Boosting Regulator Current for IC 78xx by MJ2955

Discussion in 'The Projects Forum' started by imbaine13, Oct 19, 2013.

  1. imbaine13

    Thread Starter Member

    Oct 6, 2013
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    Hello everyone. I'd like some help with a small problem. I just learnt how to use PNP transistors only a few days ago, and so far, the only building Ive done involving a PNP was turning on an LED (switching), just to familiarise myself with the basic concept. It occured to me that the only difference between the PNP and NPN as regards switching, is the NPN's base sinks current (conventional flow) while the PNP's base sources current. Basically, the PNP is turned on by powering it's base, while the PNP is turned on by grounging its base, right? I came across this current boosting cicuit, and was wondering if its possible to replace the transistor with an NPN. Could anyone please help me understand why a PNP was used in the first place?
    [​IMG]Here's the link to the full website: http://www.eleccircuit.com/boosting-regulator-current-for-ic-78xx-by-mj2955/


    Thanks a lot for your time. Have yourself a nice day.
     
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    As the current in R1 gets above 750mA, the pnp will turn on and shunt the excess current through its C/E pins, you can replace the PNP with an NPN, but then you will have to use a Negative voltage regulator like LM7905.
     
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  3. imbaine13

    Thread Starter Member

    Oct 6, 2013
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    Thanks Dodgydave,
    Unfortunately, I don't get it. Why 750mA. Could you please elaborate?
     
  4. cornishlad

    Member

    Jul 31, 2013
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    Conventional current flow ( + > - ) thru npn or pnp tansistor is in the direction of the arrow head. Hence pnp in that situation.
     
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  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    To turn-on the PNP transistor the base voltage must be 0.6V lower than voltage at the emitter voltage. So if voltage drop across R1 resistor is greater then 0.6V the PNP transistor will be ON.
    See the two examples
    For Iload = 0.2A
    [​IMG]

    And for Iload = 1A

    [​IMG]
     
  6. imbaine13

    Thread Starter Member

    Oct 6, 2013
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    Do we only consider the emitter base junction (the arrowhead junction)? >> Cornishlad
     
  7. #12

    Expert

    Nov 30, 2010
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    Only one word out of place. "Grounding" the base depends on which circuit you are using. Moving current in the right direction is what matters, and some circuits would use a different description of basically the same thing.

    Here is a concept about how to use an NPN for the booster. Not dead accurate, but it demonstrates a method.
     
    Last edited: Oct 19, 2013
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  8. imbaine13

    Thread Starter Member

    Oct 6, 2013
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    For a pnp to conduct, the emitter voltage must be higher than the base and collector voltages with respect to ground in single power rail circuits.

    Qn.I know that the emitter voltage of a pnp should be higher than the base by about 0.6v for silicon transistor. But what should the voltage difference between the base-collector (pnp) be for it to start conducting. (or does the base-collector voltage not matter and a pnp will conduct regardless of base-collector, as long as emtter-base > 0.6 and base voltage > collector voltage)


    The other thing I don't get is why a npn wouldnt work. the volatge on the right of the voltage regulator (and the emitter assuming the case of an npn) would be lower than the voltage at the base by a around 6v if the regulator is 7805 and the supply is 12v. If The 1Ω resistance is replaced by a slightly larger one, a voltage would be dropped across it and the base would have a lower voltage than the collector of the npn (which is directly connected to the power supply). Why wouldn't the npn work in this case?

    *Confused*
     
  9. imbaine13

    Thread Starter Member

    Oct 6, 2013
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    Thanks for the correction and I really appreciate the image. It helped a lot, but wouldn't excess current through the transistor's base damage it?
    Also, what is the resistor for? Can't I do without it?
     
  10. imbaine13

    Thread Starter Member

    Oct 6, 2013
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    Thanks a lot for the images. They helped a lot. But what should the base-collector voltage have to be? Is Ve > Vb (Ve - Vb = 0.6) the only condition for the transistor to turn on?
     
  11. #12

    Expert

    Nov 30, 2010
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    The collector to base ratio of current is going to operate at the gain of the transistor, thus saving the base from damage. The load resistance becomes the limiter for the base-emitter current and the current through the load is supplied mostly by the collector current. Still, if you short the output of that circuit, the transistor is going to smoke because you have forfeited the current limiting protection of the regulator. I didn't say it's a good practical circuit. I said it demonstrates a method. You can add on protection if you need it. This circuit would suck for a bench supply, but it might do just fine in a different application where the risk of shorting is nearly impossible.

    A 3 pin regulator chip sends about 50 microamps through its ground lead. That might be enough to get the diode to have a few tenths of a volt across it, but intentionally throwing a milliamp down there would get a higher voltage and it would be stable compared to depending on the waste current of the regulator chip.

    Edit: Three pin regulators always need some load current. Wasting the 5 ma or 10 ma required to stabilize the regulator makes a good current supply for the diode which is compensating for the Vbe loss in the booster transistor. Just figure the resistor for (5V - .6V) / 10 ma = 440 ohms. Use standard size: 390 ohms.
     
    Last edited: Oct 19, 2013
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Ve - Vb = 0.6V is the only condition need to tern-on the PNP transistor. But even if this condition is true,it's doesn't mean that there will be a current flow between emitter and collector.
    Try to read this
    http://forum.allaboutcircuits.com/showpost.php?p=234782&postcount=4
    http://forum.allaboutcircuits.com/showthread.php?p=165135#post165135
     
    Last edited: Oct 19, 2013
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  13. #12

    Expert

    Nov 30, 2010
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    The turn on voltage of a base-emitter junction is a continuum. In most practical circuits, the answer is .55 to .9 volts. I did a study to find the truth (for one transistor). Look here.

    http://forum.allaboutcircuits.com/blog.php?b=571
     
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  14. dsuperbad

    New Member

    Nov 3, 2013
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    two questions:

    a) is the current that enters the regulator equal to that coming out?
    b) current that flows through R1 and R2 is 0A?
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Yes. Most of the current flows out the OUT terminal; around 50uA leaves via the ADJ terminal.
    No. The regulator strives to maintain approximately 1.25v between the OUT and ADJ terminal by sourcing current from the OUT terminal. An LM317 regulator needs from 10mA current to 1.5A current flowing from its' output to provide guaranteed regulation as specified in the datasheet. For this reason, a 120 Ohm resistor is typically used as R1, as this will guarantee at least 10mA current flow at the output; 1.25v/120 Ohms ~= 10.41666...mA current. The ~= 50mA current flowing through the ADJ terminal will be added to the current flowing via R1 to equal the current flowing through R2;~= 10.4666... mA.
     
  16. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    No, you need the PNP but you can add an NPN by putting a res in the collector of the PNP and put the E-B leads of the NPN across it, then tie the collector of the NPN to the input.
     
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