Voltage regulator in a circuit

Discussion in 'General Electronics Chat' started by tekinzen, Nov 23, 2009.

  1. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Hello all. I'm new to electronics so pleas bear with me.

    AFAIK, a voltage regulator has three pins, one for the input voltage (+), one for ground (-) and the third for the output voltage that needs to be regulated.
    If the voltage regulator exists in a circuit the reference anode will become pin no. 3 of the voltage regulator while the ground (pin no 2) will become the cathode. Is my theory correct?

    I'm referring to the picture at http://forum.allaboutcircuits.com/attachment.php?attachmentid=10904&d=1247957116.

    Thanks!
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    It is safest to use Google to get a data sheet for the device in question. Fixed positive regulators (78xx)usually go In - Ground - Out. Fixed negative regulators (79xx) usually go Ground - In - Out. Adjustable regulators with 3 pins are different yet.
     
  3. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Mine is a fixed positive regulator indeed. Is my theory correct then?
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    Just to get on the same page:
    That should read "the regulated voltage output". The input is the voltage in need of regulation.

    This is a bit confusing:
    The 7808's pin 2 should connect to circuit ground, and so the output voltage is reference to that part of the circuit.

    I simply do not quite get what you mean by "reference anode". Pin 3 is the 8 volt regulated voltage source.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    One other point, if you read the fine print on these parts you will find capacitors on the input and outputs are needed, typically a 0.1µF cap on each, with a 100µF or larger cap on the input if the regulator is more than 6" away from their source.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Just to confuse you a bit more,

    You're showing a ST Microelectronics 7808 regulator in the photo.

    That particular regulator will need a minimum of 10v input to provide an 8v output, even at low current levels.

    The regulator uses the ground terminal as a 0v reference point.
    It will source current from the input terminal to the output terminal, in an attempt to keep the difference between ground and the output terminal at 8v.
     
  7. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Sorry for the confusion I may have caused with the photo. My voltage regulator is actually a 9v to 5v (low dropout).
    What I meant by reference anode is that the output pin (pin 3 of the voltage regulator) will become the anode in the circuit (i.e.: a power supply is connected to pin 1 (+5v), 2 (GND), LED anode connected to pin 3 and LED cathode connected to pin 2). Of course I would never connect the LED directly without a resistor... just an example :D
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    What is the part number of your regulator?
    A low dropout regulator implies that the voltage drop between the input and output is lower than the older standard regulators, like the 78xx series and 117/317 series.

    I'm afraid that you are using the term "anode" incorrectly.
    In the photograph of the 7808, the output pin (3) is where the regulated voltage and current exits the regulator; it is the source of the rest of the circuit's current. It would be incorrect to call it an "anode", as that would imply that the output terminal was somehow a diode. It is not a diode.

    As far as LEDs go, yes - you would want the anode connected to the more positive supply, and the cathode connected to the more negative supply.

    Just so you know, a quick method of calculating the current limiting resistor that you will need:
    Rlimit >= (Vsupply - Vf(LED)) / DesiredCurrent

    Since the output of your regulator is 5v, that replaces Vsupply in the formula.
    Let's say you had an LED that was rated for a Vf of 2.2v at 20mA
    Rlimit >= (5v - 2.2v) / 20mA
    Rlimit >= 2.8 / 0.020
    Rlimit >= 140 Ohms.
    A decade table of standard resistor values are here:
    http://www.logwell.com/tech/components/resistor_values.html
    Frequently, E24 values (green columns) are available locally.
    The other columns (E48,E96, E192) you'll usually have to order from a distributor.
    140 Ohms is not a standard E24 value. The next larger E24 value of resistance is 150.

    You could use the 150 Ohm resistor, order an E48 140 Ohm resistor, or use a combination of resistors in serial/parallel to get 140 Ohms.
    Here is an online calculator that will give multiple combinations of resistors:
    http://www.qsl.net/in3otd/parallr.html
     
  9. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Thank you all for your replies. The voltage regulator I have is LM2937ET.

