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I am using a 36VDc SMPS as a power supply. A spark is generated when I connect 36V with my load circuit.

Specification:

  • Voltage source: 36VDc
  • Current consumption: more than 3A

I want to know why a spark is generated when I attach my load circuit, and how can I reduce the spark.

I have tried some RC circuit solutions given for arc suppression circuits but I can not identify how can I use those solutions.

Is there another way to reduce the sparks?


I tested with a 1545CT Schottky diode to put at the beginning of my circuit as mentioned in this answer, but the solution isn't useful in my case.

@Manumerous, can you guide me for which type of MOSFET I have to use in my circuit with configurations to avoid a spark?

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    \$\begingroup\$ Do you have a compelling reason to quench the spark? \$\endgroup\$ – Andy aka Jul 20 at 13:17
  • \$\begingroup\$ @Andyaka Yes, because the hardware will be used in the kitchen environment, so it is necessary to quench the spark. \$\endgroup\$ – Unknown Jul 20 at 13:32
  • \$\begingroup\$ Is it an ignition hazard? If so, what gas? \$\endgroup\$ – Andy aka Jul 20 at 13:46
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    \$\begingroup\$ Connect the load before energizing the power supply. Or, if the power supply has some sort of enable/disable control, disable the output, then connect the load, then enable the power supply. \$\endgroup\$ – Adam Lawrence Jul 20 at 16:13
  • \$\begingroup\$ "A spark is generated when I connect 36V with my load circuit." - exactly where is this connection being made? \$\endgroup\$ – Bruce Abbott Jul 20 at 20:29
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This is a very common occurrence if your load contains a large capacitor bank. The problem is that if the voltage over a capacitor changes fast (as when you connect the load to the power supply), this results in very large currents. This can be seen by the following differential equation for the capacitor current:

$$I_c(t) = C \frac{dV(t)}{dt}$$

Unfortunately this is exactly the case when you connect your load to the power supply. In this case your capacitors will draw a lot of current (which is often called the inrush current)

You can reduce this by limiting the current with a large enough resistor. This approach is the simplest, but adds additional losses through the resistor and increases the input impedance. A better Option would be to use an NTC Thermistor in that case. These resistors have a negative temperature coefficient, which means that there resistance decreases when they heat up. Therefore, the resistance of the NTC Thermistor becomes very low after it is warmed up by the current passing through it. This achieves a power loss that is lower than when a fixed resistance is used. You can find some information about NTC Thermistors here: How to Use NTC Thermistors for Inrush Current Limiting

Alternatively you could add an electric switch (like a MOSFET for example), which ramps up the voltage and therefore eliminates the spark. An in depth discussion about that topic as well as schematics for how to connect the MOSFET can be found here: StackExchange Electrical Engineering: P-Channel MOSFET Inrush Current Limiting

Furthermore also the geometry of the connector can influence the stark. There are dedicated anti-spark connectors which help reduce it or ensure that the spark can not come in contact with the skin.

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  • \$\begingroup\$ The resistor may defeat the (usual) purpose for the capacitance in the first place: Low input impedance. \$\endgroup\$ – evildemonic Jul 20 at 14:46
  • \$\begingroup\$ Thanks for your input, your of course right that the input impedance increases. However the capacitors (acting as an RC-Lowpass filter) still help you filtering voltage spikes, noise and other unwanted fluctuations. The additional resistor will influence the cut-off frequency which might not be desired. But i agree its not very elegant and usually not the preferred solution. \$\endgroup\$ – Manumerous Jul 20 at 16:42
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    \$\begingroup\$ You could change the resistor to a current limiting NTC which gets bypassed after a while through a high-side Switch to reduce losses. \$\endgroup\$ – Thauer Jul 20 at 19:03
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    \$\begingroup\$ @Unknown Which MOSFET is not your primary concern here. You need to understand why first, and I have a sneaking suspicion you don't yet. \$\endgroup\$ – winny Jul 21 at 12:13
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    \$\begingroup\$ @Unknown, I adapted my answer to give you some more Information on how to implement those methods. Unfortunately there is no standard solution or standard MOSFET that i can recommend for you since this depends on your circuit and requirements. I think you should be able to get a good first understanding about the proposed solutions with the links i added. If those concepts are clear you should be able to find a MOSFET or NTC Thermistor suited to your needs. \$\endgroup\$ – Manumerous Jul 21 at 13:26
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An electric spark is an abrupt electrical discharge that occurs when a sufficiently high electric field creates an ionized, electrically conductive channel through a normally-insulating medium, often air or other gases or gas mixtures. - Courtesy Wikipedia.

A spark could be caused by breaking current flow through an inductor (resulting in a high back emf generated by a collapsing magnetic field to return its stored energy to the source). It could also be caused by breaking a high charge/discharge capacitor current. Other causes could be high surge currents through incandescent lamps etc.

Identical scenarios would be possible, when a contact is made using a switch or a relay, due to 'contact bounce' or 'chatter'.

The solution would be to use RC snubbers or VDR's across the contact or across the load.

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Your design has excessive inrush current because it has no protection for a bulk capacitive load. Thus the surge current is I=V/ESR for ESR includes cap and contact resistance.

You could consider a PTC rated for the max Joules of charge energy but that adds a minor cost and heat.

Even 2.4A USB power to Apple iPad products creates a black carbon strip in the middle for +5V and burns out one connector pad. With an extremely thin flash coating of gold, the plugs eventually wear out in a year or 2 from a daily connection.

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You can get a spark hooking up 12V jumper cables between cars because you are making and breaking a circuit in the many-millisecond time it takes to connect. The inductance in the cables or receiving circuit is surprising in its ability to keep current flowing in a collapsing magnetic field when a circuit is broken. It's nothing to worry about. If you want to reduce it, put a switch in the line and throw the switch after you connect. There will be little or no "bounce" that way.

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  • \$\begingroup\$ I can not put a physical switch over there, I have to control it with some kind of controller circuit. \$\endgroup\$ – Unknown Jul 22 at 6:01
  • \$\begingroup\$ Perhaps you could put a thermistor inline somewhere, either in the cable or in the box at either end. Thermistors are often used to stop an "inrush" of current to a device, even when the switch is electronic. \$\endgroup\$ – Roger Ellingson Jul 22 at 19:43
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My suggestions,

  1. MOSFET based inrush current limiter: https://github.com/msglazer/Anti-Spark_Switch (Generally used in octocopter)

  2. Anti-Spark XT60, XT90 connectors explanation: https://youtu.be/X71Suakve6A (Video) In Anti-spark connectors, they are using small value high power resistors approx 6 ohms. This Resistor helps in reduce inrush current due to higher capacitance in your circuit. But this resistor only contacts in a small period of time. If you connect this loose connector maybe meltdown.

Thank you for reading.

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I am using IRF3205 to switching GND, and the gate is controlled by the microcontroller for the delay, so when the connector attaches to the load circuit, the GND is already disconnected, so the spark not happens.

Solution: - When the power In signal detected by the microcontroller, it turns on the gate after a few seconds to avoid spark. (Note:- Here I can use a relay for the same purpose, but again relay is the one kind of physical switch, so spark happens inside a relay is again cause a problem after a long time so that kind of solutions are not good, so I preferred this type of solution)

Thank you all for your help.

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