DC AMPS

JESSICAJESSICA Solar Expert Posts: 289 ✭✭
Hi

Is there a formula to convert AC amps to DC amps?

In other words: If my AC load at a given moment is 20 amps, how many DC amps am I drawing from my battery (24 volts)?

Thanks.

Comments

  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: DC AMPS

    How accurate do you want to be?

    Basically 'X' Amps @ 120 VAC = 5'X' Amps @ 24 VDC.

    But (and it's a big one) in reality there is power consumed by the inverter, lost in the wires, and the actual Amperage will go up as the batteries are drained down in order to maintain the same Wattage output at the AC load.

    Metering the DC supply line is the only way to get an accurate measurement at any given time.
  • waynefromnscanadawaynefromnscanada Solar Expert Posts: 3,009 ✭✭✭✭
    Re: DC AMPS

    Amps times Volts = watts.
    10 Amps times 12 volts = 120 watts.
    5 Amps times 24 Volts = 120 Watts.
    1 Amp times 120 Volts = 120 watts.
    0.5 Amp times 240 Volts = 120 watts.
    (AC or DC doesn't matter)
    Figure out the wattage of whatever load you have in mind, then divide by the volts you're interested in. Of course if your load will be supplied by an inverter, it's a good idea to add about 15% for inverter losses / inefficiencies.
    Hope this helps.
  • BB.BB. Super Moderators, Administrators Posts: 31,090 admin
    Re: DC AMPS

    There is the Power=Volts*Current= V^2 / R = I^2 * R relationships...

    However, especially for AC circuits there is the whole complex math of phase and waveforms between Sine Wave Voltage and (ideally) in-phase Sine Wave current. Vs the real world of out of phase current and non-sine wave current (~zero amps with a very sharp current spike near the crest of the voltage sine wave, etc.)

    For motors, the current tends to "lag" the voltage wave form. You can do the math and find out that it is the Cos (current phase angle) to see how much "work" (or watts is being done).

    For Motors, the Cos (or Power Factor) can be as low as 0.6 (only 60% of the current does real work). For CFL bulbs, the Power Factor can be 0.5 or even less.

    The problem is that you have to design the AC wiring to carry the current, and the Inverters+Generators have to be large enough to supply that excessive current... For example, say you have 5x 13 watt CFL bulbs and a 120 watt old refrigerator. The current flow is probably not:
    • P=V*I; I=P/V
    • I = P/V = 5x 13 watts / 120 VAC = 0.54 amps of AC current
    The real current is:
    • Power = V * I * Power Factor
    • I = P / (V * PF)
    • I = 5x 13 watts / (120 volts * 0.50 PF) = 1.08 Amps of AC current
    So, your Inverters and Generators have to be 2x as large if you have lots of loads with poor Power Factor (mainly, motors, small electronics, many computers, etc.).

    Of course, your inverters and generators have to be large enough to start something like a refrigerator too. A standard full sized Energy Star Refrigerator which may have a 120 watt motor, may also run a 500+ watt set of heaters (defrost/ice maker) and surge current--Requiring around a 1,500 watt minimum inverter/generator any way.

    For the DC side, we have other issues to worry about... For example a 12 volt battery system will have between 10.5 to 15.5 volts... That makes a large difference in the power that an AC inverter will draw.

    Say we have that 1,500 watt inverter (hopefully True Sine Wave) to run that refrigerator and a few CFL bulbs. For a 12 volt battery bank, 10.5 volt inverter cutoff, and 85% efficient--Plus we need to size the wiring / fuses / breakers x1.25 larger that max continuous current for NEC (and to prevent fuses/breaker popping with starting loads):
    • Current = P/V * fudge factors = 1,500 watts * 1/10.5v cutoff * 1/0.85 inv eff * 1.25 NEC factor = 210 Amp rated wire + fuses
    Compared to the "nominal" calculations:
    • Current = 1,500 watts / 12.7 volts charged battery = 118 amps "ideal DC current"
    The "fudge factors" point you to almost doubling the current capacity of your DC wiring, Battery AH Rating, and fusing/breakers for a "reliable" system.

    And we did not even account for the fact that a 1,500 watt inverter may support upwards of 3,000 watts for a few seconds of surge current (fuses/wiring have thermal lag and the standard ratings do assume that you will have surge current in excess of max continuous rated current for short periods of time to start motors and such).

    For 24 and 48 volt battery systems--Just 2x or 4x the 12 volt numbers (charging, full, cutoff) and you will be fine in most cases (always read the manuals for your equipment).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • john pjohn p Solar Expert Posts: 814 ✭✭✭
    Re: DC AMPS

    As Cariboocoot wrote for measuring an inverter input current..Metering the DC supply line is the only way to get an accurate measurement at any given time. Many people talk about lots of maths to work it out but this is the only simple and accurate way As has already been said the input voltage to the inverter is not constant.
  • JESSICAJESSICA Solar Expert Posts: 289 ✭✭
    Re: DC AMPS

    Thanks to all!

    That was easier than I thought.
Sign In or Register to comment.