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Voltage Drop for Power Adaptor

Discussion in 'Electronic Basics' started by Homer, May 17, 2007.

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  1. Homer

    Homer Guest

    Hi All,

    I am planning to light up 178 LEDs (3.4V-3.8V, 25MA) using a 18V/1A
    Adapter (DC radio shack transformer) and I need a little help.

    Power Adapter says that is 18 volt but when I test it with volt meter
    it shows 22V. When I connect a string of 6 LEDs (plus small resistor)
    with it voltage is dropping to 21.4 (and shows 9.4ma current)

    I don't have problem calculating my resistors (I am using
    http://led.linear1.org/led.wiz) but it seems my 18V (or 22V) Adapter
    is not really what is says (it drops).

    My question is what should I put in that web site as my Source
    Voltage? (22V? 18V? or something else).


    Thanks in advance,

    Homer
     
  2. Chuck

    Chuck Guest

    The voltage rating on unregulated
    adapters is generally the voltage at
    full load (1 A in this case).

    If you're calling out the parts on the
    web site you could indicate both no-load
    voltage and voltage under your load.

    Chuck
     
  3. A fair approximation for your supply (based on both the
    rated voltage and your experimental results) would be a 22
    volt source that includes a 4 ohm series resistor. When you
    connect an 18 ohm resistor across it, the 4 ohm internal
    resistance will drop 4 volts and the 18 ohm resistor
    resistor will drop 18 volts.

    Include that internal 4 ohm resistance in your design
    calculations.
     
  4. Homer

    Homer Guest


    Thanks but what do you mean by unregulated adapters?
    And also if the rating is the voltage at full load then how do we
    explain the Voltage Drop with 6 LEDs (that is far less than 1A for
    sure)?
     
  5. John Fields

    John Fields Guest

    ---
    Unregulated adapters are adapters that exhibit a change in output
    voltage with a change of output current.
    ---
    ---
    As John Popelish explained earlier, your adapter can be modeled as a
    stiff 22 volt source with an internal resistor in series with its
    output, like this: (View in Courier)



    22V>---[R]----><----+
    |
    [VOLTMETER]
    |
    GND>----------><----+

    With only a typical high input impedance (10 megohms) voltmeter on
    its output the output voltage will read 22V because the voltage
    dropped across the series resistor will be very small.


    Now, with what you rigged up:


    22V
    |
    [R]
    |
    +->21.4V
    |
    [R]
    |
    [LED]
    |
    [LED]
    |
    [LED]
    |
    [LED]
    |
    [LED]
    |
    [LED]
    |
    GND

    since your whole LED string plus the resistor is dropping 21.4V and
    the current in it is 9.4mA, the string and the resistor look like:

    E 21.4V
    R = --- = --------- = 2276 ohms
    I 0.0094A

    and the circuit reduces to:



    22V E1
    |
    [R1]
    |
    +-->21.4V E2
    |
    [2276R] R2
    |
    GND

    Notice that there was 600 millivolts dropped across the internal
    resistor because of the 9.4mA flowing through it.

    That means that with that current flowing through it its resistance
    must be:

    E1 - E2 0.6V
    R = -------- = --------- = 64.52 ohms
    I 0.0093R

    and the circuit will look like:

    22V E1
    |
    [64.52R] R1
    |
    +-->21.4V E2
    |
    [2276R] R2
    |
    GND

    That means that as more and more current is pulled through it by the
    load, more and more voltage will be dropped across it until, with 1
    amp flowing through it it will drop 4 volts and the voltage across
    the load will be the specified 18V:


    22V E1
    |
    [R1]
    |
    +-->18V E2
    |
    [18R] R2
    |
    GND


    For that to happen, with 1 amp in it, its resistance must be:


    E1 - E2 4V
    R = -------- = --------- = 4 ohms
    I 1A


    and the circuit will look like:

    22V E1
    |
    [4R] R1
    |
    +-->21.4V E2
    |
    [18R] R2
    |
    GND

    Which is quite different from the 64.52 ohms with 9.3mA in it!

    The reason, of course is that the impedance of the rectifier diodes
    in the adapter decreases abruptly as the voltage increases past the
    Vf knee.
     
  6. Chuck

    Chuck Guest

    Some "wall warts" contain internal
    voltage regulators that keep the output
    voltage very close to the stated voltage
    on the adapter. If yours were regulated,
    it would deliver 18V, regardless of
    load, up to its rated current output of
    1 ampere.

    The unregulated variety have a
    non-negligible internal resistance that
    causes the output voltage to drop as the
    load increases. In a rare gesture of
    honesty, the manufacturer actually
    promises the rated voltage at full load.
    Well, the greater the external load, the
    greater the voltage that will be
    developed across the internal
    resistance. This voltage will be
    subtracted from the output voltage.

    The voltage drop follows Ohm's law,
    using the internal resistance and the
    load current.

    John has calculated the internal
    resistance to be four ohms using (22V -
    18V)/1A. That assumes a linear internal
    resistance but the 0.6V drop with a
    current of 0.0094A suggests an internal
    resistance of ~63 ohms which would never
    allow 18 volts output at 1 A. So maybe
    it is non-linear (at least over part of
    the output current range) or defective.

    It can get complicated to work with an
    unregulated supply. For less than a
    dollar, you can add a simple IC voltage
    regulator and some capacitors and have
    your output voltage stay within 50mV of
    18V at any load up to 1A.

    Chuck
     
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