multiple power rails

Discussion in 'General Electronics Chat' started by m121212, Jul 8, 2012.

  1. m121212

    Thread Starter Active Member

    Jul 24, 2011
    I'm designing the power bussing for a board that needs +20VDC, +5VDC, and -5VDC, given an input supply of +28VDC. This is a logic board that doesn't need much current, except to drive some LEDs (30mA), a 555, a reed relay (25mA), and some lightweight bjts, all on the +5V rail. The +20V and -5V are used for driving the gate of a MOSFET. The draw on those is generally zero milliamps, except for the rare 1 microsecond long pulse of 300-400 milliamps.

    For various reasons, including footprint and complexity, I am avoiding the use of switching regulation for this iteration, and will stick to using linear regulators. I think there may also be an efficiency advantage because the current draw is low.

    I have these questions:

    1) Is it a bad idea to cascade the regulators? My thinking is that the +5V regulator will be more efficient with a +20 V input, than off the +28 V supply. Also, the -5V might do better by using the +5V as the input.

    2) If there is no current draw on the output of the linear regulators, do they dissipate much? I was thinking of the 7805, 7820 and 7905, but would be glad for alternate recommendations!
  2. evilclem

    Active Member

    Dec 20, 2011
    1) No problem here that I know of. But be weary of the total current draw on the first regulator to ensure adequate headsinking.

    2) No, in theory they would dissipate nothing with no current draw; In practice, it's next to nothing.
  3. bertus


    Apr 5, 2008

    If there is no cureent drawn from the regulators, the used current will be only the quiescent current needed for the regulator itself.
    This will be not much, as you will read in the datasheet.
    This will be between 4 and 8 mA dependend on the brand and type (L, M or normal).

  4. takao21203

    AAC Fanatic!

    Apr 28, 2012
    Switchers like the MC34063 or LM2576 are very easy to use. For small currents they can work even with small RF coils if need be.

    I use linear regulators, yes, in cases where it is possible not to add capacitors, or only very small one's. If you need capacitors anyway, there is not much extra cost/complexity if you use a switcher IC.

    LM7805 makes a good voltage reference, after all.

    On the picture for instance SOIC MC34063 + large coil on top of adapter PCB. Fairly overdimensionated, but the very small SMD ferrite coil I tried caused bad efficiency (too small). Caps are just cheap 100uF. The diode is soldered directly on the adapter PCB between the pins.

    Using LM317 for 3.3V (from 12V) at nearly 100mA would be bad, in terms of cooling. And I want to stick to 12V electronic transformers only.

    If you can live with some switching noise, what many circuits actually can tolerate easily, switcher circuits can be built very simple, small and cheap.
  5. crutschow


    Mar 14, 2008
    You can cascade linear regulators to distribute the power dissipation, but the overall efficiency is unchanged. Although the dissipation of a particular regulator would be reduced, the dissipation saved on one is just transferred to a regulator further up the chain.

    But you can't use the +5V (or any positive voltage) as a source for a linear -5V regulator. A linear regulator can't change voltage polarity. You need a minus voltage source or else you must use an inverting switching regulator.
  6. m121212

    Thread Starter Active Member

    Jul 24, 2011
    Thanks for all the feedback! I didn't realize the point about the inverting switcher.

    My feeling is that I should use a linear reg for the +20, and a switcher for +5, and another switcher for -5. Does anyone know of a suitable switcher for the -5, or better yet a dual +/- 5?
  7. m121212

    Thread Starter Active Member

    Jul 24, 2011
    Looks like the MC34063 will invert as well...
  8. Rbeckett


    Sep 3, 2010
    I have an Elenco bench top PS model 750-K. It is fully adjustable and has a great diagram of how it works so that you could fab one from scratch if needed or desired. That along with a 24 VDC dual output PS and I have 6-12 AC olts, 0-20 DC volts in + or -, and a 5 VDC 1 Amp fixed voltage tap. The parts are cheap and it can be asssembled very quickly and includes a project box and nice front panel. IIRC the kit was 49 bucks and it took me about 2 hours to assemble. Just an alternative to designing and building one of your own and letting the magic smoke out of the components. You can download a PDF with all the info from Sparkfun and then order all of the parts or just get the complete kit and call it good.
    Wheelchair Bob
  9. m121212

    Thread Starter Active Member

    Jul 24, 2011
    Thanks for the recommendation - I want to put this all on one board, hopefully it will be a compact design.

    I took the previous poster's recommendations and tried building two switchers using the mc34063. The inverter, which turns -5 to 5, works like a charm. The step down is being a giant PitA!

    What's happening is that it seems to switch to the correct output voltage, for some time, between 5 and 10 seconds, then the output drives up to the input rail (+28V). I inspected all the components, and the only thing that seems to fix it is swapping out the converter chip. The new one works, but then dies again after a bit.

    I thought it might be that the input voltage is too high, but the spec sheet says +40 V is ok. t_on/t_off is 0.25, switching frequency should be about 100khz.

    I was calculating everything by hand, but I found a nice calculator online that does the same trick:

    Values I used were roughly
    Vin = 28.00 ±.1% V
    Vout = 5.00 V
    Iout = 0.20 A
    Vripple = 0.10 V

    Vin min = 27.97 V
    Ton / Toff = 0.24
    Ton + Toff = 10.00 µs (micro seconds)
    Toff = 8.06 µs (micro seconds)
    Ton = 1.94 µs (micro seconds)
    Ct = 77.50 pF (I used 100p)
    Ipk = 0.40 A
    Rsc = 0.75 Ω
    Lmin = 105.94 µH (I used 100u)
    Co = 5.00 µF (I used 6.8u)
    R1 = 10.00 kΩ
    R2 = ((Vout - 1.25) / 1.25) * R1 = 30.00 kΩ
  10. takao21203

    AAC Fanatic!

    Apr 28, 2012
    Don't rely on the calculator.

    I would recommend 270uH, and 470pF, then also try 220pF, 100pF and 50pF.
    You need a suitable diode, some are only good for 20V.
    The loss of regulation hints that the diode is not connected or does not work correctly. I have seen it sometimes.

    For low currents you can use 1n4148 or 2 or 3 of them in parallel. 1n4001 etc. will not work! The pin1 + 3 pins on the other side you can simply connect all together.

    Also instead of fixed resistors, use 20K adjustable resistor.

    Input bypass cap should be 100uF at least.
    Output cap. should be 100uF to 470uF, depends on current.

    Output must be loaded with some mA, for instance a LED.