# What calculation can be used to predict the voltage across an air gap as a plasma spark discharges over 1ms?

I am trying to determine what voltage and instantaneous current is available if using a spark gap in air or other gas. Comparable to an auto ignition. This will be DC current. There will be a main transformer like an automotive coil. It will be triggered in a CDI type layout where a capacitor will be discharged to cause the transformer primary side to have a lower voltage surge. The secondary will produce a potential of 40kv.

Discharge should last 1-1.2 milliseconds. The discharge will pass through a separate inductor coil. This coil has 14ga windings with approx 50uH inductance and 0.4ohms resistance. The circuit ends in a spark gap which allows the full voltage potential to be reached before the plasma forms and discharges the voltage.

This inductor coil just before the spark gap only needs to be powered for 1-1.2 milliseconds so its needed activation time matches the discharge time of the plasma arc.

How do I estimate the voltage drop over the arc time and the current over the same time.

simulate this circuit – Schematic created using CircuitLab

Assuming the transformer is wound 1:1000 if input is 40 volt.

The spark gap to the right of L1 is used to ensure the coil voltage rises to 40kv before discharge begins. Activation of the secondary side of the transformer occurs when the mosfet is activated and drops the voltage in the voltage trap between rectifier D1 and capacitor C1. Values in the schematic are yet to be worked out.

So once the mosfet has dropped the voltage on the (+) side of the capacitor, it will create an electron flow through the primary side of the transformer. (primary side voltage may be increased prior to the voltage trap to a couple hundred volts instead of just 40 volts) This will create the needed voltage spike from the secondary to initiate the arc at the spark gap. When this reaches near 40kv there should be a glow event at the gap. Then as voltage drops and an arc begins, the current / voltage should find the equilibrium point until the charge is exhausted and voltage drops not in inverse to current and the arc stops conducting.

So for the time which the arc event is occurring and the current / voltage is more or less stable, what current and voltage can be expected?

In this model I wish to get 1 millisecond of usable current through the inductor coil L1. I would like to be able to calculate what this current should be.

Another slightly different version

simulate this circuit

Or another option using inductive discharge instead of capacitive discharge

simulate this circuit

• Are you using the air gap to do the switching or using the circuit to drive the timing of the air gap? – Voltage Spike Jan 23 at 16:34
• Using the circuit to control the timing of the air gap. – J Hinton Jan 23 at 16:40
• Draw a diagram of the connected components. An arc has a strong negative resistance, so once struck, voltage will change to be 'whatever is needed' to keep the arc running. This generally means a certain amount of power, per mm of arc length, to keep the air hot, ionised and conducting. Less than that, the arc goes out, more than that and it collapses the power source. – Neil_UK Jan 23 at 17:16
• I don't have a drawing as yet to post but, if I can try to talk through it, at peak voltage the gap will enter the glow phase where electron transfer will begin. Then once the arc is established, voltage will drop as the current rises. The more the current rises the greater the cascade effect so more current until it lowers the voltage. The two seek balance for a time but then the supply voltage falls off and current cannot continue so the arc stops. Question is, based on the air gap, how much current flows diring the arc phase? – J Hinton Jan 23 at 20:28
• One more thought, during the arc phase, with lower voltage and Max current flow, would the current be quite high? In a theoretical example, if the transformer is 1:1000 and the circuit is complete with zero resistance, then the output side would move 1000 times the electrons of the input side. This would indicate that in a low resistance circuit situation during the arc phase the current is only limited by the total circuit resistance. – J Hinton Jan 23 at 20:47