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Parallel Vs Serial Power supplys — Parallax Forums

Parallel Vs Serial Power supplys

BasilBasil Posts: 380
edited 2007-04-23 21:46 in Propeller 1
Hi All,

The device I am designing will need both 3.3v and 5v power supplies.

I will be using the LM2937IMP 3.3V and 5V versions.

Question, is it best to connect these regulators in Serial (I.E 9V->5V->3.3V) as per the schematics for the prop demo board, or would parallel (I.e. 9V->5V & 9V->3.3V) give a better supply?

I need the supply to have as little noise as possible and efficient.

I was told that having them in serial as per the demo board produces more electrical noise, but I thought i'd ask the pro's because i'm sure you wouldn't design a board with a 'bad' power supply!

EDIT: I should note that every so often, there will be a time when up to 10A or so will be drawn from the 9v battery for < 1/2 second...I'm trying to avoid brownouts, a big cap may be needed...

Thanks,

Alec

Post Edited (Basil) : 4/23/2007 1:20:15 AM GMT

Comments

  • Phil Pilgrim (PhiPi)Phil Pilgrim (PhiPi) Posts: 23,514
    edited 2007-04-23 01:29
    It depends on how much current you need for each. On the one hand, the current drawn from your 3.3V supply will be added to the current drawn from the 5V supply if they're arranged in sequence. Also, any significant noise generatd by the 5V load might be felt on the 3.3V output. On the other hand, if each supply is derived separately from the 9V input, your 3.3V regulator will be dropping 5.7V which, when multplied by the current, may amount to a significant power dissipation.

    Short answer? There is no short answer. Like so many things, you just have to balance the tradeoffs.

    -Phil
  • BasilBasil Posts: 380
    edited 2007-04-23 02:12
    Hi Phil,

    Thanks for that, I thought it was something like that smile.gif

    FTDI chip will be powered by the USB bus so no problem with that.

    The 3.3V will be supplying the prop, boot memory and dataflash memory. (Prop = varys, boot memory = 100na standby, dataflash = 17ma max)

    5V will be supplying 12-bit ADC, ADXL78 accelerometer and MPX4115A barometer. (ADC = 400ua, ADXL = 1.3ma, MPX = 10ma)

    9V directly powers 3 x Op-Amps (Rail to rail OP777) and also 3 pyro outputs, which will be up to 15A for less than 1/2 second. (OPamp = 300ua each)

    As you can see, none of these are particulary high (If I read the datahseets right!) smile.gif

    Does this change any of your thoughts?
  • Mike GreenMike Green Posts: 23,101
    edited 2007-04-23 03:02
    Basil,
    Clearly, with the small currents involved at 5V and 3.3V, you could do it either way. With the pyro currents so high, I would carefully filter the supply to the Op-Amps, probably run the regulators in series, and put as large a capacitor as you can afford (space-wise) across the 5V regulated output so that the 1/2 second when the pyros are activated won't drop your logic supply enough to force a reset or invalid data and won't mess up whatever is driven by the Op-Amps. You might use a diode in series with the 9V power supply with a large capacitor across the output to hold the Op-Amp supply while the pyros are essentially shorting it.
  • BasilBasil Posts: 380
    edited 2007-04-23 03:25
    Mike Green said...
    Basil,
    Clearly, with the small currents involved at 5V and 3.3V, you could do it either way. With the pyro currents so high, I would carefully filter the supply to the Op-Amps, probably run the regulators in series, and put as large a capacitor as you can afford (space-wise) across the 5V regulated output so that the 1/2 second when the pyros are activated won't drop your logic supply enough to force a reset or invalid data and won't mess up whatever is driven by the Op-Amps. You might use a diode in series with the 9V power supply with a large capacitor across the output to hold the Op-Amp supply while the pyros are essentially shorting it.

    Hi Mike,

    Thanks for that.
    The op-amps aren't a critical part of the design, they are not constantly in use.
    Every 0.5 seconds or so, they will be used to detect voltage drop across the ignitors (which is where the high current is going to). This will be used as a continuity test.

    When the FET's fire the ignitors and draw all that current, I will disable the op-amps. Once the FET's are 'off' again, I will perform the continuty test with the op-amps to detect if the electric match fired correctly.

    I have attached a basic diagram of the setup for clarification.

    Would I still need the diode and cap?

    EDIT: You can ignore the resistor values, they have been changed since this was done smile.gifHi Mike,
    pyro6-1.jpg
    586 x 737 - 110K
  • Paul BakerPaul Baker Posts: 6,351
    edited 2007-04-23 03:42
    Basil said...
    Would I still need the diode and cap?

    Yes because you are planning on running the Propeller off the 9V battery too (via the regulator). 9V batteries have a high internal resistance due to the fact they are actually 6 AAAA batteries wired in series (yes quadruple A battery), when they have a high current draw that internal resistance will cause a significat voltage drop to occur and very well could brown out the Propeller causing it to reset.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Paul Baker
    Propeller Applications Engineer

    Parallax, Inc.
  • BasilBasil Posts: 380
    edited 2007-04-23 03:55
    Oh ok, so I take it the 10uf caps across each of the regulators won't be enough? Do I need caps in addition to these, or just make these ones bigger?
  • Paul BakerPaul Baker Posts: 6,351
    edited 2007-04-23 04:29
    I would start off by using 2 9V batteries, one for the electronics, another for the e-matches with the negative terminals of the two tied together. After you are done with development, if you want to try to get it all working on one battery then you can tackle that, but at least by then you know your system works in general.

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Paul Baker
    Propeller Applications Engineer

    Parallax, Inc.
  • Nick MuellerNick Mueller Posts: 815
    edited 2007-04-23 13:14
    Basil,

    You need caps **AND** a diode in front of them. Or the current drawn from your e-gnitors will discharge them.

    You could also calculate the expected voltage drop if you know the impedance of your battery and the resistance of your e-gnitors.

    Nick

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    Never use force, just go for a bigger hammer!

    The DIY Digital-Readout for mills, lathes etc.:
    YADRO
  • BasilBasil Posts: 380
    edited 2007-04-23 21:17
    Hmmm, for the cap I think ill use 1000uf. For the diode, are we talking a small zener between the +ve battery terminal and Vin on the regulator(s)?

    I will run 2x 9V in parallel for testing to be safe.

    I will design the big cap into the design which should cover everything. Ive been looking around and most other similar devices use a cap and work fine. There are even some that dont use a cap at all, but they seem to only fire 1.25A
  • Mike GreenMike Green Posts: 23,101
    edited 2007-04-23 21:22
    The diode is in series with the +9V supply (anode to the supply, cathode to the + lead of the capacitor). It can be something like a 1N4001 or a Shottky diode with a 30-50V x 1A rating, not a Zener diode. The diode prevents the capacitor from discharging through the power supply if it drops below the voltage on the capacitor.
  • BasilBasil Posts: 380
    edited 2007-04-23 21:46
    Ahhhh now I understand, will add that to my schematic too [noparse]:)[/noparse]

    I will put my 3.3V and 5V regulators in parallel to help reduce noise on the power rails. The extra power lost when reducing from 9V to the 3.3V supply over 5V to 3.3V is not as important as clean power in this case.


    Thanks guys.
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