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Voltage Regulators

Discussion in 'Electronic Basics' started by Abstract Dissonance, Feb 20, 2006.

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  1. I'm tried to do my ps but I'm having some problems. I ended up burning out
    one of the regulators for some reason but the real issue is that the
    regulators don't seem to be regulating.

    If I setup a very basic transformer->rectifier->filter->regulator

    then with no load I get the right voltage but with a 2.2k ohm + LED load the
    voltage drops from 8 to 4. Its a MC78M08 regulator and surely it shoudln't
    drop 4 volts at that load? I don't use any of the "extra" stuff such as
    protection diodes and ripple reducing capacitors and the voltage I'm putting
    in is about 35 volts.

    The transformer is a 25.2VCT and I'm using the full secondary. The bridge is
    [email protected] and the filter cap is a 200V @ 1mF(I know I could use a larger one
    but surely it is enough).

    As far as I can tell I'm not getting any.............


    WTF. I just switched my DMM down to 20V cause I was measuring it on 200 and
    now its reading 7.74V with variations going from +- 1V about every second.

    I didn't know I had to choose the right voltage range on my DMM to get an
    accurate reading ;/ Is this true? That really sucks if so ;/

    i.e., turning the switch to 200V gives me a reading of 4.4V and switching it
    to 20V gives a reading of 7.74V. Any reason why it would do this? Low
    battery? Messed up meter? Or do I just need to make sure to select the
    proper range?(which seems kinda odd cause you would have to know it ~ before

  2. Pooh Bear

    Pooh Bear Guest

    What do you mean by 1mF ? It *should* mean 1000uF but popular US usage ( why
    are Americans *always* so bloody minded ? ) was to use mF = uF.

    If you have 1uF, that's your problem.

  3. cause it makes complete sense. m = *m*illi, u = micro cause u looks like
    the greek symbol from micro that is often used. I don't see a problem with
    it? Just cause someone started a bad trend should we keep on perpetuating
    well, its not.
  4. Capacitance units aside ;-) I would not expect that difference in reading
    between 200v and 20v range from any worth while VOM.
    It's not unusual to have the regulator wrongly connected and what you are
    saying sounds like it to me. Have another look :)
  5. redbelly

    redbelly Guest

    Once in a while, I get flaky readings on a DMM. It usually means the
    battery is low.

    Try replacing the battery in your DMM.


  6. Chris

    Chris Guest

    Hi, Jon. Take a deep breath, and look at things one at a time.

    Before anything else, check your meter. It shouldn't change readings
    between the 20V and 200V range. Get a known good 9V battery, and
    measure 9V on the 20V and 200V ranges. If they don't agree, replace
    the battery on the meter and try it again. If the meter still doesn't
    agree between ranges with a fresh battery, junk it and get another one
    for $10 or $15 at the hardware store. You need a known good meter to

    First, you want to look at your unregulated, filtered power supply. To
    see if it's working properly, try putting a load on the supply _before_
    the regulator. As a rule of thumb, you should figure that, for a 60Hz
    supply (120Hz ripple), 8,000uF will give you 1V of ripple per amp of
    current. That means a 1,000uF cap will give you 8V of ripple per amp.
    On your DVM, that 8V will look like a 4VDC reduction per amp, because
    the ripple is roughly a triangle wave. So, get a power resistor, load
    up the power supply, and see what's going on.

    Let's assume you have a 100 ohm, 20 watt resistor handy. Put it across
    the 35V peak supply. You'll have about 1/3 amp load, which means 1/3 *
    8 or about 2.7V of ripple. On your DVM, that will look like your 35V
    will go down half that, or down 1.4V to about 33.6V. If your DC
    voltage measured across the cap doesn't go down by about 1.4V +/- 0.25V
    with that 100 ohm resistor load, then something's wrong. Your bridge
    rectifier may be hooked up wrong. Your cap may not be 1000uF. Your
    cap may be old and bad. Your cap may actually be 1uF (which would
    cause the problem you describe). Your transformer may be messed up.
    But you will _know_ that there's a problem in that part of the power
    supply. And if you have another value power resistor, work it out for
    that load. But if that part of the power supply performs properly,
    then you know that's where the problem _ain't_.

