Linear regulators (LM317, LM317AHV, LT 783) Current handling.

Discussion in 'The Projects Forum' started by dondempsey, May 13, 2016.

  1. dondempsey

    Thread Starter New Member

    May 10, 2016
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    I have only just joined All about circuits as a result of attempting to construct a new Regulated 60V 3Amps supply for my refurbished Bailey 30Watt amplifier (it had been viciously attacked in my workshop by wood lice, slugs and damp and smoked quite profusely when powered up! ).
    The original voltage regulator used discrete components (now obselete ) and I thought it would be nice to use more modern technology (not so modern as an SMPSU!!).

    I have built a number of supplies and bread boarded them, but they all seem to be somewhat noisy. The initial challenge was that the output from the Reservoir capacitor was in the region of 95Volts which only the LT783 could handle. I replaced the transformer such that the voltage is now a more modest 75 Volts, which is still a tad high for the LM317AVH BUT it seems to cope OK.

    My query is related to the inclusion of the pass transistor in the circuit. Most of the application data I have read (e.g TI, LT,) tend to place the transistor such that it is switched on by the current flowing INTO the regulator through a resistor connected across the emitter and base and involve the regulator in supplying a smaller part of the current to the load. My most successful attempt was actually using the output from the regulator by connecting it directly to base of the pass transistor (a Toshiba 2SC5200) with the load connected to the emitter.

    I have simulated both approaches using LTspice and they both appear to work OK but I don't know how to measure ripple and noise using LTspice.

    If someone could please comment on my approach I would be most grateful, I have been around the houses quite a few times and am very befuddled about this - maybe I should not have dismissed a buck converter, though I would have problem soldering the very small technology we have now (my 60W Solon iron is a bit on the large size).

    Regards,

    Don
     
  2. crutschow

    Expert

    Mar 14, 2008
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    3,227
    The downside of putting the transistor on the output of the regulator is that the regulation is degraded by the base-emitter voltage drop of the transistor (which varies with current and temperature), but that probably isn't a significant factor for your amplifier application.

    You can measure ripple and noise in LTspice by doing a transient analysis on the regulator with the maximum load current.
    But I would expect it to be quite low.
    If you've had previous problems with noise then that may be the result of layout or poor decoupling and possible oscillations.
    What type of noise have you seen?

    And buy yourself a much smaller iron for any electronic circuit soldering.
    A 60W iron is a definite no-no for that purpose. ;)
     
  3. BobTPH

    Active Member

    Jun 5, 2013
    782
    114
    A linear supply like you are building is probably better for audio gear.

    What was the problem with using the pass transistor in the usual way? Perhaps you did not choose the resistor correctly.

    Also, how large a filter capacitor are you using after the bridge?

    Bob
     
  4. dondempsey

    Thread Starter New Member

    May 10, 2016
    4
    0
    Hi, Thank you for your prompt response, do you mean that the regulated output will be different than calculated as a result of the base emitter voltage drop? As this is an adjustable regulator can I not simply adjust for this?
    My concern is that the regulator efficiency would be reduced but I came to the conclusion that as long as I loaded the regulator output with a resistor that ensured it always consumed the minimum recommended current for "good" regulation (say 20mA), this would be ok. The drop across base and Emitter is very low and won't change to much as a proportion of the output voltage. I had thought that the output from regulator input directly into the pass transistor would also provide a much tighter control over the output transistor than it reflecting the changes into Vin on regulator IC.
    I had a look a the tabs on ltspice and there is a tab called noise but this is mutually exclusive with the transient tab??
    Based on your excellent advice i have obtained a much smaller soldering iron and it does seem much easier soldering the strip board.
    Once again many thanks for the prompt response,

    Don
     
  5. dondempsey

    Thread Starter New Member

    May 10, 2016
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    0
    HI Bob,
    Thanks for the input, yes I agree removing the noise from an SMPS could be difficult and is definitely not wanted! Your suggestion regarding the resistor selection is absolutely true, the problem I had was my inability to calculate the resistor. The output transistor obviously needed to "switch on" such that it took over the current feed from the LM317HV but I was running out of measuring devices (ammeters) so I could not adopt a completely empirical approach. If you are able to clarify the calculation for me I would appreciate it.
    The reservoir capacitor is 15000ufd/100V Aluminium Electrolytic shunted by a 100N as this is being breadborded I also have a 1 ufd 100v cap close to the regulator Vin.
    In closing do you have any views as to why I should not drive the pass transistor from the Vout of the regulator rather than the input current to Vin?

    Regards

    Don
     
  6. dl324

    Distinguished Member

    Mar 30, 2015
    3,235
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    The major issues:
    1. Output regulation will be affected by the added BE junction that's (nonlinear and) outside of the feedback loop.
    2. The external pass transistor can't be protected by the over current and thermal protection builtin to the regulator.
    Was there a problem with the classical approach to extending regulator current? This example is from the 1980 NatSemi Regulator Handbook:
    upload_2016-5-14_6-50-32.png
    This variant of the external pass transistor circuit protects the transistor by setting up a current divider between the transistor and regulator; sharing is determined by the ratio of R1 and R2. You want the regulator current to be at it's maximum under full load. D tries to match the BE drop of Q1.
     
    Last edited: May 14, 2016
  7. dondempsey

    Thread Starter New Member

    May 10, 2016
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    0
    HI, thank you for the input. Unfortunately I am not sure I agree with your second point - if the regulator shuts down, that is no current flows, the Vout will drop and the pass transistor will surely switch off?

    With regard to the into from NATSEMI (1980), Linear Technology, TI and ON all have later info in their dataheets. The problem is they are using pass transistors which are now OBSOLETE. I have managed to find equivalents and actually ordered these such that I can try their design and then compare it with the approach I was using with the transistor on the output of the regulator chip.

    Many thanks for your input, if my response seems ungracious it is not intended. My problem is that what started as a need to build, what should have been a straightforward regulated supply, has turned into a major distraction of what actually is happening with me having difficulty finding spice models for "old/obsolete" devices.

    Once again thanks for bothering☺☺

    Regards,

    Don
     
  8. dl324

    Distinguished Member

    Mar 30, 2015
    3,235
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    It's your prerogative to agree or disagree with anything you're told.

    A transistor connected as an emitter follower will amplify current based on the transistor beta. If you assume 10, the pass transistor will dissipate 10X more power than the regulator. At 3A, the transistor would be dissipating P = IV = 3A * 15V = 45W. The regulator would be dissipating 0.3A * 15V = 4.5W. If the regulator isn't being stressed, it won't shutdown. It's worse if you assume a higher beta.
    You don't have to use the exact components. It's the circuit topology that matters.
     
  9. dannyf

    Well-Known Member

    Sep 13, 2015
    1,775
    360
    that approach is to allow the regulator to handle large current;

    In your case, a tracking pre-regulator (in the form of a transistor or another regulator) is used - its configuration is similar to the one above.
     
  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    Note that the voltage rating for the 317, 350, etc. is the maximum *differential* voltage across the input-output pins, not the voltage from the input pin to GND. This is because the entire regulator circuit floats on the shunt resistor from the ADJ pin to GND. If the input is 150 Vdc and the output is 130 Vdc, a standard LM317 will work just fine because the differential voltage is only 20 Vdc.

    ak
     
  11. crutschow

    Expert

    Mar 14, 2008
    12,991
    3,227
    As long as the voltage isn't exceed during startup or transient events.
    For safety, a zener diode could be connected across the regulator to conduct before its maximum voltage is exceeded.
     
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