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Simple Voltage Reduction

  1. Dec 27, 2009 #1
    I plan on using two solar cells that can each generate 3 volts and ~80mA each and wiring them in series to get an array that generates 6 volts and ~80mA.

    However, I need ~4.5 volts from the array. What would be the best way to decrease the voltage to this level with the given array?
     
  2. jcsd
  3. Dec 27, 2009 #2

    berkeman

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    Probably a buck DC-DC voltage regulator. Search National Semiconductor's website for datasheets and details.
     
  4. Dec 27, 2009 #3
    Is there no normal component that can bring something down 1.5 volts?

    EDIT: a single solar cell would generate ~3 volts. Is it easier to raise the voltage of one solar panel or lower the voltage of two wired in series (at ~6 volts) with the addition of extra circuitry?
     
    Last edited: Dec 28, 2009
  5. Dec 28, 2009 #4

    vk6kro

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    If you know the current will not vary, you could use a series resistor. Calculate it as follows:
    resistance (in ohms) = 1.5 volts / current in amps.
    eg resistance (in ohms) = 1.5 volts / 0.08 A = 18.75 ohms

    A better way would be to look for a low dropout adjustable regulator on Google, or at an electronics store.
    Look for one that will give 4.5 volts out and will allow the input to drop to within 0.5 volts of the output voltage (ie <5 volts).

    Don't get a LM317 as these need at least 6 volts to give 4.5 volts out so your output would drop each time the sun went behind a cloud.

    Another alternative is to use a shunt regulator. These are also available and may be better for this application. You wouldn't even need a series resistor because the solar panel would limit the current itself. One type is the TL431.

    Another way would be to get two lead acid cells and charge them up in series from the solar panel. This would give you a very steady 4.5 volts that would last for some time after the sun had set.
    You would need to put a diode in series with the battery to avoid it discharging back into the solar panels. You probably wouldn't need a series resistor as the solar panel would not be able to deliver more than 80 mA anyway.
     
  6. Dec 28, 2009 #5
    Are you saying that there would be no need to bring the current down if the panels were charging a battery and the battery powering the device? Because that is what I have planned. The batteries are the typical AA NiCad rechargeable type meant to be charged by solar panels; ~900mAh, 1.2 volts each (I will be using two, wired in series).
     
  7. Dec 28, 2009 #6

    vk6kro

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    Yes, that is right. A solar panel acts as a current generator and limits its own current if short circuited.

    Since the 80 mA is a very modest current, it should be OK for battery charging as it is.

    You probably will need a diode, though, as the battery could discharge into the panel at night.

    I wondered how you were going to get 4.5 volts from NiCd batteries, though. They give 1.2 volts each so you could get 1.2, 2.4, 3.6, 4.8 but no 4.5 volts.
    That is why I suggested lead acid which would give 2.2, 4.4 etc.
     
  8. Dec 28, 2009 #7
    Yes, I will be using a diode. Here are the schematics I'm using:

    http://www.anybodyburns.com/pathlight/images/schematic1led.gif"

    Sorry about the confusion. To clarify; the solar array has ~80mA and ~4.5 volts.

    Referring to the schematic, then, will everything be okay, considering that I will be using a ~6 volt solar array instead of a ~4.5 volt one?

    EDIT: Crap. I said "currrent" in my last post instead of "voltage." I meant voltage.
     
    Last edited by a moderator: Apr 24, 2017
  9. Dec 29, 2009 #8

    vk6kro

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    Are you using a 4.5 volt array or a 6 volt one? You seemed to be saying both.

    You might be able to use just one of your 3 volt arrays if you used a Schottky diode such as the 1N5822. There are lots of them, so you could get almost anything for this.
    These have a lower voltage drop than ordinary silicon diodes.
    That would give you about the same current as two arrays and you could use the other array for something else.

    The circuit looks OK, but you would have to avoid overcharging the batteries. Measure the current when you first install it to see how many hours you will need to use it to charge the batteries.

    I would use a resistor in series with the LED. About 100 ohms to start with and see if that gives you enough brightness. You can probably add more LED/ resistor combinations in parallel if the battery can cope with it.

    Just for interest, here is another similar circuit:
    solar light 2.JPG
    It apparently gives good light from an almost flat battery.
     
  10. Dec 29, 2009 #9
    The schematic I posted calls for a 4.5 volt array, but my only options are a 3 volt or 6 volt array, so I need to know if I must modify the circuit to get 4.5 volts or not, and if so, would it be easier to increase by 1.5 volts or decrease by 1.5 volts?

    But what about the schematic that says I need ~4.5 volts out of the array to charge the batteries properly? Or would only 3 volts be fine?

    I plan on it.

