Calculating combined feedback (feedback from multiple rails)

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  • Th3_uN1Qu3
    Believe in
    • Jul 2010
    • 6031
    • Romania

    #1

    Calculating combined feedback (feedback from multiple rails)

    First of all, if you didn't know it already: I suck at math. But i know basic electronics theory, and when something bugs me i take out the pen & paper and stay on it till i've figured it out. If only i'd be treating homework like this as well...

    Anyway. I've been wondering for a long while how the values of the resistors in the combined 5v/12v feedback in group regulated ATX supplies were selected. When only one rail is in the feedback loop, you have a simple voltage divider. When two or more rails are involved however, you can't just calculate two voltage dividers and stick them together: it won't give the expected results because currents are also involved.

    So, i decided to look into it a bit. And i quickly discovered it's nothing more than plain ole Ohm's Law, nothing fancy.

    Say we have a power supply with one 5v and one 12v rail, coming from the same transformer, and with the output inductor on a common core. Typical ATX PSU situation. We want to include both rails in the feedback loop. The voltage we need to present to the error amplifier is 2.5v.

    The first step is to fix R3, the bottom resistor in the feedback divider. Its value should be selected so that the current flowing thru it is at least 100 times the input bias current of the error amplifier. For the TL494 this is 1uA, so we would need at least 100uA flowing thru R3. I set R3 to 1k to keep things simple, i could have gone quite a bit higher in value - as a matter of fact, i could have used 10k. Anyway, moving on.

    Once you have set R3, you need to find out the current flowing thru it. If you selected it based on the current, you have that already, but chances are if you did that, you're gonna end up with an ugly value, so best pick a standard resistor first and then calculate the current. Apply Ohm's Law, I= U / R. 2.5 volts over 1kOhm is 2.5mA.

    Now, we need to fix R1 and R2. This is actually a lot simpler than it looks. Let's see. First, since both the current thru R1 and the current thru R2 will be going thru R3, it occurs that
    • I(R3) = I(R1) + I(R2).

    To keep things simple, i assumed that the input rails will be loaded equally, ie the feedback is split 50-50 between 5v and 12v. Thus,
    • I(R1) = I(R2) = I(R3) / 2 = 1.25mA.

    We know the currents. Now, we also need the voltages to calculate the resistance of R1 and R2. But look at it - we have those already! Since we want to have 2.5v at the junction of the feedback resistors, the voltage drop of the upper resistors must be the rail voltage minus 2.5v. So we have:
    • U(R1) = 5 - 2.5 = 2.5v,
    • R1 = U1 / I1 = 2.5v / 1.25mA = 2k
    • U(R2) = 12 - 2.5 = 9.5v
    • R2 = U2 / I2 = 9.5v / 1.25mA = 7k6.

    That's all folks! I hope that clears up some confusion that some of you might have had when looking at such a feedback divider.

    If for example, you wanted to have 60% of the power on 12v, and 40% on 5v, you would make I(R2) = 0.6*I(R3) and I(R1) = 0.4*I(R3). You can add more rails as well - the current thru the bottom resistor of the divider will always be the sum of the currents flowing thru the resistors coming from the rails, so once you've picked your ratios it's the same calculation.
    Attached Files
    Last edited by Th3_uN1Qu3; 09-14-2012, 04:23 PM. Reason: forgot the attachment... duh!
    Originally posted by PeteS in CA
    Remember that by the time consequences of a short-sighted decision are experienced, the idiot who made the bad decision may have already been promoted or moved on to a better job at another company.
    A working TV? How boring!
  • ben7
    Capaholic
    • Jan 2011
    • 4059
    • USA

    #2
    Re: Calculating combined feedback (feedback from multiple rails)

    Good job!

    And, LOL, I was thinking about this last night! xD Thinking of making a multiple output supply (1.5v, 3v, 6v, 12v, and 24v) for a model railroad (to power lights and things).

    -Ben
    Muh-soggy-knee

    Comment

    • Th3_uN1Qu3
      Believe in
      • Jul 2010
      • 6031
      • Romania

      #3
      Re: Calculating combined feedback (feedback from multiple rails)

      If you don't need too much power on those rails, i'd reg the 3v off the 6v using a LDO, and then take the 1.5v from the 3v with another one, having only 6, 12 and 24v on the main transformer. Else you're going to run into trouble, both with the voltage reference (that can be solved, you can divide the reference down) and with the turns ratio of the transformer - you can't have fractionary turns.
      Originally posted by PeteS in CA
      Remember that by the time consequences of a short-sighted decision are experienced, the idiot who made the bad decision may have already been promoted or moved on to a better job at another company.
      A working TV? How boring!

      Comment

      • ben7
        Capaholic
        • Jan 2011
        • 4059
        • USA

        #4
        Re: Calculating combined feedback (feedback from multiple rails)

        Originally posted by Th3_uN1Qu3
        If you don't need too much power on those rails, i'd reg the 3v off the 6v using a LDO, and then take the 1.5v from the 3v with another one, having only 6, 12 and 24v on the main transformer. Else you're going to run into trouble, both with the voltage reference (that can be solved, you can divide the reference down) and with the turns ratio of the transformer - you can't have fractionary turns.
        I'd say probably a max of 1.5A on each 1.5v, 3v, and 6v rail. 12v would have a max of 3A. 24v, a max of 500mA. The total output power would be about 75 watts. (Actually it would be less, the LDO regs will dissipate some of the power)
        Muh-soggy-knee

        Comment

        • Th3_uN1Qu3
          Believe in
          • Jul 2010
          • 6031
          • Romania

          #5
          Re: Calculating combined feedback (feedback from multiple rails)

          Now for a few real-world examples. The ATX power supply here has 40/60% 5v/12v feedback. If you calculate the power distribution, it's almost the same as the feedback ratio. Actually it's 44% 5v/56% 12v, but it's close enough.

          Whereas this one here has 56% feedback on 5v and 44% on 12v. No power rating is given on this schematic, except the total power of 200W - but this is clearly an older design.
          Originally posted by PeteS in CA
          Remember that by the time consequences of a short-sighted decision are experienced, the idiot who made the bad decision may have already been promoted or moved on to a better job at another company.
          A working TV? How boring!

          Comment

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