Connect with us

Voltage regulator puzzle - any ideas?

Discussion in 'Electronic Basics' started by beatbox, Jul 13, 2007.

Scroll to continue with content
  1. beatbox

    beatbox Guest

    Hi,
    I built this PSU circuit:
    http://modular.fonik.de/soundlab/sl_psu.pdf

    All is well, except I'm getting -21v from the output of the 7912
    voltage regulator instead of -12v. The +12, and +/-9 outputs are all
    fine.
    I tried removing the 7909, then the 7809. I was still getting -21v
    from the 7912.
    I then tried the 7909 in the 7912 slot, and I got -9v as expected, so
    the circuit seems sound.
    I have tried substituting the 7912 twice. Same result.
    The AC power supply is 9V, 500mA.
    I'm measuring the voltage going into the positive voltage regulators
    as 30v, and -30 into the negative voltage regulators.
    Could I have a bad batch of 7912s, or can anyone suggest something
    else I might be doing wrong?

    Thanks!
     
  2. Phil Allison

    Phil Allison Guest

    "beatbox"

    ** Suspect your 7912s are mislabelled or fakes.

    May really be 7812s or just some cheap BJT.



    ........ Phil
     
  3. Tom Biasi

    Tom Biasi Guest

    Where are you getting the 30 volts with 9 volts in?
    Check the orientation of the diodes.

    Tom
     
  4. Hi,

    The 7912 needs a minimum load, then it works fine.
    If you understand german you can read the thread

    "7912 braucht Mindestlast?"

    in d.s.e. There the same problem is discussed.

    HTH
    Falko Rudolph
     
  5. Phil Allison

    Phil Allison Guest

    "Falko Rudolph" <


    ** TOTAL BOLLOCKS !!!




    ** Mein Kampf is damn good page turner too.






    ........ Phil
     
  6. Guest

    Hi Falko,
    Ja, ich kann ein Bisschen Deutch verstehen, und das problem scheint
    mir aehnlich. Ich bin aber nicht sicher wie ich ein loesung finden
    koennte.
    Wie aendert man die Last (load?).
    Danke,
    Mike
     
  7. beatbox

    beatbox Guest

    I'm posting from a couple of accounts here, just to confuse things ;)
    My (very basic) question, after reading that thread, is how might I
    change the load in my circuit to see if that has any effect?
    Thanks.
     
  8. Hammy

    Hammy Guest


    If you're saying you want to test the PSU's response to a fast load
    transient? Use a Mosfet to switch from no load to full load. If you
    don't have a function generator to generate your gate pulses just use
    a pwm like a 555 timer and a driver IC.
     
  9. Hammy

    Hammy Guest

    Sorry here is a schematic.

    http://i9.tinypic.com/61x3ep3.png

    I just did this a couple of days ago so you can see a scope image
    here.

    http://i11.tinypic.com/4m8428g.gif

    A 7808 going from no load to 1 amp (2.3uS pulse per 26 uS).Channel 1
    (yellow) is the 7808 output voltage dropping by 1.5V. Channel two is
    the gate signal to a logic FET the driver gets its Vcc from the output
    of the 7808.Thats why the peak gate voltage drops.
     
  10. Phil Allison

    Phil Allison Guest

    "beatbox"


    ** No load is needed on a 7912 to get -12 volts out.




    ......... Phil
     
  11. Just add a Z-Diode >12V from Output to ground. When they are right in the
    german thread this will create a load current whenever the output voltage is
    too high. If it works after that you have found the problem. IIRC the
    input/output capacitors had an influence, too.

    Gern geschehen.

