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Current passing for regulator IC's

Discussion in 'General Electronics Discussion' started by sureshot, Jan 24, 2018.

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  1. sureshot

    sureshot

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    Jul 7, 2012
    Looking for some help. I'm wanting to try and do a current pass circuit for a TS7812 L CT 100mA voltage regulator. My aim would be 2 to 3 Amps by way of a single pnp pass transistor as an emitter follower. My problem is keeping the regulators work to under 100mA and preferably around 80mA under load conditions. Would i be right in thinking my current limiting resistor would be the following: 12 Volts - 0.6 Volts / 0.080 mA = 142.5 ohms ,so the nearest 0.5 watt rsistor for limiting current through the 100mA voltage regulator
    Thanks for reading, any help much appreciated thank you.
     
    Last edited by a moderator: Jan 25, 2018
  2. WHONOES

    WHONOES

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    Why not use a voltage regulator that is capable of delivering the current you require. They are freely available.
     
  3. Audioguru

    Audioguru

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    An emitter follower ruins the excellent voltage regulation of an IC regulator IC because its base-emitter voltage changes when the current changes.
    A PNP pass transistor added to a positive IC voltage regulator is not an emitter follower and produces excellent voltage regulation because it is inside the negative feedback loop. Like this:
     

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  4. AnalogKid

    AnalogKid

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    Jun 10, 2015
    You didn't do us the courtesy of including the datasheet for the part you are using, so this is only guesswork.

    Assuming your part is similar to the standard LM7812, using a PNP pass transistor will get you additional output current while limiting the current through the regulator, plus maintain the regulation performance of the chip. See page 20, Figures 13 and 14.

    ak
     

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  5. sureshot

    sureshot

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    Sorry for not including the data sheet, i have used L7812 1 Amp regulators before. And the higher current 78S12CV 2 Amp version, both as stand alone and in a circuit with a pass transistor. I wanted to try and see if i could amplify the current using the 7812 100mA version of this linear regulator.

    I will try and post the pdf data sheet, its the TO92 package i was hoping to use. As my opening post, i was wandering if i had the maths right for the current limiting resistor for the regulator. That was 12 Volts - 0.6 Transistor drop 11.40 Volts / 0.080 mA = 142.5 ohms resistor. So the closest value to that figure. That is if i've got the maths right. Thank you again for any help with this.
     

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  6. sureshot

    sureshot

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    Jul 7, 2012
    I was using a popular circuit that uses a pnp pass transistor to boost regulator current. a few web sites have this circuit diagram. I first saw it in a fairchild data sheet, for boosting 7812 regulator current. But that was for the L7812 1 Amp regulator, and not the 7812 100mA version of this regulator.
     
  7. sureshot

    sureshot

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    Jul 7, 2012
    Thank you for the two circuit diagrams. Iv'e got the TIP36C transistors, but would still like to pair one up with the 78L12 100mA voltage regulator. I was just looking for some help to see if i got the maths right for my current limit resistor on the regulators input. I only want it to do 80mA of the work, the rest of the work being done by the transistor. My working out was 12 Volts - 0.6 Volts drop for the transistor, so 11.40 Volts / 0.080 mA current for the regulator = 142.5 ohm resistor. I'm not entirely sure I've got the math right.
     
  8. Audioguru

    Audioguru

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    Sep 24, 2016
    The resistor does not limit the current, instead it senses the current then turns on the transistor.
    You want the transistor to turn on with a base-emitter voltage across the resistor of 0.6V and a current of 80mA so the resistor value is 0.6V/80mA= 7.5 ohms. But why not 50mA then the resistor is 0.6V/50mA= 12 ohms.
     
  9. sureshot

    sureshot

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    This is where I'm unsure, in the picture below that circuits current says it conducts current above 600 mA , below that value the regulator is doing the work. So i don't understand how a 12 ohm resistor could get the transistor to switch on at 50 mA I'm not saying its wrong, more that i don't get how a 10 ohm resistor switches the transistor on at above 600 mA. But a 12 ohm resistor switces on the transistor at 50 mA. I'm unsure. Thank you for the help with this.
     

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  10. sureshot

    sureshot

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    Jul 7, 2012
    I think I'm getting some of it now. So its the voltage across the resistor that turns the transistor on ? So if the transistors base maximum voltage is say 5 Volts, i would won't it to switch on below that. Maybe 1 or 2 Volts to turn on the transistor for a current limit on the regulator of 50 mA would be, say 2 Volts / 0.050 mA = 40 ohm resistor. If I've understood you correctly.
     
  11. Audioguru

    Audioguru

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    Sep 24, 2016
    The base-emitter voltage turns on the transistor. It begins to turn on when the base voltage is about 0.6V less than its emitter voltage and is about 1.8V when the collector to emitter current is typically 10A.
    Then for a base-emitter voltage of 0.6V the current in the transistor is about 50mA. Then the resistor is 0.6V/50mA= 12 ohms.
    The picture you found has a 1 ohm resistor and if the transistor conducts 600mA then its base-emitter voltage is 1 ohm x 600mA= 0.6V which is wrong because the datasheet of the 2N2955 shows about 0.78V for 600mA.
    The picture I posted has a 3 ohm resistor and a TIP42 transistor that begins to turn on when its base-emitter voltage is 0.6V.
     
