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Homework Help: Can i use a 1mA source to charge a 3V rechargeable battery?

  1. Mar 7, 2008 #1
    Can i use a 1mA source to charge a 3V rechargeable battery?
    (1mA source is the output from Photodiode. I can't put an amplifier because of some reason. Note that, the photodiode in this case is used to transfer power only.)

    Got one engineer told me that i can't do that. Because instead of charging the battery, actually, i'm discharge the battery. Both of the battery and source will try to reach a balancing between them, so the current from high voltage battery will flow out to the source.

    But, there is another answer for that question, said i can charge the battery. Because, the 1mA source will“pushes” the charge in battery from + end to – end.

    I'm confusing which one is correct?
  2. jcsd
  3. Mar 8, 2008 #2


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    If the resistance of the photodiode is small enough that the battery is discharging more current through the diode than the diode is pushing in, then you should probably listen to the engineer. What's the resistance?
  4. Mar 8, 2008 #3
    How to know the resistance in photodiode? I can't found from the spec sheet.
  5. Mar 8, 2008 #4


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    Current flows from high potential to low potential. If the voltage developed across the photodiode is less than that of the battery, it cannot charge the battery.

    - Warren
  6. Mar 8, 2008 #5
    Means that i have to know the voltage of photodiode?

    The problems is that i only manage to calculate the output current from photodiode by using formula: Ip=RPin, where Ip is output current from photodiode; R is responsivity; and Pin is the output power from laser source.

    May i know how to get the voltage of photodiode?
  7. Mar 9, 2008 #6
    Anybody help???
  8. Mar 9, 2008 #7
    The current is not important to charge a battery. It is more the voltage. The photodiode has to create a larger voltage, in the opposite direction of the battery trying to push the current around the circuit, in order for the battery to be charged.

    You will not discharge the battery, if the photodiode has a large enough reverse break down voltage (the maximum voltage you can put across the diode 'backwards', the way round it stops current flowing, before the component catastrophically fails and becomes practically 0Ohm resistance). The reverse break down voltage must be larger than the battery voltage.

    The easiest way to get to know the photodiodes' voltage is either look on the specification sheet or get a scope/DMM out and measure it. Personally I would do the second.
  9. Mar 9, 2008 #8
    From the specification sheet, i found that the reverse voltage is 30V and maximum satuaration input power is 10dBm.

    However, availabled laser source in my school is 8dBm. So that supplied 8dBm into photodiode, will only get the voltage of 0.4V. Why like this happen? Why the voltage is so low?
  10. Mar 9, 2008 #9
    Just incase you have misunderstood, the reverse voltage is not the voltage that it will 'produce' when the maximum input power is provided. A photodiode is essentially a diode with some bits bolted on to make it sensitive to light. At school level it is suitable to see a diode as a short circuit in one direction of current flow and a open circuit in the other. In the case that you place a voltage across the diode so you create the open circuit case (called reverse biasing the diode) if you increase the voltage, in this case above 30V, then the diode will fail as an effective open circuit, the structure of the silicone inside will become damaged and will make it an effective open circuit.

    As long as the battery you are trying to charge is below 30V then no power will be dissipated, as there is no current, as the diode is acting as a open circuit.

    Photodiodes are not actually use to generate electricity but more as a switch or sensory element. They are designed to detect light and alter their resistance accordingly not to generate electricity so the voltage you get out of these things is small as the amount of energy they can convert from light into an electrical signal is very low.

    You could, in theory, use a transformer to step up the voltage however then the current will get smaller and smaller which means that it will take a very very very long time to charge the battery, it would almost definitely very impractical.

    Also - In your original post you talk about using an amplifier - amplifiers, in the cases that I can think of which would step up the current or voltage, require some kind of input power supply which must be greater than or equal to the output you want - so you would expend more energy in stepping up the voltage/current than you got by charging up the battery.
  11. Mar 9, 2008 #10
    No, you can't charge a battery with 1 mA. The engineer was correct.
  12. Mar 9, 2008 #11
    Why? What is your reasoning TVP45?
  13. Mar 9, 2008 #12
    Tell me the characteristics of the battery and I'll give you a detailed answer.
  14. Mar 9, 2008 #13
    I don't know them, I'm not the original poster. I just don't see any theoretical reason why it won't charge the battery - it would be massively impractical for any real application but still possible.
  15. Mar 9, 2008 #14
    I know that photodiodes is not used to generate electricity. Due to some reason, i have no choice but only can use photodiodes. I can't even use other additional device such as transformer. The only things that i have is a laser source, photodiode, and rechargeable battery.

