How To Make A 5V Regulated Power Supply For Micro-Controllers?

Discussion in 'Embedded Systems and Microcontrollers' started by Amlan Panigrahi, Oct 26, 2015.

  1. Amlan Panigrahi

    Thread Starter New Member

    Jul 9, 2015
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  2. JohnInTX

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    Jun 26, 2012
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    It looks OK. Your power LED is shown backwards. Good use of the caps and the 1N5400.
    You should do a thermal analysis to ensure that the heatsink will be enough for high currents at the higher input voltages. Yours looks a bit small.
     
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  3. Amlan Panigrahi

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    Jul 9, 2015
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    Thanks a lot...much appreciated! Will be working on it!
     
  4. Amlan Panigrahi

    Thread Starter New Member

    Jul 9, 2015
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    Yeah :) right in the schematic LED's terminals are reversed. :) Thank you!

    Your help will be appreciated regarding the heating problem. However, I thought of using a switching regulator with more than 70% efficiency. Or else the classic way to optimize the heat when dealing with high current would be by increasing the size of the heatsink.


    You suggest...
     
  5. JohnInTX

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    Jun 26, 2012
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    Well, they are two different things of course.
    The linear 7805 is simple to build and understand but is not particularly efficient.
    A switcher (for this kind of thing I like TI's simple switcher stuff) is much more efficient but more expensive, both the chip and the extra components around it. But design tools are free and easy to use.

    EDIT: as a personal rule of thumb -I'll use a linear for no-brainer stuff like a local supply for a PIC or something like that. For anything that has to do enough power management to be a heat problem with big sinks, I use a switcher. I don't like hot things and these days the switchers are cheap and easy to use. YMMV.

    Here's how to do the thermal analysis for the 7805.
    First you need to know how much heat the junction (internal chip of the regulator) can take and how fast it can get rid of the heat (that's what the heatsink is for but that's getting ahead).
    If you look at the datasheet for a TO-220 package 7805, you will find that the Thermal Resistance from Junction to Case is 5 degrees C per Watt of dissipation. That means that if you bolt it to an infinite heatsink, the junction temperature will rise 5degC above ambient if it is dissipating 1 watt of heat.
    The WATTs come from the familiar W=V*I where V is the difference between the input and output voltage and I is the load current.
    So.. if you have 10V in, 5V out and 1amp load you have 5Watts of heat to dissipate. (10-5)volts/1amp.
    At 5degC/Watt, the junction temperature will rise 25DecG above ambient - with our infinite heat sink. If ambient is 30degC, the junction temperature will be 30+25=55degC junction temp. Referring to the datasheet we find that this is well within the 125degC absolute maximum rating.

    BUT. You don't have an infinite heat sink. Yours is much less. I would guess maybe 35degC/Watt (but find a datasheet). Temperature resistances ADD like other resistances in series so let's see what we have:
    (Junction_to_case + case_to_heatsink + heatsink_to_ambient) *degC/Watt total thermal resistance (Rtt for now)
    Rtt for this example is:
    5degC/Watt + 35degC/Watt = 40degC/Watt. (we will consider case_to_heatsink as 0 for now).

    Junction_Temperature = (Rtt degC/Watt * Watts ) + Ambient Temp degC.
    Example:
    For 12Vin, 5Vout, 500ma load current.
    Watts dissipated = (12-5)/.5 = 3.5Watts.
    3.5Watts * 40degC/Watt + 30degC ambient temp = 170degC!!!! This is way more than the 125degC rating and will cause the regulator to go into thermal shutdown. Told you the heatsink looked a bit small.
    What to do?
    Lower the input voltage and/or load current
    Increase the size of the heatsink (to lower its degC/Watt rating - a big finned TO-220 will be maybe 5-10degC/Watt IIRC).
    Add airflow - see the heatsink datasheet.
    Realize that TO-220 is limited by the size of its package. That's why bigger regulators like the LM350K use a TO-3 package - lower junction-case thermal resistance and can bolt to a bigger heat sink.

    I did not touch much on the case-heatsink interface. That has to be good as well. A bit of thermal grease or a silicone pad is usually used. Either has a thermal resistance of something like 1/2degC/Watt. Don't just bolt the tab to the heatsink. You might consider a 7805FA package from NJR. All plastic to bolt right to the sink.

