Calculating Current from DC to AC

Discussion in 'General Electronics Chat' started by AutoNub, Oct 31, 2011.

  1. AutoNub

    AutoNub Thread Starter Member

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    Hello,

    We have a 115VAC supply going to two 24DC supplies. One 24DC supply is 1A and the other is 2.1A. How do I figure out the total alternating current going from the 115VAC supply to the two parallel connected 24DC supplies? It can't be as simple as just adding the currents of the DC supplies (1+2.1=3.1A), or can it? Is there a square root of 3 calculation in there somewhere?

    I understand Kirchhoff's current law states that current in has to equal current out, but I've never actually done a hand calculation from DC to AC before for calculating current. Thanks in advance for an clarifications!
    Last edited: Oct 31, 2011
  2. ErnieM

    ErnieM AAC Fanatic!

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    Perhaps the best way to approach is to do a power calculation, where the power in is the power out plus losses.

    Let's guess supply 1 runs 24V @ 1A out with an 80% efficiency, that is:

    P = (24*1) / 0.8 = 30W

    And supply 2 runs 24V @ 2.1A out with an 75% efficiency, that is:

    P = (24*2.1) / 0.75 = 67.2W


    So the total power in is 30W + 67.2W = 97.2 W

    The current is then just power over voltage:

    97.2W / 115V = .85 A

    That's a rough order of magnitude calculation for an average current. Power supplies, especially linear output stages, are notorious non-linear, requiring high current spikes for short periods of time.

    Kirchhoff's current law does apply, but it yields nothing useful to finding the current. A power supply has 4 leads, 2 for AC, 2 for DC. Each pair of leads has the same current going in as going out, so they all sum to zero.
    AutoNub likes this.
  3. AutoNub

    AutoNub Thread Starter Member

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    So if I'm rating the required minimum source current for the system in question, is there a general percentage used as a rule of thumb for adding to the calculated currents to account for these required spikes? I understand the inrush current only lasts for milliseconds, but if the source is literally incapable of delivering that high current for any duration of time, would this be a problem?

    Forgive me for the questions. I'm only a recent graduate with very little real world experience. I deeply appreciate you sharing your knowledge. Thank you.
  4. bountyhunter

    bountyhunter Well-Known Member

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    Actually, you forgot power factor. For a single phase 115VAC design, it's usually about 0.5 - 0.6. That means the RMS current flowing into the AC lines will be close to double what you calculate based on power calculations alone.
  5. #12

    #12 AAC Fanatic!

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    Linear or switching power supplies start by filling a capacitor as a reservoir. As long as the output power is less than the supply limitations, it will eventually get there. Maybe a dozen cycles or a 5th of a second. It depends on things you haven't told yet.

    When the supply fails to have enough current, the tops of the sine waves get clipped off. A linear regulator will simply deliver lower (and very noisy) voltage until the supply catches up. A switching regulator might get irritated and go into self-protect because it is struggling to throw maximum amps which are not available.

    That's as far as I can go with what I have to work with.
  6. Adjuster

    Adjuster Well-Known Member

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    Many power sources, such as batteries or transformers may supply a moderate short-term current excess with just a slight fall in voltage, provided that it's not too excessive or too prolonged. Take things too far though, and something might overheat, or at least a fuse or breaker might blow.

    In a simple situation, turning the system on in sections rather than all at once may be enough to mitigate the surge. Devices such as larger motors are routinely provided with current-limiting starting systems for their own benefit as much as that of the supply. Similarly, regulated power supplies especially switched - mode types may be made to "soft start", so that the inrush is less.

    Some specialised power sources like solar batteries and (very rare) series fed supplies have inherent current limits: try to take too much current from these and their output voltage will collapse. This is an especially delicate situation where some types of regulated SMPS are used, as they can have a negative input voltage/input current slope. This can lead to a regenerative collapse of the input voltage unless things are arranged properly.

    If you are in such a situation, it is necessary to manage inrush currents carefully. Starting up in a programmed sequence and the use of "soft-start" systems may be sufficient, but extra reservoir capacitors or even storage batteries may also be required, especially if previously un-powered devices may be suddenly connected to the system.
  7. ErnieM

    ErnieM AAC Fanatic!

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    The inrush starting current is larger, but I mean during normal operation an old school transformer/bridge/cap input scheme draws a very non-linear non-sinusoidal set of current slugs. A better more expensive supply should smooth these out some.

    Perhaps you should let us know how you are getting this AC voltage. Sometimes the source can be a problem, I have some friends who claim to have blown out their microwave ovens when they tried to run them off a DC battery to AC inverter. (Or was it the inverter that failed? Not 100% sure but I think it was the oven).

    Hey, it's what we do. ;)
  8. AutoNub

    AutoNub Thread Starter Member

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    Can anyone confirm this assessment? Did the efficiencies (0.8 and 0.75) already account for the power factors or do we need to add to our calculations? I believe efficiency and power factor are related (efficiency might be determined by using power factor), but are the two terms interchangeable in our calculations?

    In college, we used the simple efficiency formula η=(Pout/Pin)*100%. The only time we ever got more detailed in calculating efficiency was in reference to generators and motors (electrical or copper losses, brush losses, core losses, mechanical losses, and stray load losses). In these cases we'd represent our losses by drawing a power-flow diagram. In contrast, PF is simply cos(θ)=P/S (refering back to the power triangle, [(real power)/(apparent power)]). Unfortunately, the data sheets corresponding to the power supplies in question don't appear to provide enough information.

    The 115VAC single phase supply is coming from large ships. This is a marine application. EMIs are not an issue (everything is very thoroughly shielded and extensively tested). I can't be more specific due to confidentiality issues. The DC power modules described are Acopian power supplies. I have data sheets which show efficiencies, but don't see power factors. I could contact the manufacturer if necessary.
    Last edited: Nov 3, 2011
  9. Adjuster

    Adjuster Well-Known Member

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    Power Factor is the ratio of real power (W) to apparent power (VA) in a circuit. It is a separate issue to efficiency, and only really affects it in the sense that a low power factor increases the losses in the distribution system relative to the amount of actual power delivered.

    http://en.wikipedia.org/wiki/Power_factor
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