    The rechargeable battery in this experiment is a Samsung handphone's battery. The battery was labelled as 3.7V.
  16. Mar 9, 2008 #15
    This is a Lithium ion battery. You normally recharge them at a constant current until they reach some portion of full charge and then switch to constant voltage for the rest.

    In the simplest case, 1 mA requires several hundred hours.

    And, even if the photodiode put out the correct voltage, the "smart" charging circuit would require much more than 1 mA to run it.

    It may also be strictly a point of view. I am a retired engineer; to me, extremely impractical is impossible.
  17. Mar 9, 2008 #16
    @ TVP45 - Ok, I understand what your saying and do agree with your points, but while still extremely impractical, and to be honest completely useless, I don't agree that it can be said to be impossible - impossible to market, sell, obtain funding for or even find a useful application for but not impossible to create. But then were going round in circles, I understand what you mean, and yes, your right. I bow down to superior knowledge in this case. =]
  18. Mar 9, 2008 #17
    Does it the reason that I can't charge the battery? How much current does "smart" charging circuit require?

    Actually, I should consider the V or I when measure a battery? Why?

    When charging a battery, I should consider V or I?

    I understand that this is an impractical case. But, it still can work right? Just take a long time right?
  19. Mar 9, 2008 #18
    A summary for you:
    1. - You need the photodiode to create a larger voltage than the fully charged voltage of the battery.
    2. - Some current will be required to drive the 'smart' charger circuitry, 1mA is probably not enough as said by TVP45.
    3. As the current would be tiny the time to charge a battery, even a small one, would be very very large.
  20. Mar 9, 2008 #19
    A photodiode used for generating power can be modeled as a current source parallel with a forward biased diode. The size of the current depends on the light intensity. If no load is connected, all of the current will go through the diode and this will produce a voltage according to the normal diode equation. for a silicon diode this means smaller than 0.7V.
    For a gallium Arsenide diode < 1.7V
    If you connect a battery to this, the battery will just discharge through the diode, and damage it.
  21. Mar 10, 2008 #20
    Bacially, how to measure a battery? Measure directly with multimeter?
    I need to consider voltage or current when measuring the battery?
    What is the different between fully charged battery and fully discharged battery?

    I had done an experiment to measure the voltage of fully charged handphone's battery and fully discharged handphone's battery by using multimeter. But i found that the voltage for both of them is same. Why?
  22. Mar 10, 2008 #21
    Keep in mind that a Lithium ion battery is never fully discharged (unless of course it has been damaged or has sat unused for several years). I don't know where to find a voltage curve just now, but you should see a voltage between (maybe) 3 V ("needs recharged") and 4 V ("charged"). Those numbers vary a little depending on the amp-hour rating of the battery and the manufacturer.

    But, you need to always check a battery under load. Since your battery has an amp-hour rating of (I think) 800 mAh, you'll want to put a load of about 50 mA on it, so use a 100 Ohm, 1/4 W resistor. That should show the difference.
  23. Mar 14, 2008 #22
    How to increse the voltage of output from photodiode, so that the voltage is higher than battery's voltage?
    Remarks: Only can add-on passive components; active components are not allow.

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  24. Mar 14, 2008 #23
    Why do you want to do this? The chances of damaging the battery are so great and the chances for any success so small.
  25. Mar 14, 2008 #24
    This is my school project. I have no choice.
    Got any way to increse the voltage coming out from photodetector?

    Why the chances of damaging the battery are so great? How the battery gets damage?
  26. Mar 15, 2008 #25
    Excuse me if I sound critical. I understand your plight and sympathize. But, what school do you attend that would give this as a project?

    Leaving aside the question of whether a photodiode is active or passive (the heuristic definition of active is that current is controlled by an electrical signal and then we all get to argue over whether light constitutes an electrical signal), the only way I can imagine doing something like this is a group of photodiodes. Or, you could use modulated light (a chopped beam would probably work) and then feed the output of the photodiode into a transformer, rectify it, filter it, and try that. Even Rube Goldberg would roll his eyes at that one.:rolleyes:
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