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First time Voltage Regulator questions — Parallax Forums

First time Voltage Regulator questions

UghaUgha Posts: 543
edited 2009-01-28 09:32 in General Discussion
I'm working on a project that uses a voltage regulator for the first time so I have a couple quick questions...

My voltage regulator has a 1amp limit. This is the same as on the BOE so I'm assuming that's plenty of juice... if I were to heat sink it, would it be able
to go a bit higher?

How do you know when to use a heat sink and when its not needed?

The caps used in the circuit on the input and output... the ratings stated are 0.33uf and 0.1uf... would using a much higher rating (10uf) cause any problems?

Another cap question... I'm assuming the schematic calls for ceramic caps... would using electrolytics cause any problems?


I'm pretty sure I know most of this but I don't want to take any chances... any feedback would be great.


Thanks!

Comments

  • davejamesdavejames Posts: 4,047
    edited 2009-01-27 19:54
    Ugha,

    Most single chip voltage regulators self-limit the amount of current they pass.

    The regulator's spec sheet will list the maximum free air power dissipation into a stated amibient air temperature. Going beyond that would require a heat sink on the device.

    The regulator's spec sheet will have info on suggested input/output decoupling caps. Going below those values may introduce too much ripple and going over them may cause oscillation issues.

    One of the top rules in electronics is, never assume anything. Educated guesses are much better. I'd stick with whatever the spec sheet states.

    Regards,

    DJ
  • Erik FriesenErik Friesen Posts: 1,071
    edited 2009-01-27 20:18
    It is a little more complicated than that there is a 1 amp limit. Very likely that is with a minimum voltage drop like .5 or whatever the regulator is rated at. Understand that if you are pulling 1 amp at 3.3v for example, and you are feeding it with 10v, that you are dissipating (10-3.3)=6.7v x 1a = 6.7w, and it will take over 1 amp at 10v. If you look at the data sheet it will give a w/c heat rise. So it will depend on a few factors but if you are dissipating 5w I am guessing you will need a heatsink or risk bumping against the limiter.

    Another thing is that a given size ceramic cap typically has much quicker response than a larger cap (ESR). You might get away with a 1uf but I doubt a 10uf will work well in a critical application
  • Phil Pilgrim (PhiPi)Phil Pilgrim (PhiPi) Posts: 23,514
    edited 2009-01-27 20:26
    Ugha,

    Which voltage regulator are you using? The caps issue is entirely dependent on that. Some regulators allow the output capacitance to increase beyond the minimum specs without limit. Others impose both upper and lower ESR limits on the caps, as well as capacitance limits. This will determine what type of cap you have to use. It goes on and on. So give us a part number, and we can provide more incisive advice.

    -Phil
  • LeonLeon Posts: 7,620
    edited 2009-01-27 22:42
    Regulators that use 330nF and 100nF capacitors require ceramic ones.

    Leon

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  • UghaUgha Posts: 543
    edited 2009-01-28 00:15
    I never realized it was so complicated...

    It's a 7805a voltage regulator from radioshack... I had a hard time finding a datasheet for it but I finally found the
    attached... although it seems slightly incomplete.
    I may have missed it... but I also can't figure out the voltage drop from the datasheet.

    I'm using a 6v 2.8AH lead acid battery to feed the 7805a... I'll be drawing 5v through the regulator for some
    electronics and 6v straight from the battery for 4 motors.

    I connected it up with two 10uf electrolytics and ran an LED and a motor for about 2 hours with no heating what
    so ever on the regulator... this is before I read Leon's post.

    Are the caps for smoothing the power so sensitive electronics like a microprocessor or other ICs don't fry or
    is there some other reason?

    I don't get why this regulator requires ceramics of a specific size and the BOE has a 5v regulator with two 1000uf
    caps connected.

    Maybe I should just connect the battery to a BOE and use it to regulate the power... I didn't do this originally
    because I read that its better not to have a microprocessor on the same power source as a motor (or in this case,
    4 motors).

    Any opinions or advice is absolutely welcome.

    Edit: I'm using the second example in the datasheet, except with the 10uf caps.
    pdf
    pdf
    7805a voltage regulator.pdf
    285K
  • Mike GreenMike Green Posts: 23,101
    edited 2009-01-28 00:36
    The 7805 has a fairly high voltage drop (about 2.5V). You can't feed it from a 6V battery. It can't regulate with only a 1V drop. Advantages of the 7805: It's cheap. It's available almost anywhere. It's pretty rugged. It doesn't require much in the way of input or output capacitors. 1uF or almost anything more on the output will work and 0.1uF·or more or even nothing (if close to the input power source)·on the input is more than adequate.

    For a 5V output (7805), I'd suggest at least 8V for an input voltage.

    Post Edited (Mike Green) : 1/28/2009 12:42:06 AM GMT
  • UghaUgha Posts: 543
    edited 2009-01-28 00:59
    So the 4.8ish volts I'm getting after regulation is either incorrect or unstable?

    I'm assuming that this is a case where an O-scope would be very useful.
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2009-01-28 01:23
    No, this is a matter of not enough V_IN, which should be greater than or equal to 7.5V (min.), call it 8V.· 6V is not enough.
  • UghaUgha Posts: 543
    edited 2009-01-28 01:55
    Alright... I'll just use the battery unregulated for my motor supply.

    I also have a couple regulators from a surplus store's random packages... anyone know much about a LX8386-05ip? I know its a 5 volt regulator and I've yet to look at the datasheet (was curious
    if you guys had any experience with it)

    Same deal with a PLX8386-05ip. I'm assuming its pretty much the same as the other.
  • Brian218Brian218 Posts: 92
    edited 2009-01-28 05:54
    It's a low drop out regulator; it should need 6.5 volts input voltage for 5v output. Current rating is 1.5 Amps..

    I've never used it but here's a link for a datasheet
    http://www.datasheetcatalog.org/datasheet/microsemi/LX8386.pdf

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    This post is a work of art. Variations in spelling and grammar are intentional, artistic endeavors that add value to all of mankind.
  • RDL2004RDL2004 Posts: 2,554
    edited 2009-01-28 06:07
    In case it wasn't clear from earlier posts, the power dissipation in linear regulators is calculated by (V in - V out) x I out, the result is in Watts.
    The voltage differential is usually the cause when there is overheating problems.

    8 in - 5 out x 1 amp = 3 watts
    12 in - 5 out x 1 amp = 7 watts

    18 in - 5 out x 1 amp = 13 watts

    Note that in each case the current is the same.
    When thinking of how much heat this represents, consider holding the equivalent size light bulb in your hand while it's on.

    Here is some good basic info on regulators.


    www.national.com/appinfo/power/files/f4.pdf

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    - Rick
  • LeonLeon Posts: 7,620
    edited 2009-01-28 09:32
    The capacitors are there primarily to ensure that the regulator is stable. Without them, or if they are the wrong type or value, it might oscillate. A friend of mine copied a power supply that I'd designed using the 7812 regulator, omitted them and wondered why it didn't produce any output. I found it was oscillating at VHF, added the capacitors, and it worked.

    Leon

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