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When Measuring Watts...

Discussion in 'General Electronics Discussion' started by MrClamperSir, Mar 27, 2016.

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  1. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    Hello!!

    If I have 10vin from the transformer but capable of 14vout with the help of my regulator, 460Ω of fixed resistance and a 5k pot, how would I calculate the various watts I can expect from this circuit? Also do I measure from the 10vin or from 14v?
     
  2. davenn

    davenn Moderator

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    a little unclear on the setup

    please show a schematic for your project/thoughts
     
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  3. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    image.jpg
    Does this help?
     
    Last edited: Mar 27, 2016
  4. dorke

    dorke

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    1st thing that will limit any "Watts" you can get is the transformer itself.
    i.e you need to know what is it's watt rating(you can't get more than that).

    2nd thing is the voltage drop on the regulator.
    i.e Vreg=14 - vout ; PDreg=Vreg * iload .
    the lower Vout is the more power is lost in the reg.

    3rd thing to consider is the maximum theoretical Vout you can get .
    It is dependent on the regulator's Vdrop min.(that has to be looked-up in the datasheet)
    For a "common" reg. Vdropmin is 3V so Vout max=14-3=11V
     
    Last edited: Mar 27, 2016
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  5. Anon_LG

    Anon_LG

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    Jun 24, 2014
    I do hope your bridge rectifier has some actual diodes in it, rather than just wiring connections!!!
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Also note that as a rule of thumb, your transformer should be rated for 1.414A RMS for each Amp you wish to draw from a bridge rectifier.

    Transformers are often rated in VA, so a 15VA 10VAC transformer if driving a bridge rectifier would only be able to supply 1A to the load.
     
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  7. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    The data sheet does not list the watt rating. It lists the voltages and amps along with Hz.

    This is where it gets confusing for me. There's about 12vac I'm getting from the 10vac (2a) secondary side of the transformer. After the bridge rectifier it's 10.6vdc. I'm using the lm 338 reg which boosts the voltage to 14.72vdc.
    So I guessing when I calculate my load minus the the vout, I would multiply it by the amps?
    If this is correct, would I multiply it by the amps capable or what the load is drawing?
     
  8. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    I appreciate your concern. You'd be happy to know the diodes are in place!!
     
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  9. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    The transformer is rated 20VA. I only need 0.5A-1.5A (it's more like 1A max but I'm trying to leaving a little room in there) on the load.
    Am I able to draw this current with the 20VA rating?
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    A 10V 20VA transformer driving a bridge rectifier and a filter capacitor is capable of 20VA / (10V * 1.414) = 1.414A

    And a 338 cannot boost the voltage.

    It would be reasonably safe to ensure your input voltage to the 338 will be about 12V under load, leading to a reasonable range of output voltages being 1.25V to 9.5V at up to 1.4A Thus, you can deliver about 13.5W into a load. (whilst dissipating about 6.5W in heat)
     
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  11. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    How do you calculate the dissipating watts? Because all my tools run at different voltages from 3.5V to 7.5V. I want to be able calculate Vin and Vout in Watts dissipated amongst the varying loads.

    Also if the 338 isn't boosting the voltage why do I have around 10.6V into the reg and 14.72V on the output?

    Thank you for your help
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    My guess is that it is due to an error in measurement. (I'm assuming it is an LM338 or similar and not a switching regulator of some sort).

    Given that you can get 1.4A out, you can calculate the watts out as voltage * 1.4

    You mat be able to draw more current but you will be exceeding the rating of the transformer if you do.
     
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  13. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    Thank you. I will remeasure tomorrow and post the results.
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    note on your circuit exactly where you take the measurements.
     
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  15. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    image.jpg

    So it appears I'm getting 12.5VAC before the bridge rectifier and 16.1VDC after. Which means I've got 16V into the regulator. So I'm guessing this is the number I use to calculate the wattages in dissipated heat?
     
    Last edited: Mar 28, 2016
  16. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    So the formula would be (16V-VariableVoltage) x 1.4A?
     
  17. dorke

    dorke

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    If you wont to get a more realistic result do this:

    1. Connect a load that will draw 1A(or 1.4A) from the unregulated DC voltage(measurable with a series Ampere-Meter) .

    2. With this load measure the unregulated DC voltage and use that instead of the 16V no-load voltage in your formulas .
     
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  18. MrClamperSir

    MrClamperSir

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    Feb 3, 2016
    Ok thanks. I'll give this a try tomorrow and post the results.
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    you should find that the output of the transformer drops, and the voltage drop across the rectifier rises.

    The end result will be a lower voltage across the filter capacitors.
     
  20. dorke

    dorke

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    Jun 20, 2015
    P.S
    Another thing to take into account is the AC Line voltage tolerance.
    If I am not mistaken,in the US it should be 120VAC±5% so 114-126VAC.

    while measuring the above ,
    measure the line voltage ACV(on a high scale voltage) and compensate for the worst higher case of 126V.
    e.g say you measure 117V then the compensation value should be
    (126-117) / 117 =0.077 (7.7%)
    The unregulated DCV in the max case is thus 1.077 higher than what you have measured.

    In the same way,
    you have a compensation on the maximum voltage output of the Regulator.
    This time from the lower 114V side, like so:
    (114-117) /117 = -0.025 (-2.5%)
    The unregulated DCV (and max regulated voltage) are lower by that amount in the worst case of lowest input AC-line voltage.
     
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