Can anyone clarify the purpose of transistor in this LM317 regulator circuit.

Discussion in 'The Projects Forum' started by Parrot, Mar 22, 2014.

  1. Parrot

    Thread Starter New Member

    Mar 22, 2014
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    Hi all,

    I wonder if anyone can help me ?

    I've been bending my brain all evening trying to work out what the transistor does in the following variable power supply circuit.

    As far as I can figure, the circuit would work perfectly well without the transistor. Not only that, but I'm not convinced that the transistor will ever turn on.

    This is an actual circuit that I purchase for a couple of euros from China. I only wanted the LM317 and a heat-sink to contribute to making a little variable power supply, but I got this little circuit delivered to me at a cheaper price than I could of purchased the heat sink and LM317 alone. Hey, I view that as free components added !

    Out of curiosity, and as a learning exercise, I produced a schematic of it. In fact I wish I hadn't, because now I'm stuck thinking about it. Can anyone out there enlighten me on why it's designed with the transistor this way and just how it works ? Far as I can see, the transistor never turns on, because the base can never be at a higher voltage than the emitter as the wiper of the pot returns to ground. I can't be right though .. because then, why have the transistor ? Hence, I'm coming to you folks to see where I'm going wrong.

    Incidentally, D5 is to protect the voltage regulator from inductive spikes from a load ?

    I was also amused to find that the little PCB has two different capacitors identified as C4 (corrected by me in the schematic). You really do get what you pay for. But if I learn anything from this, it'll be two euros well spent.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    When the pot VR1 gets worn/dirty/noisy and the wiper breaks contact with ground, the current will flow through VR1 from one end to the other through the base of the transistor, and pull the output down to roughly 2v. That's better than having the output float high.
     
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  3. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    My guess is that the pot terminals 2 and 3 should connect to ground, leaving the transistor base free for applying a separate control voltage to shut down the regulator (pull the output low) if required. But SgtWookie's explanation is more plausible.
    D5 prevents reverse current flow through the regulator at input switch-off, when C5 is still charged.
     
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  4. canonfodder

    New Member

    Mar 22, 2014
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    I think you better check the pot connections carefully. I think you may have mixed them up. Make sure that the connections of the pot go where your schematic shows. You are correct in saying that the transistor doesn't seem to make sense as you show it.
     
  5. wayneh

    Expert

    Sep 9, 2010
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    It does make sense. It's a failsafe as already noted. Pots aren't perfect.
     
  6. #12

    Expert

    Nov 30, 2010
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    It makes sense to me. R1 and VR1 have about 9.6 ma of current to deal with. What seems wrong is the idea that the 10K pot is capable of adjusting the output voltage to 97.4 volts. You might have read the label wrong or you might need to change the potentiometer. A resistor in parallel with the pot would just keep Q1 on all the time, so that won't fix anything.

    The datasheet says C2 and C3 should be .1 uf ceramic and 1 uf (tantalum or aluminum), in that order.

    Wookie and Alec covered the rest of this.
    If the wiper connection inside the pot goes bad, Q1 will slap the output to as low as it can get, and D5 protects the 317 against backward current when you turn the power off.
     
  7. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    I like SgtWookie's explanation.

    To add to that, the transistor means that the pot does not have to conduct the full 10mA current required in the LM317 voltage divider.

    That means that a standard small 10k pot can be used instead of a special value (low resistance) high dissipation pot.
     
  8. #12

    Expert

    Nov 30, 2010
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    I think you stayed up too late tonight. If the wiper attaches a ground in the middle of the pot, there is no current left to go to the base of the transistor.
     
    Last edited: Mar 23, 2014
  9. Parrot

    Thread Starter New Member

    Mar 22, 2014
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    Yep, that is what was puzzling me. I thought I must of been going nuts. But it does indeed go to ground. Had me really scratching my head as I didn't see the use of the transistor in that case.

    So a massive thanks to SgtWookie,

    He seems to have hit the nail on the head. In effect, the only time there is every any current going towards the base of Q1 is in an error state, where the wiper breaks contact and can go open circuit. The transistor will only ever turn on once VR1 is in this fault condition, and as SgtWookie says, it'll drop the voltage, instead of letting it float high hence saving any little project attached from getting fried.

    Thanks for that. Something else I've learnt that I would not of thought of.



    And a general "Thank you", to all those that replied. It seems my two euro investment (including delivery no less) paid off rather well. I got a voltage regulator with two safety protections (Q1 and D5) that I never would of thought of; and I learnt some things as well.
     
