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Typically, you apply a known source voltage. Apparently not in your case, so you'd better explain what this circuit applies to.lepp1892 said:1) I want to find out the source voltage.
lepp1892 said:hmm...
How about knowing the max voltage that will be across both parallel branches seperatly? Would that give me enough knowledge to then find the source voltage?
This isn't exactly a circuit equation. I am actually trying to find breakdown voltages across multiple layers of materials and I came across an article that had this as a way to solve it but it didnt go into much detail about solving it.
lepp1892 said:Ok well the first branch has a voltage of 30 kV and the second branch has a voltage of 4 kV.
lepp1892 said:Its really just adding the 2 voltages together?
lepp1892 said:I got the 30 and 4 from the dielectric strength of the materials. The dielectric strength of the dielectric layer is 30 kV/mm and it is going through 1 mm and the air gap is 4 mm and the dielectric strength of air is about 1 kV
lepp1892 said:hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.
lepp1892 said:hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.
DragonPetter said:What is the purpose of the voltage divider and two different dielectrics? Are you trying to compare one to another?
If you just want to test what voltage it takes to break through the materials, can't you just place the materials in series with a variable voltage source and increase it til they break down and write down that voltage?
psparky said:Hmmmm...I don't think you are looking at things correctly.
Incidentally...why do you want to find the voltage across the loads? Why does that interest you? If you tell us why you are interested in voltage across load...that might help us solve your question.
Your breakdown tests appear to be spot on. If the dielectric were rated with a breakdown of 30kV then you'd expect this to be conservative, so finding it can reliably withstand at least 33kV would confirm this.lepp1892 said:hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.
We are trying to test our parts dielectric strength and we need to have an air gap so we can use less tooling so we don't have to have a different test fixture for every part.
NascentOxygen said:Your breakdown tests appear to be spot on. If the dielectric were rated with a breakdown of 30kV then you'd expect this to be conservative, so finding it can reliably withstand at least 33kV would confirm this.
Examining your RC model at 50Hz, we can overlook the presence of both resistances because they have little effect in comparison with capacitances of the values you give. Your air gap capacitance is so much smaller than the dielectric capacitance, that the air gap will carry 98% of the applied voltage until the air breaks down and conducts. This will occur at around 4kV.
For applied voltages greater than 4kV the air gap is effectively conducting, leaving all of the applied voltage impressed across your dielectric. With your dielectric's rated breakdown strength of 30kV then for applied voltages exceeding 30kV you will find the dielectric is subject to breaking down.
https://www.physicsforums.com/images/icons/icon2.gif Essentially, the air gap is not interfering with your voltage tests.
When the applied voltage is removed, the air gap capacitor repairs itself; alas, the plastic dielectric does not.
Some, but negligible here.DragonPetter said:Won't the air gap have some resistance at this point when it is ionized, rather than acting strictly like a short?
An RC circuit is a type of electrical circuit that contains both a resistor (R) and a capacitor (C). It is commonly used in electronic devices to control the flow of current and store electrical energy.
The voltage in an RC circuit can be calculated using the equation V = V₀ * (1 - e^(-t/RC)), where V₀ is the initial voltage, t is the time, R is the resistance, and C is the capacitance.
The time constant in an RC circuit is the product of the resistance (R) and capacitance (C) in the circuit. It represents the amount of time it takes for the capacitor to charge to 63% of its maximum voltage.
In an RC circuit, the voltage initially increases rapidly as the capacitor charges, then slows down and approaches a maximum voltage as the capacitor becomes fully charged. The rate at which the voltage changes is determined by the time constant.
The resistance (R) affects the rate at which the capacitor charges and discharges, while the capacitance (C) determines how much charge the capacitor can hold. Higher resistance will result in a slower charging rate and lower capacitance will result in a lower maximum voltage in the circuit.