Hello everybody,

I have been analysing a system where the power supply is a 1000 V battery. This battery has a contactor that controls the discharge circuit. This circuit connects to an inverter, which in turn connects to a motor. The cable between the battery and the inverter is of considerable length, and simulating the opening of the contactor (necessary in an emergency situation) produces very high voltage peaks, in the order of tens of kV for a very short period of time. My question is, is this possible or is my simulation wrong? What can cause a voltage spike of this magnitude during such a short period of time?

Best regards and thanks,

 

Inductance in the Motor.
If You are not aware of this,
You should not be working in the same building that contains any ~1000-Volt Equipment.
A ~1000-Volt Battery is just Death waiting to happen.
.
.
.

 

Hi, post a schematic to get the best responses. Include your calculations as well otherwise no one knows your thought process.

 

The magnitude of the voltage spike from and inductor is proportional to the inductance and the time rate of change of the current in the inductor. These can be LETHAL.

 

The magnitude of the voltage spike from and inductor is proportional to the inductance and the time rate of change of the current in the inductor. These can be LETHAL.

And simulators cannot accurately handle this because it depends so much on behavior outside normal operating characteristics and on stray resistance, capacitance and inductances.

 

The use of a snubber diode, effectively presenting a short circuit to any back emf, together with a suitably rated varistor across the supply should reduce the contact arcing.

 

Hi, post a schematic to get the best responses. Include your calculations as well otherwise no one knows your thought process.

Hi, thank you for your response. You are right, I should have uploaded an image. I do it now, with the values of all components. I add the parasitic inductance of the cable and the parasitic capacitances of the cable:

With this inductance, the peak voltage that should be produced (VL = L * dI/dt), taking into account that the contactor used (according to tests carried out) takes 50 ns to cut off the current completely and the rated current is 500 A, is 17.2 kV. Is this possible?

Furthermore, in the simulations performed, the voltage peaks are even higher. Maybe it is a problem with the software I use (Matlab/Simulink), can anyone help me with this problem?

 

The use of a snubber diode, effectively presenting a short circuit to any back emf, together with a suitably rated varistor across the supply should reduce the contact arcing.

Hi, thank you for your help. I had also thought about a dv/dt filter or a damping capacitor. Best regards

 

Add a reverse biased diode after the contactor and see what the simulation does – you could also add a capacitor that would limit the voltage, through being charged by the voltage spike.

 

Add a reverse biased diode after the contactor and see what the simulation does – you could also add a capacitor that would limit the voltage, through being charged by the voltage spike.

Hi, thanks for the reply. As for the solution you propose, I will look for a diode for that voltage level, but my main question is whether such high voltage peaks (around 17.2 kV) are really possible. What do toy think? Thanks and best regards.

 

In such a circuit with a considerable length of cable, where the capacitance and inductance is spread along the cable length, it is difficult to determine the back emf of the collapsing inductive field – and its interaction with the cable capacitance.

But if you are measuring such a voltage, and seeing the effect (switch arcing), I’d believe it.

Bear in mind that if you place a capacitor post the switch in an attempt to limit the switch-off voltage spike, that it will be subject to a very fast 1,000V voltage rise (at switch on) – and may require some series resistance to limit the peak charge current.

 

In such a circuit with a considerable length of cable, where the capacitance and inductance is spread along the cable length, it is difficult to determine the back emf of the collapsing inductive field – and its interaction with the cable capacitance.

But if you are measuring such a voltage, and seeing the effect (switch arcing), I’d believe it.

Bear in mind that if you place a capacitor post the switch in an attempt to limit the switch-off voltage spike, that it will be subject to a very fast 1,000V voltage rise (at switch on) – and may require some series resistance to limit the peak charge current.

Yes, the next step is to select the capacitor and the series resistor. Anyway, I will continue to investigate the voltage spike that occurs when the contactor opens, because I am still unclear. Thanks for your help.

 

There is also the possibility that the voltage spike is produced by the inverter (when de-energised), therefore you could try fitting a reverse bias diode across the supply at the inverter input.

 

There is also the possibility that the voltage spike is produced by the inverter (when de-energised), therefore you could try fitting a reverse bias diode across the supply at the inverter input.

No, this is not possible because in the simulation the inverter is still in conduction mode. This simulation is made to simulate the behaviour in case of a battery emergency, where the battery does not have time to ask the inverter to open the circuit. And it has to open the battery.

 

This may seem wacky idea, but if the back emf is generated over the positive conductor; connect a reverse bias diode between the switch contact and the inverter plus input. This will necessitate an additional cable (between the switch and inverter) but it need not be of large cross sectional area.

In theory, with the conductor shorting the diode, it need not have a high voltage breakdown rating, only being sufficient to withstand the back emf current/energy.

 

dV=L*di/dt

 

dV=L*di/dt

Yes, I have use this formula to calculate the voltage peak analytically. With the inductance of this circuit (1.72e-6 H), the peak voltage that should be produced (dV = L * dI/dt), taking into account that the contactor used (according to tests carried out) takes 50 ns to cut off the current completely and the rated current is 500 A, is 17.2 kV. Is this possible? What do you think?

 

This may seem wacky idea, but if the back emf is generated over the positive conductor; connect a reverse bias diode between the switch contact and the inverter plus input. This will necessitate an additional cable (between the switch and inverter) but it need not be of large cross sectional area.

In theory, with the conductor shorting the diode, it need not have a high voltage breakdown rating, only being sufficient to withstand the back emf current/energy.

I have try the idea in the simulation, it helps to reduce it a bit. Thank you for your idea.

 

The fact is that shutting off a rather high power inverter via an "E-STOP" disconnect certainly can produce quite high voltage spikes that contain enough energy to cause damage. Suddenly disrupting the operation at a random part of a cycle may not cause damage every time, but certainly at some points of the inverter cycle damage can happen. The long cables to the motor, and the motor's momentary generation condition add to the lack of specific analysis.
My suggestion is adding gas-tube over-voltage transient protectors across the motor leads. And assure adequate ARC flash protection for that battery contactor section. The warning sign on the enclosure should announce "Hazard of Death". It does keep most folks from opening the door to take a look.

 

With regards my wacky idea reducing the voltage spike slightly, not all diodes are equal. Try the simulation using a fast acting diode to see if this improves the effect.

With regards MisterBill2’s suggestion of using GDTs to limit voltage spikes, bear in mind that once ‘fired’ due to a voltage spike, GDTs remain conducting until the current falls below a holding level, even when used on AC systems. To prevent this resulting in a hazard or causing a fuse to operate, they are normally used with an MOV in series to cut the current flow once the circuit voltage falls to normal levels. So unless you can guarantee that any GDTs will only operate on a de-energised circuit, some additional circuit protection is required with their use.