dc capacitor divider?

Hi all,

Suppose, I have two capacitors connected in series to act as a divider (same value caps) and i test with a DC input say 4V, what would be the output at the top plate of the cap C2 ? What I thought was (and what i suppose to be right) is that since the capacitors are equi-valued, charge stored across each cap will be the same; so to the left of C1 if positive charges are developed, an equal amount of negative charges will be available on its right plate. At the same time an equal amount of positive charge will be on the top plate of C2, and the bottom plate as is grounded, will have no charge. So effectively, at the output node (which is the top plate of C2), the charges cancel out, 0V will be available at the output.

But, the transfer function of the system (Vout(s)/Vin(s)) = 1/2. So, this states that , output must be 2V
Also, if i apply charge conservation axiom at the output node, initial charge is 0C = Vout*C + (Vout-4)*C [final charge];
This again give me 2V at the output.

So, which of the answers 0V or 2V is right? In either case, what is the wrong with the contradicting proof(s)?
Plz help me out...

Hello rsashwinkumar !

You are thinking too complicated.

You are connecting 2 caps [equal values] in series across a 4V voltage source.

So each cap. will be charge equally with 2V across it.

rsashwinkumar said:

the bottom plate as is grounded, will have no charge.

Click to expand...

That is incorrect. Just because the plate is grounded does not mean that it has no charge; it must have a charge if the opposite plate is charged. The charge will, like the other capacitor, be the opposite sign, equal magnitude so that the charge overall on the capacitor is zero.

So, we have two capacitors in series, each with identical charge across their plates. Each capacitor will have a potential across it according to Q=CV. It can be simply seen that if Q is the same, and each C is the same, then each V must be the same too.

Since the sum of the two potentials must add up to the total potential, they must be half each of the total potential.

The considerations bringing to 2V are right.
The failure of the first thinking (from which you wuold conclude that you have 0V on C2) is to consider that there are no charges un the bottom plate of C2. There are negative charges that are provided by the - terminal of the battery that equilibrate the positive charges provided to C1.
Maybe you get confused because of the ground connections. Imagine that there is a connection between the - terminal and the bottom of C2 but they are not at ground.

Regards

Z

Thanks FoxyRick,

Going by your argument, suppose if i have say three capacitors (same value) in series, then both the top and bottom plates of the middle capacitor, wont have charges as they would have been cancelled out by the equal opposite charges present in the first and last capacitors, so, effectively, there would not be any voltage across the middle cap.
In that case, too transfer function method suggest 1/3rd of the voltage must be present across middle cap so does charge conservation? What is the missing point here?

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Thanks zorro,

https://en.wikipedia.org/wiki/Voltage_divider#Capacitive_divider

This page describes capacitor divider as not applicable to DC inputs. Also my simulations show that zero volt is only available at the output. So, what is the missing detail ?

Every plate will have a charge, if the top one (or indeed any other) plate does. For three in series, with the top connected to the positive, the order of charges on each plate will be:

+
-
+
-
+
-

If you look at a single capacitor, its net charge will be zero because the charges on its plates are equal magnitude, opposite charge. Again, this must be. But, the plates are charged. Nothing is cancelled out as such; the sum of charges for the whole system is simply zero because we have equal positives and negatives on the plates

Every capacitor will therefore have a potential difference across it according to Q=CV.

The same process applies then as I did above and yes, each capacitor will have one third of the total supply voltage across it (if they are all equal C of course).

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rsashwinkumar said:

This page describes capacitor divider as not applicable to DC inputs. Also my simulations show that zero volt is only available at the output. So, what is the missing detail ?

Click to expand...

That is correct. The reason is that as soon as you connect anything to the junction between the capacitors, you alter the whole system and disturb the simple charge balancing that the capacitors do because charge can then flow somewhere else. All of the previous discussion assumes that nothing is connected to the junctions.