    I think someone else here told me to do that, and recommended a 220 ohm resistor. Here is the schematic I drew that implements that; tell me what you think of it? The added resistor is labled "R4."

    Picture005-2.jpg

    How does it accomplish that?
     
  11. Dec 29, 2009 #10

    vk6kro

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    But what about the schematic that says I need ~4.5 volts out of the array to charge the batteries properly? Or would only 3 volts be fine?

    The solar panel voltage has to be higher than the battery voltage plus the series diode voltage.
    That is 2.4 volts plus 0.15 volts or 2.55 volts. So, 3 volts will charge the batteries if you use a series Schottky diode instead of the silicon one in the circuit.
    Worth a try if it saves one solar panel.

    It apparently gives good light from an almost flat battery.
    How does it accomplish that?

    That circuit is an oscillator which generates a higher voltage than the battery voltage to run the LED. White LEDs need 3.5 volts to run them, so you can't run them on 1 or 2 NiCd batteries.
     
  12. Dec 29, 2009 #11
    Ah... that makes sense.

    Definitely. But wait, are you saying there is a chance it won't work?

    Okay. Hey, do you know how many volts blue LEDs need to run?
     
  13. Dec 30, 2009 #12

    vk6kro

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    Definitely. But wait, are you saying there is a chance it won't work?

    That's right, but you get a money back guarantee. .........:-)
    Don't believe anything will work until you build it and see it working.


    Okay. Hey, do you know how many volts blue LEDs need to run?

    Have a look at this site:
    http://en.wikipedia.org/wiki/Led
    There is a nice chart of LED voltages there.
    Blue LEDs work between 2.48 and 3.7 volts, so you would be very lucky to get one working on just two NiCds. That's why that other circuit I sent would be worth trying. I have seen similar setups in solar path lights. They run a white LED off one NiCd and white LEDs need 3.5 volts.
     
    Last edited: Dec 30, 2009
  14. Dec 30, 2009 #13
    Ah, of course.

    This product (the blue version): http://www.solarbotics.com/products/sbled/" doesn't even say how much voltage they need... though it does say "rated for 30mA continuous current"; will my batteries of ~550mAh (according to schematic) work alright with that?

    Anyways, regarding the solar array charging... accounting for the batteries voltage and the voltage drop of http://www.solarbotics.com/products/d3/": 2.4 + 0.22 = 2.62 so I can use a 6V solar array (6 volts in full sunlight) just fine. Technically I could use a 3V array, but that would only have to drop 0.38 volts before the LED turned on, thanks to a rouge cloud or something. So it's probably safer to use a 6V array, so it would have to drop 3.38 volts before the LED turned on. Correct?
     
    Last edited by a moderator: Apr 24, 2017
  15. Dec 30, 2009 #14

    vk6kro

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    I just did a little experiment. I set up a red LED with 330 ohms in series to a power supply and a 1000 ohm pot (connected as a rheostat) across the LED. This way I could check the actual light output of the LED for different voltages, without destroying the LED.
    Below 1.5 volts, there was nothing from the LED. At 1.52 volts I could JUST see a faint trace of red light. At 1.72 volts it was at full brightness.
    I tried it with a yellow LED. It started to glow at 1.7 volts and was at full brightness at 1.9 volts.

    So, voltage is extremely important. If it is less than the minimum value, you won't see anything.

    Yes, you have it OK.
    You want the solar panel to keep charging even if the panel isn't in full sunlight. So, it is better to have some voltage in reserve.
    NiCd batteries are very vulnerable to overcharging, though, so you will need to watch out for that possibility. It is a bit of a juggling act because the exact behaviour of the solar panel is not known.
    I have a set of wires with crocodile clips on each end and I can quickly set up an experimental setup to find out what really works. This is better than any simulator.
     
    Last edited by a moderator: Apr 24, 2017
  16. Dec 30, 2009 #15
    Oh cool!

    So, if a yellow LED needs ~2 volts to be at full brightness, how long do you think it would take the batteries, at 2.4 volts (full charge) to run down to below 2 volts? And if the answer is "too fast", then would three batteries instead of two at 3.6 volts be too much for the LED?

    Also, once that is decided, I believe on a different thread you recommended a resistor in series with the LED to prevent damage to the batteries, as NiCad batteries don't take kindly to being fully discharged. I think you said that stopping the batteries from being discharged further than 0.875 volts would prevent damage. That would be 1.75 for two batteries, 2.625 for three, etc... correct? In that case, what ohm (for example; 2.2K) resistor would be used for this purpose?

    Yes, I will take precautions regarding battery overcharging. I guess some experimentation will be required to figure out how far they have been charged in how long.
     
