Battery ?

Discussion in 'General Electronics Chat' started by Mathematics!, Dec 19, 2009.

  1. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    I found this in a previous post

    Code ( (Unknown Language)):
    2. Battery capacity in mA*hr
    4.                 Alkaline    C-Zn    NiCd        NiMH        Lithium
    5.                 --------    ----    --------    ---------   -------
    6. AA               2700       1100    600-1000    1700-2900     3000
    7. AAA              1200        540                800-1000
    8. AAAA              625
    9. C                8000       3800                4500-6000
    10. D               19500       8000                900-11500
    11. 9 volt            565        400    120         175           1200
    12. CR123                                                         1500
    13. CR2025   dia=0.787, height=0.098                               160
    14. CR2032   dia=0.787, height=0.126                               225
    15. CR2450   dia=0.965, height=0.197                               610
    18. Silver Oxide   Volts mA*hr     Alkaline  Volts  mA*hr   Dia, in   Height, in
    19. ------------   ----- -----     --------  -----  -----   -------   ----------
    20.     SR41        1.55   42        LR41      1.5    32     0.311      0.142
    21.     SR44        1.55  200        LR44      1.5   150     0.457      0.213
    My question is this

    I rigged up enough alkaline D in seires to make 9 volt dc approx.
    The camera I have plugs into the house with an adapter cord that says
    input 120vac 60hz
    output 9volts dc max current drawn 500mA
    And in the manual I read that the camera runs on 9volt dc with 100mA to max 150mA.

    Also in the manual you can send away for a battery so you can use the camera without haveing it plugged into the house with the 6 ft adapter

    My question is why won't my D cell batteries work to power the camera?

    Going by the chart seems it should?

    Here is the links to the camera and battery pack for it.

    Curious why the D cells won't work 19500mAh but maybe when you but them in series it is not good enough to make the equivalant of Lithium 800mah.

    The only difference is this battery is rechargeable mine is not..... but it still should work.
    Or maybe the mAh are based on different current per hour that screws up the powering of the device .....etc

    Thanks for anybody that clears this up...

    If I by the battery pack will it last long for recharging purposes....
    Last edited: Dec 19, 2009
  2. k7elp60

    Active Member

    Nov 4, 2008
    The D cells(six in series) should work fine.
  3. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    6 in series will not equal 9volts dc ?
  4. Mike33

    AAC Fanatic!

    Feb 4, 2005
    Yes, 6 in series will equal 9V. 6*1.5v = 9v
  5. Audioguru

    New Member

    Dec 20, 2007
    Six brand new alkaline cells are about 9.6V. When used the voltage drops quickly to about 7.2V. Then the voltage drops slowly to 6V when the battery is dead.
    We don't know what is the minimum voltage for the system.
    The AC adapter is not fully loaded so its voltage is probably higher than 9V, maybe 11V.

    The Ni-MH battery uses a number of 850mAh AAA cells but we don't know how many (eight will make about 9.6V) and we don't know the voltage of its charger.

    Of course you must connect your battery with the correct polarity.
  6. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    By this it means I need to supply the camera with 800mAh/5 = 160mA or
    800mAh/3 = 267mA approx. Or something between...

    The output of the cord for the camera say's 9volts dc 400mA
    The ac adapter also say's Class 2 Power Unit
    input: 120vac 60hz 120mA
    Efficiency level: IV

    I have figured out the polarity of the camera already.

    Either way If I get the exact current/voltage ratings it should work.

    My only question is how much voltage/current is this Lithium 800mah rechargeable battery outputing to the camera to power it.

    It is rated for 800mAh so by the calculation above it should need steady current in the range of 160mA to 267mA but I am unsure of the voltage rating it needs for this current?

    If anybody knows the power this battery pack device supplies that would help?

    Also in the specifications on page 14/15 it say's max current consumption by the camera is 150mA. but from the above calculation range of 160mA to 267mA seems to be what the current should be at to power the camera for 3 to 5 hours??? The math is just not adding up ???

    Also from the adapters output voltage/current the power will be 3.6watts

    So I am only assuming that the Lithium 800mah must supply 3.6watts of power but the question is what current or voltage it is supplying the camera with.....?

    If it is 9volt then that imples 3.6/9volts = .4 = 400mA but then ofcourse that exceeds the 150mA current max consumption?

    say we have it supplying 100mA then if power is 3.6watts => 3.6/.1 = 36volts dc?

    Anybody shed some light on this....
    This is a perfect example of knowing the math doesn't equal knowing electronics ?????

    Thanks for any help I am more curious on making the math work then building the damn thing now ahhhhhhhhhhhhhhhhhhh

    I can even calculate the energy in joules using E = P*T assuming power is constant which I think is a vaild assumption...

