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LM338: Laboratory Power Supply design and Variable current adjust knob(potentiometer)???

Gurumurthy J
Prodigy 140 points
Part Number: LM338

Hi to all !!!

I want to design  laboratory Power Supply circuit(30V @ 5 ampere )using LM338. I should be able to control current from minimum to maximum Level.

I can put a potentiometer between output terminal and adjust pin , but the problem is it get too hot and burns. Hence i want design current adjustable circuit with out magic smoke because of potentiometer. help me by giving by various techniques with schematics.

  • 0
    TI__Guru 55940 points
    Hi Gurumurthy,

    For us to help you to improve your current design, it would be helpful to see the schematic that you are using. The reason for heat is going to be the amount of power being dissipated. Since Pd = V x I the way to reduce the heat is to reduce the voltage drop and/or current. You can spread the heat by using a series resistance; however, you will then need to account for the additional voltage drop.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Prodigy 140 points

    Dear Ryan,

              Here is the schematic taken from LM338 Data Sheet.

    LM338.doc

  • 0 in reply to Gurumurthy J
    TI__Guru 55940 points
    Hi Gurumurthy,

    Thank you for indicating which schematic from the datasheet that you are basing your design on. As you can see the schematic is literally made up of the LM338 and one external resistor by which the output current is set. As stated in my last post, you can spread the heat by adding series resistance; however, you would need to account for the additional voltage drop.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Prodigy 140 points
    Dear sir,
    it would give insight , if schematic is given. No interconnectdetails is present. So provide schematic and thanks a lot sir.
  • 0 in reply to Gurumurthy J
    TI__Guru 55940 points
    Hi Gurumurthy,

    This circuit uses the LM338 in unity configuration. As such the output voltage of the LM338 is Vref. The resistor is used to set the output current by Ohm's Law. The example schematic uses a potentiometer rather than a fixed resistor in order to indicate that the resistance is the variable to adjust the output current. For your application it is up to you to decide the best way to set the resistance in order to achieve the your desired output current. Since you want an adjustable output current, you will need to find a way to change the resistance.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Prodigy 140 points
    Dear Ryan,

    Words cannot explain content. Please give some schematic and explain....
  • 0 in reply to Gurumurthy J
    TI__Guru 55940 points
    Hi Gurumurthy,

    The circuit you are attempting to build is one linear regulator and a resistor. The schematic in the datasheet is all that is required to build the circuit. Any method to change the resistance (in order to adjust the current) will be specific to your application and is your responsibility. Please note that the principle that this circuit is built on is Ohm's Law. As stated in the datasheet the voltage is fixed at Vref for this design; therefore, any adjustment that you choose to make to the resistance will directly impact the output current.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Prodigy 140 points

    Thanks Ryan,

                          I can understand this ,but problem is i need put Potentiometer of high wattage for 5 ampere current.  which is much costly.

    example:  

    load current =5 ampere 

    voltage =1.25V

    power= Voltage *Load Current= 1.25*5=6.25Watts (This much wattage will cost more Than Regulator IC)

    I want to avoid this. Is there any other technique to build Current control adjust knob.