Capacitor selection for MCP1827 LDO

Discussion in 'The Projects Forum' started by moocrow, Oct 12, 2013.

  1. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    0
    Hi,

    Firstly, before posting I have searched on Google and here on the subject of input capacitors and found most of what I need but have one or two remaining questions.

    I am using an MCP1827 LDO regulator, the fixed 5V model with 5 pins, datasheet here:

    http://www.google.co.uk/url?sa=t&rc...IiLgKgOCZFz-oqYj5odbQGg&bvm=bv.53899372,d.d2k

    I put the circuit together, using the recommended resistor and capacitors, except that lacking ceramic capacitors of the right size (4.7uF and 1uF for input and output respectively) I substituted them for electrolytics. From a position of significant ignorance, I couldn't see how they could be reverse biased by the circuit and so assumed the fact they are polarised wouldn't matter, as long as they were connected the right way. I assumed, for both input and output, that they were just there to compensate for a drop in output from either the original voltage source or the regulator, should the load suddenly increase.

    The power supply circuit worked for some initial testing, several times, with virtually no load, and then a couple of times running a Raspberry Pi Model A, so up to 500mA. I disconnected everything to add a switch then on reconnecting it failed. It turned out that the LDO was outputting around 1.2V, though within around 30 seconds this climbed to around 1.6V. At this point I tested the temperature of the LDO with my finger and it was extremely hot (I was using a DC power supply at 6v as input, the plan being to use batteries when the circuit is finished).

    I did some further reading and now understand that the polarity does seem to matter.

    My questions are these:

    1. Much discussion of voltage regulators refers to regulators oscillating. However, it's not clear whether this means that the LDO has a tendency to output AC, which would explain why polarised capacitors would not work, or merely fluctuating DC, in which case they ought to be fine. Which is it?

    2. The schematics in the datasheet for the LDO show non-polarised capacitors and refer specifically to a 1uF ceramic capacitor on the output. This is easy enough to do. However, the datasheet shows a non-polarised capacitor of 4.7uF on the input and I have struggled to find a ceramic capacitor of such a large value online. Rs components have them, but only as surface mounted and not through hole. So, does the input capacitor really need to be non-polarised, in which case what is the right sort of capacitor to use? The datasheet says it's needed if the circuit will be run from batteries or from a DC supply with several inches of wire, which is the case with my circuit.

    Thank you,

    Jim.

    EDIT: The datasheet is really very helpful and includes significant discussion on the appropriate size and type of the output capacitor, i.e. ceramic and 1uF, and of the *size* of the input capacitor, i.e. 1uF - 10uF. The schematics clearly show a non-polarised capacitor on the input (as well as the output) but give no clue as to what sort of capacitor the input should be.
     
    Last edited: Oct 12, 2013
  2. wayneh

    Expert

    Sep 9, 2010
    12,137
    3,054
    I would not hesitate to use a polarized electrolytic on the input or the output. Certainly they will not be destroyed by depolarization. Other types such as ceramics may provide different responses to transients. But it sounds like you want a big, low impedance current source on the input and a low ESR electrolytic should be fine for that.
     
  3. MrChips

    Moderator

    Oct 2, 2009
    12,446
    3,362
    The purpose of the capacitors is not to provide compensation from varying loads. That is the purpose of the regulator itself.
    The purpose is to suppress high frequency oscillation of the regulator.
    You can use 4.7-10μF tantalum electrolytic on the input and 0.1-1μF ceramic on the output. The important point is that the capacitors must be placed at the terminals of the regulator, not at any distance away from the chip.
     
  4. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    0
    Thank you for the helpful answers. I will try the combination of tantalum and ceramic capacitors suggested by Mr Chips.

    I'm a bit puzzled as to what the problem is now however, given wayneh's point that using electrolytics ought to be OK. The voltage across the regulator output and ground fell from 5V to around 1.2V, which made me assume that the capacitor between the output and ground was broken and creating a short circuit (albeit one with enough resistance to produce a voltage of 1.2V).

    Are there any typical problems which can cause this type of behaviour? The reading I did afterwards pointed to the capacitors which is why I came up with the theory that my use of electrolytics was to blame.

    Thanks.
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,392
    1,605
    It's my experience hard cap failures are either completely open (rare) or completely shorted (more common). 1.2 to 1.6V on the output sounds like something else: cap is not impossible, but see either what is also getting warm or what makes problem go away when removed.
     
  6. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,291
    1,255
    It sounds like the regulator is in current limit so it doesn't burn itself up. Make sure you hooked everything back up right and make sure there are no near shorts on the output.
     
  7. wayneh

    Expert

    Sep 9, 2010
    12,137
    3,054
    Per the data sheet:
    "For applications that have output step load requirements, the input capacitance of the LDO is very important. The input capacitance provides the LDO with a good local low-impedance source to pull the transient currents from in order to respond quickly to the output load step."

    Am I wrong to see a contradiction?
     
  8. MrChips

    Moderator

    Oct 2, 2009
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    There is no contradiction. The purpose of the regulator as a linear feedback controller is to provide the proper compensation despite any transients from the load. Of course, in order for the regulator to do its job it requires a low impedance source, which is what the input capacitor does.
     
  9. DickCappels

    Moderator

    Aug 21, 2008
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    Mr. Moocrow,

    It might not be a capacitor problem and could be the chip is simply overheating (as noted by Donv) because it is not adequately heatsunk. Did you adequately heatsink the regulator?

    500 ma means that your regulator would dissipate half a watt for each volt difference between the input and output voltages.
     
  10. moocrow

    Thread Starter New Member

    Apr 3, 2011
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    0
    Hi,

    Thank you for all the replies, I'm really grateful.

    In answer to DickCappels' question, I am not using a heatsink. The input voltage is 6V. I may have overestimated the current drawn by the Raspberry Pi in my original post. I just re-read this article:

    http://www.daveakerman.com/?page_id=1294

    (although this article is about running the Pi from 3.3V I am running it from 5V)

    ... which I consulted when I was planning the circuit. I'm using a Raspberry Pi Model A which it says may be pulling as little as 115mA in addition to around 150mA for the WiFi dongle, rather than the Model B I am used to which pulls around 500mA without the dongle.

    Accepting that the voltages are different (this sounds like a stupid thing to say in hindsight) the project in the above article doesn't use a heatsink so I assumed it would be OK. I have seen general recommendations about use of heatsinks which refer to the difference between input and output voltages and as in this case the difference is only 1V, though that is the device's maximum input voltage.

    After the fault occurred I tested the circuit without a load shortly after (several times within the next hour) , so according to the equation in the datasheet (which matches DickCappels' calculation) the power dissipation ought to have been minimal. As the LDO has over-temperature protection, I thought it was the case that if it had overheated under load and was allowed to cool, then tested without a load, it ought to work again pretty much immediately (i.e. as soon as the temperature drops below 140C, page 19)?

    Cheers.
     
  11. moocrow

    Thread Starter New Member

    Apr 3, 2011
    21
    0
    Hi,

    Thank you to all of those who replied to my problem, above.

    As per Mr Chips' advice I replaced the capacitors, double checked that there were no stripboard shorts or any resulting from the casing etc and everything worked. I have worked on it for two full evenings with the Raspberry Pi running and there have been no problems. I have periodically checked the temperature of the regulator and it seems fine, so all good.

    Inevitably I have now encountered another issue which I'll probably post on soon (interfacing a digital output with an optocoupler).

    Thanks again,

    Jim.
     
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