Putting capacitors in series in a DC circuit is only useful in order to increase the total working voltage, not to act as a potential divider. Remember, while charge can move into and out of a capacitor's plates, it cannot flow through the capacitor from one plate to the other. The charge cannot cross the dielectric. No current can flow beyond the brief current to charge the plate. AC circuits are a different matter though, since they work on moving the charge backwards and forwards anyway...

...

Just a suggestion: try to visualise what is happening in terms of the movement of electrons. When the top plate is connected to positive, electrons are attracted towards the positive source and are pulled off that plate. So, the plate now has a positive charge because it is missing some negative electrons.

The other plate of that capacitor now experiences a positive charge close by, and so electrons are attracted to the second plate. These come from the wire and subsequently from the plate of the second capacitor. Effectively, the electrons in the plate-wire-plate junction are shifted upwards. This then leaves a negative charge on the first capacitor's bottom plate (because it has an excess of these 'shifted' electrons) and a positive charge on the top plate of the second capacitor (because its electrons have moved away)

This continues downwards for however many capacitors are in series.

At the bottom plate of the bottom capacitor, the electrons near that plate will experience a pull towards its top plate, just as above, and electrons will shift from the ground (or negative supply potential, whatever you want to call it) into that bottom plate, making it more negative.

So, every capacitor gets equal and opposite charges on its plates.

rsashwinkumar,

Shouldn't this question be in the "Elementary Electronics" section, and not in the "Analog Circuit Design" section?

Suppose, I have two capacitors connected in series to act as a divider (same value caps) and i test with a DC input say 4V, what would be the output at the top plate of the cap C2 ?

Click to expand...

The English language capitalizes the personal pronoun "I". You have not marked the capacitors with a "1" or "2" in your schematic, so I don't know to which capacitor you are referring.

What I thought was (and what i suppose to be right) is that since the capacitors are equi-valued, charge stored across each cap will be the same; so to the left of C1 if positive charges are developed, an equal amount of negative charges will be available on its right plate.

Click to expand...

Two things are wrong with your above statement.

1) Capacitors do not charge, they energize. If you apply 100 volts across a capacitor, it will have the same net charge it had at zero volts, namely zero net charge. Any charge put on one plate will be equally removed from the opposite plate for a net change of zero. The plates will be imbalanced, and this will cause a voltage to form across the plates. This represents a storage of energy in an electrostatic field. Therefore, the capacitor is "energized" when a voltage is applied, or the plates are imbalanced with respect to charge.

2) There are no positive charges involved. All the charge carriers in the circuit are electrons, which have a negative charge. If this were a bipolar junction transistor (BJT), then we could talk about holes which do have a positive charge.

At the same time an equal amount of positive charge will be on the top plate of C2, and the bottom plate as is grounded, will have no charge. So effectively, at the output node (which is the top plate of C2), the charges cancel out, 0V will be available at the output.

But, the transfer function of the system (Vout(s)/Vin(s)) = 1/2. So, this states that , output must be 2V
Also, if i apply charge conservation axiom at the output node, initial charge is 0C = Vout*C + (Vout-4)*C [final charge];
This again give me 2V at the output.

So, which of the answers 0V or 2V is right? In either case, what is the wrong with the contradicting proof(s)?
Plz help me out...

Click to expand...

Your reasoning is too convoluted for me to understand, so let me show you how I would solve the problem of determining the voltage across one or more capacitors connected in series.

Applying a voltage (E) across a capacitor string will imbalance the charge across each pair of plates by a value of (Q). Using the familiar relationship C*E = Q, or E = Q/C, we get the following voltage divider

(Q/Cx)/(Q/C1+Q/C2 ... Q/Cn), where Cx is any capacitor in the string. All the Q's cancel out. For only two capacitors, the voltage divider for C2 becomes C1/(C1+C2), assuming C2 is the last capacitor in the string.

