Voltage regulator question

Discussion in 'General Electronics Chat' started by Bryan47, Mar 4, 2011.

  1. Bryan47

    Thread Starter New Member

    Dec 1, 2010
    I need to order some voltage regulators and I will probably get them from Jameco. I will need them to handle about 2A current. Here are the 3 they have for 7812 https://www.jameco.com/webapp/wcs/s...gId=10001&freeText=7812&search_type=jamecoall

    The TO220 package ones say 2.2A but the data sheet only says they are 1A and I've always seen them to be 1A so I don't see why it says 2.2A. On the other hand the TO-3 package one says 1A?

    I'm confused as to what regulator to order. Thanks a bunch.
  2. Audioguru

    New Member

    Dec 20, 2007
    Why not read about the details on the datasheet?
    A 7812 is guaranteed to work well with an output current of 1.0A if you keep it cool enough. But some of them will have an output current of 2.2A but it might not regulate very well at such a high current.
    You can't order the "good" ones and they might not have any.
    Jameco should not lie about spec's.
  3. tracecom

    AAC Fanatic!

    Apr 16, 2010
    Datasheets can be confusing as well as contradictory. There are so many numbers, charts, and graphs that it is hard to be sure what they all mean. This problem is compounded when the datasheet listed is from a manufacturer that is different from the part that you get. (You should be aware that Jameco is not an authorized distributor for National Semiconductor parts even though they offer some NS parts for sale.)

    If you are going to be trying to squeeze maximum power from the LM7812, your best bet is to look at the datasheets from the manufacturers (NS, Fairchild, ST Micro, e.g.) to see if any of them have any real advantage(s) over the others. You can contact the manufacturer with specific questions although it may take a few days to get answers. Then, order the exact part you want from an authorized distributor.

    ETA: Or, you can do what many of us do. Order one, build it up, test it, and then you'll really know.
    Last edited: Mar 4, 2011
  4. JMac3108

    Active Member

    Aug 16, 2010

    Current ratings on linear voltage regulators are max limits that the part will put out before it shuts down. This does not mean that in your application you can get the maximum out of it. You have to look at your input and output voltages, load current, and the thermal resistance to make sure that the part will not get too hot. Let me give a quick example with made-up numbers ... you can follow this procedure for your design.

    Input Voltage = 12V
    Output voltage = 5V
    Load current = 200mA
    Thermal resistance of TO-220 package = 50 degrees C per Watt (from datasheet)
    Maximum die temperature = 125C (from datasheet)

    Calculate voltage drop accross the regulator = Vin-Vout = 12V-5V=7V
    Calculate power dissipation in reglator = VI = 7V*500mA=3.5w
    Calculate temperature rise = 3.5W * 50 C/W = 175C
    Calculate device temperature = ambient temp + temp rise = 25C + 175C = 200C

    Your part will get to 200C which is well above the rated 125C and will either shut down or burn up depending on if your particular device has thermal protection. This tells you that you will need a way to cool the part. In this case it means a heatsink and fan.

    Hope this helps.
  5. Jaguarjoe

    Active Member

    Apr 7, 2010
    The LM338K is a 5 amp adjustable 3 pin regulator. In the TO-3 (K) package, its thermal resistance (J to C) is only 1 deg C per watt. This is about the easiest regulator to keep cool. 5 amps gives you a nice cushion for the 2.5 you need. It requires 2 small resistors to operate at any given voltage from 1.25 to 37. With a variable resistor in one leg it becomes a variable output regulator.

  6. JMac3108

    Active Member

    Aug 16, 2010
    You have to use the Junction-to-ambient number, not the junction-to-case number to perform your calculation. For the LM338K the J-A thermal resistance is 35C/W. You perform this calculation with J-A to see if you need a heatsink and external cooling in your application. I
  7. Jaguarjoe

    Active Member

    Apr 7, 2010
    Right you are!