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How to find max amperes

Discussion in 'General Electronics Discussion' started by keraynopoylos, Sep 1, 2012.

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  1. keraynopoylos

    keraynopoylos

    9
    0
    Sep 1, 2012
    Hi all.

    I have this PSU:
    http://cnc4you.co.uk/resources/PSU.pdf

    So I know the Output(DC) Volts, Amps(max) and Watts(max).

    What I'm trying to calculate, is the max current(amps) of the INPUT (AC).

    I want to know that so I can properly fuse my mains power cord and also buy a power switch and Estop switch of the appropriate specs.

    I would be grateful if the method is briefly explained too, as I'll have to recalculate(after adding a few more loads.

    I have found that the formula should be:
    Imax = P / Vp x cosφ

    Is P the Real power(and if so how to I calculate it?) or the Apparent power(Vdc x Imax = 36 x 11)?

    I guess Vp would be the input voltage(AC), 240V?

    And how do I calculate the cosφ, when the specs only give me the efficincy(85%)? Is the power factor the same as the efficiency(just not expressed in %), so would it be 0.85?

    Thanks in advance.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    Simple way is to use the power in watts to determine the current as if it were a purely resistive load.

    Then multiply it by a magic number between 2 and 10.

    If I had a 350W power supply, I might fit a 5A fuse.

    Remember that a switchmode power supply will draw heavy currents when it is switched on (50A is not unusual even for a small power supply). Larger power supplies may incorporate additional circuitry to limit the peak current. Your fuse needs to be rugged enough to survive this.

    Failures which blow fuses are typically those which place a short across the incoming mains (in a SMPS) so you're going to get a pretty dramatic spike in current that will blow any reasonably rated fuse very quickly.

    The only condition where these power supplies will gradually increase current is where you have a brownout. If you knew the ratings of the switching devices you could fuse it so that this (and perhaps the inductor) was not subject to abnormal currents.
     
  3. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    The datasheet you linked to clearly states...

    "Inrush Current ~ cold start, 40A/230V"

    So in theory that will be the max pull of the device, fuse it for 50A and you should be good... And as Steve pointed out almost any real failure will still pop a 50A fuse in no time... If you fuse it lower than say 40A you risk the fuse not being able to handle the inrush for any sustained duration, a 50A fuse gives you a little buffer room to compensate for variances, and will still fail upon a legit short...
     
  4. keraynopoylos

    keraynopoylos

    9
    0
    Sep 1, 2012
    I see your point.

    This raises another matter though...

    I've browsed through numerous Estop switches and on/off switches and haven't found any with rating above 20A...
    (I think I've checked the whole farnell.com catalogue)
     
  5. john monks

    john monks

    693
    1
    Mar 9, 2012
    Code:
    [CODE]
    [/CODE]Not enough information.
    I cannot tell what the fuse should be because I don't know if the input is inductive, capacitive, or if the PSU only looks at the input peak voltage like into a bridge rectifier or what. So you are dependent on what the manufacture tells you. And because the input is spect for AC or DC I assume the input is a rectifier. If there is RFI filtering that may affect it. Frankly I would probably take the input power, divide it by the input voltage, multiply that result by 2 or a little more like Steve suggests, and pick a slow-blow fuse of that value.

    Power factor or PF is the real power divided by the apparent power.
    The number is between 0 and 1.

    So a perfect capacitor with a reactance of 220 ohms will have an apparent power 220 watts if connected to a 220 volt line. But the real power is 0 watts. This is because the power from the line is going right back into the line resulting in a net zero power going into the capacitor.
    This is because when the Line voltage is going from 0 to the peak voltage the capacitor is charging absorbing energy. When the line is going from peak voltage back to zero the capacitor is discharging putting energy back into the line. The maximum current is when the voltage is changing the most, around 0 volts, that is when the voltage is going from negative to positive or positive to negative.

    In this case the PF is 0 because real power divided by apparent power or 0/220 is 0.

    I don't know where you got your formula from but I work in electronic manufacturing and I do not use formulas That I cannot derive. Not even ohms laws. And your formula is just another reason why I don't use them. They often times tend to block understand and cause unnecessary confusion. I focus on a scientific analysis of capacitors, resistors, and inductors, and definitions. The rest is just math.

    If you want a mathematical analysis of power factor PF I will be happy to do it.
     
    Last edited: Sep 2, 2012
  6. keraynopoylos

    keraynopoylos

    9
    0
    Sep 1, 2012
    First of all thanks for all your replies.

    I didn't get much of that. Sorry, very limited knowledge(just what I remember form highschool and a bit of wikipedia reaseach). You mean that due to lack of specification we can't get an accurate calculation so will be using the following rule of thumb.

    Input voltage=240V AC? Or should it be the RMS Volts?(and how would I calculate that? Would it be 240*0.707?)

    Input power=?
    Would we calculate it from:
    Efficiency/100=EnergyOutput/Energy Input=>
    Energy Input=100*Energy Output/Efficiency=100*Vdc*Idc/Efficiency=100*36*11/85=466W
    ?

    Which would give us(assuming Input Voltage=240v): Input Power/Input Voltage=466/240=1.95A?
    And then multiply by 2 or 3? (So Imax ac=4-6A?)
    Why do we do the multiplication(by 2 or more -or between 2 and 10 as (*steve*) has suggested?