    I have it connected to a 9V battery which I want regulate at 5v like this:

    Pin 1 (+) connected to battery anode, Pin 2 (GND) connected to battery cathode. When I measure the output voltage with my multimeter I still read 8V+ when I connect Pin 2 (GND) and Pin 3 (output) to my multimeter.

    I know I need a capacitor too, but I need to know why I still get unregulated voltage. This is a positive voltage regulator.
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    I tend to draw simple picture of my regulators because I've wired them wrong too.

    For example, for the LM317 I have this.

    [​IMG]

    Basically it keeps it easy to track.
     
  11. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Strange, If I wire mine according to your schematic I get a regulated voltage but it's only 1.8V. Maybe that's why this needs a cap?
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    Bill_Marsden did not mean for you to connect up your regulator as he indicated for the LM317; you have a different regulator. If you connect it up as shown for an LM317, you may damage your regulator or the battery.

    You have a TO-220 package. With the letters/numbers on the package facing you so that you can read them, pins down, the pins are numbered left to right, 1, 2, 3.
    1 is IN, 2 is GND, 3 is OUT. The tab is also GND.
    There are numbers following the LM2937ET.
    It will look like one of the following:
    LM2937ET-5.0
    LM2937ET-8.0
    LM2937ET-10.0
    LM2937ET-12.0
    LM2937ET-15.0
    OK, instead of referring to a battery as having "anodes" and "cathodes", you should instead refer to them as "positive battery terminal" and "negative battery terminal", otherwise confusion on the part of the readers may result.
    I suspect that your part number has a -8.0 at the end of it.

    You must use a capacitor of at least 10uF from the OUT to GND terminals.
    This is a requirement that is stated in the datasheet:
    http://cache.national.com/ds/LM/LM2937.pdf
    See page 8.
    If you do not have a capacitor of at least 10uF across the OUT and GND terminals, do not expect for the regulator to work properly.

    Also, if the wiring from the battery is more than 3" long, you will need a 0.1uF capacitor between the GND and IN terminals.
     
    Last edited: Nov 27, 2009
  13. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Thanks for correcting me. I have yet to learn how to use the right terminology.

    It is strange indeed for my regulator has 5.0 and not 8.0 at the end. It gets more confusing now...
     
  14. SgtWookie

    Expert

    Jul 17, 2007
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    If you add a 10uF capacitor across the OUT and GND terminals, you will have a 5v output.
     
  15. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    That is if I connect the terminals like in the data sheet (pin 1 In, pin 2 GND, pin 3 out), correct?
    I don't have a 10μF cap, can I use a 100μF instead?

    On another note, just touched the regulator and burned myself :D

    I "lied". I have a 10μF cap. I get 4.96V when I put it in my circuit but that wouldn't be sufficient to power a PIC micro-controller which requires 5V, correct?
     
    Last edited: Nov 27, 2009
  16. SgtWookie

    Expert

    Jul 17, 2007
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    Then you need a heat sink on the regulator. Make certain that it is connected as shown in the datasheet. What you have described above is correct.

    Well, let's see... 4.96V is 99.2% of 5v.

    Have you checked the datasheet for the PIC microcontroller that you are using to see what are the acceptable ranges are for it's proper operation?

    Look in the Electrical Characteristics section.
     
  17. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    Should have said that in a separate reply... When I have the 10μF in the circuit, the voltage is 4.96V and the voltage regulator is lukewarm not hot. It was hot when I had it incorrectly wired and was getting 1.8V regulated voltage.

    The micro-controller I'll be using is PIC12F629. Looks like the voltage range is 2-5.5V (http://www.microchip.com/wwwproducts/Devices.aspx?dDocName=en010113)
     
  18. SgtWookie

    Expert

    Jul 17, 2007
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    Incorrect wiring frequently causes heat - and smoke. :eek:

    Your PIC12F629 will operate very happily on 4.96V. So will other TTL and CMOS logic ICs.
     
  19. tekinzen

    Thread Starter Member

    Nov 22, 2009
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    I have learned it the hard way :D
    Thanks again for your helpful guidance!
     
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