Regulator problem

Discussion in 'The Projects Forum' started by Dalaran, Dec 4, 2009.

  1. Dalaran

    Thread Starter Active Member

    Dec 3, 2009
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    Bringing back an old post. But I have used a schematic very similar to the one given in the datasheet to create a power supply. The problem is as I add load to the output pins (my circuit) the voltage drops, and it seems not to be regulating correctly. Any idea what I might be doing wrong as the pot and everything else seems to be working fine.

    Thanks.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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  3. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Be sure to clearly indicate the input voltage, the output voltage you desire, and the resistance of the load (or amount of current it should draw) on the schematic that you post.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    Maybe the input voltage to the regulator drops lower than its dropout voltage.
    Maybe because the transformer cannot supply enough current or because the main filter capacitor is too small.
     
  5. Dalaran

    Thread Starter Active Member

    Dec 3, 2009
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    Hey all I apologize for the silly mistake. I ran it by my buddy that was over yesterday and he knew right away I had connected the positive post to the adj terminal and not Vout. I should have realized this too since I was getting down to 0V.

    Thanks again for the help all.
     
  6. ELECTRONERD

    Senior Member

    May 26, 2009
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    Another common mistake is connecting the Vout back panel to ground. Some of back metal panels are designed to connect to ground and others aren't, so make sure you define which is which. The designers of the regulators typically choose what pin the back panel should be based on the best thermal dissipation.

    Austin
     
  7. Dalaran

    Thread Starter Active Member

    Dec 3, 2009
    168
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    Thanks again.

    Now that this is working I would like to add on to it by using a microcontroller to display both the voltage and current outputs.

    To get the current is it possible just to use a very small resistor in series and measure the voltage drop across this using the ADC of the microcontroller? This could then be converted into a current based on ohms law and displayed on a screen.

    Am I right with my thinking or will I need to take other precautions when putting this resistor in an measuring the voltage.

    Thanks.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    If you use a resistor in series with the load, your regulation will lose precision.

    You really should start a new thread with this new twist; otherwise the beginning of the thread will get confused with this new subject material.

    I suggest something like "Monitor power supply current/voltage with uC?" for a title.
     
  9. Dalaran

    Thread Starter Active Member

    Dec 3, 2009
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    If I want to use a simple 6V wall-wart and a fixed 5V regulator do I need anything other then caps on the input an output? What sizes are recommended? 100nF on the input and 10-100uF on the output? Would be used for 1-2A.

    Thanks.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    You will be very disappointed in the performance of a power supply constructed like that.

    A fixed-output 7805 regulator has a minimum voltage drop from the input to the output of about 2v; that goes higher when drawing maximum current (around 1A).
    You will need a large filter capacitor on the input. Generally, DC wall warts do have rectifiers in them, but if they have filter caps, they are very small in size. Figure on roughly 3,000uF per 1A of output current. You will still have some ripple voltage on the filter cap.

    If you want to have any chance of your supply working halfway decently, you will need to use a LDO (low dropout) type regulator. If you want to use a standard type fixed voltage regulator, you will need more input voltage; around 8v-9v. If you are wanting up to 2A output current, you will need a large heat sink on the regulator.
     
  11. Dalaran

    Thread Starter Active Member

    Dec 3, 2009
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    Thanks. I can go with a larger wall-wart that is not a problem. Why do I need such a large capacitor on the input here compared to a variable supply using the lm317? There a small 100nF cap is suggested on the input and a larger on the output. Is there an ideal size on the output then?
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    OK. Keep in mind that whatever linear voltage regulator you select, it will have some "dropout" voltage; the dropout voltage will almost always increase as current flow increases. Refer to the datasheet of the regulator under consideration for the numbers. Some include charts/graphs of dropout vs temp vs output current.
    The small 100nF cap on the input of the LM317 is in addition to whatever filter caps you might have after the rectifier/bridge. It's just there to help take care of any inductance in the wiring between the filter cap(s) and the regulator itself. It aids in rapid transient response, and helps to prevent high frequency oscillation.

    10uF + 100nF (0.1uF) is a good combination as a general "rule of thumb" for general use. If you are using certain types of ICs (particularly 555 timers) you will want larger caps right near the ICs; for 555 timers use 220uF + 100nF in parallel.

    If you are using the LM317 as a current regulator, there will be a minimum 3V dropout between the IN and ADJ terminals, and you do not want a cap on the output.
     
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