do it than an LM317. (And I'm the person who suggested the LM317! ) Also, if you do use the LM317, I think you'd need to go for a 12V raw input to ensure the regulator has enough headroom. (Output voltage is too close to input voltage with a 9v in, 8.4V out.)
posted by Larry David Syndrome at 6:55 PM on October 17, 2018
+9v ----->]----- +8.35v out . It wouldn't be a bad idea to also tack a 100uf/16V electrolytic capacitor (pay attention to polarity!) between the 8.35V output and ground.
Agree with flabdablet that a switching converter module is a much more elegant way to do it than an LM317. (And I'm the person who suggested the LM317! ) Also, if you do use the LM317, I think you'd need to go for a 12V raw input to ensure the regulator has enough headroom. (Output voltage is too close to input voltage with a 9v in, 8.4V out.)
posted by Larry David Syndrome at 6:55 PM on October 17, 2018
I agree with Larry David Syndrome about the unlikelihood of an LM317 being happy to deliver 8.4V out with 9V in and the consequent desirability of using 12V in. This was the assumption behind my efficiency argument for the switching regulator.
The vendor claims efficiencies of "up to 92%" for the switcher, so let's assume that's a bit generous and that it would be good for 80% instead. That means that if your projector is consuming 11W, the switching converter would need to dissipate 11W * ((100% - 80%) / 80%) = 2.75W, which is OK for a mostly-groundplane PC board a couple of inches square.
If you're running a linear regulator to drop a 12V input to 8.4V, then the voltage drop across the regulator will be 12V - 8.4V = 3.6V. In a linear regulator, unlike a switcher, the output current and input currents are equal. Power dissipation is proportional to voltage drop for elements carrying the same current, so if the load is dissipating 11W then the regulator will dissipate 11W / 8.4V * 3.6V = 4.7W which is starting to be an uncomfortable amount of heat for a small package to get rid of.
Larry David Syndrome's idea of using a simple series diode as a voltage dropper from a 9V supply beats both of these options though. Again, the common-current linear model is applicable, but in this case the voltage dropping element only needs to get rid of 0.7V, so it will dissipate 11W / 8.3V * 0.7V = 0.9W, which is enough to make a typical rectifier diode package run a bit warm but not to cook it, and handily beats the efficiency of the switching converter too.
So if you do indeed have access to a 9V 2A supply whose voltage is regulated at least as well as that of a discharging battery (i.e. not particularly well), the series diode option is probably your best way to make a simulated 7.8V - 8.4V battery out of it.
posted by flabdablet at 10:18 PM on October 17, 2018 [2 favorites]
The vendor claims efficiencies of "up to 92%" for the switcher, so let's assume that's a bit generous and that it would be good for 80% instead. That means that if your projector is consuming 11W, the switching converter would need to dissipate 11W * ((100% - 80%) / 80%) = 2.75W, which is OK for a mostly-groundplane PC board a couple of inches square.
If you're running a linear regulator to drop a 12V input to 8.4V, then the voltage drop across the regulator will be 12V - 8.4V = 3.6V. In a linear regulator, unlike a switcher, the output current and input currents are equal. Power dissipation is proportional to voltage drop for elements carrying the same current, so if the load is dissipating 11W then the regulator will dissipate 11W / 8.4V * 3.6V = 4.7W which is starting to be an uncomfortable amount of heat for a small package to get rid of.
Larry David Syndrome's idea of using a simple series diode as a voltage dropper from a 9V supply beats both of these options though. Again, the common-current linear model is applicable, but in this case the voltage dropping element only needs to get rid of 0.7V, so it will dissipate 11W / 8.3V * 0.7V = 0.9W, which is enough to make a typical rectifier diode package run a bit warm but not to cook it, and handily beats the efficiency of the switching converter too.
So if you do indeed have access to a 9V 2A supply whose voltage is regulated at least as well as that of a discharging battery (i.e. not particularly well), the series diode option is probably your best way to make a simulated 7.8V - 8.4V battery out of it.
posted by flabdablet at 10:18 PM on October 17, 2018 [2 favorites]
Probably better still, though, if you're not willing to wait for Aaxa to ship you their official 12V power supply and you can't source one of your own with a compatible connector, is just to open the thing up and solder a connector you can use to the points where the OEM power socket supplies 12V to the circuit board.
posted by flabdablet at 1:21 AM on October 18, 2018 [1 favorite]
posted by flabdablet at 1:21 AM on October 18, 2018 [1 favorite]
This thread is closed to new comments.
posted by Larry David Syndrome at 6:45 PM on October 8, 2018 [3 favorites]