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voltage regulator question

Discussion in 'Electronic Basics' started by tempus fugit, Dec 1, 2003.

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  1. Rich Grise

    Rich Grise Guest

    True.

    and adding a little more wouldn't
    Not necessarily true, as you've noticed by now. :)

    particularly when it was much smaller than the circuit's
    Well, there's still X volts at the output of the regulator,
    but what you did was create a voltage divider.

    Also, the voltage drop varied
    Use Ohm's law. The only way to change the load current with
    a voltage-regulated source is to change the load resistance.
    I haven't seen the circuit, but it sounds like you've got
    16 ohms in series with whatever the load is, and the load
    itself changes. So, if you're drawing 100 mA at 10 V, the
    total load on the supply is 100 ohms. That's the sum of your
    16 Ohm series resistor and the load itself, giving 84 ohms;
    the load sees 8.4V. When you turn down the load so the total
    is 10 mA, the total load is 1000 ohms, and the 16 is fixed,
    so the adjustable part of the load is now presenting 984 ohms.
    (please double-check my arithmetic. %-} ) So it sees 984 *
    ..01 = 9.84V, almost 1.4V difference.

    HTH!
    Rich
     
  2. tempus fugit

    tempus fugit Guest

    Thanks Rich

    How? I've got the resistance in series with the output.


    it sounds like you've got
    I do have 16 ohms in series with the load, but the load doesn't change. All
    I did was connect a 100 ohm resistor from the regulated out (after the 16
    ohms) to ground. I did sweep the voltage output of the PS from 5-15v to see
    what the effect on the regulation was.
    Defaut has pretty much convinced me not to worry about the short circuit
    protection, but I'd still like to figure out why this made such a
    difference.

    Thanks again


    So, if you're drawing 100 mA at 10 V, the
     
  3. tempus fugit

    tempus fugit Guest

    Thanks again. The whole supply is fused at the AC input, so I should be OK
    at the xformer - I was more worried about the regulator.
     
  4. tempus fugit

    tempus fugit Guest

    Sorry, that's what I meant.
     

  5. As far as the voltage regulator is concerned, the 16 ohms is part of
    the load. The regulator will keep the voltage at its own output
    terminal constant. The 16 ohms and the load resistance are apparently
    in series across the regulator output, so the current drawn by your
    intended load also flows through the 16 ohms, and, according to Ohm's
    Law, produces a voltage drop in that resistance. That voltage drop
    will vary with the current drawn by your intended load.
     
  6. default

    default Guest

    Like Peter wrote, the 16 ohms is part of your total series load
    resistor. The regulator only cares about what its output pin voltage
    is with respect to its "adjustment" terminal. If the numbers don't
    calculate out - then you have a circuitry or part malfunction.

    With the "16 ohms in series with your load resistor," (forgetting the
    accuracy of that statement for the moment) and a load resistor of 100
    ohms I calculate the following drops:

    With the supply voltage set to 15 volts your 100 ohms will have 12.94
    volts across it.

    With the supply voltage set to 12 volts your 100 ohms will have 10.35
    volts across it.

    With the supply voltage set to 10 volts your 100 ohms will have 8.72
    volts across it.

    With the supply voltage set to 8 volts your 100 ohms will have 6.88
    volts across it.

    Assuming I didn't make a mistake, the laws of Ohm still prevail and
    allowing for the ±5% tolerance in your meter, and ±5% tolerance in
    your 16 and 100 ohm resistor, etc.. The actual reading you get should
    fall within ±10% of those voltages at your 100 ohm resistor)
     
  7. Rich Grise

    Rich Grise Guest

    OK - that's what I meant by the "load" changing. The part of the
    circuit _after_ the 16 oms in series is its "load." You've changed
    the load from the point of view of the 16 Ohm. In a way, you could
    say that you've added an additional load. Draw a schematic, and
    trace current paths. Preferably on paper, with a pencil. Imagine
    the water flow model - I've found that helps me a lot.

    And maybe we should all be clear, in a situation like this,
    what we're referring to as the "load." From the point of view
    of the supply alone, the whole thing from the 16R out is the
    load. From the point of view of the supply+16R circuit,
    your DUT and the parallel resistor is the load. From the
    point of view of a circuit consisting of the PS, the 16R
    and the 100R, the load on the 16-100 point is your DUT.

    Or something like that - I think the chronic is kicking in. %-}

    Cheers!
    Rich



    I did sweep the voltage output of the PS from 5-15v to see
     
  8. tempus fugit

    tempus fugit Guest

    ahhh. I think I see now. I don't think I measured the voltage at the output
    pin of the regulator, but only at the output of the PS itself (which of
    course is after the 16 ohms). So what you're saying is that the voltage I'm
    reading is actually like measuring the voltage part way through into a
    circuit? Hence the voltage drop.


    That makes sense now. Simple thing - I shouldn't have overlooked it.
     
  9. default

    default Guest

    I think you've got it.
     
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