Ratch

Hello,

I am afraid, none of the above answers is correct.
At first we have to consider two different cases: IDEAL or REAL (lossy) capacitors:

* IDEAL: There is no final answer because each distribution of charges resp. voltages is possible. For example: Why not V=V1+V2=Q1/C+Q2/C with unequal voltages?

* REAL: The voltage distribution is determined by the (parallel) loss resistances only.

LvW,

* IDEAL: There is no final answer because each distribution of charges resp. voltages is possible. For example: Why not V=V1+V2=Q1/C+Q2/C

Click to expand...

The voltages across each capacitor connected in series are determined solely by the source voltage, and the capacitance value of each capacitor in conjunction with the values of the remaining capacitors. Q1 and Q2 in the above equation are equal for series capacitors.

* REAL: The voltage distribution is determined by the (parallel) loss resistances only.

Click to expand...

The losses due to leaky capacitors are usually small with respect to the transient current need to energize the capacitors. If the leakage current becomes a problem, the capacitor is usually replaced.

Ratch

Hi Ratch

Ratch said:

The voltages across each capacitor connected in series are determined solely by the source voltage, and the capacitance value of each capacitor in conjunction with the values of the remaining capacitors. Q1 and Q2 in the above equation are equal for series capacitors.

Is this a claim without proof? WHY do you think they are equal?

The losses due to leaky capacitors are usually small with respect to the transient current need to energize the capacitors. If the leakage current becomes a problem, the capacitor is usually replaced.

Click to expand...

Te effect is as follows: After all transients are disappeared the losses allow a continuous dc current, which solely determines the steady-state voltage distribution (Ohms law).
The question whether or not to replace such capacitors is not subject of the discussion.

The following links may be helpful (I hope they work):

https://www.google.de/url?sa=t&rct=...soCgAg&usg=AFQjCNGg7gPI8lMZvThqoeh7dRzp57kOgw


https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.167.3034&rep=rep1&type=pdf

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Here is another indication for the fact that the problem has no solution (ideal capacitors):

No simulation program does allow such a series connection of capacitors because no dc bias point can be calculated. There is no solution for ideal capacitors.
Each node must have a (perhaps high-resistive) connection to ground.

LvW,

Is this a claim without proof? WHY do you think they are equal?

Click to expand...

Because of Kirchoff's current law. Since the capacitors are in series, they all must have an equal charge flow through them.

Te effect is as follows: After all transients are disappeared the losses allow a continuous dc current, which solely determines the steady-state voltage distribution (Ohms law).

Click to expand...

For ideal capacitors, the losses are not a factor. For nonideal capacitors, it must be assumed that the losses are minimal enough such the voltage supply can supply a continuous small transient current to make up for the current leakage. So if a capacitor slowly loses voltage due to leakage, the voltage supply will supply the current to keep the capacitor up to its previous voltage. If the leakage becomes excessive, then we are not talking about a plain string of capacitors anymore.

Neither of the links you sent me pertain to the OP's question.

The first link makes a simple problem hard with the addition of a third capacitor. C1 will increase in voltage as the switching sequence progresses. Its voltage will be equal to a nonconvergent series.

The second link makes a big deal out of figuring out the energy distribution of one capacitor de-energizing into another capacitor. Usually the problem is done by assuming a resistance in the conduction path between the capacitors. But if they want to say that the energy is also used in moving around molecules in the dielectric, that is no problem with me. The bottom line is that the total energy before and after has to be the same and its distribution explained.

Ratch

I think the subject of this thread is an interesting one - but it is not as simple as it seems.
For example, none of the previous contributions in this thread have considered the peak current at t=0 (voltage switch closed), which approaches infinite if all ohmic resistances in the circuit are neglected.
I know that this and similar "Two-capacitor problems" have already caused several papers and written contributions in the past.

