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Need help about a 7805 regulator circuit !

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by Zarkabul, Jun 7, 2005.

  1. Roff

    Roff Well-Known Member

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    I cobbled together a little simulation in SwitcherCAD for this problem. The values for R1 and Rsc are shown as .params. As was pointed out, this is hypothetical, and doesn't use real parts that you can buy or steal.

    Ron
     

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  2. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Hi Ron,
    It looks like your sim is using 0.228 ohms for R2{R1} and R3. Then with 1A through the 7805 regulator the voltage drop across R2{R1} will be only 0.228V and the booster transistor Q4 won't be anywhere near being turned on.
    The 7805 must pass 0.6V/0.228 ohms = 2.63A for the booster transistor to begin conducting, and when the booster transistor reaches a collector current of 4A with a Vbe of 1V, the 7805 will have 1.912V/0.228 ohms = 8.39A through it! :(
     
  3. Roff

    Roff Well-Known Member

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    I should have numbered the resistors differently.
    {R1}=3 ohms=the value of R2. (.param R1=3)
    {Rsc}=0.228=the value of R3. (.param Rsc=0.228)

    Ron
     
  4. dave

    Dave New Member

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  5. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Hi Ron,
    Don't you argee that if {R1} is 3 ohms, and if the regulator has 1A thru it, then the voltage drop across {R1} is slightly less than 3V and your booster transistor Q4 would be turned-on so hard that the other transistor Q3 would current-limit? Then the Rsc would have about 0.9V across it and wouldn't you know, have nearly 4A through it! :shock:

    But it doesn't work that way because Vbe of Q4 plus the voltage across Rsc reduces the voltage across the 3 ohms resistor, reducing the current in the regulator to about only 633mA. :lol:
     
  6. Roff

    Roff Well-Known Member

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    Hi Len,

    The additional current (1A-633mA=367mA) will flow through Q3, if Q3 is robust enough to carry that current. It might be better to make R2=2.2 ohms, which (in my sim) increases its max current to about 856mA, and reduces the max current through Q3 to about 104mA. You can't cut this too thin without running into component tolerance problems, but then this is just an academic exercise.
    If anyone wants to try this, here is the SwitcherCAD file. I have posted it as a text file, because the forum doesn't accept .ASC files. Save it to C:\Program Files\LTC\SwCADIII (or wherever you keep your schematic files) and rename it with the .ASC extension.
     

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  7. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Hi Ron,
    I changed the file type of your sim file before downloading it and got SwCADIII to open your schematic. I watched it run a few things but it is different with R1 = 2.2 ohms.

    Do you think the Current limiter transistor is active with a 5A load?
    If this is the case then I am giving my appologies to Russ and Morgen. :oops:

    Do you think it looks like this:
     

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  8. Roff

    Roff Well-Known Member

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    Yeah, I think that you were missing the fact that Q2 will conduct when it limits the voltage across Rsc. It will supply the difference between The regulator's current and the current through R1.
     
  9. amazing_ang

    amazing_ang New Member

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    hi
    i need help with designing a power supply of 5v-5a(high voltage,high current).
    cant get the current to boost by using a current booster.if u have the circuit diagram even that will do...but not by using a MJ power transistor only current booster
     
  10. Roff

    Roff Well-Known Member

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    Sounds like you need a switching regulator.
     
  11. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi there,


    Here's how it all works......


    Since Ir1 depends on the two Vbe drops of the transistors just before cutoff:

    R1=(Vbe1+Vbe2)/(Ir1-Ic1/B1)

    but since Vbe changes we want to allow a 50 percent
    change in the two Vbe so we would get:

    R1=1.5*(Vbe1+Vbe2)/(Ir1-Ic1/B1)

    or

    R1=1.5*(Vbe1+Vbe2)/(RegIlimit-Io/B1+RegIlimit/B1)

    Since Q2 begins conducting at Vbe2:

    Rsc=Vbe2/((Io-RegIlimit)*(1+1/B1))

    where

    Vbe1 is base emitter voltage of Q1, and
    Vbe2 is base emitter voltage of Q2, and
    RegIlimit=Regulator I limit
    Ic1=collector current of Q1
    Io is total output current
    B1 is beta of Q1
    Rsc is Q2's base emitter resistor
    Ir1 is current through R1


    Not too difficult?


    Numerical example 1:

    Io=5 amps
    RegIlimit=1 amp
    B1=100
    Vbe1=1 volt
    Vbe2=1 volt

    Calculate R1:
    R1=1.5*(Vbe1+Vbe2)/(RegIlimit-Io/B1+RegIlimit/B1)
    so, R1=3.125 ohms.

    Calculate Rsc:
    Rsc=Vbe2/((Io-RegIlimit)*(1+1/B1))
    so, Rsc=0.248 ohms


    Numerical example 2:

    Io=5 amps
    RegIlimit=1 amp
    B1=10 (much lower than before)
    Vbe1=1 volt
    Vbe2=1 volt

    Calculate R1:
    R1=1.5*(Vbe1+Vbe2)/(RegIlimit-Io/B1+RegIlimit/B1)
    so, R1=5 ohms.

    Calculate Rsc:
    Rsc=Vbe2/((Io-RegIlimit)*(1+1/B1))
    so, Rsc=0.227 ohms


    Another point of view is that if R1 goes too low the regulator will cut back too soon, and if R1 is too high Q2 has to do more work.
    A good value would be one that always allows Q2 to work, but as little as possible.
     
    Last edited: Oct 13, 2008
  12. Roff

    Roff Well-Known Member

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    Mr Al, this thread is over 3 years old. I don't think amazing ang wants to use the circuit that was being discussed. I could be wrong.
     
  13. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Roff,

    Well, if he does now he has a good idea how it works :)
    Someone else might find this useful too.
    I thought it was a little interesting too, seeing as how it would work with other regulators too.
     

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