Need some help with a project

Discussion in 'The Projects Forum' started by pityocamptes, Jul 25, 2012.

  1. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
    Been a while since I delved into electronics, but trying my hand at it again. I have a project, simple in nature, and was hoping someone could help me, with regards to the circuit/schematic. I am looking at making a regulated voltage supply. The power supply at max will be about 5 volts at 3 amps. Basically, what I need is a regulated power supply at a fixed 2.5 volts and whatever amps the initial power supply is providing. I only need to ensure that the output voltage stays at 2.5 volts no matter what the initial voltage is and the amps are not changed. If anyone could help me out on something like this I would greatly appreciate it. Thanks!
     
  2. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    through hole or SMT?
    Whats the minimum input voltage?
     
  3. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
    Through hole probably for now. Min input voltage I would say ~3.5 (just to be safe), max should be at 5v - 5.5v. Thanks!
     
  4. mcasale

    Member

    Jul 18, 2011
    210
    12
    So, the input to your regulator is the 3.5v to 5.5v?
    The output is a fixed 2.5v at a maximum load of 3 amps?

    Can't you use a regulator IC that lets you adjust the output to whatever you want?
    Or do you need to design something from scratch?

    Be sure to check the maximum power dissipation in your regulator. It could be as high as (3v)(3a)=9Watts. That's a lot. You'll need a heat sink.
     
  5. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
    If anyone has a schematic that would be great. Looking for something that may not have a high dissipation, if possible. Thanks again!
     
  6. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
  7. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
    Anyone? Thanks.
     
  8. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    You just need a LDO (low drop out) voltage regulator that is capable of 3.5-5.5V in and 2.5V fixed output. (and up to 3A of current..do you really need that much?)
    I've been looking for one for you that was through hole but am not having any luck finding one that is actually in stock.
     
  9. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
  10. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
  11. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    I believe all you need is a 1uF cap on the input to ground and a 47uF cap on the output to ground.. If you buy the fixed 2.5V one the resistors are internal and not required.
     
  12. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0

    OK, so you are saying I do not need the resistor on the schematic (on page two of the pdf) between legs 4&5? Thanks. Also, what exactly is PG? I see the Vo but not sure what is meant by the PG (coming off leg 5).
     
  13. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
  14. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
    Hey thanks. After looking through the net, and some previous PS on previous projects, way in the past, it looks like I used the LM150. However, given my amperage load I have looked at the LM338T(8Amps) or LM350(3Amps). For 2.5 volt output I will need 240ohm resisitors for R1 and R2. My question is, since my input/output voltage is a small differential, what wattage do I need? I calculated at least 3.75 watts. Is this correct? So I guess I could go with a 5 watt on R1 and R2. Now if I use a potentiometer for R2 to dial in exact voltage due to 5-10%+- resistance differential for carbon resistors, do I have to find a 5 watt pot for R2? Thanks again. Here is the schematic for the LM350 I was looking at on page 8.

    http://www.ti.com/lit/ds/symlink/lm350-n.pdf
     
  15. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    No..You need a LDO regulator... The LM350 needs a Vin of Vout+3V
     
  16. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0
    Hey, thanks again! i really appreciate your help. I will stick with the LDO. Also as far as a heat sink, I was having a friend help me out, and they indicated I should have a heat sink between 3-4 C/W - which is pretty big. I just wanted to make sure that that sounded right to you? Thanks again!
     
  17. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    No.. (well maybe depending on your environment,etc..) here is the basic math
    Assuming 5.5Vin max (and 3A load)
    (5.5-2.5)*3A=9W (So it will be dissipating 9W max)
    Rjc=2 degC/W (junction to case)
    Rcs=1degC/W (thermal grease resistance)
    Rsa is what we need to figure out (heat sink to air/ambient)

    if Tjmax is 125deg C (max junction temp)
    and if ambient is 25deg C
    that means we have 100deg headroom
    now
    100/9W=11.1 (total Rja)
    So 11-2-1=8 (where 2 is the Rjc and 1 is Rcs)
    So in effect you need a heatsink of 8 degC/W or better.
     
  18. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0

    Pretty hot where I am at. Ambient, during those max temp times of the year, would be around 43C.
     
  19. pityocamptes

    Thread Starter Member

    Jul 25, 2012
    75
    0

    Also, just to clarify, when you mean "better" that would mean lower, as in <=8? Could you explain in laymans terms what exactly is meant by the 8degC/W? Does that mean that the heatsink dissipates 8degC for every watt? If so, then why are higher C/W rated heatsinks smaller, ie 26degC/W? Or is this kind of like metal gauges, with the lower number being thicker, ie better? Thanks again!
     
  20. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,769
    969
    Yes less than 8.

    Basically when you see 10deg C/W it means the part will heat up 10degC for every watt it dissipates.
    So for example a chip with no heat sink with a Rja (junction to ambient) of 50degC/W (and assuming a 25 deg ambient) will be running at 75 deg C if its dissipating 1 watt or 125 deg C if its dissipating 2 watts.
     
Loading...