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Battery needed for POV display project?

  1. Aug 21, 2013 #1

    perplexabot

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    Hello all. I have been working on a side project, a Persistence of vision display. I have had and still have a bunch of problems I must deal with. I feel as though it has something to do with the power needed by the components and the battery I am using. Can someone help with this matter?

    For example, the components are (other than the battery):
    8LEDs with a shift register
    atmeg328
    a hall effect sensor and another LED

    is there a way to know what type of battery I need?
    Also a bigger question once I get the right power source, would I need to "partition" the power source in order to satisfy each component? maybe with a voltage divider? Or am I wrong? Any help would be appreciated, Thank you.
     
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  3. Aug 21, 2013 #2

    davenn

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    well your ATmega will have a max supply voltage that it needs to be kept under
    the datasheet will tell you its standard supply voltage

    what is your shift register ? what sort of IC(s)
    again the datasheet will tell you its standard supply voltage

    there's some starting points

    do you actually have a circuit diagram for this project?
    if so did you draw it or did you get it off the net ... post the cct diagram

    Dave
     
  4. Aug 21, 2013 #3

    perplexabot

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    Hey, thanks for the reply. Yes, I know that the max voltage can be found in the datasheets, but once these voltages are known, what then? I built this thing myself, so I didnt follow a schematic, I will upload my schematic soon. Thank you.
     
  5. Aug 21, 2013 #4

    meBigGuy

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    All of your components have their own voltage requirements. One of the steps of design is selecting components that allow a simple power system. It's generally a balancing act. Often you need a regulator to condition the batteries for proper operation. This might be a switching boost regulator to convert from ~1.5 batteries to 3.3 or 5V. Or maybe you can select components with a wide enough range to operate directly from multiple cells. Or, you can regulate multiple cells down to the voltages you need (and maybe drive the LED's directly off the battery).

    You need to choose a battery technology and a battery size and then understand it's voltage range and discharge characteristics. Then you determine how to condition that to supply the components you have chosen. It is actually the hardest part of many designs.

    A computer mouse usually has a single alkaline cell or paralleled cells and a SMPS to convert to the voltages required by the processor and Laser Module. That would be a 1.5V Boost regulator.
     
    Last edited: Aug 21, 2013
  6. Aug 22, 2013 #5

    perplexabot

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    Interesting. So your saying choose a battery first then figure out how to "distribute" power? So say I have three components, each with a max Voltage of V1, V2 and V3. If i know that the device I am building/designing will at one point use all three components, then the battery should be at least V1+V2+V3, is that correct? But that's only voltage, what about the current needed by the components (in other words, should I be looking at power instead of voltage)?

    Yes, I forgot to mention that I am using a voltage regulator for the atmeg, but all the other components are connected to unregulated voltage (directly to the 9v battery that I randomly chose). Once again, I will post a schematic soon, sorry for the inconvenience.
     
  7. Aug 22, 2013 #6

    perplexabot

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    Hey guys, here is the schematic. I just drew this now, hopefully with no errors.
    I still don't know how to tackle my problem.

    Thank you all.
     

    Attached Files:

  8. Aug 22, 2013 #7

    meBigGuy

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    Nice start.

    One thing you are missing is that you need to avoid anything that is 9 volts from leaking to the ATMega. Anything higher than it's supply voltage will damage the outputs/inputs. Let's assume you are running at 3.3V.

    For the LED shift register, use a 3.3V logic part and drive transistors that sink the LED current to ground.

    For the hall effect, also use a transistor to isolate the processor from 9V.

    Maybe you can just run everything at 5V in which case it would all run off the +5V regulator and your lack of level conversion would no longer be an issue.
     
  9. Aug 22, 2013 #8

    perplexabot

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    Is that not achieved with the voltage regulator (vol reg)? Since the output of my vol reg is the input voltage of the atmeg (where the vol reg output voltage is programmed to be 5v)

    Are you assuming the atmeg voltage is 3.3v or my battery (voltage source) is 3.3v?

    By "3.3v logic part" do you mean another vol reg with a 3.3v output? Are you saying I need to have the input voltage for the shift register regulated to 3.3v? And are you also saying instead of having the leds '-' terminal connect straight to ground, have them connected to a voltage buffer?

    Hmmm, this makes sense, basically just insert a voltage buffer between the hall sensor and the atmeg in order to reduce the "cascading effect," is that true?

    This solution seems to be a simple one and best suits my situation since I have already done most of the soldering (lesson learned: never solder until you know you have a working design. Should have known that from the start).
    So are you implying that after I have inserted the buffers, I can try to distribute the regulated voltage to all the components? Wouldn't I then need to worry how much current each component needs?

    I want to say thanks again for your invaluable help and your time. I apologize for my many questions.
     
  10. Aug 23, 2013 #9

    meBigGuy

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    The voltage problem I was referring to is that the IO pins of the processor generally cannot be subjected to voltages higher than its power supply pin. So you have to use level translation to interface to the 9V circuits. There are various ways to do that.

    Also, assuming you find a 9V shift register, the output voltage of the processor is barely enough to cause it's input to switch (unless it is a specialty part). Generally you don't want to connect 9V IO directly to a 5V processor. There are better ways to do that (generally the shift register drives transistors that sink the LED current from the 9V)

    EDIT: I just found a neat part (TLC6C598-Q1) that deals with all this nicely: http://www.ti.com/lit/ds/slis142b/slis142b.pdf [Broken]. It would work perfectly for you, and nicely illustrates the output drive transistors.

