Understanding 78XX high current voltage regulator

Discussion in 'General Electronics Chat' started by drbenne, Apr 7, 2015.

  1. drbenne

    Thread Starter Member

    Jul 30, 2013
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    Hi guys,

    I've been looking over the typical applications in the datasheet for various 78XX voltage regulators trying to build and learn more about each configuration when I came across the two designs for high current and high current with short circuit protection.

    My confusion arises because of the use of PNP transitors and the term "B" in the formulas for determining both
    R1 and Io in the high current configuration.

    I understand that the PNP's function is to pass the higher current. What I don't understand is what role R1 has in biasing the transistor. From what I can see there is no base resistor resulting in a high emitter/base current. How does this not blow the transistor?

    For the short circuit protection, what is the logic in providing short circuit protection based on the transistor configuration?

    I've been trying to study PNP basics but can't seem to wrap my head around them. NPNs haven't been too much of a problem.

    Thanks,

    Dave
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Say the load current is 4A (too much for the LM78xx alone). How much current flows through the PNP and how much flows through the regulator?

    The Vbe of the PNP when it is turned on is about 0.7 to 0.9V. How much current has to flow through R1 (3Ω) to create a base bias of 0.9V? I=E/R = 0.9/3 = 0.3A.

    If ~3.5A out of 4A of the load current goes through the PNP (bypassing the LM78xx), what would be the base current of the PNP? Say that β is 50, so 3.5/50 = 0.07A, so that adds to the 0.3A we calculated above, so the current that flows through the regulator is .37A (which goes to the load) , so the total load current is 3.5+0.37 = ~3.9A which is close to the postulated 4A.

    So you can see that the regulator carries about 10% of the load current while the PNP carries 90%. That mix is effected by R1.

    Note that the regulator is still regulating. If the load current decreases, the regulator reacts by passing less current, which reduces the base current of the PNP, which adjusts the total current to the load...

    p.s. B=β=current gain of the PNP.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is a sim. Independent variable is load current, from 5mA to 5A. Dependent plots of V(out), the current out of the regulator, the collector current of the PNP, the current through R1. Note that at load current = 4A, about 3.25A from the PNP and about 0.7A from the regulator.

    49.gif
     
  4. drbenne

    Thread Starter Member

    Jul 30, 2013
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    Thanks Mike,

    I thought that B=β, just wasn't totally certain. I guess my next question is about the current flowing into and out of the node of the PNP base and R1. What directions are the currents flowing? I am confused as to how the PNP turns on. Is the base current being fed back in the emitter through R1?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    The idea behind the NPN is directly analogous to the PNP. The base-emitter junction of a NPN is and PN diode so base current flows into the base pin and out the emitter pin. The base-emitter junction of a PNP is a NP diode so base current flows into the emitter pin and out the base pin. As a consequence the "on" base-emitter voltage for a NPN is about 0.6V while it is -0.6V for a PNP. In both cases collector, in the active region, the collector current is β times the base current. But for a NPN this is the current flowing into the collector and out the emitter while for a PNP it is the current flowing into the emitter and out the collector.
     
  6. drbenne

    Thread Starter Member

    Jul 30, 2013
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    All of the explanations have made sense so far but could someone explain the function of the short circuit protection of the 2nd circuit? I have an idea but I don't want to sound too stupid describing my explanation...

    thanks,

    dave
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    As more current flows through Rsc (guess what 'sc' stands for), Q2 is turned on more and more, right? What happens to the voltage base-emitter voltage of Q1 if Q2 is turned on hard?
     
  8. imraneesa

    Member

    Dec 18, 2014
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    Can you explain me this. does not PNP need negetive current near base. how the PNP starts without negetive current near base? can you explain me this please.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    It does have negative current at the base (if you are assigning positive current to be current flowing into the base).

    As soon as enough current is flowing through the 3Ω resistor to develop about 0.6V to 0.7V across it, then current can start to flow from the emitter to the base of the transistor, which means that many times that amount of current can flow from the emitter to the collector.
     
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