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What am I missing in this voltage regulator design?

  1. Jul 18, 2013 #1
    DUJ4pxZ.jpg

    The above (top) design is one for a adjustable regulated voltage using an IC. The bottom schematic is the actual design of the IC regulator.

    My question is, am I missing an obvious way to derive the relationship for Vo? How does this design work, anyway? I figured it had something to do with the feedback to the ICs reference voltage (ground, middle pin).

    If it were me, I would just use a fixed voltage regulator, like the 7805, ground the middle pin and take the output between R1 and R2 (with R2 being shunted by a cap). In this way, the output is simply a voltage divider. Although, perhaps I am just not seeing the merit of using a true adjustable IC like this one.
     
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  3. Jul 18, 2013 #2

    nsaspook

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    The 7805 connected as a variable regulator will have a larger VDO than the LM317 and the regulator standby current (several mA) will flow into R2 causing the circuit to lose voltage stability. With a 12 output you will have 5 vdc on R1 and 7 vdc on R2 (common R values for 1&2 470ohm,510ohm for ~12 vdc). With a LM317 the regulator terminal adjustment current will be much lower(100uA).

    http://www.ti.com/lit/ds/symlink/lm117.pdf
     
  4. Jul 23, 2013 #3

    Averagesupernova

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    Sherrellbc, I am not 100% sure what you are trying to say. Do you mean to say you would simply take the node between R1 and R2 as a voltage source? You must know this can't work since a voltage source needs a low output impedance.
    -
    The way all 3 terminal regulators work is they try to maintain a constant voltage between the adj pin and the output pin. They don't care about anything else other than overcurrent.
     
  5. Jul 31, 2013 #4

    Baluncore

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    It is clear from the internal diagram that the output of the LM317 will rise or fall so as to make the Vout pin one voltage reference above the Vadj pin. That requires a current through R1 of Iadj which must then flow on through R2 to ground.

    So; Vout = (Vref / R1) * (R1 + R2) = Vref * ( 1 + R2/R1 )
    I cannot see where your extra component “+ Iadj * R2” comes from.
    Maybe you have counted the reference voltage twice.

    Another view is as a simple potential divider driven by Iadj. The ratios are;
    (Vref / R1) = Vout / (R1 + R2) which obviously gives the same result.

    By drawing the R1:R2 divider with a bend in it you are hiding the intuitive solution.
    Instead draw the R1:R2 divider straight from Vout to ground. Bend the connection to Vadj.
     
    Last edited: Jul 31, 2013
  6. Aug 10, 2013 #5

    meBigGuy

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    Are you putting a resistor in series with the output? (maybe I'm misunderstanding what you are describing)

    That means you have no load regulation and have no need for a regulator in the first place.

    The advantage is that the LM317 has "zero" output resistance at whatever current you draw. The output voltage is not load dependent as it is with a divider. You use a divider to create the reference, not to directly create the output voltage.

    Think of it kind of like an amplifier driven by a reference. You can draw all you want from the amp without affecting the reference. In the LM317 there is feedback to make the output even more stable.

    Because of the feedback the internal amp in the LM317 will do whatever is required for there to be essentially 0 voltage difference between its internal inputs (marked with --).
     
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