Ratch said:

Because of Kirchoff's current law. Since the capacitors are in series, they all must have an equal charge flow through them.

yes, no doubt about this. But does this necessarily mean, that the voltages (charges) of both (ideal) capacitors are equal? The answer is: No.
The only requirement is that the SUM of the capacitor voltages equals Vdc. And this can be fulfilled with each combination V1+V2=Vdc with V1=Q1/C and V2=Q2/C .

For ideal capacitors, the losses are not a factor. For nonideal capacitors, it must be assumed that the losses are minimal enough such the voltage supply can supply a continuous small transient current to make up for the current leakage. So if a capacitor slowly loses voltage due to leakage, the voltage supply will supply the current to keep the capacitor up to its previous voltage.

Sorry, I cannot agree. A simple model can support my claim: Two lossy capacitors (each with a parallel resistor) are connected in series. Under steady state conditions there will be a continuous dc current through the resistive chain. For each resistor Ohms law applies. That`s all - because there is no other solution. Thus, only the losses determine the voltages across the capacitors.

Click to expand...

I think, the problem can be qualitatively answered without long and involved calculations. Let me try it:

There is a general rule: If an electronic circuit (with known and fixed parts values) has only one single solution for the voltage and current distribution within the circuit, it does not matter if you
* calculate all voltages and currents as a function of the driving voltage source, or
* calculate the driving voltage source as a function of known node voltages and path currents.
In contrary - if the circuit has more than one single solution this rule does not apply.

Therefore the question: Is there only one single voltage distribution between both capacitors possible, which leads to the fixed Vdc value? Answer: No - there are multiple solutions as long as the SUM equals Vdc.
I think, nobody will argue against the fact that two equal capacitors can carry different charges (with corresponding voltages V1 and V2 across its plates) and that the series connection of these elements leads to a net voltage of V1+V2.
Thus, the above rule is violated in the present case - which means: Because of multiple possible solutions there is none !

LvW is right. The voltage across each capacitor after charging has stabilized is determined solely by the leakage resistance of each capacitor. That is why, when capacitors are placed in series to achieve a higher voltage rating you will always see a resistor added around each capacitor. The idea is that no matter what the leakage resistances are the explicit resistors will dominate the leakage and cause the voltage drop to be evenly distributed across the several capacitors in series. That is the only safe way to parallel capacitors if the supply voltage is greater than the voltage rating of the individual capacitors.

LvW,

OK, I calculated set up a model consisting of a 1 volt source with 2 series capacitors, each in parallel with a resistor of R ohms, which represent the losses or equalizing resistance. Capacitor C1 has a value of C and C2 has a value of 2C. The voltage across C2 is easily calculated to be 1/2-1/6(exp(-(2/3)*t/(R*C)) . When R===> ∞, which represents an ideal capacitor, the voltage across C2 is 1/3 volt. This is to be expected, because twice the capacitance means half the voltage of C1 which is 2/3 volts. When R is finite and t===> ∞ the voltage across C2 becomes 1/2 volt. This shows that if we wait long enough, the voltages across each capacitor will be proportional to the losses in parallel with the equalizing resistors. In either case, there is only one steady state value for the voltages across each capacitor.

yes, no doubt about this. But does this necessarily mean, that the voltages (charges) of both (ideal) capacitors are equal? The answer is: No.
The only requirement is that the SUM of the capacitor voltages equals Vdc. And this can be fulfilled with each combination V1+V2=Vdc with V1=Q1/C and V2=Q2/C .

Click to expand...

The losses and equalizing resistors will eventually determine the voltages and thereby the charge imbalance of each capacitor. There is only one voltage and charge imbalance value for each capacitor in the steady state. Yes, the voltage sums do equal the source voltage.

Sorry, I cannot agree. A simple model can support my claim: Two lossy capacitors (each with a parallel resistor) are connected in series. Under steady state conditions there will be a continuous dc current through the resistive chain. For each resistor Ohms law applies. That`s all - because there is no other solution. Thus, only the losses determine the voltages across the capacitors.

Click to expand...

