Assistance with circuit modification - DC to DC inverter

Discussion in 'The Projects Forum' started by letsbully, Mar 23, 2012.

  1. letsbully

    Thread Starter Member

    Mar 23, 2012
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    Is there a way this circuit can be easily modified to output a higher voltage? Currently the circuit outputs 4.5v from an input of approximately 1v (measured with my multimeter). Can any of the components be substituted/swapped to output 7 (or even 9) volts? Thanks.

    [​IMG]
     
  2. #12

    Expert

    Nov 30, 2010
    16,252
    6,751
    Increase the size of the inductor.
     
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  3. letsbully

    Thread Starter Member

    Mar 23, 2012
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    Increasing it from 470uH to a higher inductance would work? (I tried using a lower inductance (150uH) to see if that would make a difference but the voltage output stayed the same at 4.5v)
     
  4. KJ6EAD

    Senior Member

    Apr 30, 2011
    1,425
    363
    You need to increase the voltage rating of C2 to handle the higher voltage. It's currently 3.3V, too low even for the 4.5V it's seeing!
     
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  5. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    Thanks! I actually forgot to mention that I replaced the capacitors in the schematic with 16V capacitors.

    Here's the curious thing. I did a little experimenting and this is what I'm getting:

    Current source: 1.2v AA NiMH battery. Here are the voltages I am getting using various inductors that I tried. My battery is connected to where the 1.5v battery goes (of course) and my multimeter is connected to the "out" of the circuit.

    I get 17.4v using a 150uH inductor, 10.1v with a 470uH inductor, 11.5v with an inductor I found (not sure the rating) that has 25 turns of what looks like 14 awg wire around 2" diamater core, and 6.0v with an unknown inductor that has 7 turns of 16 awg wire around a 1/2" core.

    Does that sound right? I'm trying to learn the relationship between the output voltage and inductance with this "experiment" (without getting too technical and math intensive). It seems that it's turning out that the lower the inductance the higher the output voltage in this circuit.

    Also, is the inductor the only part responsible for "creating" the output voltage? What are the resistors doing in the circuit? Are they affecting the voltage too? Thanks for the info.
     
  6. ifixit

    Distinguished Member

    Nov 20, 2008
    638
    108
    Almost everything will change the output voltage. Without feedback it will be difficult to get a predictable voltage out. If the load current is known and doesn't change, then that would help.

    What is the load current requirements?

    With a fixed load, you can tweak R2 a little to get the voltage you want. However, the output will change as the battery dies.

    Regards,
    Ifixit
     
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  7. #12

    Expert

    Nov 30, 2010
    16,252
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    Looks counter intuitive. Instead of a larger inductor delivering larger bites of current, it works better with a smaller inductor because it runs faster. Now try it with a load and see which inductor holds a higher voltage under load.
     
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  8. letsbully

    Thread Starter Member

    Mar 23, 2012
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    The current requirement will be 1A (eventually), but for now the current requirement is pretty much "any". I know/suspect that for 1A I will need an inductor capable of that and resistors beefy enough to handle the current.

    What's curious to me is that under load, the voltage output is 2.6v no matter what inductor is used. I had 1 LED connected at first and then had 6 LEDs connected in parallel in the circuit and tested inductors I listed previously and the output was measured at 2.6v in both cases. Without LEDs the voltage output is what I listed in my other post, depending on which inductor is used. Do I need a larger inductor to an output to stay at 7v? Different resistors?
     
  9. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    That circuit can not possibly supply anywhere near an amp of current. Analyze the power in/out and then calculate the currents through the transistors (and the base drive currents). You will see what I mean. That is basically a low current novelty circuit.
     
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  10. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
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    A flyback converter stores energy in the inductor when the transistor is on in the form of:

    1/2 L (I) squared

    Then releases the energy when the switch is off.

    But in your design the base drive to the switch transistor is so weak it basically starves the switch so you are fine tuning the L value to get the most energy stored based on the transistor collector current available.
     
