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Ask an Expert Forum Voltage regulator questions - capacitor on the input, difference between different regulators
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Voltage regulator questions - capacitor on the input, difference between different regulators

sawik123
sawik123 over 9 years ago

Hello everyone,

 

I have some questions about voltage regulators (it's the first time I'm dealing with them, so the question are *really* basic - but I wasn't able to find answers anywhere, so I've figured I'll ask).

 

I have a 6 AA battery pack that I want to get 5 volts out of. First I got this voltage regulator and it seems to be working: https://www.adafruit.com/products/2164

However, it's dropdown can be as high as 2.5V, meaning that once the batteries go below 1.25 V each (which as far as I know is not the end of their lifetime) it may stop working (if I understand correctly. Or will this one do fine?).

 

Anyway, I've decided to get a regulator with a lower dropdown. I got these two (I expect the output current to be around 1A, so I got ones that support >= 1.5A just in case)

MIC29150-5.0WT - MICREL SEMICONDUCTOR - LDO VOLTAGE REGULATOR, 5V, 1.5A, TO-220 | Newark element14

MIC29300-5.0WT - MICREL SEMICONDUCTOR - Fixed LDO Voltage Regulator, 2.5V to 26V, 370mV Dropout, 5Vout, 3Aout, TO-220-3 …

 

Now basically my questions are:

  1. what capacitor should I use on the input? The spec doesn't really specify that. It only marks it as "C In". Does the value of that really matter? Will 10uF work fine (so I'd have 10uF on both input & output)?
  2. just out of curiosity - what really is the difference between them? Seems like the second one supports higher current and has slightly higher dropdown. Is that really the only thing? Will they pretty much operate the same, or will the second one like get more warm or something? Basically what I'm trying ot ask is - why would anyone want to use the first one over the second one?

 

Thanks,

Tom

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  • shabaz
    shabaz over 9 years ago +2 suggested
    Hi Tom, The spec does specify that you may need a 100nF cap on the input (see page 20), you can additionally add the 10uF if you find you need it (it depends on what the impedance of your source is; there…
  • shabaz
    shabaz over 9 years ago in reply to sawik123 +2 verified
    Hi Tom, The 100nF and 10uF are used together, the reason is that they perform differently due to the material used in their construction (ceramic vs electrolytic). Both are advised. Figure 1 in this PDF…
  • jw0752
    jw0752 over 9 years ago +2 suggested
    Here is a link to Recom's DC to DC Book of Knowledge just in case you want more information on using switchers for converting voltages and some other really neat things. http://www.recom-power.com/de/americas…
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  • jw0752
    0 jw0752 over 9 years ago

    Hi Tomasz,

    If you can get by with one Amp you may want to check out these Recom Convetors. Since they are switching regulators they do not waste as much energy in IR loss. Further there is only a 1.5 volt drop out and probably no need for heat sinking. With efficiency up to 94% more of your battery power will go to your application.

     

    http://www.newark.com/recom-power/r-785-0-1-0/dc-dc-converter-5w-5v-1a-sip/dp/62X9542

     

    Incidently if you need more than one amp these devices can be paralleled.

     

    John

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  • sawik123
    0 sawik123 over 9 years ago in reply to jw0752

    Thanks! That's really helpful, I'm learning a lot. I wasn't aware of the fact that the linear regulators have such low efficiency. It would drain my batteries in no time. I really need to buy some books and learn about this stuff...

     

    Anyway, I guess I'll go with the switching regulator you've suggested. I'm pretty sure that I'm not going to drain more than 1A - I have to power up a small LCD screen that's supposed to draw 500mA, and a Raspberry Pi with no peripherals (I think it's supposed to draw around 300mA).

    Worst case I'll connect them in parallel (one for the screen, one for the raspberry). But I think just one should be fine. I'll measure that later.

     

    However, there still is a question of what capacitors to use... neither the spec nor the "book of knowledge" really specify that (or I can't find it again). Maybe it's some common knowledge that I'm lacking. Can you by any chance give me some advice on that?