    Falko
     
  12. Hi,

    If you were to start with an input of 9Vac rms, in theory you would have an
    AC input signal that was about 12.7 Vac, peak-to-peak, which you want to
    rectify to a useable source for your +/- DC inputs to the voltage regulator
    chips. By the time you subtract off the forward bias drop across the diodes,
    and the voltage lag from the numerous capacitors in the circuit, you may end
    up with an input voltage 10 or 11 Vdc (+/-) , maybe .... For the 7912
    regulator, you are supposed to have an input voltage is between (-14.5Vdc)
    to (-30Vdc), which may even change to (-16Vdc) to (-22Vdc), depending on the
    how the load effects the operating temperature. Now the dropout voltage is
    2V, which means that you need keep the input voltage at least 2V greater
    than the expected -12Vdc output, or the chip will not conduct. You should
    check to see that you really are supplying -13Vdc (or a value more negative)
    to pin 2 of the 7912, and the ground (0V) on pin 1. If you don't have at
    least a negative 14 volt differential between these two pins, then the
    regulator will stop conducting, and I'm not sure what you would then read as
    your output voltage, but if the input voltage should read something close
    to -9Vdc, then the -21Vdc you have read on the output seems suspiciously
    close to the sum o (-9Vdc actual input) + (-12Vdc expected output), with no
    output current to available... I got cross-eyed trying to follow the leads
    on the PC drawing. An equivalent schematic would have been easier, or
    quicker to digest. Plus, I'm not sure what you meant by the statement: "I'm
    measuring the voltage going into the positive voltage regulators as 30v,
    and -30 into the negative voltage regulators." If all you had for your
    input voltages were the rectified DC voltages you got from a 9Vac source,
    you'd have a pretty hard time coming up with either a plus or minus 30Vdc as
    an input to anything in that circuit. If you were lucky enought to squeeze
    out 11-12Vdc (plus and minus) after rectifying the AC input, that would
    probably be enough over the 2Vdc cutout voltage for everything up through
    the 9Vdc outputs (plus and minus), but may not have been enough to stay
    above the extra 2Vdc needed to allow the 12Vdc output to stay active. The
    absence (or presence) of a load can also effect how much "extra" input
    voltage is needed to maintain the expected output voltage.

    I hope this helps. Maybe it's all smoke.

    -DW
     
  13. Hammy

    Hammy Guest

    The circuit is a voltage multiplier thats how he's getting the 30V. ;)
     
  14. neon

    neon

    1,325
    0
    Oct 21, 2006
    7912 means 12 v output but you must insure that the adjustment pin is not tied to a voltage source of some kind. It may be a fried 7912 if you have a very large capacitor > 100ufd across the output install a reverse diode across the input to the output. to protect the regulator . consider this 12 + 9= 21v ring a bell
     
    Last edited: Jul 17, 2007
  15. beatbox

    beatbox Guest

    Hi,
    Thanks for that useful explanation. I'm learning as I go along here.
    There's a schematic here:
    http://electro-music.com/forum/phpbb-files/sl_psu_204.gif
    The voltage differential I'm getting between the input pin and the
    ground pin of the 7912 is -30v.
    The voltage between the input pin of the 7812 and its ground pin is
    30v. The 7812 outputs 12v quite happily. Would that fit with your
    theory?
    (The same voltages are measured between the inputs and grounds of the
    7809 and 7909 by the way, both of which are working OK.)
    If you could explain what a "load" means in this context that would be
    really useful. (Is that basic enough for a "basics" group?)
    I'm not sure where to start looking for a fix, but I have started by
    ordering some new 7912s by a different manufacturer. Will report back
    on that...
     
  16. beatbox

    beatbox Guest

    Thanks,
    I'll try it.
    Mike
     
  17. I don't want to insult your intelligence but are you positive you're
    measuring from the ground to the 1912s? If you happen to measure from
    +12 to -9 (or +9 to -12) then you'd get your odd 21volts. check for
    shorts between the regulators and the ground rail.
     
  18. beatbox

    beatbox Guest

    Hi,
    Yes, sure as can be. I'm testing from the designated ground and output
    points in the circuit, as well as from the regulator pins to double
    check.
    Mike
     
  19. beatbox

    beatbox Guest

    I replaced the 7912 with one from a different manufacturer. It
    worked.
    So I got there in the end, though my nice PCB is now in a mess from
    all that soldering and desoldering.
    Thanks all.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-