    I got the formula "Current (single phase): I = P / Vp×cos φ" from this website and I found from this document(from TDK-Lambda) that IL = Pav / VL×PF×Eff
    where:
    IL - RMS line current in Amps
    Pav - average output power in Watts
    VL - AC line voltage in Volts
    PF - input power factor
    Eff - efficiency of the supply
    (so they add the Efficiency to the equation)


    Finally one more thing.
    Assuming we calculate the max input ac current to be for example 6 amps, would it be for some reason wrong to use 2 fuses in series, one slow-blow with a rating just above of the 6 amps(lets say 7 or 8) and a fast-blow wiath a 50 volt rating, so that the slow blow will cut off the supply in case we get a constant current(not a spike like the Inrush current) of above the 6A that is our limit whithin a few seconds and the fast blow will provide instant protection from classic short-circuits where we would get a very high current?
    (yes, safety is very important for me as I am inexperienced and don't want to harm anyone)

    I apologise for the far too many questions :)
     
    Last edited: Sep 2, 2012
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
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    Jan 21, 2010
    240V AC *is* the RMS value, so power divided by that, yep.

    Ideally calculate the input power from the output power an efficiency, but the fudge factors are there to take this sort of stuff into account anyway.

    You don't want 2 fuses in series.

    The fuses are there to provide protection if a failure occurs, not to prevent you from overloading the PSU. If it fails, you're going to see some pretty high currents (It's the nature of faults in SMPS power supplies)
     
  8. john monks

    john monks

    693
    1
    Mar 9, 2012
    Now I see what's happening.
    I am assuming that the efficiency is about 85%
    So your formula will be input current = output power divided by (input voltage X PF X efficiency)
    for example you might have 200W / (230V X .85 X .65) = 1.57 amps.
    You should multiply this by 2 to get 3.15 amps.
    Choose the next slow-blow fuse up, maybe 5 amps.

    Now that I saw the datasheet I can figure it out. But even here this is not precise but is close enough most of the time.

    If you have trouble seeing this I will explain it better.
     
  9. Timescope

    Timescope

    43
    0
    Aug 30, 2012
    If we take the characteristics of BS1362 fuses into account, a 5 amp fuse should be adequate :
    " The standards specifies breaking time versus current characteristics only for 3 A or 13 A fuses.[citation needed]
    For 3 A fuses: 0.02–80 s at 9 A, < 0.1 s at 20 A and < 0.03 s at 30 A.
    For 13 A fuses: 1–400 s at 30 A, 0.1–20 s at 50 A[verification needed] and 0.01–0.2 s at 100 A "
    http://en.wikipedia.org/wiki/BS_1363
    Timescope.
     
  10. keraynopoylos

    keraynopoylos

    9
    0
    Sep 1, 2012
    @(*steve*) yes, you're absolutely right. I thought +-240V were the peak values(which are actually 400V) but 240v is the RMS.

    (*steve*)&CocaCola and john monks&Timescope you have a different approach concerning the fuse. (*steve*)&CocaCola propose a high rating(sth like 50A) fast-blow fuse that would withstand the Inrush current but would instantly open with a short and john monks&Timescope prefer an only-double-the-caclculated-normal-input-current rated slowblow fuse.

    Sorry to keep coming back to the same issue and I respect your opinions, but I have to ask why you think one solution is preferable over the other.

    (and me being stubborn: why an in-series solution that combines both the fuses(and the advantages) is not advisable? - and would it be the same as using a dual-element fuse?)

    Thanks once again!
     
  11. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    For me it goes like this...

    I don't know the duration of the inrush for the power supply, or any other fluctuations in current it might have when running...
    I don't know the duration and or limits that any particular slow-blow fuse has off hand, to compensate any fluctuations...

    Without those variables one can only take an educated guess at what slow-blow fuse might work...

    For me I grab a 50A and I know it works, regardless of not knowing the variables, and I know that as a 'safety' device it's still going to pop on a short due to failure...

    IMO neither approach is more right or wrong, they are just different thoughts... From my experience if there is one thing that bugs, annoys and drives crazy is when a fused item is too tightly rated for the device resulting in a roll of the dice if it works on any given day or blows the fuse...
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    However, you might raise a few eyebrows with a 50A fuse on a device connected to your 10 or 15 amp circuit. :D
     
  13. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    Yep, you might...

    But if the 10A or 15A circuit can't handle the inrush you are back at square one...

    Also using underrated switches in such applications can pose a problem due to excessive arching...
     
    Last edited: Sep 3, 2012
  14. keraynopoylos

    keraynopoylos

    9
    0
    Sep 1, 2012
    There's one thing I forgot to ask about regarding john monks' last reply:
    How did you calculate that PF=0.65..? It's not given at the specs .pdf...


    Other than that, after all we conclude that if it wasn't for the inrush current, a 6A fuse would probably be a good choice, right?

    From the same mains lead, I will be powering a router too. The router's rated input is 1350W and rated voltage 230V. Would I find the current by dividing 1350W/230V=5.9A?
    And would that be the peak current(is rated input power=max input power?)

    And if so, the fit-for-purpose fuse would be 6+5.9=12A?(again, disregarding the inrush current)
    (sorry for asking the obvious, but I'd rather not assume, given the high current and my ignorance)

    So when selecting a "master" switch, I should go for a rating above the inrush current?
     
  15. john monks

    john monks

    693
    1
    Mar 9, 2012
    PF of .65 was take from the application note 502 that you posted.
    This is only an example. You will have to recalculate based on the output power and the line voltage.
     
  16. keraynopoylos

    keraynopoylos

    9
    0
    Sep 1, 2012
    Thank you all for your answers
     
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