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  11. #12

    Expert

    Nov 30, 2010
    16,252
    6,751
    Uh...did you put resistors in series with the LEDs or just connect them to the output?
    Without resistors to limit their current they would act like zener diodes on a feeble amount of current and lock all your output voltages to the LEDs internal voltage... 2.6V

    I think you should stop now and look up a REAL boost converter on the www.national.com website. Their calculators will show you more stuff than you could ever get out of a AA battery, get you valid answers, and do it faster.
     
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  12. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    Thanks for the info. It's basically the first circuit I've done from scratch, start to finish.

    Got it.

    I just connected them to the output. Makes sense now.
    Thanks for the site. It looks interesting.
     
  13. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    What about this circuit? How can it be modified to output 5v?

    [​IMG]
     
  14. #12

    Expert

    Nov 30, 2010
    16,252
    6,751
    Follow the instructions on the datasheet about what resistance the Rfb resistors should have.
     
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  15. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    Here's the overview page of LM2622 from National and the datasheet directly from Ti.

    The formula they give is

    Rfb1 = Rfb2 x (Vout - 1.26 / 1.26)

    So would Rfb1 be 22.2K ohms then? I did Rfb1 = 7.5 x (5 - 1.26 / 1.26) = 22.2xxx. Is that correct?

    Thanks for guiding me. Most of this stuff is like reading a language that I don't know.

    Edit: They also mention a different setup in the datasheet. The image I linked to in the last post shows a setup for 600kHz. They list the components to a 1.3Mhz schematic. Is there any advantage to the 1.3Mhz to a 600kHz setup?
     
    Last edited: Mar 26, 2012
  16. #12

    Expert

    Nov 30, 2010
    16,252
    6,751
    It's 22.2k
    You have to carry the labels when you do math.
    1.3 MHz uses a smaller inductor but is pickier about layout details. Good thing the National website provides drawings of a lot of their chips on a properly designed circuit board.
     
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  17. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    Awesome. Thanks a lot! :D

    Right before I came back to check if anyone replied to my question I went back to the datasheet realizing that I could verify whether or not my answer was correct by using the values for Rfb2 and the Vout that were already given in the sheet to see if I get the same value (40.2K) for Rfb1. Makes total sense to me now.

    Another question: I can't find a 22.2K resistor. If I use a 22K and 200ohm resistor in series, it should add up to 22.2K and be the same as using one resistor because when resistors are connected in series their combined resistance is equal to the individual resistances added together. Is there a "real world" caveat to this? I shouldn't have to add anything to compensate for "losses" of anything right?
     
    Last edited: Mar 26, 2012
  18. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    One other thing. Do wattage ratings of resistors matter in this application? What if I use, say, a 3W resistor instead of a 1/4W? Would that affect the current output?
     
  19. #12

    Expert

    Nov 30, 2010
    16,252
    6,751
    Yes, 22k + 200 = 22.2k
    There are also 1% resistors you can buy with much closer values to what you calculate.
    Then you find out you still haven't got it perfect because of the tolerance of the reference voltage in the chip or the current in the feedback path and you wind up settling for 5.1 volts or 4.9 volts because it doesn't hurt anything.

    Another way to look at it is, if 22200 ohms will get you 5.0 volts, 22000 ohms is 1% off and you will get 5 volts less 1% = 4.95 volts. Not a problem...except the 22k resistor is not guaranteed to 1% accuracy. It's guaranteed to 5% or 10% accuracy.

    It's never perfect the first time. Most of the time, it isn't perfect on the second try. Take hold of that perfection gene of yours and suppress it!

    and yes, wattage always matters. Do the math or get familiar with the smell of burning parts.
     
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  20. letsbully

    Thread Starter Member

    Mar 23, 2012
    34
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    Sounds good. I'll make note of that. Thanks!

    As far as watt ratings go, I guess buying a beefier resistor than needed would not adversely affect the circuit after all then. I didn't know whether or not it would somehow affect the current output if I used a resistor that was rated at 2W instead of 1/4W as in the previous example (as if the resistor would eat up more current if it was rated higher).
     
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