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I Calculating pre-charge of capacitors on electric vehicle

  1. Nov 27, 2016 #1
    Hello!

    I found how to calculate here: http://liionbms.com/php/precharge.php
    But the problem is, that when I used online calculator, it gives me that with resistance 531, my charging time to 288V will be about 2.3 seconds instead 5 seconds, that I need.
    Online calculator here: http://mustcalculate.com/electronic...p?vfrom=0&vto=288&vs=320&c=1880u&r=531&time=5
    We have:
    • Capacity that we need to charge in controller: C=1880uF or 0.00188F
    • Battery nominal voltage: U=320V
    • Time, that is needed to charge controller capacitors to 90% of the nominal battery voltage: t=5s
    We need:
    1. 90% from 320V
    2. Needed resistance R in ohms
    3. Maximum current during precharge
    4. Energy E in joules
    5. Power P in watts
    6. Peak power PPeak in watts
    1. U1=(320V/100)*90%=288 V
    2. R=t/C/S=5s/0.00188/5=531 ohms
    3. I=U/R=320/531=0.6 A
    4. E=(C*V2)/2=(0.00188*2882)/2=78 Joules
    5. P=E2/T=782/5=15,6 W
    6. PPeak=U2/R=2882/531=156 W

    Is everyithing correct? What is that 5 (S) in 2nd calculation?

    Thank you!
     
  2. jcsd
  3. Nov 27, 2016 #2

    BvU

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    The 531 ##\Omega## is not right.
    Charging a capacitor over a resistance goes according to $$V_{\rm cap}(t) = V_{\rm batt}(1-e^{-{t\over RC}})$$ so of you want ##V_{\rm cap} = 0.9 \,V_{\rm batt} ## you need $$ 0.9 = (1-e^{-{t\over RC}})
    \Leftrightarrow e^{-{t\over RC} }= 0.1 \Leftrightarrow -{t\over RC} = \ln 0.1\Leftrightarrow R = {-\ln 0.1 \over tC}$$
    [edit] mistake (cut&paste while typesetting, see below -- well spotted)
    $$ -{t\over RC} = \ln 0.1\Leftrightarrow R = {-t \over C \ln 0.1}$$


    For the peak power dissipated in the resistance you need ##V_{\rm batt}##, not the 288 V.
     
    Last edited: Nov 27, 2016
  4. Nov 27, 2016 #3
    Hello!

    Thank you for your fast reply!
    I think there is a small mistake in your equasion at the end, because R=-t/(C*ln0.1) or am I wrong?
    R=-t/(C*ln0.1)=5/(0.00188*ln0.1)=1155 Ω
    The graph is also now correct: http://mustcalculate.com/electronic...?vfrom=0&vto=288&vs=320&c=1880u&r=1155&time=5

    4. E=(C*V2)/2=(0.00188*2882)/2=78 Joules - should I use here also battery back voltage?

    The needed resistande for 5s charging time is 1155 Ω.

    Thank you!
     
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