    Let's assume you run everything at 5V through the regulator for your first attempt. The regulator will have to supply all of the current, so it might get warm. Say you run each LED at 14ma. That's 8x.014 = 112ma. Dropping 4V across the regulator (9V to 5V) means you have added about 1/2 watt to the required dissipation of the regulator.

    Depending on the LM317 regulator package and the power requirements of the processor and sensor, it may get warm. For example a TO-220 package LM317T has a Thermal Resistance of 80degreesC/Watt. So with no heat sink, if you dissipate 1 watt the junction will heat to 80C + Ambient, or about 105C or more, which is about its limit (125C is the limit, so you are pushing it) Add a small heat sink and it drops accordingly (look at the heat sink specification to determine thermal resistance).

    Regardless of the power architecture you choose, you need to consider the dissipation in any regulators.
     
    Last edited by a moderator: May 6, 2017
  11. Aug 23, 2013 #10

    perplexabot

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    Wow, amazing reply. So much information! Thank you so much. That part you found would have worked perfectly, I wouldnt have to worry about the 8LEDs. It turns out my 8bit register can take -.5v up to 7v.

    Ok so I think I know what to do now.
    I will regulate my 9v battery down to 5v, so that the atmeg, hallsensor and shift register all have 5v voltage supply. Then I will need to make sure the inputs to the atmeg do not exceed 5v (which I assume they wont since the components' power supply is 5v). I will also need to make sure that the atmeg does not output signals that are over 5v (which I also assume wont since the atmeg's power supply is at 5v).

    I then need to place a buffer amp between the hallsensor and the atmeg. Finally, if I do not use the part (the shift reg) you found, I need to add mosfets (acting as switches) to turn on the 8Leds. If I do use the part you found, then I can connect the 8Leds as shown in my first schematic.

    What do you think?
     
  12. Aug 23, 2013 #11

    meBigGuy

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    Sound like you got it.

    Notice that your first schematic shows the shift register sourcing the power, and the part I showed would sink the current.

    Post the hall effect input buffer when you get it done.
     
  13. Aug 24, 2013 #12

    perplexabot

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    Hello. Your saying in the initial schematic, the shift register sources power, I think this makes sense since current is emitted from the shift register to the LEDs. You also say that if mosfets are introduced (between the shift register and the LEDs) to turn on the LEDs the shift register will sink current rather than source, I am having a problem wrapping my head around that. I realize that the shift register does not supply current in this case but only switches on the power supply to the LEDs. I can't rationalize how current sinks into the shift register. May you please elaborate?

    I attached my buffer amp. It is a degenerate common drain mosfet configuration. Is this the correct approach?

    Once again thank you.
     

    Attached Files:

  14. Aug 24, 2013 #13

    meBigGuy

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    I'm confused about what you are doing.
    You need for the shift register to drive into N channel FETs that sink 9V supplied current to ground through the LEDs (there is now a logical inversion). This isolates the shift register from 9V while allowing the 5V from the shift register to turn on the fet. That's how the TI part works. Sinks LED current to ground.

    As for your hall effect sensor, I don't know what part you are using so I don't know how to interface it. If it is running from the regulated supply you probably don't need a FET. If it is running from 9V what you drew might have issues. Generally to level shift from 9V to 5V you drive the 9V signal into a source grounded N channel fet (possibly through a voltage divider) and the FET drain has a pullup to 5V. The drain connects to the uP.
     
  15. Aug 24, 2013 #14

    perplexabot

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    Thank you for your very informative reply. I will run my hall sensor using the 5v regulated voltage, so according to your last reply, I don't need a buffer. I have one more question.

    Q: Is it possible to connect the LEDs to the 5v regulated voltage and NOT use fets to drive them?

    Is it because 5v is not enough for 8LEDs?

    Thanks again for all your help, I definitely learned a lot.
     
  16. Aug 25, 2013 #15

    meBigGuy

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    If you connect the shift register to 5V you can do exactly what you showed in the first drawing if the shift register can source enough current (which I doubt). Or, you can turn them around and sink current (some drivers can sink more than they can source). Either way is fine. You need only consider the drive capability of the shift register, the LED's requirements, and the heating of the regulator (as I described in post 9).

    I don't know what the LEDs require. Let's say they drop 2.6V at 14ma. The problem is finding a shift register that can supply that. For example the 74HC595 can only source 4.5 volts at 8ma. Some can only supply 4ma.

    But, it depends on the LED. How much current you need for the brightness you want, and how much voltage drop occurs at that current for that LED.

    The basic calculation goes like this. Say you can settle for 8ma, assume a 0.5V shift register register voltage drop, assume 2.3V LED voltage drop => 2.8V drop total. Leaves 3.2V across the resistor. 8ma at 3.2v is 400 ohms. So if you use 400 ohms you will get near to 8ma through the LED if my assumetions were close. If the LED drops 3V, then there will only be 1.5V across the resistor so you use about 200 ohms.
     
  17. Aug 25, 2013 #16

    perplexabot

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    Hmmm. Ok, I will need to do a little more thinking/reading. Thanks again for all your wonderful help.
     
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