  17. Dec 30, 2009 #16

    vk6kro

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    So, if a yellow LED needs ~2 volts to be at full brightness, how long do you think it would take the batteries, at 2.4 volts (full charge) to run down to below 2 volts? And if the answer is "too fast", then would three batteries instead of two at 3.6 volts be too much for the LED?

    This depends on a few things.
    assume 2000 mAH for the batteries and 20 mA current for the LEDs.
    On those figures, the LED would last 100 hours.

    It doesn't help to add extra batteries because when a NiCd or NiMH battery goes flat the voltage drops rapidly.

    Also, once that is decided, I believe on a different thread you recommended a resistor in series with the LED to prevent damage to the batteries, as NiCad batteries don't take kindly to being fully discharged. I think you said that stopping the batteries from being discharged further than 0.875 volts would prevent damage. That would be 1.75 for two batteries, 2.625 for three, etc... correct? In that case, what ohm (for example; 2.2K) resistor would be used for this purpose?

    I hope I didn't say that.
    The resistor in series with the LED is to protect the LED. Above the turn-on voltage of a LED it behaves like a Zener diode and is hell-bent on self destruction. So, you have to limit the current with a resistor.
    It is essential that you don't discharge NiCd or NiMH batteries below 0.8 V (or 0.6 V isn't too bad) and leave them like that.
    If the load is LEDs, they will stop conducting below their turn-on voltage, so they tend to protect the batteries anyway.
     
  18. Dec 30, 2009 #17
    Ah!

    How come?

    So is there a need for a resistor in series with the LED? What kind of resistor (2.2k ohm for example)?
     
  19. Dec 30, 2009 #18

    vk6kro

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    Because it would be wrong to say that.

    You can easily calculate it.

    Suppose you have one LED that turns on at 2 volts. Assume it is on and the supply is 6 volts, so a series resistor must have 4 volts across it.

    Now, if the LED needs to have 20 mA flowing in it, this must also flow in the resistor, because they are in series.

    So to work out the resistor size:
    R = V (resistor) / I (in amps)
    R = 4 volts / 0.02 amps = 200 ohms

    You need to do this calculation to suit your own LEDs and voltages, though.
     
  20. Dec 30, 2009 #19
    Oh, cool! So if the yellow LED uses 2.7 volts and 30 mA, then this is the answer...

    R = 2.7 / 0.03 = 90 ohms

    Am I doing that right?
     
  21. Dec 31, 2009 #20

    vk6kro

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    No.

    It is the voltage across the resistor that matters for this calculation.

    So, if the supply was 12 volts, the resistor would be calculated like this:

    R = (12 - 2.7) / 0.03 = 310 ohms
     
  22. Dec 31, 2009 #21
    Ohh. So the voltage of the batteries, then, right?

    R = (battery voltage) - (LED voltage) / (LED amperage) = (needed ohms of resistor) ?

    Sheesh, in the case that the LED voltage is 2.7 and the battery voltage is 2.4, I need more batteries don't I?
     
  23. Dec 31, 2009 #22

    vk6kro

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    Yes, or you could have a look at that circuit I sent you.

    It uses the voltage developed across the inductor to generate more voltage than the battery supplies and this is then used to power the LED.

    It is a very clever circuit.
     
  24. Dec 31, 2009 #23
    I can't really read that schematic. Is that coil-looking thing the oscillator? How does that work, and what kind is needed? How many batteries does it need?
     
  25. Dec 31, 2009 #24

    vk6kro

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    I have redrawn the oscillator part of the circuit to look more conventional.

    Because the original contains a copyright message, I have kept it on the page.
    This circuit should not be used for commercial purposes.

    solar light 3.JPG

    The 1 M resistor is fed from the solar panel, so when this is grounded the circuit will oscillate. When the panel is operating, the oscillator is not oscillating and is also not drawing any current. This seems to be the function of the diode in the emitter of the PNP transistor.
     
  26. Dec 31, 2009 #25
    I don't have much experience with reading electrical schematics, so I am little confused about a few things.

    1. What is the curly-looking thing labeled "22uH"?

    2. There are no labels on the transistors in either circuit. I can tell that one is PNP and one is NPN but I can't tell which, nor what type is needed.

    3. What's a "1 M" resistor?

    4. The original schematic shows the solar panel and D1 and D2 diodes. Yours only shows one diode and no solar panel.

    EDIT: Oh, wait, you had only redrawn the oscillator part. I'm still rather confused though, because I'm not sure how yours fits into the big picture. If it's not too much trouble, could you draw the full circuit with labels? That would be great.

    Also, this oscillator only gives 1.2 volts to the LED. If I needed more, could I add batteries in series with the one shown without any further alteration of the circuit?
     
    Last edited: Dec 31, 2009
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