    That's how you do it. Its the same procedure for most devices, transistors, MOSFETS etc. Keep in mind that ambient temps can be considerably higher when this is put in a case. Also, even though the regulator may be happy at 100degC, the temp on the heatsink will easily cause painful burns.

    Have fun.
     
    Last edited: Oct 27, 2015
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  6. MikeML

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    Oct 2, 2009
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    Arduino doesn't want 5V, it wants about 6 to 12V because it has an on-board, low-dropout regulator that makes 5V. Likely, off-board stuff (shields) are needing 12V at more current than your supply can deliver...
     
  7. Amlan Panigrahi

    Thread Starter New Member

    Jul 9, 2015
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    In that case, my friend replacing the IC7805 with IC7809 will give enough potential to operate Arduino likewise you can make another power supply sing the same setup but only replacing the IC7805 by IC7812. That's all. If you need more amps then parallel IC7812 may be considered with bigger sinks but they have to be protected from current leakage by using Diodes (which will drop the current by .7V) but you'll get the required current.

    Also, if you will do the math with the caps that I have used in my circuit It will give you nearly 5.86V, which is enough to operate the Arduino if you will talk practically. I've checked it myself. If in your opinion it would not work... it can be then powered by using IC7809. What Say?
     
  8. Amlan Panigrahi

    Thread Starter New Member

    Jul 9, 2015
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    Thank you Very much for your kind and detailed opinion. It is really helpful, the ambient temperature part was well said. That has to be kept in mind when doing the heat analysis. However, if we operate under low current, let's say 200mA and the input voltage is 12V then

    Power wasted = (12V – 5V) * (0.5A + 0.005A+0) = 3.465W

    Which will suck the battery for no reason and in the worst case it will melt the soldering. To this problem what modification the circuit will need. if you can guide me.

    See, I could have used, DE-SW0XX family of switch-mode voltage regulators but they will bankrupt you (Kidding)... So something that is in the budget and easily available.


    I thought of using LM-317 but it has the same problem of heat dissipation and the output voltage will not be fixed for different input voltages.
     
  9. DickCappels

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    Aug 21, 2008
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    Below is the schematic referenced at the start of the thread, just in case it is not available to all who view this thread now or later. In the future, please post your designs at allaboutcircuits.com rather than merely referring readers to another site.

    [​IMG]
     
  10. nerdegutta

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    Dec 15, 2009
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    why aren't C1 and C2 labeled the same way on the schematic?

    The LM7805 will be really warm, when powering a RPI.
     
  11. MikeML

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    You asked for an opinion, so you are going to get it: I suspect that you are not really looking for criticism, but are just trying to get traffic to that web site which is typical of the kind I hate. Glitzy graphics, nags about not allowing pop-ups or cookies.

    Just like the web site, your supply is full of unnecessary things.

    C4 does nothing. If the supply is fed from a 12V battery, C4 is not needed, because the output impedance of the battery is orders of magnitude lower than the reactance of C4 at any useful frequency. If the battery were to be replaced with a unfiltered wall-wart, then it is nowhere near big enough to be a proper filter capacitor at 100 or 120Hz.

    According to the TI datasheet, C1 should be 0.33uF, not 0.1uF

    C3 does nothing. According to the TI datasheet, C2 should be 0.1uF, and that is sufficient to stabilize the 7805. This capacitor is required for stability (prevent oscillation), and to hold up the output voltage for transient loads that change faster than the internal regulation circuit can follow. C2 was chosen by the chip maker to satisfy both of these requirements, so your C3 is unnecessary. More is not better...

    Frankly, why do you publish a schematic, that you claim is your invention, when the correct circuit is on the manufacturer's data sheet? You would do much better to just refer a prospective builder to the data sheet, and show a photo of a completed board that complies with Section 9.1, Applications Section of the data sheet.


    78.gif
     
  12. JohnInTX

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    Jun 26, 2012
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    Sure it will. That's what voltage regulators do. The LM317 (like the others) has an internal voltage reference that it uses to control the output voltage under varying load currents AND varying inputs (within reason).
    Actually at 200ma its 1.4035W (adding the quiescent current as you did). So, while it runs cooler (Tj ~86degC) its still plenty hot. And there you see why a linear regulator is not the best choice for battery-operated equipment when it has to cope with high input voltages and/or load currents.
     
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