I agree with the above statement, but it appears to contradict your first statement which says "The only requirement is that the SUM of the capacitor voltages equals Vdc. And this can be fulfilled with each combination V1+V2=Vdc with V1=Q1/C and V2=Q2/C." Surely you mean that the resistance across each capacitor has to change to accomodate that statement.

Therefore the question: Is there only one single voltage distribution between both capacitors possible, which leads to the fixed Vdc value? Answer: No - there are multiple solutions as long as the SUM equals Vdc.

Click to expand...

There is only one solution for a fixed set of resistance values.

tunelabguy,

LvW is right. The voltage across each capacitor after charging has stabilized is determined solely by the leakage resistance of each capacitor.

Click to expand...

Is that what he means? I am not quite sure.

Ratch

Hi Ratch,

as I have mentioned - it is an interesting problem. Not new - but not a simple question.

I think we are not allowed to neglect connecting wire resistances if we use an IDEAL voltage source.
Otherwise we face a problem. Let me explain: What happens if we at t=0 connect one single capacitor (with or without parallel losses) to an ideal voltage source of 1volt?
According to the circuit simulator the voltage across the capacitor immediately goes to 1volt.
Is this possible? Is this realistic as this is equivalent to a charging current of infinite value?
Therefore, I think that the described ideal configuration without any losses not only is unrealistic, but it is a circuit configuration that cannot be treated mathematically.

And the same applies to the configuration described by you (two lossy capacitors in series). If both capacitors are empty at t=0 they represent a short at t=0 leading to a current of infinity that does not allow us to try a mathematical description of the charging process. And therefore, I do not agree to the formula you have given in your last reply.
I know that the circuit simulator - in case of an ideal voltage source without any series losses - seems to confirm your formula (1/3volt across C2 for t=0). But because of my former the explanation I am sure the simulator fails (must fail) in this case because of the current problem. Remember that each simulator completely refuses to accept two (ideal) capacitors in series.
As soon as we put another resistor in series with the lossy capacitors the voltages across the capacitors is zero for t=0 (as expected)..

I consider this as another justification for the claim that it makes no sense to ask for a solution where there is no solution.

In summary:
When at t=0 a dc voltage is connected to two IDEAL capacitors in series (without losses) it makes no sense to ask for the voltage distribution between both parts because there is no single solution (due to totally unrealistic assumptions and a computational problem with a current approaching infinity).

In case of realistic lossy capacitors we can calculate
(a) the voltages across the capacitors for t===> infinite (charging stabilized) in case of an IDEAL voltage source (zero series losses), or
(b) the voltages across the capacitors as a function of time in case of a real voltage source with series losses. In this case, all capacitor voltages are zero for t=0.

Thank you
LvW

I don't believe that the problem is really complicated. Only referring to the "ideal capacitor" scenario:

1. there's no meaningful solution for the pure DC case. The "middle" node can expose any voltage, you need to define either the voltage, a charge ratio, or an equivalent quantity.

2. if the voltage source is switched, e.g. at t=0, the initial voltage or charge must be known.

LvW,

I think we need to simplify the problem further. Let's see what happens when we put an ideal step voltage (E) across an ideal capacitor (C). I am going to use Laplace transforms to calculate the charge on this capacitor after the ideal step voltage is applied for an infinite time.

The Laplace transform of the ideal step voltage is E/s, and the impedance of the capacitor is 1/Cs. Dividing the impedance into the voltage gives us the Laplace current CE. The inverse Laplace gives us the current in the time domain of CE*Dirac(t). Now, as you must know, the Dirac function is defined to be infinite in value and infinitesimal in duration. Oh, and did I mention it is defined to have an area of 1? Now I will integrate the current from 0 to infinite time to get the total charge imbalance (Q) of the capacitor, and it is easily seen to be Q = CE.

