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enter image description hereI am building my e-motorcycle. I am planning to use only one contactor instead of using three contactors. In my plan, pre-charge resistor and contactor circuit will be main contactor too. I am curious about how much power will be dissipated in this setup. I will share schmematics of my system.enter image description here

I am asking what is power dissipation when motor drain current. My battery voltage 76 volts it drains 50 amps. How much power will be dissipated on resistor?

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  • \$\begingroup\$ Peak power in the pre-charge resistor will be given by \$ P = \frac {V^2 } R \$. Voltage across the reistor will decay by 63% in one time constant, \$ \tau = RC \$ where \$ C \$ is the on-board capacitance. \$\endgroup\$
    – Transistor
    Commented Jan 12, 2022 at 8:28
  • \$\begingroup\$ Why are the fuses on the load side as opposed to the battery side? What protects the battery and its wiring? You have a wire that can deliver 1000’s of amps going to a keyswitch. That’s just wrong. \$\endgroup\$
    – Kartman
    Commented Jan 12, 2022 at 10:44
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    \$\begingroup\$ "I am asking what is power dissipation when motor drain current." It will be zero when the relay is energised. \$\endgroup\$
    – Transistor
    Commented Jan 12, 2022 at 14:28
  • \$\begingroup\$ " It will be zero when the relay is energised." However, we put resistor and contactor between battery and inverter. When motor pulls the power, there will be current on that path. So, resistance should dissipate some power in my view. I would like to hear your ideas if I am wrong. Thanks \$\endgroup\$
    – Burak Bozdoğan
    Commented Jan 13, 2022 at 5:10

1 Answer 1

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It seems like the controller will be connected through the charging resistor whenever the battery is connected. The power dissipated will be whatever is required to operate the controller without the motor energized plus the dissipation of the charging resistor carrying the controller current. That assumes that the voltage drop in the charging resistor is not too high from preventing the controller from operating. That may not be very much power, but it will eventually discharge the battery unless you have a manual switch that disconnects the battery.

The initial capacitor charging current will occur When the battery is connected. The peak current will be battery voltage divided by the 1000 ohm charging resistance. That will drop to 63% of that in one time constant as described in a comment by @Transistor.

What is shown, does not seem to be a very satisfactory chargine arrangement.

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  • \$\begingroup\$ Thanks for your kind comment. I agree with your idea. One thing is not clear for me. In the circuit there is key switch which connects to contactor. Is this key switch has controlled the contactor or not? \$\endgroup\$
    – Burak Bozdoğan
    Commented Jan 13, 2022 at 5:33
  • \$\begingroup\$ The key switch must control the contactor, but it must also power the low-power circuits of the controller. There must be an internal connection between B- and GND. The low-power circuits must be powered from PWR while the motor power comes from B+. Both the low power and motor power circuits have a common connection that is the connection between B- and GND. \$\endgroup\$
    – user80875
    Commented Jan 13, 2022 at 14:44
  • \$\begingroup\$ You appear to have expanded the question by revising it. That is generally not a proper thing to do, but since it may help with the original question, I won't complain very strongly. Using two contactors avoids the problem that I pointed out with the battery draining through the charging resistor. I assume that the 3rd contactor is redundant, operating simultaneously with the 2nd, but is, in effect, in series with the 2nd contactor reducing the chance of excessive arcing when opening under load. \$\endgroup\$
    – user80875
    Commented Jan 14, 2022 at 2:11

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