    Once you get that taken care of, look at your regulator. Just get a
    12VDC wall wart. Put the + at the input pin, with the - of the wall
    wart goijng to the middle pin of the 7808. Measure the output voltage.
    Yes, there might be some high frequency oscillations, but it _will_
    give you 8V +/-10%, or it doesn't work. Then put a 220 ohm resistor
    between the output pin and GND, and see if the output voltage changes.
    If it changes more than several mV with a 36 mA load, it's busted.

    If you look at the data sheet, you'll see that the Absolute Maximum
    input voltage to that device is 35V, and you've already come right up
    against that limit. You need to think this one out a little better, or
    lay in a good supply of spare parts. I'm figuring your 7808 is damaged
    (they act that way sometimes when they've been overstressed).

    Putting 35VDC at the input of an LM7808 isn't doing it right, Jon. Not
    only that, but let's say you're pulling 1/2 amp from the 7808. With a
    1000uF cap, you should have an average DC of 33V. Your power
    dissipation will be (33V - 8V) * 0.5A = 12.5 watts. Lotta extra heat,

    The point of troubleshooting is to make good decisions based on
    available information. Read the datasheet, Jon. That's always
    important. And Absolute Maximum means that, if you exceed that, even
    by a millivolt, the manufacturer says you will have problems. But when
    you're doing something new, or even troubleshooting something that's
    known to have worked, you have to be open to the available facts. Try
    to make the observations that will make it apparent where the problem
    is, but more importantly, where it is _not_.

    Make good observations, and believe your observations, even if they
    don't seem to make sense. The things you're seeing and measuring are
    trying to tell you something, if you can take the time to learn their
    language and are prepared to listen to them.

    You also may want to take a minute and actually think out what you're
    doing here, and see if you can get there with the parts you have.
    Again, I'd really recommend just buying one or two open-frame linears,
    combine them with an LM317 for the variable voltage, and be done with

    (And by the way, people stopped caling 1000uF "1 mF" about 35 years
    ago. If your cap says "1mF" on it, the cap is almost certainly a
    fossil, and shouldn't be used, anyway.)

    Good luck
  7. Pooh Bear

    Pooh Bear Guest

    It simply isn't common practice to quote capacitance as milliFarads though. As
    I explained it's prone to being misunderstood due to prior bad US practice.

  8. Pooh Bear

    Pooh Bear Guest

    1000uF @ 200V is actually physically quite large !

  9. A 1 milliFarad (1000 uF) 200 volt capacitor would be very large.

    If the capacitor is marked "1mF", then I suspect that it is almost
    certainly 1 uF - you need a much larger value for your primary filter
    capacitor - 1000uF at 50 V would be a good start.

    For historical reasons, "mF" = "uF" - in ancient times, a 1000 uF (or
    1 milliFarad) capacitor would have been impossibly large - no-one
    would have a use for such a thing, so, as the Greek alphabet had not
    yet been invented, we used "mF" to mean "microfarad" - that practice
    is so deeply ingrained in the culture, that it is very undesirable to
    use "mF" for anything other than microfarads.

    Peter Bennett, VE7CEI
    peterbb4 (at)
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  10. ehsjr

    ehsjr Guest

    When you finally get it sorted out, be aware that you will
    be pounding that chip pretty hard if you draw appreciable
    current through it. For example, at 200 mA (40% of the chip's
    rating) it will need to dissipate 6 watts. That's not a factor
    with your present load, but you really don't want to go over
    about 30 mA if the chip is not mounted on a heat sink. The
    problem is that the 30 volt Vin-Vout means the chip has
    to dissipate 30 * Iout.

  11. John Fields

    John Fields Guest

  12. Mike

    Mike Guest

    Hello Jon

    First thing, Use the center tap of the transformer and 2 diodes to get
    the DC output of the base supply down to a reasonable level. 1000uf is
    a bit on the low side. Use a lower voltage cap, it'll be MUCH smaller.

