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Increasing voltage regulator current

  1. May 3, 2017 #1
    I'm looking to draw about 6A. I have an LM338 regulator that can supply up to 5A. Searching the internet I came across this schematic,
    boosti.gif

    So basically when the 1R resistor drops 0.6V then the transistor conducts the remainder of the current. I calculated my own values for those two resistors. Was wondering if my values are fine,

    I want 3A to pass through the regulator, so I got R1 to be,
    R1 = 0.6V / 3A = 0.2 ohm

    PR1 = 3^2 * 0.2 = 1.8 W

    The remaining 3A can pass through the transistor, using the 2N6491 PNP transistor, this one
    http://www.onsemi.com/pub/Collateral/2N6487-D.PDF

    From the datasheet hfe at 3A is 50
    upload_2017-5-3_23-57-36.png

    So, Ib = Ic/hfe = 3A / 50 = 0.06 A

    Vb = Vin - 0.6 (Vin = input voltage to the regulator)
    = 13.85 - 0.6 = 13.25 V

    Rb = Vb/Ib
    = 13.25 / 0.06 = 221 ohm
     
  2. jcsd
  3. May 4, 2017 #2

    PhysicoRaj

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    I had done this before and almost similar to your steps except for this part:
    Now verify:
    Place that 221 ohms in its place and apply KVL to the base - emitter loop.
     
  4. May 4, 2017 #3

    Averagesupernova

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    That's an awful lot of capacitance to be hanging on the output.
     
  5. May 4, 2017 #4
    I think it's OK for six amps.
    However, the same might be a bit short for the input side.
     
  6. May 4, 2017 #5

    Averagesupernova

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    The reason I wonder about it being to much capacitance is the amount of current draw on power up. The 3 terminal regulator on its own is quite likely able to handle a very very large capacitance. However, when we have the current boost transistor added on and we power up, it is like a short circuit until that cap on the output is charged. I know from experience that this type of current boost configuration WILL smoke when its output is shorted.
     
  7. May 4, 2017 #6

    PhysicoRaj

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    Agreed. There's no point in filtering the output of a regulator. The output voltage will stay constant as long as the input is greater than or equal to a minimum voltage, which will be provided by the input capacitor for as long as possible.
    The datasheet specifies a small cap at the output, just in case the output leads catch up EMI or transient noises.
     
  8. May 4, 2017 #7

    Tom.G

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    Depends on the usage. For instance at the upper end of the audio frequencies, the output impedance of the regulator is equivalent to a 50uF cap. That's not much for a 6A supply. edit: see post two down. end edit

    Use current limiting as shown in https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf, pg 20, fig 14. It uses one transistor and one resistor.
     
    Last edited: May 4, 2017
  9. May 4, 2017 #8

    PhysicoRaj

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    I did not understand.. Or you might be talking for example about a leaky audio amp which requires supply bypass caps?
     
  10. May 4, 2017 #9

    Tom.G

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    You're right , I'm off by a factor of two. The 50uF should have been 120uF.
    If you look at the output impedance vs frequency of the LM338 you see that at 20kHz the impedance is a about 135 milliOhm. Capacitive reactance of 135 milliOhm at 20kHz requires 118uF. Unless I messed it up again, a 6A load will give 810mVpp ripple from the bare regulator.The below image is from: http://www.ti.com/lit/ds/symlink/lm338.pdf, pg7, fig8

    blob.jpg
     

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    Last edited: May 4, 2017
  11. May 4, 2017 #10

    PhysicoRaj

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    Correct.. I hadn't considered that. Thanks for pointing it out.
    So the current limit should compensate for the cap transients for high current application.
     
  12. May 5, 2017 #11
    For this low voltages, and for this kind of typical circuit it's not a big deal at this target current: the energy involved is still low, so unless you make a really big output capacitor bank (over a few thousand uF) it won't harm the components.
    At this level it is typically the fuse which you have to be careful with.
    Also, it worth to note that for output filtering adding a smaller ceramic cap can spare a lot on the bigger electrolytic one.
     
  13. May 5, 2017 #12

    Baluncore

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    TheRedDevil18.
    You should put a 10uF tantalum and a 0u1F ceramic capacitor in parallel with the output of the regulator.

    With higher current regulators the cost of heatsinks becomes prohibitive. Linear regulators are being progressively replaced with switching converters. Switching 3 amp adjustable regulator modules now cost a couple of dollars including postage. 12 amp modules cost about $5 complete, including postage.
    For example;
    http://www.ebay.com.au/itm/14225835...49&var=441318526436&ssPageName=STRK:MEBIDX:IT
     
  14. May 6, 2017 #13

    davenn

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    just a note ..... if you put an @ sign in front of a nickname, the member will be notified of your response :smile:
     
  15. May 6, 2017 #14

    Baluncore

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    Yes, if I needed OP attention I would have put an @ in front of the name.
    TheRedDevil18 has been away for a couple of days, so probably has notifications turned off anyhow, there is no hurry.
     
  16. May 6, 2017 #15

    sophiecentaur

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    I feel so out of touch at times. I'll have to try it.
     
  17. May 6, 2017 #16
    That value seems a bit high. I simulated it and I get a very low current across the transistor. When I remove it, it works fine. Not sure if I really need that base resistor
     
  18. May 6, 2017 #17

    PhysicoRaj

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    Yes. That is because you are assuming the current through the base resistance is because of the voltage across the input of the regulator, which is not the case.
    Base current will be the resultant of the voltage between the free end of base resistance and the 'emitter'.
    That according to me will be the potential across the 0.2 ohm current sense resistor. Call this Vb.

    Then another important parameter is the base-emitter drop (Vbe) which you will have to subtract from Vb. I checked the datasheet and I think for your current the Vbe will be around 1 volt.
    So you see the current sense resistor needs to drop more than 1v to force a good base current and turn it on.
    That changes the 0.2 ohm resistance.

    After that, you'll get an appreciable value of Rb if you apply KVL to the base-emitter loop.. do not omit it.

    Good luck.
     
    Last edited: May 6, 2017
  19. May 6, 2017 #18

    Baluncore

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    The current sense resistor 1R sets the maximum current through the regulator by starting to turn on the PNP pass transistor when the regulator has over about 0.6 amp current.
    So what is the purpose of the Rb = 10R in series with the base of the PNP ?
     
  20. May 6, 2017 #19

    PhysicoRaj

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    It will be very small. But without the base resistor the transistor will turn completely on and the base current could be well over the limit.
     
  21. May 6, 2017 #20

    PhysicoRaj

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    Looking at the datasheet, it seems that this transistor can handle upto 5A of base current.
    So maybe it won't get busted without Rb.
     
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