Connect with us

Basic Qs on LM3886 power supply design

Discussion in 'Electronic Design' started by Mahendra Reddy, Jul 21, 2003.

Scroll to continue with content
  1. I have some bsic questions in designing power supply for LM3886
    amplifier for 4ohm speakers. Please help me in designing this. Not
    only for this requirement but in general I need to understand these
    things. Sorry for asking so many basic questions.

    LM3886 datasheet page 16 says it requires +/-21V at 4.5A for 40W into
    4ohm load.

    1.) Does it mean it dissipates 42V(+/-21V dual supply) * 4.5A = 189
    watts? i.e Pd = V * I

    2.) To calculate power dissipation in the above equation is it correct
    to consider +/-21V(dual supply) as 42V? If I need to consider 21V in P
    = V*I then what should be the voltage if I use 42V single supply
    configuration of LM3886? Power disspation and current requirement
    should be more or less same in +/-21V dual supply and 42V single
    supply. Am I correct?

    3.) Is 4.5A at +/-21V equal to 2.25A on +21V supply and 2.25A on -21V
    supply? If I use ammeter to measure the current drawn by the amplifier
    I see some amps drawn on +V supply and some amps drawn on -V supply.
    So is it sufficient to use a 18V-0V-18V transformer supplying 2.5 amps
    to get approx +/-21V and 4.5amps(2.25 at +21V and another 2.25 at
    -21V) required for 40W into 4 ohms in dual supply configuration?

    4.)What is continuous output power and what is Instantaneous peak
    output power of an amplifier?

    5.)What if I give required voltage but the trnasformer delivering less
    current(for example 1A against the required 4.5amps) and turn up the
    volume to maximum? Will it distort the signal? or simply doesn't sound

    Thanks in advance
    Mahendra Reddy
  2. CFoley1064

    CFoley1064 Guest

  3. R.Legg

    R.Legg Guest

    4.47 Apk from the supply into the load to produce 40W RMS.
    3.16A/2 = 1.58Arms from each rail. ' Irms = Ipk x (D/2)^.5 '

    18V peak is required into the load, so they are telling you that the
    output saturation voltage of the amplifier is 3V @ 4.5A.
    1.58A x 21V x 2rails
    66.4W total from the supply
    The amplifier dissipates 66.4 - 40 = 26.4W
    1.58A x 42V x 1rail = 66.4W
    66.4W total from the supply
    If the supply is not current limited below 4.5A and filter caps are
    large enough, (assume 10% droop), 18-0-18V transformer will produce
    roughly 27V average.

    This will increase power consumption to
    1.58A x 27V x 2rails = 86W
    For the same 40W into 4R output.

    Power dissipation in the amplifier would be 86 - 40 = 46W
    40Wrms output, as you are requesting, and 4.5^2 x 4R = 80W peak
    LM3886 is thermally limited below around 90W output, using the
    specified supply voltages, according to the data sheet.

    The transformer will not current limit at it's rating, it will simply
    overheat if overloaded, unless the LM3886 overheats first and limits
    at a lower output power level.

    Full power output is not the worst case power dissipation condition
    for a linear amplifier.

Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day