I Calculating pre-charge of capacitors on electric vehicle

  • Thread starter t00mas
  • Start date
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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/electronics/capacitorchargeanddischarge.php?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!
 

BvU

Science Advisor
Homework Helper
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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:
2
1
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 $$ 0.9 = (1-e^{-{t\over RC}})
\Leftrightarrow e^{-{t\over RC} }= 0.1 \Leftrightarrow -{t\over RC} = \ln 0.1\Leftrightarrow R = {-\ln 0.1 \over tC}$$

For the peak power dissipated in the resistance you need ##V_{\rm batt}##, not the 288 V.
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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