What is the breakdown voltage for this two dielectric capacitor (BaTiO3,LPDE)?

In summary, BaTiO3 has a thickness of 7mm and a dielectric constant of 1260, with a dielectric breakdown of 97MV/m. LDPE has a thickness of 7mm and a dielectric constant of 2.3, with a dielectric breakdown of 3kV/m. In a two dielectric capacitor with BaTiO3 and LDPE, the maximum voltage that can be applied is 52MV for BaTiO3 and 472kV for LDPE. At a source voltage of 3776V and a plate separation of 0.008m, the electric field will be equal to 472kV. Since the breakdown voltage of BaTiO3 is
  • #1
HelloCthulhu
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Homework Statement
A copper parallel plate capacitor has an area of 0.049m. A 1mm thick low-density polyethylene (LDPE) film is placed in between two 3.5mm thick BaTiO3 plates and placed inside the parallel plate capacitor. Both dielectrics have an area of 0.049m. Find the maximum capacitance, voltage and charge for the capacitor. If this two dielectric capacitor were to exceed its maximum voltage, which dielectric will breakdown first?
Relevant Equations
Q=CV

V = Ed

electric constant = $ \epsilon_{0}=8.854\times 10^{-12}$

Capacitance:
$$C=K\epsilon_{0}\frac{A}{d}$$

Capacitance for 2 capacitors in series:
$$C_{T}=\frac{C_{1}\times C_{2}}{C_{1}+C_{2}}$$

Breakdown voltage the capacitor:
$$ V_{max}=\frac{V_{bd}}{\frac{C_{T}}{C_{1}}}$$
BaTiO3
thickness = 7mm = 0.007m
dielectric constant = K = 1260
dielectric breakdown = E = 97MV/m (97kv/mm)

(Structural, Dielectric and Ferroelectric Properties of Tungsten Substituted Barium Titanate Ceramics, 2009)

LDPE
thickness = d = 0.007m
dielectric constant = K = 2.3
dielectric breakdown = 3kV/m (300V/mm)

https://www.azom.com/article.aspx?ArticleID=428

In a two dielectric capacitor, if one dielectric breaks down the capacitor fails. To solve for the voltage across each capacitor in series, first find the capacitance:

Max voltage for BaTiO3:
(97e6 MV)× 0.007m = 679000 = 680kV

Capacitance:
$$C=\frac{(8.85\times 10^{-12})\times1200\times0.049}{0.007}=0.00000007434F=7.434\times10^{-10} F=74340pF$$

Charge for BaTiO3:
(680e3V)*(0.00000007434e-10F) = 5.05512e-12C = 5.05512e-6μC

Max voltage for Low Density Polyethylene - LDPE:
(27e6V)×0.001m = 27000V = 27kV

Capacitance for LDPE:
$$ C=\frac{(8.85\times 10^{-12})\times2.3\times0.049}{0.007}=0.000000000997395F=9.97395\times10^{-10}F=997.395 pF$$

Charge for LDPE:
(27e3V)×( 9.97395e-10F) = 2.6929665e-5C = 26.929665μC

Total capacitance:
$$C_{T}=\frac{(7.434\times10^{-10})\times(9.97395\times10^{-10})}{(7.434\times10^{-10})+(9.97395\times10^{-10})}=9.84190445\times10^{-11}F=984.190445pF$$

BaTiO3:
$$V_{max}=\frac{680\times10^{3}}{\frac{984.190445pF}{74,340pF}}=\frac{680\times10^{3}}{0.0132390428436911}=51363229.8V=51MV$$

LDPE:
$$V_{max}=\frac{460\times10^{3}V}{\frac{984.190445pF}{997.395 pF}}=\frac{460\times10^{3}V}{0.986760957293750}=466171.6491V=466kV$$
 
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  • #2
I assume that the plate area is 0.049 m2 not 0.049 m.
I am not sure why you say you have 2 capacitors in series when you have 3. You have a sandwich in which the "bread" is BaTiO3 and the "filling" is the plastic.
I disagree with your value of the total capacitance. For trouble shooting your solution, it would be easier if you found an algebraic expression for the equivalent capacitance starting from the well known expression $$\frac{1}{C_{eq}}=\frac{1}{C_B}+\frac{1}{C_L}+\frac{1}{C_B}$$ where subscript B stands for BaTiO3 and L for LDPE. Solve for ##C_{eq}## and then substitute the numbers.
 
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  • #4
HelloCthulhu said:
I've done a few these before, but now I'm worried I've been using the wrong equation. Could you take a look at the first problem I posted here?

https://www.physicsforums.com/threads/max-values-for-a-2-dielectric-capacitor.941359/#post-5955152
I looked. That problem has two dielectrics. You can use that solution here if you consider double thickness for the LDPE dielectric. The equation in post #2 can be written as $$\frac{1}{C_{eq}}=\frac{1}{C_B}+\frac{1}{C_L}+\frac{1}{C_B}=\frac{1}{C_L}+\frac{2}{C_B}.$$
 
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  • #5
Ok, let's see if I can get this right...

