# What is the spark, caused by opening/closing a circuit, dependent on?

I have a DC circuit that runs 1 A current in it when it is closed. When I close the circuit, i notice that there is a spark between the terminals of connection when they're brought close enough, just before completing the connection. I understand that this is due to the electrical breakdown of air but the air is such an excellent insulator (breakdown voltage=3000V/mm).

1. Is this sparking dependent on the closed circuit current (higher current, higher probability of sparking) or the potential difference between the terminals or some other factor?
2. How would there be a difference in potential if it's just two broken ends of a circuit which are going to be joined together to make it live?
3. If it is dependent on the closed circuit current, what is the threshold current to cause the sparking?
• Knowing the voltage and load characteristics would help improve the already good answers Apr 20, 2021 at 22:11

Most switches exhibit a phenomenon called switch bounce. The action of bounce can cause rapid, millisecond-level pulsing to the switched device. If the circuit being switched has any inductance and capacitance (generally the case with a power switch), these bounce pulses will cause rapid changes in current and voltage, leading to transients that can arc as the switch opens and closes until it settles down.

How much arcing happens has a lot to do with the amount of current and voltage being switched, and the reactive nature of the load. But there will always be some. And, also, whether DC or AC is being switched. More about this below.

Once the arc happens, a local plasma (ionized gas) cloud is formed across the contacts, reducing the insulation resistance of the gap. The arc flows through this plasma.

Now here's where DC and AC differ: with DC, once the arc lights, it will stay lit until the circuit stabilizes as long as there is enough voltage and current available to sustain the arc through the plasma cloud. With AC, the current through the plasma will extinguish at or near the next zero-cross point in the AC cycle.

This is why the switch voltage ratings are lower for DC than for AC on the same switch: for a given voltage, greater contact-open distance is required to extinguish the DC arc. More here: https://hackaday.com/2018/07/24/a-dramatic-demo-of-ac-versus-dc-switching/

Sparking is normal. Every switch will do this. It's why switches wear out.

You don't need much current to get a spark. Ever get a static-charge zap after walking across a carpet and touching a doorknob?? That's nano or picoamperes. (At thousands of volts...)

The amount of spark, and the resulting "damage" to the switch contacts, is a function of both voltage and current. i.e. "power".

There's also a component of inductance. If you cut the power to a spinning motor, you get a huge kickback voltage and a big spark. THis is why (non-electronic-controlled) power tools have a switch that goes "click" when you use it... It's designed to make and break that contact very very fast.

V= L * dI /dt

So you have 3 variables Inductance, L Current, I at the time of opening switch and speed of dry contact switch, dt.

The speed of a dry contact switch is so fast that it will always arc with the same current when it was closed. The air ionization-time may be in microseconds so the actual dx/dt in mm/us millimetres per microsecond determines how much voltage occurs in latency T for ionization in say 10 us. So once the air is ionized you can stretch the arc much greater than it can bridge to start the arc.

Usually, for AC, RC snubbers are used to suppress relay, or transfer switch arc voltages each custom designed depending on the energy to be absorbed. In very high power high voltage transfer switches, there will be a long travel to extinguish the air or it will have contacts sealed in a vacuum or perhaps SF6 gas or maybe have a jet of air to extinguish the arc.

For DC loads, diodes are used to connect the kickback voltage to the opposite supply rail being switched so they are reverse biased when the switch is closed the become forward biased when the switch opens until the current collapses.

• Maybe add increasing arc length paths and magnetic arc control for completeness / interest. Apr 20, 2021 at 22:09

The spark is caused by the dielectric breakdown of air due to a electric-potential difference between the contacts of the switch. Current has nothing to do with it directly, however it's likely that a higher closed-circuit current for a given circuit would mean the circuit itself has a higher potential source within it, implying a higher closed circuit current and the probability of sparking at a larger gap.

To answer your question 2: does your circuit have any form of voltage source in it? If so, that's likely where your potential between the two "broken ends of the circuit" comes from. If there was no potential while the circuit was open, then why would current start flowing when the circuit is closed? With no potential source, no current will flow.

Try using a volt meter to measure the voltage across the two ends of the open circuit before you connect them. If you're getting sparks when connecting the wires, there should be a potential there before the circuit is closed.

• While I am not sure of the exact mechanics but for example a 12V car battery will spark ferociously if a low-ohm connection is made, but there is absolutely no sparks for a high-ohm load, so current must surely have something to do with it. Could it be a high-voltage during contact bounce driven by cable inductance? Mar 17, 2021 at 14:30
• A high-impedance material will more easily allow a large potential across it. In your car battery case, if you get a high-valued resistor connected to the (-) terminal, as you bring the other end closer to the (+) terminal, the potential at that end will float closer to the battery's (+) terminal, resulting in a lower potential to break down the air. With the same experiment with with a low-valued resistor, as you bring the unconnected end towards the (+) terminal, it's potential will be lower because the lower value of the resistor will help pin it to the potential of the (-) terminal. Mar 17, 2021 at 15:09
• @Arcatus until a current begins to flow, there will be no voltage drop across a resistor, so the switch contacts will have 12V between them regardless of the load, and will therefore always arc at the same distance. However, with a low series resistance a large current will flow producing a visible spark, but with a high resistance the current will be low and the voltage across the arc will collapse.
– Frog
Apr 18, 2021 at 4:35

DC will always spark - it's much easier to switch AC, because the voltage is zero twice every cycle. DC switch contact tips need to be bigger than AC for the same current. Silver tips are often used because the oxides of silver are also electrically conductive. Motor vehicle ignition, prior to electronic controls, used mechanical switching ('Kettering' or 'points' Ignition). A capacitor across the switch terminals extended the life of contact tips ('points').

Sparking or arcing is caused by the conduction continuing through the air after the contacts open at least a bit. Air is an insulator up to a dielectric breakdown voltage (which depends on electrode shape, air pressure etc), but the conduction through the air can also be aided by metallic ions ripped or boiled off from the contact surfaces. More refractory metals are usually better in this regard. Once an arc or spark forms, the ionized air and other ions lower the threshold for conduction, so it tends to continue until the voltage and current drop far below what was required to initiate the discharge.

Here you can see a crude Jacob's Ladder where the operator induces a spark at the bottom and the heated air from the spark itself causes the discharge to rise to the top where the wires are too far separated for the arc to continue. This is likely with an AC transformer, so the ionization is substantial enough for the arc to survive the zero crossings .

Inductance, even stray inductance from a few meters of wire that is not twisted or side-by-side, causes the voltage across the gap to increase as the contacts open.