# ZVS driver, calculate current for final voltage...?

#### Hamlet

Joined Jun 10, 2015
448
I'm inputing 12v at 8A, and generating a 7mm spark at the output, so should be about 7kv...

12v * 8 = 96w primary, 96w/7kv = .0137A, or 13.7mA

Did I do that right?

#### MisterBill2

Joined Jan 23, 2018
8,709
Were missing all of the context here, at least i do not see any context and so how can I provide a useful answer? Or is this an ongoing thread that somehow got turned into a new one and I missed all of the 59 remarks previously made.

#### Hamlet

Joined Jun 10, 2015
448
Sorry, here's the schema:

I'm trying to calculate the final current, at the left side. I believe the voltage is 7kV...
My current at the right side, current in, is 8A @ 12v, or 96watts.

#### DickCappels

Joined Aug 21, 2008
7,418
Close. You need to figure in an efficiency factor, which will decrease the output power with respect to the measured input power.

#### Mouthpear

Joined Dec 24, 2018
25
Close. You need to figure in an efficiency factor, which will decrease the output power with respect to the measured input power.
How do you do that?

#### DickCappels

Joined Aug 21, 2008
7,418
It starts with being able to measure the current and voltage in and out of the circuit. I guess this is another way of saying that you cannot know what you cannot measure.

#### MisterBill2

Joined Jan 23, 2018
8,709
yes because without context you can't leave a comment that actually pertains to the subject. Otherwise it's just leaving comments to rack up comment counts. Right?
I have never considered "comment counts" as related to comment accuracy or usefulness. Not every posting individual always provides the needed information so that relevant answers can be provided.
For this circuit with the load being an arc the turns ratio can help to figure the output open circuit voltage. The voltage across an arc is much less, and usually limited by the source impedance.
In this case the source impedance is the sum of the secondary winding resistance, (easy to measure) and the secondary winding reactance, which depends on both the inductance and the frequency. So that method gets complicated because of needing to measure the inductance and the frequency.

#### Hamlet

Joined Jun 10, 2015
448
With the arc extinguished, the current draw on the system is 2.5A. I don't know if that is useful.
Using the Hz setting on my multi-meter, it shows 12kHz. I lost count on the turns ratio of the
secondary, but an aprox. guess is 133:1.

At any rate, I was looking for an outcome with current rating
on the final being greater than that using an automotive ignition coil, but at a lower voltages,
so my standoff insulators don't become overwhelmed, as this is
going to be for an electric fence, abet pulsed at a few milliseconds/second.

#### MisterBill2

Joined Jan 23, 2018
8,709
OK, now there is material for some additional calculation. The no-load current is 2.5 amps and the loaded current is 8 amps then the extra power is what is spent in the load portion of the system 8.0-2.5=5.5A. so if the no load power is 30 watts then the load power is the rest of the total 96 watts, 66 watts. The voltage across the arc is not very high because an arc is a fairly good conductor. If the resistance of the secondary is known, and power= current (squared) times resistance, then current is the square root of the term Power/resistance. That will give an approximate value, because the actual resistance of the arc is not zero.

Now the normal operation of all the electric fence chargers that I have worked on is discharging a capacitor across a coil that is the step-up transformer primary. So it is a single high voltage spike. But I have not worked on any of those "weed-burning" fence chargers that are able to deliver a lot more energy. So it is important to be aware of the hazard potential of a maintained voltage fence charger. That may matter in the TS application of fence charging. Keeping people out is different than intimidating cattle.