HIGH VOLTAGE TECHNIQUES Basic Electrode Systems

HIGH VOLTAGE TECHNIQUES Basic Electrode Systems

Basic electrode systems Different configurations Parallel plate electrodes Co-axial cylinders Concentric spheres

Parallel plate electrodes Plane-plane electrode system U: applied voltage E: electic field d: electrode seperation (the distance between electrodes)

Electric field Which coordinate system we should use? Cartesian coordinates!!! What does it depend on? x-coordinate? y-coordinate? E (hence V) only depends on the y-coordinate!!!

Electric field and potential equations Electrical charges We can use Gauss Law Laplace s equation In cartesian coordinates

By Laplace s equation Laplace s equation in Cartesian coordinates (dependent variable is y) General solution for this differential equation is Constants A and B is determined by the data of the problem which is called boundary conditions.

Boundary conditions y = 0 y = d V = V 2 = 0 0 = A.0 + B B = 0 V = V 1 = U U = A.d + B U = A.d + 0 A = U/d

Solution Potential Equation for Parallel Plate Electrode System V = V y = Ay + B = U d y

Potential change by y V=V(y) U U/2 V=(U/d)y d/2 d y

Equipotential lines

Solution Electric Field Equation for Parallel Plate Electrode System E = dv dy = d U d y dy = U d What is the meaning of minus sign?

Electric field lines

Electric field lines ALWAYS perpendicular to equipotential lines (ORTHOGONAL)

Electric field change by y What is the meaning of this graph? E is constant (U/d). It has the same intensity everywhere! Electric field is HOMOGENOUS!!! (Uniform)

Electrical discharge U DISCHARGE Discharge can be in a form of Breakdown Flashover Partial discharge

When does it breakdown? U U b : Breakdown voltage E = U d E b = U b d : Breakdown electric field, breakdown strength E Eb Discharge in the insulation!!!

Electric field equation by electrical charges Gauss s Law: For uniform symmetrical surfaces: (S: surface of the electrode)

Potential equation by electrical charges 2 U = E ds = E. d 1 E = U d V = E dy + c = E. y + c For y = 0 V = V 2 = 0, hence c = 0 V = E. y + 0 E = V y

Potential and Electric Field Equations for Parallel Plate Electrode System: V = V y = U d y E = U d

Capacitance Parallel plate capacitor

Capacitance Q = C. U charge voltage capacitance C = Q U = D. S U εe. S = U = ε U d. S U C = ε. S d = ε rε 0. S d

Capacitance F/m m 2 C = ε rε 0. S d F m : Dielectric constant of the insulation = 0 r [F/m] r : relative dielectric constant (no unit), relative permittivity 0 = 8.854 10-12 F/m dielectric constant for space, permittivity S : Area of electrode surface [m 2 ] d : electrode separation, insulation thickness [m]

Capacitive resistance the resistance of a system to current going through because of its capacitance X c = 1 ωc = 1 2πf C I = U R I = U X c C X c I C X c I

Multilayer dielectrics

This dielectric configuration is connected in SERIES! I = I 1 = I 2 = I 3 = = I n Q = Q 1 = Q 2 = Q 3 = = Q n C U = C 1 U 1 = C 2 U 2 = C 3 U 3 = = C n U n U is the voltage applied to the entire system U 1, U 2,, U n are voltages across the corresponding dielectric layers U = U 1 + U 2 + U 3 + + U n

Voltages across each dielectric layer: C U = C 1 U 1 = C 2 U 2 = C 3 U 3 = = C n U n U 1 = U 2 = C U C 1 C U C 2 U n = C U C n

Equivalent capacitance of the system X C = X C1 + X C2 + X C3 + + X Cn 1 ωc = 1 + 1 + 1 + + 1 ωc 1 ωc 2 ωc 3 ωc n 1 C = 1 C 1 + 1 C 2 + 1 C 3 + + 1 C n C : equivalent capacitance of the system, total capacitance of the system U : voltage applied to the entire system C i : capacitance of the i th dielectric layer U i : voltage across i th dielectric layer

Calculating capacitance C = ε. S d, C 1 = ε 1. S d 1, C 2 = ε 2. S d 2,, C n = ε n. S d n 1 C = 1 + ε 1. S d 1 1 ε 2. S d 2 + + 1 ε n. S d n = = 1 S n d i ε i i=1 1 S d 1 ε 1 + d 2 ε 2 + + d n ε n A

Total capacitance, equivalent capacitance A = n i=1 d i ε i = d 1 ε 1 + d 2 ε 2 + + d n ε n 1 C = A S C = S A

Potential and electric field across 1 st dielectric layer U 1 = C U C 1 = S A U ε 1. S d 1 = U A d 1 ε 1 E 1 = U 1 d 1 = U d 1 A ε 1 = U d 1 A. ε 1

Potential and electric field across 2 nd dielectric layer U 2 = C U C 2 = S A U ε 2. S d 2 = U A ε 2 d 2 E 2 = U 2 d 2 = U ε 2 A d 2 = U d 2 A. ε 2

Potential and electric field across i th dielectric layer U i = C U C i = S A U ε i. S d i = U A ε i d i E i = U i d i = U ε i A d i = U d i A. ε i

Electric field strength across any dielectric layer, E i, becomes greater than breakdown strength of that dielectric, there will be an electrical discharge. E i E bi Discharge in that layer of the insulation!!! Breakdown, short circuit in i th dielectric layer

Questions What happens in the other layers? What changes? What happens to the entire system?

Equivalent of capacitance connected in parallel: U = U 1 = U 2 = U 3 = = U n E = E 1 = E 2 = E 3 = = E n = U d If E E bi BREAKDOWN C = C 1 + C 2 + C 3 + + C n

Non-parallel configuration What happens to electric field if the plates are NOT parallel? In order to evaluate that, we have to know the behaviour of field lines at boundary surfaces.

Next topic... Refraction of electric fields