    Next, Make absolutely sure that the regulator is connected properly.

    Front View
    | o |
    | |
    | |
    input | | | Output

    Next, Put a 1-10uf capacitor between the output and ground pins of the
    regulator and if the regulator is more than a few inches from the main
    filter cap, put a .1-1uf on the input pin also.

    Next mount the regulator on a small-medium sized heatsink.

    Btw, An MC78M08 has a "peak" output current rating of 700mA.
    The data sheet says "Maximum output current with adequate heatsinking
    is 500mA".

  13. John Fields

    John Fields Guest


    I dt
    C = ------

    rearranging to get the ripple voltage due to a 500mA load gives us:

    I dt 0.5A * 8.3E-3s
    dV = ------ = ---------------- = 4.15V
    C 1E-3F

    Using the supply you suggest:

    17VDC 8VDC
    / +-------+ /
    MAINS>---+ +-------[DIODE>]--+----| 78MO8 |----+
    P||S | +---+---+ |
    R||E-CT-+--[1000µF]--+ | [16R]
    I||C | | GND |
    MAINS>---+ +----|--[DIODE>]--+ GND

    would put the input of the regulator at 17V. With a 500mA load on
    the output of the regulator, the ripple would cause the input
    voltage to vary from 17V to about 12.8V.

    The regulator needs about 2 volts of headroom (typical, I couldn't
    find worst case) to stay in regulation, so with a worst case
    input-to-output differential of 12.8V - 8V = 3.2V there should be no
    problem at all with a 1000µF cap.

    But what about accounting for the cap's tolerance?

    If it's + whatever, -20%, then we've got 800µF, minimum, and we
    wind up with:

    I dt 0.5A * 8.3E-3s
    dV = ------ = ---------------- ~ 5.2V
    C 8E-4F

    of ripple, so with an 8V output and a 17V - 5.2V = 11.8V input
    valley voltage, we wind up with an input-to-output differential of
    11.8V - 8V = 3.8V.

    Still no problem.
    No. None is required for stability, although 0.1µF will improve the
    transient response time.
    No, 0.3µF is the recommended capacitance if the regulator is "some
    distance" from the filter cap.

    With an input voltage of 17V, an output voltage of 8V and a 500mA
    load, the regulator will be dissipating:

    Pd = (Vin - Vout) Il = (17V - 8V) * 0.5A = 4.5 watts

    The regulator has an R theta junction-to-case of 2.5°C/W, so for 4.5
    watts that comes to 11.25°. If we want to keep the junction at 70°C
    or less with an ambient air temp of 25°C, then, we'll need to get
    something which will keep the case at 70°C - 11.25° = 58.75 ~ 60°C

    With an ambient temp of 25°C, a case temp of 60°C, and 4.5 watts to
    get rid of, that means the heat sink needs to have a thermal
    resistance of:

    60°C - 25°C
    Rth = ------------- = 7.778 ~ 8°C/W

    Going to Thermalloy's site at:

    And searching around a bit for TO-220 type heat sinks yields:

    At 7.7°C/W, damn near perfect!
  14. John Fields

    John Fields Guest


    The regulator needs about 2V of headroom to stay in regulation,
    which means that for a fixed 8V output, the input _always_ has to
    stay at 10V or higher.

    With a 500mA load and 1000µF filter, the lowest the input will drop
    to will be 12.8V, so that's no problem and 1000µF will be OK
    More rewrite:

    so the lowest the input voltage will drop to will be
    17V - 5.2V = 11V
  15. It seems like it was the battery as I'm getting a consistant reading across
    the different voltage ranges.

    Thanks for the help
  16. ray13

    ray13 Guest

    With no capacitance on the output of a 3 teminal regulator you run the
    risk of parasitic oscillations. Oscillations upwards of 1Mega Hertz.
    Detected by your DMM because it's not rated for that frequency.
    Detected by little pn junctions and filtered by stray capacitance,
    present a DC voltage to the DMM circuit. Switch ranges and the stray
    capacitance changes changing the reading. No? Get a good ole VTVM out
    preform a DMM in my opinion.
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