  12. sureshot

    sureshot

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    Thank you for getting back to me. So what value would be a good starting point, if I'm using a 78L12 100mA voltage regulator ? I don't really won't to keep burning the regulator out, where by its drawing to much current. In short I'm trying to find the math formula to get the resistor value so I'm only drawing around 50 mA to 80 mA from the 78L12 regulator. And the transistor take the rest of the current up to 2 to 3 Amps maximum, typically 2 Amps. After reading up some, turn on transistor voltage was 0.6 - 1.2 Volts. What I'm looking for is the formula to determine a maximum current of 50 - 80 mA of work from the 78L12 regulator.
    Thanks again for the help.
     
  13. Audioguru

    Audioguru

    2,744
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    Sep 24, 2016
    The transistor also needs base current. With a collector current of 2A then the minimum current gain of a TIP42 is 25 then its base current is 2000/25= 80mA which will flow through the regulator in addition to the 50mA in the resistor.Then the total current in the regulator is more than the maximum allowed current. Reduce the current in the resistor or test many then select a high gain transistor or select a regulator with a higher maximum allowed current.
     
  14. sureshot

    sureshot

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    Jul 7, 2012
    Thanks for your reply. So it might not be possible to achieve a power output of 2 Amps with a 12L78 100mA regulator ? Just wandering if a darlington like a TIP147 would allow a better scenario for less current through the base emitter resistor. Would it be possible to increase the resistor value to reduce current through the base resistor, but still get it to switch on ? Although i know the regulator would be doing less work than 50mA. Would that be possible.
     
  15. sureshot

    sureshot

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    Jul 7, 2012
    Maybe the only way is to breadboard this to see, putting a couple of multimeters in circuit to read current under load conditions. As a novice I'm not sure if i can obtain 2 Amps output current using a 78L12 and a single pnp pass transistor. I look at the current pass circuit for LM723 circuits, and its achieved there with a pre driving transistor to the main power transistors. Maybe that two transistor where the first drives the second might work.
     
    Last edited by a moderator: Jan 25, 2018
  16. davenn

    davenn Moderator

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    @sureshot

    I have edited your thread title and a number of your posts

    you have to STOP using the term "amplifying current" it is incorrect .... nothing is being amplified
    Rather refer to "current pass" circuits for voltage regulators

    Dave
     
  17. sureshot

    sureshot

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    I know any opposing is futile. I'm only saying as I've seen it mentioned in many other electronic sites. A quick Google does refer to current amplified. I would hop you wouldn't lock the thread at this point. I'm not quite there yet with a conclusion. I'm mostly a forum watcher from the side lines. Many threads get locked, some look justified for saftey or stupidity, this thread is neither of those.
     

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  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    If you google for "current pass regulator" I believe you get a better quality set of results than if you google for "current amplifying regulator".

    In fact, if you do it now, you might find an answer to your question in the first couple of hits. :)

    But let me assist...

    We'll start with a circuit suggested by @Audioguru

    high current voltage regulator.png

    The problem with this circuit is that using a 78XXL it is hard to get your 2A output current whilst keeping the current through the regulator low enough.

    If we want Q1 to start to turn on at 10mA, R1 needs to be about 33Ω. Because of Rsc, that current will rise to about 20mA as the current limit starts to kick in. So, we have 60mA of possible base current for Q1. This means it will need to have a current gain greater than 33. A transistor with an HFE of 40 to 50 should be fine.

    A NJW1302GOS-ND seems to fit the bill (as do many others)

    Using this, assuming an HFE of 75 to 150 at 2A, the current through the regulator will be between about 35mA and 55mA.

    Under these conditions, you might have as much as 5V across Q1, plus another 0.7V across Rsc, so for an output of 12V you will need an input of at least 18V, more if it has significant ripple.

    At lower input voltages, the current through the regulator will increase for a given output current.

    An alternative is to replace Q1 with a darlington. If you assume an HFE of 1000, you can keep the current through the regulator much lower (and more consistent), however R1 needs to be adjusted to drop 1.4V at 10mA (68Ω should be fine) and it will only rise to 15mA as the current limit starts to kick in. The current through the regulator at this point will be approximately 17mA.

    Note that the transistor Q2 prevents the current through Q1 from exceeding 0.7/Rsc A. If you attempt to draw more, the current will rise until the regulator is also at its limit. 78XX series regulators are protected to some extent against overcurrent and will shut down if badly overheated. However this shouldn't be considered a method of providing fool-proof short circuit protection.
     
  19. sureshot

    sureshot

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    Jul 7, 2012
    Thank you Steve for your reply. I can see there's no single one answer to this resistor value. Looks like many variables, dependent on transistor type, current drawn, and input voltage after rectified and filtered. I do understand what you have explained. For me its probably best if i breadboard this, trying a few transistor and resistor combinations.

    The reason behind the 78L12 was space saving on the stripboard, but if i can keep the current through the regulator low, with a minimalist heatsink, that would work. I've got a few graphics card and chip set heatsinks of old IT equipment. My goal was try and keep everything on a 150 x 100 prototype board. With fan for cooling with N/O thermal switch to turn the fan on and off.

    If i bread board this, i can see from a couple of multimeters what is going on. Both idle, and under load. I've got a few transistor types to try out.
    Thank you for the breakdown explanation, it has helped. And thank you again to everyone for your help.
     
  20. davenn

    davenn Moderator

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    I had no intention locking the thread, and I never mentioned such

    I just want you to learn correct terminology, it would be a wise thing to do :)

    and that file you attached is NOT the correct terminology
     
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