Voltage Regulator

Discussion in 'General Electronics Chat' started by roberto.drago, Apr 20, 2010.

  1. roberto.drago

    Thread Starter New Member

    Apr 20, 2010
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    Hi,

    We would like to convert a 9V voltage to a 5V voltage using a voltage regulator as discussed in http://forum.allaboutcircuits.com/showthread.php?t=25761. Yet out concern is that we need about 2A of current. Could the regulators be simply connected in parallel? or should there be addition circuitry in the case that the regulators do not balance the current amongst themselves.

    Thanks,
    Roberto
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Are you considering using a 9v "transistor" battery as your source of power?

    If so, look for a different kind of battery. A 2A load on a 9v battery will kill it within a few minutes, and it's likely that the battery will explode.
     
  3. roberto.drago

    Thread Starter New Member

    Apr 20, 2010
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    Hi,

    Thanks for you reply. No actually we will be using a power supply which can handle this current.

    Thanks again,
    Roberto
     
  4. ELECTRONERD

    Senior Member

    May 26, 2009
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    kingdano likes this.
  5. roberto.drago

    Thread Starter New Member

    Apr 20, 2010
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    Hi,

    This is what we need!! Thanks a lot. Just one last question, what is the purpose of the NPN transistor and TTL logic base input shown on the datasheet? (see the example circuit : 5V Logic Regulator with Electronic Shutdown)

    Thanks,
    Roberto
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    There are a lot of applications schematics in the datasheet.

    The transistor shown causes the output to go to almost zero volts.

    You will need a large heat sink for the regulator, as 4/9ths of the total power, or 8 Watts will be dissipated in the regulator itself.
     
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  7. kingdano

    Member

    Apr 14, 2010
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    This. Definitely.

    Please go through the calculations in the datasheet to properly size your heatsink.
     
  8. roberto.drago

    Thread Starter New Member

    Apr 20, 2010
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    So should I leave the base TTL connector floating (or connected to ground) if I want a constant 5V output?
     
  9. ELECTRONERD

    Senior Member

    May 26, 2009
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    That's for a logic regulator, if you want it. To simply get 5V out, you use the equation:

    V_O_U_T = V_R_E_F(1 + \frac{R_2}{R_1}) + I_A_D_JR_2

    So since you have a 9V input, and you want a 5V output, we can find out the R2 resistor value:

    Parameters:

    • V_O_U_T = 5V
    • V_I_N = 9V
    • V_R_E_F = 1.25V
    • I_A_D_J = 50\mu A
    • R_1 = 120\Omega
    Substitute and solve Algebraically:

    5V = 1.25V(1 + \frac{R_2}{120\Omega}) + 50\mu AR_2

    Now we may conclude that R2 = 358.3Ω ≈ 360Ω

    Your circuit schematic for the voltage regulator is shown on page 5.

    Austin
     
    Last edited: Apr 20, 2010
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  10. kingdano

    Member

    Apr 14, 2010
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    that's well done.


    *applause*
     
  11. roberto.drago

    Thread Starter New Member

    Apr 20, 2010
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    Hi

    We found out that our power supply gives 18V, what values change in order to get the 5V from the same LM338?

    Thanks,
    Roberto
     
  12. kingdano

    Member

    Apr 14, 2010
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    You have seen the equations done - no one here is interested in doing your work for you.

    I think the point of the forums is to help you figure it out, not to do it for you.
     
  13. roberto.drago

    Thread Starter New Member

    Apr 20, 2010
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    Thanks for all your help kingdano. Appreciate it, yet we are still students and learning electronics.

    The thing is that the input voltage does not seem to be related to the output voltage. With our given supply, we are not getting the required output voltage.
     
  14. ELECTRONERD

    Senior Member

    May 26, 2009
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    According to the equation It isn't. It has to do with the voltage divider resistors. As long as you have a voltage supply within the constraint limits, and 18V does, you'll be fine. The regulator will have a voltage drop across it's input and output in order to power itself (Vref = 1.25V).

    Austin
     
  15. Jaguarjoe

    Active Member

    Apr 7, 2010
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    With an 18 volt supply and a 5 volt output, the regulator will be dissipating 26 watts. I don't think it will survive.
     
  16. kingdano

    Member

    Apr 14, 2010
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    this is why you should go through the datasheet equations for calculating an appropriate heatsink for your application.

    i have recently used an LM317T to regulate 36V down to 8V and found a heatsink which dissipated enough heat to satisfy the device.
     
  17. ELECTRONERD

    Senior Member

    May 26, 2009
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    With an adequate heatsink and probably using the TO-3 package, the LM338 should work with 18V. Also keep in mind that this regulator is capable of 5A, while the op is only using 2A.

    Austin
     
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  18. ifixit

    Distinguished Member

    Nov 20, 2008
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    Hi,

    Alternately, you could put a power resistor in series with the regulator to dissapate some of the un-wanted heat in the resistor instead of the regulator.

    Can you calculate what value and power rating you might need?

    Regards,
    Ifixit
     
  19. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
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    Thats right. The power has got to go away somewhere.. It doesn't all have to be in the regulator. You can continually drop before the regulator as long as you are still in range for proper regulator operation.
     
  20. ELECTRONERD

    Senior Member

    May 26, 2009
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    Indeed, that would work. He could have power dissipated through a resistor or a series of resistors in which they create an overall voltage drop. This in turn will make the LM338 voltage drop very low.

    Excellent Idea! ;)

    Austin
     
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