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capacitors in series with DC voltage

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eecs4ever

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So if i have 2 ideal capacitors in series, suppose the values are equal.
Assume theres no parasitic resistances.

----||-----||--------

and i apply 6v DC voltage accross them.

what is the voltage in the middle?

1. Its 3V by the capacitor divide? C/(C+C) . Is this valid for DC voltages?
Does this work because of charge sharing?

OR

2. Since the capacitors block off DC completely, the middle node remains
floating, and thus the voltage is undefined. ?

Which is right?
I'm not sure. I hope someone knows =)
 

japholix

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The first is the correct one :

The total series capacitance is C/2, the charge stored in each capacitor is then 6*C/2 = 3C and now the voltage across either one is just 3C/C = 3 because of the symmetry
 

    eecs4ever

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Naveed Alam

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eecs4ever said:
2. Since the capacitors block off DC completely, the middle node remains floating, and thus the voltage is undefined. ?
this point is right... b/c suppose u put 2 reversed bias diodes in place of these 2 capacitors, the situation is same and for that case ur second option is valid..(if u practically see this, u might get zero voltages across each capacitor, and with some volt-meters it will fluctuate)
so, ur second option is valid, for sure

Regards
 

sharas

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Option 2 is correct. Every node must have a DC path to ground.

I simulated it in SPICE, and it wouldn't compile.
 

v_c

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That is because spice is having trouble trying to invert a singular matrix (which has a determinant equal to zero). In reality, all capacitors have series and parallel resistances and there will not be any floating notes. In theory, as I mentioned earlier, an impulse of current will flow to charge the capacitors instantaneously. So this is a case of spice lying to us. Just put a large resistance across the capacitors (say \[ 10M \Omega=10^7 \Omega\]) to fix this issue. Putting this capacitor across the capacitor is not cheating as the real system will have such parasitic elements.

Best regards,
\[v_c\]
 
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    eecs4ever

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sharas

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Thanks, V_c,

You are right, it will actually be a voltage divider.
 

rami.masadeh

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i dont think that it can work as a voltage devider because the capacitors will reject the DC voltage to pass throw it, and it has a very high resistenace to the DC voltages, so it will be treated as open circuit.
 

verilog_coder

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well i have a feeling that both capacitors will burn out or battery will burn out.
 

v_c

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I think the question that was posed by the original poster was a conceptual problem. Using this as a voltage divider in practice might not be realistic. Although, you will see 2 capacitors used with dc a voltage sources and 2 transistor switches to make a half-bridge topology. If you do a google search on "half bridge" you will see how a dc voltage of , say 100V, is made into two 50V sources using 2 capacitors across the 100V source.

So to summarize ... by itself this circuit might not do much, but used in other circuits it does serve a purpose as a type of voltage divider.

Best regards,
v_c
 
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    eecs4ever

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Kral

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eecs4ever,
The equivalent capacitance is C/2.
Let Vs = the supply voltage.
The total charge =Qt =CVs/2
Let Qc be the charge on each capacitor
1/2 the total charge is stored in each capacitor,
so Qc =Qt/2 = CVs/4
Let Vc = voltage across a single capacitor.
Qt/2 = CVc/2 = CVs/4
Solve for Vc
Vc = 2Vs/4
Vc = Vs/2
~
The same reasoning can be applied to unequal capacitances.
~
So, for your example, the voltage across each capacitor = 3V.
Regards,
Kral

Added after 27 seconds:

eecs4ever,
The equivalent capacitance is C/2.
Let Vs = the supply voltage.
The total charge =Qt =CVs/2
Let Qc be the charge on each capacitor
1/2 the total charge is stored in each capacitor,
so Qc =Qt/2 = CVs/4
Let Vc = voltage across a single capacitor.
Qt/2 = CVc/2 = CVs/4
Solve for Vc
Vc = 2Vs/4
Vc = Vs/2
Regards,
Kral
 

    eecs4ever

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hani51

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hi
i think theoritically the capacitors considered with infinity resistance so the two adjacent plates and the conductor in between will be considered as insulator between the outer plates of both capacitors and the voltage is floating and the two capcitors will behave like one capacitor with a conductor in the electric feild inside

practically as mentioned before there is a very high impeadance which can be considered as parallel resistance
this is for steady state operation

for transient operation the DC voltage will be rise exponentially and very fast depend on conductor resistance and capacitor resistance and capacitance which will cause some current to flow till the voltage build up on the capacitors till they reach the steady state then the voltage in between the two capacitors will be floating (undefined)

correct me if i am wrong please
 

artem

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you can try to find answer either in time domain or in frequency domain - because applying DC votage can be replaced by sum of sine cosine functions .
Not going into each particular frequency constituting
DC pulse we can instead calculate voltage drop per each frequncy constituting pulse.
Then voltage across each capacitor will be calculated via normal serries connection impedance where impedance for C will be 1/wC . Per each frequency we have a impedance
c/(c+c) which is irrespective of frequency . That means voltage on capacitor connection point will be proprotional only to capacitor values and option 1 is in place.
Quite good example on Fourier series justification).

Or you can try to go into differential equations and solve bit longer math there .
 

v_c

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I thought I already answered this question last week . Check out my responses here
Article: Big Door Handle

I think there are two topics with the same subject line! I was getting confused since I did not see my reply with the pictures on this thread.

Best regards,
v_c
 
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    eecs4ever

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Sal

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Hi

If the capacitors are ideal, the voltage you get at the middle point should be zero after the transient time is over.
If initially the 2 capacitors are discharged and you apply the power supply voltage a short circuit will occur , and after a while the middle point of the capacitors should be like any other point outside the circuit , if measured by an ideal multimeter(infinite input impedance)

cheers
 

eecs4ever

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Sal,

I don't think you are right. v_c answered this question already in a different place. I accidentally posted twice.

h**p://www.edaboard.com/viewtopic.php?p=556441#556441
 

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