    So 5hours would mean the Lithium 800mah battery would have energy E = 3.6 * 5hrs = 3.6watts * 18000s = 64800 J.

    Can anybody put the math together for me. I understand internal resistance , tempature ,...etc can effect the calculations but I should beable to have a reasonable approximation.....
    Last edited: Dec 20, 2009
  7. beenthere

    Retired Moderator

    Apr 20, 2004
    Those are the magic numbers.
  8. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    Ya , but if that is the case why does it say max current consumption for the camera is 150mA shouldn't it be greater then 400mA if the output of the adapter is giving 9volt dc with 400mA??? I would think since 400mA of current would kill the camera because it exceeds the max current consumption of 150mA?

    What I was think is that the 9volts dc with 400mA output of the adapter is with no load or maybe the max current/voltage/power that this adapter can supply not the actual current being supplied when the camera is pluged in.... maybe that is it , maybe not but still confused because the math is not working out???

    400mA > 150mA max

    Either way the math doesn't work out here either (with the mAH /h should give you the steady current needed by the battery pack but if this is true the current rating would well exceed the max current consumption?

    Still looking for clarity?
    I know tempatures , and internal battery resistance can play a roll but I definitely know it can't effect it by this much !
    Can somebody figure out the math because this mathematician just cann't do the math
    Last edited: Dec 21, 2009
  9. kubeek

    AAC Fanatic!

    Sep 20, 2005
    Ok if the camera says 150mA max, then it shouldn´t consume more than that.

    Another question is, isn´t the camera checking for the presence of the batteries (presumably recharchable), and won´t turn on without them?
    Does the original battery have three or more contacts?
  10. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    Then why does the output of the power adapter for the camera say 9volts 400mA if that where the case I would think the 400mA of current would exceed the max rating of 150mA and damage the camera ????

    As for your other statements read pervious posts. Right now I am just concerned about making the math work out and why the 400mA > 150mA won't kill the camera, and why the mAH / H => a current of 800mAh/5 = 160mA or 800mAh/3 = 267mA approx. still much greater then 150mA max

    Math still doesn't add up???????????????????????????????????????????????????????????????????????????????????????
    Last edited: Dec 21, 2009
  11. SgtWookie


    Jul 17, 2007
    If your adapter is rated for 9v @ 400mA, and it's an unregulated "wall wart" type plug supply, it may have an output voltage that is significantly higher with only a 150mA load on it - perhaps in the range of 11v-13v.

    I'm not suggesting that you try using more voltage. I am, however, suggesting that it's not cut and dried exactly what voltage your camera is seeing.

    Why don't you measure your adapters' voltage output when there is no load on it?

    I have a feeling you'll read between 13v and 15v.
  12. kubeek

    AAC Fanatic!

    Sep 20, 2005
    maybe you should try to read AAC books again, especially the Ohms law. When the wallwart says 450mA, that means that the maximal current you can draw is 450mA, not that it will allways produce 9V and 450mA regardless of load (how would it produce 450mA with the load disconnected?).
    Because of the ohms law, it is not possible to create a device that makes 9V and 450mA at all times regardless of the load applied.
  13. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    Well the adapter that plugs into the power outlet has a sticker on it that says input 120vac 60hz , 120mA , output 9volts dc 400mA

    So then what does this 9volts dc 400mA mean """I would think it means the voltage/current that is being supplied to the camera when it is plugged in"""? Really if that is not it is it a max rating because I don't really see the point in saying this output values if they are not true????
    Is it the max voltage/current when you short the adapter output or something?

    Mathematically this is not true since a steady 9volt battery with the device being just a 20ohm resistor would produce a steady 450mA.
    Leaving out resistor tolerances , internal battery resistance , wire resistance this would be exact.... don't really know where you came up with that?

    If the 9vdc 400mA output is just a max rating for the output to a particular load then I would think any 9vdc battery with more the ability to supply a steady 100mA of current should power the camera.... (since the cameras max current consumption is 150mA I would think any battery that can supply around or alittle below this max value would be perfect to use? Correct me if I am wrong? Other then the fact that the battery would have to have enough mAh to supply it steady with this 9vdc 100mA,...etc for enough time to be relatively useful i.e a few hours or more would be good ??
    Last edited: Dec 22, 2009
  14. SgtWookie


    Jul 17, 2007
    Did you even bother to read my post?

    Measure the output of your adapter with no load on it.

    It's going to be higher than 9v with no load.
  15. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    ya , your right it is about 11.5 volts or so with no load.
    But that still doesn't answer my question on what the hell the 9vdc , 400mA output means on the adapter? Because it obviously is not a max rating.

