LM338 Power Supply: Help with Schematic. Thanks

Discussion in 'The Projects Forum' started by TheLaw, Oct 2, 2010.

  1. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    Hi,

    I was originally set on building an amp but seeing that I really don't have a true lab power supply, I thought I'd build one seeing that some pretty simple ones can be made. Now I could have gone with an LM317 but I thought I might want a little more juice for future projects. Also, it would open more doors of opportunity as I won't be limited to circuits that use <2A of current.

    So for all of the projects I plan on doing, I hope to be using one of these.

    Here's what I am basing it off of: http://worldtechnical.blogspot.com/2...l#comment-form

    I have two questions.

    1.) Does that transformer need to be 24V? (24V seems a bit overkill for what I will ever need it for.) I was hoping I could use something like a 12V transformer which will bring down the cost substantially. Is it just a drop in replacement?

    2.) Can that trimpot (I am assuming its a trimmer), be replaced with a regular non-trimmer potentiometer? And does 2K5 stand for 2.5K? or is that like a dual ganged pot?

    Thanks very much.

    -TheLaw
     
    Last edited: Oct 2, 2010
  2. wayneh

    Expert

    Sep 9, 2010
    12,088
    3,027
    The max voltage of the transformer just needs to be a bit more than your highest DC voltage. Any "extra" voltage just ends up being burned off as heat in the regulator anyway.

    Depending on the load, you'll gain voltage going from the AC voltage, measured as rms, to the DC voltage, which can be as high as the peak-peak minus the approx. 1.4 volts of diode drop at the rectifier.
     
    TheLaw likes this.
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Well, the author of that article was a bit overly optimistic on how much current they would be able to get out of that supply as pictured; the heat sink is just not very adequate, and the input capacitor is too small.

    If the output current were 5A and the input was 60Hz, there would be over 8v ripple on the 4700uF cap. Even at 3A, there would be 5v ripple. I'd use 22,000uF instead.

    Aluminum heat sinks are not as efficient at cooling as copper is; only about 59%.

    If you used a 2.5k pot with only a 12v input, about half the pot would not be usable.

    R1 is shown as the standard value; 120 Ohms. This causes a nominal 10.417mA current to flow through it, as Vref (the voltage measured at the OUT terminal respective to the ADJ terminal) is nominally 1.25v. 1.25v/120 Ohms ~= 10.417mA

    The 10mA number is important, as it's required in order for the regulator to provide guaranteed regulation. If there is less than 10mA current flowing from the OUT terminal, the output voltage may be significantly higher than desired.

    Vref sets the minimum output voltage that you can achieve with this circuit. As I mentioned previously, Vref is nominally 1.25v, but it may be anywhere from 1.2v to 1.3v and still be within the manufacturer's specification.

    There is also Iadj to consider. This is a small amount of current that flows out of the ADJ terminal; it can be as high as 100uA (0.1mA), but usually is more in the 50uA to 70uA range. With low values for R1 and R3, this doesn't come into play very much, because 60uA x 2.5k = 150mV. However, if you use larger values of resistance for R1 & R3, it makes much more of a difference.

    1k and 2k pots are easy to find; as they are standard values of resistance.

    Let's say you want your highest output voltage to be around 12v.

    12v-1.25v = 10.75v.
    10.75v/1000 Ohms = 10.75mA current required through R1.
    1.25v/10.75mA = 116.2 Ohms for R1.
    If you wanted to make certain that you would have up to 12v output, you could decrease R1 to 110 Ohms.
     
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  4. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    SgtWookie, you are the best. You've been very helpful in all of my endeavorers. I will have to reread your post and hopefully fully grasp what you are saying. Thank you to all the responses. There's a chance I will still have a question.

    Like....should the pot be somewhere around 5k instead?

    And, regarding power supplies and safety. A lot of other people on different forums and in tutorials, and in videos have always said that it could be dangerous. But what could really happen besides the circuit getting blown out or maybe a little bit of smoke?

    Thanks to everyone.
     
    Last edited: Oct 3, 2010
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Simple Ohm's Law shows that the current in the 120 ohm resistor is 10.4mA.
    The same current would try to be in a 5k pot so Ohm's Law again shows a voltage of 52V across a 5k pot which is impossible.
     
  6. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    Oh good point...
     
  7. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    What does he mean by only have of it would be "useable"?
     
  8. wayneh

    Expert

    Sep 9, 2010
    12,088
    3,027
    If your output is 12v, then you're dropping 1.25 across R1 and 10.75 across R3, which happens at R3=1,152 ohms. So at any higher ohms setting at R3, you're maxed out relative to your supply voltage. Half of the turning range of the pot makes no difference to Vout.
     
  9. TheLaw

    Thread Starter Member

    Sep 2, 2010
    228
    2
    Thanks. =D
     
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