Contactor compatibility with the BMS
Quote from by Mail on 21/10/2021, 8:10 amHi again, got another question, is this contactor compatible with the bms for chargers and consumers disconnect (separate busses so two will be used)? I'm having a hard time sourcing the ones people recommend here in Sweden. As far as I can see it should work and the current draw is acceptable, but it's not my main area of expertise. I realise this might be a bit outside the scope of questions that you could be expected to answer, so no hard feelings if you don't.
Hi again, got another question, is this contactor compatible with the bms for chargers and consumers disconnect (separate busses so two will be used)? I'm having a hard time sourcing the ones people recommend here in Sweden. As far as I can see it should work and the current draw is acceptable, but it's not my main area of expertise. I realise this might be a bit outside the scope of questions that you could be expected to answer, so no hard feelings if you don't.
Quote from philtao on 21/10/2021, 8:12 amAlways a pleasure to share knowledge so that you get the best possible experience when going to lithium. I never heard about this contactor before, but it looks serious with complete datasheet. I do not see anything that would make it not work with the BMS. The key aspects to consider in relation to the BMS driving circuit are:
- the coil circuit inrush current of 3.8A - which should be OK for the BMS as it is very short (a few milliseconds?) - the BMS can support up to 7A for 100 ms
- the datasheet does not say if there is a protection diode in the coil circuit to avoid very high voltage spike when the relay is switched off. these diodes are called "flyback", "freewheeling" or "snubber". If the contactor does not have such diode, then you should install one (1N4007 for example).
Philippe
Always a pleasure to share knowledge so that you get the best possible experience when going to lithium. I never heard about this contactor before, but it looks serious with complete datasheet. I do not see anything that would make it not work with the BMS. The key aspects to consider in relation to the BMS driving circuit are:
- the coil circuit inrush current of 3.8A - which should be OK for the BMS as it is very short (a few milliseconds?) - the BMS can support up to 7A for 100 ms
- the datasheet does not say if there is a protection diode in the coil circuit to avoid very high voltage spike when the relay is switched off. these diodes are called "flyback", "freewheeling" or "snubber". If the contactor does not have such diode, then you should install one (1N4007 for example).
Philippe
Quote from Daniel on 21/10/2021, 10:13 amThanks for the answer, I thought that since it has positive and negative terminals it is with protection diode, can it be any other reason for that(the N version has non polar loaded terminals)?
Thanks for the answer, I thought that since it has positive and negative terminals it is with protection diode, can it be any other reason for that(the N version has non polar loaded terminals)?
Quote from philtao on 21/10/2021, 10:56 amThe polarity option of the contactor is for the power terminals (+A1 / -A2).
The need for a protection diode is on the red and black coil wires.
The coil is an inductance:
- when it is powered ON, some current flows through the inductance and there is energy stored in the inductance
- when it is powered OFF, it takes some time for the current in the inductance to go down (the time to dissipate the energy it has stored) and, as the command switch (the BMS output) is open, there is nowhere for that current to go to... and that creates a huge voltage build up
The protection diode is mounted between the red and black coil wires to give a path for that current.
I found an image on Internet showing the contactor coil (L and R), the BMS output (switch) and the protection wire (D).
- the contactor red coil wire is connected to the BMS output
- the contactor black coil wire is connected to ground
The polarity option of the contactor is for the power terminals (+A1 / -A2).
The need for a protection diode is on the red and black coil wires.
The coil is an inductance:
- when it is powered ON, some current flows through the inductance and there is energy stored in the inductance
- when it is powered OFF, it takes some time for the current in the inductance to go down (the time to dissipate the energy it has stored) and, as the command switch (the BMS output) is open, there is nowhere for that current to go to... and that creates a huge voltage build up
The protection diode is mounted between the red and black coil wires to give a path for that current.
I found an image on Internet showing the contactor coil (L and R), the BMS output (switch) and the protection wire (D).
- the contactor red coil wire is connected to the BMS output
- the contactor black coil wire is connected to ground