Power Supply Using LM350

Discussion in 'General Electronics Chat' started by sureshparanjape, Mar 11, 2012.

  1. sureshparanjape

    Thread Starter Active Member

    Feb 10, 2012
    I am a total novice in the wonder land of electronics hobby.
    I am following instructions given in "http://pic16f628a.blogspot.in/2010/07/variable-dc-power-supply-125-18v-using.html" for a variable DC power supply proect. The project gives a mathematical formula for Vout as follows:
    Vout = 1.25 V ( 1 + R2/R1) + Iadj R2
    As per advice I have downloaded LM350 data sheet and it also has the same formula. I wish to know what V stands for in the formula.
    One can observe that 1.25 V (1+R2/R1) is always bigger than 1.25 V and Vout would be maximum when R2 is maximum, in the present project 5 K.
    My different inputs for Vin I got, as expected, different maximum values for Vout. Lowering of R2 resulted in smaller Vout (also smaller than Vin).
    Hence my query is :What does V in the above formula stand for?
    Thanking an answer in advance.
    Last edited: Mar 11, 2012
  2. Wendy


    Mar 24, 2008
    This was neither feedback nor a suggestion, so I have moved it to a more appropriate forum.
  3. Markd77

    Senior Member

    Sep 7, 2009
    Just ignore it, it just means volts and isn't really part of the formula.
  4. sureshparanjape

    Thread Starter Active Member

    Feb 10, 2012
    I feel then there is a problem. Vout doesn't depend on Vin;however it appears to be as was observed in an experimentation.
  5. SgtWookie


    Jul 17, 2007
    The formula is simplified, and assumes that you will have the input voltage at least as high as the desired output voltage plus the regulator's dropout voltage. The dropout voltage median is approximately 1.7v, but will vary widely over Iout and temperature.

    See the plot in the National Semiconductor/TI datasheet for the LM117/LM317:
    Page 6, 2nd down on the right, "Dropout Voltage".

    If (Vin-Vdropout) < Vout, then the output voltage will not be regulated.

    Additionally, I will caution you that if you are considering attempting to use a digital potentiometer, you will likely destroy it, as digital pots usually have a voltage supply limit of around 5v.

    Rather than using a digital pot, consider using several different values of fixed resistors for R2, and a fixed 120 Ohm resistor for R1. You will need to use open collector or open drain switches controlled by the PIC output(s), otherwise you risk damaging the PIC I/O pins, as they have internal clamping diodes.
  6. sureshparanjape

    Thread Starter Active Member

    Feb 10, 2012
    Thanks for sparing time and giving important informations.
    I am(rather was) using standard POT, which I seem to have damaged since it got over heated while experimenting. I was thinking of using digital POT but as you advise I will not( I would have very likely not observed 5 v restriction!).
    I am puzzled by seeing the graph of Vout against Vin. It is only for the range 0 to 5 for Vin, when it is said that it can be used for maximum of 30 v for Vin.
    Looking at the graph of Vout against Vin, it appears that the relation is linear. Is there a linear relation ?
    Thanking you again for your time and advice.