Total capacitance:

$$C_{eq}=\frac{2}{9.97395\times10^{-10}}+\frac{1}{7.434\times10^{-10}}=3350394444.19$$

$$C_{eq}=\frac{1}{3350394444.19}=0.0000000002984723F=298.4723pF$$

BaTiO3:

$$V_{max}=\frac{680\times10^{3}}{\frac{298.4723pF}{74340pF}}=\frac{680\times10^{3}}{0.0132390428436911}=169366470.3V=169MV$$

LDPE::

$$V_{max}=\frac{460\times10^{3}V}{\frac{298.4723pF}{997.395 pF}}=\frac{460\times10^{3}V}{0.299251851}=1537166.76V=1.54MV$$In a capacitor with a source voltage of 12.3kV and a plate separation of 0.008m, the electric field will be equal to 1.54MV (12320V/0.008m = 1540000V). Since the breakdown voltage of BaTiO3 is 169MV, LDPE will be the first dielectric to break down.
 
  • #6
Can you please fix your equations? Use two $ signs to bracket them instead of one. Also, your answer would be much more legible and easier to troubleshoot if you used symbols instead of numbers. Less stuff to write and you can always substitute at the very end. It's a good habit to acquire.
 
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  • #7
kuruman said:
Can you please fix your equations? Use two $ signs to bracket them instead of one. Also, your answer would be much more legible and easier to troubleshoot if you used symbols instead of numbers. Less stuff to write and you can always substitute at the very end. It's a good habit to acquire.

Thank you for the posting correction! Not sure how I forgot that. However, I'm still confused about what you mean by replacing numbers with symbols.
 
  • #8
Thank you for fixing the equations. Strictly speaking, your first equation is post #5 should have ##\dfrac{1}{C_{eq}}=\dots##. You can fix that at some later time.

For BaTiO3 you have $$C=\frac{(8.85\times 10^{-12})\times1200\times0.049}{0.007}=0.00000007434F=7.434\times10^{-10} F$$ That should be ##7.434\times10^{-8}##F. Check your other numbers.

By symbols I mean no numbers. I would write the equivalent capacitance as ##C_{eq}=\dfrac{C_LC_B}{2C_L+C_B}## and write the charge on the capacitor at a given voltage as $$Q=C_{eq}V=\frac{C_LC_B}{2C_L+C_B}V$$and the voltage across just the LDPE as $$V_L=\frac{Q}{C_L}=\frac{C_B}{2C_L+C_B}V.$$ Since ##V_L+2V_B=V## you can easily find ##V_B## and so on for whatever you do next.

Note that if you happen to make a mistake in calculating the value of one symbol (as you have in this case), it would be much easier to recalculate the final answer without having to redo the intermediate calculations.
 
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  • #9
I'm still a little confused (still need practice with the derivations), but let's see how I do...

BaTiO3 capacitance:

$$C=K\epsilon_{0}\frac{A}{d}=7.434\times10^{-8} F$$

LDPE capacitance:

$$C=K\epsilon_{0}\frac{A}{d}=9.97395\times10^{-10}F$$

Total Capacitance:

$$C_{eq}=\dfrac{C_LC_B}{2C_L+C_B}=971.33 pF$$

BaTiO3 max voltage:

$$V_{max}=\frac{V_{bd}}{\frac{C_{T}}{C_{1}}}=52MV$$

LDPE max voltage:

$$V_{max}=\frac{V_{bd}}{\frac{C_{T}}{C_{1}}}=472kV$$

In a capacitor with a source voltage of 3776V and a plate separation of 0.008m, the electric field will be equal to 472kV (3776V/0.008m = 473kV). Since the breakdown voltage of BaTiO3 is 52MV, LDPE will be the first dielectric to break down.
 

1. What is breakdown voltage?

Breakdown voltage is the voltage at which an insulating material, such as a dielectric capacitor, experiences a sudden increase in electric current and becomes conductive.

2. How is breakdown voltage determined for a two dielectric capacitor?

The breakdown voltage for a two dielectric capacitor is determined by measuring the voltage at which the electric current suddenly increases through the capacitor. This can be done through various testing methods, such as using a high voltage source and monitoring the current through the capacitor.

3. What is the significance of the two dielectric materials, BaTiO3 and LPDE, in determining the breakdown voltage?

The two dielectric materials, BaTiO3 and LPDE, play a crucial role in determining the breakdown voltage as they have different dielectric strengths. The dielectric strength is the ability of a material to withstand high electric fields before experiencing breakdown. Therefore, the combination of these two materials in a capacitor will affect the overall breakdown voltage.

4. How does temperature affect the breakdown voltage of a two dielectric capacitor?

Temperature can have a significant impact on the breakdown voltage of a two dielectric capacitor. As the temperature increases, the dielectric strength of the materials may decrease, leading to a lower breakdown voltage. Therefore, it is essential to consider the operating temperature when determining the breakdown voltage of a capacitor.

5. Can the breakdown voltage of a two dielectric capacitor be improved?

Yes, the breakdown voltage of a two dielectric capacitor can be improved by using materials with higher dielectric strengths or by increasing the distance between the two dielectric layers. Additionally, proper design and manufacturing techniques can also help improve the breakdown voltage of a capacitor.

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