    I mean the input of the adapter is correct it say's 120vac 60hz so I thought the output of the adapter would be 9vdc 400mA ?

    So is this 9vdc 400mA occur when the camera is plugged in i.e the cameras load???? When does this 9vdc 400mA occur as output from the adapter?

    At least I am assuming the power is constant regardless of the voltage/current? Mean if I measured 11.5vdc the current would just be 0.31304347826086956521739130434783 A = 313mA approx.

    but when it is 9 volts => 400mA because the power always has to be 3.6watts
    Correct me if I am wrong and the power can fluctuate as well?
  16. SgtWookie


    Jul 17, 2007
    If you connect a 22.5 Ohm 5 Watt resistor across the output of the adapter, that will draw 400mA current at 9V.

    With that load, your adapter output will likely be somewhere around 8.5-9v.

    However, your load is much less than that; only 150mA. That's about 72 Ohms with 10.7v across it.

    Since the adapter's output is not regulated, with the lesser load of 150mA, the actual voltage of the adapter will probably be more like 10.5v-11v.
  17. BMorse

    Senior Member

    Sep 26, 2009
    Your camera can use up to 150mA, your adapter can supply up to 400mA, this doesnt mean it is feeding the whole 400mA into the camera! Your power supply can say 9 VDC at 10000 mA and your camera will still only draw 150mA....

    On the other hand if the power supply said 150mA, and the camera consumed 400mA then that would be bad, since the camera would draw more than the power supply can provide. So higher numbers from power source in output current capabilities is better when the device draws less.
  18. Mathematics!

    Thread Starter Senior Member

    Jul 21, 2008
    Ok , so any battery that is 9volts and can provide a steady 150mA > should work?

    Obviously I am a little shaky on this reasoning since their are many batteries for example that supply 12volts but a 12volt car battery would have a current that could damage the in circuitary of some sensitive devices that run on 12volts.

    Won't to be completely safe so you don't kill the device you must match the voltage/current "power ratings" ??? To much power could kill the camera I would think regardless of if you match the correct voltage?

    Is this output of 9vdc , 400mA just a reference output because this really doesn't have any meaning anymore? Can any device that needs 3.6watts of power use this adapter ? I wouldn't think so if it had a comparator circuit that compared 5volts with something else you would at least need
    5 volt supply but you could make 3.6watts with a 2 volt and alot more current....?

    If the camera can only consume 150mA max that would mean the cameras resistance would be 9vdc/.150 = 60ohms
    a 60ohm resistor rated at 3.6watts or greater. This would be the camera equivalent of a resistor.

    For example if we had 3 9volt batteries one that delievered 9watts , one that delievered 18watts , and one that delivered 27watts of power. And you had the camera that was an equivalent of a 60ohm 2.5 watt resistor I would think the camera would work with the first 2 batteries but would be damaged by the third since the third exceeds the amount of power that the camera can dissipate....

    But that is just me.
    I would also think knowing the equivalent resistance of the camera you would also need to know what the power rating for the resistor. Since a 1kohm resistor could not be used to the main house outlet 120vac if it was rated for a 1/2watt but if it was 1kohm and rated at 100watts then this would be far beyond what you need...

    So the question is if I any of the 9volt batteries in the list would exceed the power ratings for my camera. Or will the current from the batteries be in the correct range to not damage the camera???

    Last edited: Dec 23, 2009
  19. Paulo540


    Nov 23, 2009
    Like Bmorse said, You could hook up a 9v battery rated at 2,000,000,000,000,000,000,000,000,000,000 mAh and it won't damage your circuitry, think of mAh as the capacity and ability to discharge, not the actual output. the output current is determined by the circuitry of the device, it won't take more than it needs unless of course something were shorted. Also, if the camera uses a rechargeable li-ion, then most of the current it is referencing is charging current. I have never heard of a 9v li-ion battery either, are you sure its not a 3.6 or 7.2v?
  20. SgtWookie


    Jul 17, 2007
    OK then...

    Let's say you had a DC supply that had an internal resistance of 6.25 Ohms, that measured 11.5v with no load.

    What would you get with varying loads on it?

    Why don't you try to see what happens in a simulation?

    Set up a voltage device, with a voltage of 11.5 and internal resistance of 6.25 Ohms. Ground the negative side of it.

    Then set up a current device, and try sinking 150mA of current to ground from the positive side of the voltage device.

    Then run the simulation, and see what voltage you're measuring at the output of the voltage device.

    Or you could just ask how much voltage would be dropped across 6.25 Ohms (the internal resistance of the voltage source) with a 150mA current through it, and then subtract that result from 11.5v.

    Then try the same thing with a 400mA load.
    Last edited: Dec 23, 2009