     

    Also in the spec it mentions to add a diode to prevent current going back through the regulator, but since I'm only powering a small LCD and a Pi I probably don't need that, right?

     

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  • shabaz
    0 shabaz over 9 years ago in reply to sawik123

    Hi Tom,

     

    Now that there is a bit more information!! beware that you can't assume 1A is sufficient if you're using the Pi 3.

    According to this link the current consumption (I expect with nothing attached) during a reboot reaches 720mA.

    If you make the assumption 1A is enough for the Pi 3 I think this is just waiting to fail once your parts arrive.

    If your LCD screen really does need 500mA as you say, then you definitely need a higher power DC-DC converter

    Anyway it is unclear what Pi you're using.

     

    Also, there are LiPo USB power bricks which would power the Pi and display (save on needing 6xAA batteries).

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  • sawik123
    0 sawik123 over 9 years ago in reply to shabaz

    I'm using Rasperry Pi Zero - from what I've read, it's got a really low current draw, ~200mA. I've read on some blog how someone was testing it's draw and it peaked at around 450mA under really heavy load. So that's why I hope to fit in the 1A with both (worst case I can dim the backlight on the LCD screen by 50%, that's supposed to drop it's draw to around 370mA).

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  • jw0752
    0 jw0752 over 9 years ago in reply to sawik123

    Hi Tomasz,

    The Data sheets are not real clear on this I agree. The problem seems to be that the need is very application dependent. I would recommend a 4.7 uF to a 10 uF on the input of the Converter. While they say it is optional I would also put a 47 uF on the output as well as a 0.1 uF ceramic. This is in line with the recommendations from shabaz. The diode is also optional but if you have one why not afford protection if it is available.

    John

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  • jw0752
    0 jw0752 over 9 years ago in reply to sawik123

    Hi Tomasz,

    For the fun of it I put 8 used AA batteries in a pack and powered a 5 V Recom Converter. Here is a picture of the set up. The little voltmeters each draw 20 mA.

    image

    I did not put any capacitors on the input or output in this case.

    John

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  • jw0752
    0 jw0752 over 9 years ago in reply to sawik123

    Hi Tomasz,

    For the fun of it I put 8 used AA batteries in a pack and powered a 5 V Recom Converter. Here is a picture of the set up. The little voltmeters each draw 20 mA.

    image

    I did not put any capacitors on the input or output in this case.

    John

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  • sawik123
    0 sawik123 over 9 years ago in reply to jw0752

    Interesting. I guess the capacitors help with filtering and stuff? Which is something you would not see on a voltmeter?

    Anyway, thanks a lot for the help! I'll put the the capacitors there anyway, because why not image and the diode.

    shabaz - thanks a lot for the help as well!

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  • Robert Peter Oakes
    0 Robert Peter Oakes over 9 years ago in reply to sawik123

    The larger the capacitor the lower the frequency it tends to filter efficiently

    Many people question why most schematics show a 10 or 100uF capacitor in parallel with 10 to 100nF capacitors, they think simplistically about capacitors in parallel and series (Much like resistors in parallel and series) but they do not consider the parasitic impedance / reactance etc at different frequencies

     

    For instance, a Large electrolytic at high frequencies can have a very high reactance (Think Resistance) due to the inductance of the capacitor (Its made up of a coil of metal after all) and is therefore useless at filtering the noise from all the digital switching that can be into the Mhz or even higher. Putting a smaller nF or pF capacitors in parallel to the electrolytic solves this problem, so they compliment each other

     

    so looking at this view

    image

    you can see how this happens, the frequency increases, the capacitor impedance lowers but the internal inductor impedance increases, at a certain point the impedance from the inductor becomes significantly more than that of the capacitor

     

    In a power regulator like the one we're talking about in this thread, the larger capacitor is for smoothing of the typical mains voltages and to reduce low frequency oscillations, the lower value caps reduce the high frequency noise from the load and other sources.

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