Adjustable voltage regulator help

Discussion in 'The Projects Forum' started by pilotmatt, May 11, 2010.

  1. pilotmatt

    Thread Starter New Member

    May 11, 2010
    First of all, let's start by saying that my electronics knowledge is very old and incredibly rusty. This is not my area of expertise... so please be kind!


    A colleague and I are trying to design an adjustable linear voltage regulator. We are using the LM338T regulator in a circuit where the output is controlled by a variable resistor/potentiometer.

    The datasheet for the LM338 is here :

    The circuit diagram we're using is given as an attachment; effectively the same as the 1.2V-25V Adjustable Regulator shown on page 7 of the datasheet, less the input capacitor C1.

    V-IN for the circuit is ~28VDC, and V-OUT would generally be 12-14VDC, although from time to time it may need changing to suit the load (8V - 24V). Based on these figures and the information in the datasheet, we've arrived at figures for R1 of 120Ω and R2 of 2.2kΩ (variable).

    Our issue is how to choose precisely the correct resistor and potentiometer with respect to tolerances, voltage ratings and power ratings. I recall that P=I²R but how could we find I in order to choose the correctly power-rated components? Is it as simple as P=Vref²/R?

    The guy assembling this (luckily, that's my colleague) prefers axial resistors as they're easier to solder into the peg board.

    Elsewhere in the datasheet it explains Iadj and I1 and quotes figures for Iadj, but not I1. Clearly we want to avoid choosing an inappropriate component and having it flashover.

    Any advice welcome!


    p.s. we've bought a heatsink for the regulator, but will we need additional cooling too?
  2. Norfindel

    AAC Fanatic!

    Mar 6, 2008
    This regulators work by forcing aprox 1.2v over R1, that sets the current thru R2, as no current can go inside the ADJ terminal. With that information, you can calculate the power rating. It's I^2/R, and isn't going to be large, as the current with a 120 ohm resistor is going to be 10mA (Note: the real reference voltage is 1.24v typical, not 1.2v, i think they use 1.2 to make calculations easier, but isn't as precise).

    You will be a lot more interested into calculating the power dissipation, which i suppose is roughly (Vin-Vout)*Iout (you better check in the datasheet, or application notes) and required heatsink. I don't have any specific documents to point you at, but you can get some results in a google search:

    The maximum allowable thermal resistance of the heatsink is:

    Rth_{(s-a)} = \frac{Tjmax - Tamb}{Pd} - Rth_{(j-c)} - Rth_{(c-s)}

    Rth_{(s-a)} means Thermal Resistance heatSink to Air (heatsink manufacturer supplies this number).
    Rth_{(j-c)} means Thermal Resistance Junction to Case (component manufacturer supplies this number).
    Rth_{(c-s)} means Thermal Resistance Case to heatSink (depends on the thermal compound used, and isolator, like mica or silpads).
    Tjmax is the maximum junction temperature (supplied by manufacturer, but better you try to keep this lower than 90^oC).
    Tamb is the maximum allowed ambient temperature that the device is going to work. You decide this.

    Also, take a look at figure 3, that shows the protection diodes, and reference bypass capacitor for added ripple rejection.

    EDIT: here, i found an online calculator for heatsinks. If you have any doubts, please ask:
    Last edited: May 11, 2010