How to Plot Current vs. Time for a 2.2 kΩ Precharge Resistor?

  • Thread starter Thread starter nikita
  • Start date Start date
  • Tags Tags
    Circuit Resistor
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 6K views
nikita
Messages
11
Reaction score
0
please help me with plotting of graph between "current versus time" for precharge resistor- 2.2 kΩ 20W.
this is placed across a contactor,which is linked to a motor controller with capacitors of value:6768μF.
closing time of resistor is calculated as 2.97 sec!
 
Engineering news on Phys.org
i have calculated time constant using formula 5T=R*C, and it is coming 2.97 sec! but when i am plotting the graph of voltage versus time using formula v= 1-e^(-t/rc) graph is coming linear instead of exponential. where m i going wrong?
i have also used the formula v=e0(1-e^(-t/rc)), from this graph is coming exponential but time constant is coming very large. around 150 it is showing steady state!
battery voltage is i.e e0 =148v
r=2200 ohm,c=6768μF

i m really confused.what to do?
i have attached the graph which i have got.
 

Attachments

  • rc graph.jpg
    rc graph.jpg
    12.9 KB · Views: 814
  • rc1 graph.jpg
    rc1 graph.jpg
    9.6 KB · Views: 838
Last edited:
T = RC == 14.88. 5 time constants represents settling time, which gives 74.45 seconds.

Where did you get 5T=R*C
 
meBigGuy said:
T = RC == 14.88. 5 time constants represents settling time, which gives 74.45 seconds.

Where did you get 5T=R*C
5T is the time for current to reduce to a manageable value.
and you are saying correct that it will come 74.45 sec,but since our application is electric vehicle,it is practicaly impossible to wait for that much time before vehicle can start.so, I am apprehensive about it.
 
For the precharge circuit - the number of Time Constants needed is to get to within ??% of the full battery voltage, and then allow the main contactor to close. 3 TC = 95% of the voltage... with 2.2K Resistor ~ 3*TC=45S, double up the resistor ( parallel) ~23 Sec, or 2 TC = 85% of the Full voltage. --
This issue is not to get the current down to a manageable value during precharge, it is how much current in-rush to the capacitors can you accept at the end of the precharge cycle - when the main contactor closes? Look at the contactor specs - it may provide some info on number of operations at Current = I -
The better you can define what you need to do ( with real numbers) the better you can find the best technical solution.
 
How much time do you want to spend to precharge to 5T? If the answer is 3 sec, then tau= 0.6 so R must be 0.6/6768uF = 88 ohms.

You need to decide:
1. How much settling do you need. Perhaps 5T is overkill (as Windadct said)
2. How large a precharge current you want to deal with (not sure what the limitation is here).