Notice I was able to calculate the charge imbalance without any reference to resistance. The infinite value of the impulse is counterbalanced by its infinitesimal duration to give a finite charge that can be easily calculated without any reference to internal or external resistance. And with a true step function and ideal capacitor, the theoretical time it takes to fully energize the capacitor is infinitesimal.

Of course, if real time values of resistance and a less than perfect step function are considered, then we should expect a finite time to energize the capacitor.

Ratch

FvM said:

2. if the voltage source is switched, e.g. at t=0, the initial voltage or charge must be known.

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Yes, of course - but up to now I think we have assumed an initial voltage of zero. Thus, let`s assume it is known. What do you think does this mean with respect to the problem under discussion?

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Ratch said:

the impedance of the capacitor is 1/Cs.

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Hi Ratch,
to me this is the impedance of a capacitor only in case it is connected with a sinusoidal source. Therefore, your calculation could not fully convince me.

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More than that, I have some problems with the expression you gave in post#14 : 1/2-1/6(exp(-(2/3)*t/(R*C)).
What kind of charging current did you assume during calculation?

Hi Ratch,

here comes another test which can support my claim that there is no single solution:

An ideal dc voltage source is connected to a series connection of two ideal capacitors C1=1uF and C2=2uF - however, the second (grounded) capacitor C2 has a (closed) switch in parallel which is opened after the first capacitor has completely charged, for example at t=1E-3 sec.
Switch parameter: On resistance Ron=1E-3 Ohm, off resistance Roff=1E7 Ohms.
This circuit can be analyzed by a simulation program because of the finite series resistance of 1E-3 Ohms.

After start of a TRAN analysis the first capacitor C1 is charged to 1 Volt very quickly (due to the small time constant Ron*C1=1E-9 sec). The voltage across C2 of course is zero.
If this configuration (voltage source connected to a series combination of two capacitors) would have only one single solution showing a voltage distribution between both capacitors due to the capacitance ratio, the capacitor C2 (after opening the switch at t=1E-e sec) now should accumulate charge and show a rising voltage.
But - this is not the case. The voltage remains constant, which is no surprise because - according to Kirchhoffs voltage law - there is no charging voltage across C2.

Summary: The same configuration as used in your description now shows a completely different voltage distribution between the capacitors - just because of another switching sequence.
That means: The configuration as described by the questioner in post#1 (dc voltage at two caps in series) has no single unequivocal solution (in contrast, for example, to a resistive divider that always has the same divider ratio - independent on any switching events).

LvW,

to me this is the impedance of a capacitor only in case it is connected with a sinusoidal source. Therefore, your calculation could not fully convince me.

Click to expand...

OK, let's call it an immittance then. That won't stop a Laplace calculation as long as the circuit is linear. And any waveform can be represented by a collection of sine waves.

More than that, I have some problems with the expression you gave in post#14 : 1/2-1/6(exp(-(2/3)*t/(R*C)).
What kind of charging current did you assume during calculation?

Click to expand...

I used a 1 volt step simulating a switch turning on at t=0. The expression above is for the voltage of the second capacitor. I can calculate the current or charge if you wish.

Ratch

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Lvw,

here comes another test which can support my claim that there is no single solution:

Click to expand...

Bring it on.

Summary: The same configuration as used in your description now shows a completely different voltage distribution between the capacitors - just because of another switching sequence.
That means: The configuration as described by the questioner in post#1 (dc voltage at two caps in series) has no single unequivocal solution (in contrast, for example, to a resistive divider that always has the same divider ratio - independent on any switching events).

Click to expand...

The results are not unexpected, and do not contradict what I showed before. You are putting a voltage source in opposition with a capacitor that has been energized to the same voltage. Their voltage sum will be zero, so no current will exist in the circuit until the voltage of either the power supply or the capacitor changes. I already shows the voltage distribution of two capacitors in series with a resistor connected across each, and it had a distinct solution. It you put switches in the circuit to set up initial conditions, a Laplace calculation can find the solution to that also.

Ratch