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  10. THE_RB

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    I just assumed the circuit was mis-drawn, obviously they swapped pot pins 2 and 3.
     
  11. #12

    Expert

    Nov 30, 2010
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    I didn't even read the pin numbers.:confused:
    I just use the symbol.
    Maybe I should consider the pin numbers in the future.
     
  12. Parrot

    Thread Starter New Member

    Mar 22, 2014
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    Just for clarification, as there seems to be question about the pot's pinout, the circuit diagram did not come with the circuit. I made it myself. But here in this photo, you can see that the wiper is indeed directly soldered to the ground plane as portrayed in the schematic. Hope this helps to clarify.

    By the way, please forgive the mouse drawn annotations.

    Cheers all,
    Parrot.
     
  13. DickCappels

    Moderator

    Aug 21, 2008
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    The purpose of D5 is to protect the output transistor in the LM317 from C5 discharging "backwards" through the LM317's output when power is switched off.

    Q1 is connected as a shunt regulator. This configuration is also used as a "VBe Multiplier". It is the error amplifier in the system and the LM317 merely serves as a current and thermally limited power stage.

    Using the VBe of Q1 results in the output voltage drifting by 0.3%/°C or more -very poor compared with the LM317, unless the original application, such as a bias supply, requires such drift.

    Also, the open loop gain of this single transistor stage is very low, so the huge output capacitor of 470 uf is no surprise and most likely included to filter hum.
     
  14. muanpuia

    New Member

    Jul 1, 2013
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    If your schematic is right, i think youd have far better control of the output voltage if you swap the VR1 pin 2 and 3. But in that way, 10k maybe too much ohms to get a linear swing voltage output
     
  15. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Wow the pot wiper does look to be connected to ground!

    That circuit starts to make sense if Q1 is a PNP, and you have it's C and E reversed. Then the pot acts a variable resistor to ground and will bias the base of the PNP on.

    As it stands if Q1 is NPN something looks VERY funky.

    Do you have a clear photo of the face of Q1?

    And have you got any voltage readings on the 3 pins of Q1 with the pot in different positions?
    :)
     
  16. Parrot

    Thread Starter New Member

    Mar 22, 2014
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    Thanks DickCappels, for your reply about D5 and Q1.

    With regards to Q1, your thought is very close to my own initial one when I first saw a transistor in the circuit near the adjust pin of the regulator. Indeed, that is how I thought it should work, and the reason for my confusion and the original post; as the wiper of the pot does not seem correctly wired for this to actually be a VBe multiplier. If Q1 were used in such a configuration, would not the base be connected to the wiper and the bottom of VR1 to GND ?

    This is not the case in this circuit. As it is wired, and so far as I can understand it, no current ever flows to Q1 under normal operating conditions as the wiper of VR1 is grounded and all current will follow that path no matter what variable resistance is set by the wiper. ie, Q1, under normal conditions, has no effect on the biasing of the voltage regulator. It is an open circuit.

    I couldn't understand why this was so, as it seemed a waste of a transistor! But SgtWookie in post #2 pointed to the fact that if the pot fails or gets noisy with age, Q1 will immediately start to conduct and pull the adjust voltage down. And I have to agree, that this error mode protection does seem to be the role of Q1.

    I just wanted to reply, as your thoughts closely resembled my own before I noted that the wiper was grounded.

    Cheers,
    Parrot.
     
  17. Parrot

    Thread Starter New Member

    Mar 22, 2014
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    Sorry The_RB, a clear photo of Q1 would be hard. I'll see if I can nab the wife's camera when she's done with it. My webcam would just show a grainy blob. And I've not even powered the circuit up yet. I gave away my bench power supply years ago, but am restarting in electronics so I'm building myself a new little power supply and this circuit is going to be tagged onto that.

    However, using a magnifying glass, a torch, and with severe facial contortions, it can be clearly identified as a 2N5551 (NPN). So yes, very funky indeed and the reason for my original post.

    I think post #2 by SgtWookie has the solution to the problem though.

    Cheers,
    Parrot.
     
  18. crutschow

    Expert

    Mar 14, 2008
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    I agree with the Sgt. The transistor is to prevent the output voltage from rising if the pot wiper opens momentarily due to dirty wiper or continuously due to a failure.
     
  19. DickCappels

    Moderator

    Aug 21, 2008
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    Ha! You're right. The wipe and one leg would have to be swapped.

     
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