# Can DC capacitor charging energy be made more efficient by adding an inductor?

#### Ele1

Joined Dec 3, 2014
17
Where did half of the capacitor charging energy go?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4

When a capacitor is charged from a battery in a DC circuit, half the energy is lost.
QV is the energy from the battery and (1/2)QV is the energy stored on the capacitor.
The other (1/2)QV is lost to resistance.

Question:
If an inductor is placed in series with the capacitor then would (some of the) energy
lost to resistance be instead stored in the inductor and further charge the capacitor
passed (1/2)QV as the capacitor charging current decreased?

As the capacitor charging current decreases, would the inductor respond by becoming
a source in series with the battery, increasing the circuit voltage to resist the decreasing
capacitor charging current?

#### MikeML

Joined Oct 2, 2009
5,444
Adding an inductor will do nothing but make it take longer for the capacitor to reach its final voltage, which will be the same with or without the inductor. The final charge on the capacitor will be the same; 1/2CV*V

#### JDT

Joined Feb 12, 2009
657
When a capacitor is charged from a battery in a DC circuit, half the energy is lost.
QV is the energy from the battery and (1/2)QV is the energy stored on the capacitor.
The other (1/2)QV is lost to resistance.
The energy is lost to resistance. The non-mathematical answer is this:
When the capacitor is first connected to the battery the charging current is infinite. Well, it would be, but in fact is limited by the internal resistance of the battery and the connecting circuit, and the series internal resistance of the capacitor. This is where the energy losses occur.

Inserting an inductor will control the current. So in many cases the energy loss will be lower.

You are right, that the inductance will try to keep the current flowing even after the capacitor is fully charged. This causes overshoot or ringing.

Basically, the LC series network will oscillate at its resonant frequency, damped by the resistive losses in the circuit.

#### MikeML

Joined Oct 2, 2009
5,444
...
Inserting an inductor will control the current. So in many cases the energy loss will be lower.
...
I agree that adding an inductor might reduce the peak current. I disagree that the energy loss will any less. The charge (and energy) in the capacitor at the end will no more or less with or without the inductor.

Last edited:

#### shortbus

Joined Sep 30, 2009
8,951
Probably should start my own thread but... What about discharging a capacitor through an inductor? Does it then make the energy stored in the cap last a slight bit longer?

#### MikeML

Joined Oct 2, 2009
5,444
Probably should start my own thread but... What about discharging a capacitor through an inductor? Does it then make the energy stored in the cap last a slight bit longer?
Since an ideal inductor is lossless, then discharging a lossless capacitor through a lossless inductor would take forever. #### Bernard

Joined Aug 7, 2008
5,788
Only if L is infinite.

• #12

#### crutschow

Joined Mar 14, 2008
28,153
If you connect an ideal inductor in series with an ideal capacitor and apply a step voltage, V to the inductor, then the inductor will store energy during the capacitor charge and transfer it to the capacitor (resonant charging).
If you terminate the charging at the point the inductor current drops to zero then all the energy is transferred to the capacitor and the final capacitor voltage is twice the source voltage, giving a final energy on the capacitor of 1/2 C(2V)^2 = 2CV^2, with no energy loss in the transfer.

• Mark Flint

#### MikeML

Joined Oct 2, 2009
5,444
Only if L is infinite.
No, only if R=0. L only controls the frequency of the the oscillating circuit, not the rate of energy dissipation.

#### #12

Joined Nov 30, 2010
18,223
I'd give you a, "like" but you didn't make me laugh out loud (as Bernard did).

#### ian field

Joined Oct 27, 2012
6,539
Where did half of the capacitor charging energy go?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4

When a capacitor is charged from a battery in a DC circuit, half the energy is lost.
QV is the energy from the battery and (1/2)QV is the energy stored on the capacitor.
The other (1/2)QV is lost to resistance.

Question:
If an inductor is placed in series with the capacitor then would (some of the) energy
lost to resistance be instead stored in the inductor and further charge the capacitor
passed (1/2)QV as the capacitor charging current decreased?

As the capacitor charging current decreases, would the inductor respond by becoming
a source in series with the battery, increasing the circuit voltage to resist the decreasing
capacitor charging current?
The inductor will have winding resistance, and probably more of it than just the wires.

The nearest thing I can think of is a SMPS circuit with soft start - but you'd still have to work hard for any efficiency gain.

#### WBahn

Joined Mar 31, 2012
26,398
Only if L is infinite.
Nope. The value of L and C merely set the frequency at which the energy will oscillate back and forth between the two. As long as R=0 there is no way for the energy to leave the circuit (except possibly as RF radiated energy) and so it will ring forever.

#### WBahn

Joined Mar 31, 2012
26,398
Where did half of the capacitor charging energy go?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4

When a capacitor is charged from a battery in a DC circuit, half the energy is lost.
QV is the energy from the battery and (1/2)QV is the energy stored on the capacitor.
The other (1/2)QV is lost to resistance.

Question:
If an inductor is placed in series with the capacitor then would (some of the) energy
lost to resistance be instead stored in the inductor and further charge the capacitor
passed (1/2)QV as the capacitor charging current decreased?

As the capacitor charging current decreases, would the inductor respond by becoming
a source in series with the battery, increasing the circuit voltage to resist the decreasing
capacitor charging current?
In theory, yes. But it's not as simple as that.

What you have to do is throw the switch to connect the source, inductor, and capacitor in series. Then at just the right moment when the inductor current reaches a maximum, you have to throw the switch to put the inductor and the capacitor in series (or parallel -- same thing with only two components) and then, just as the current dies in the inductor, open the switch to prevent the capacitor from discharging back through the inductor. When you throw the switch to remove the source you will have an inductive kick that will cause arcing across the switch contacts unless you snub them out with a diode or capacitor across the contacts and there will be losses associated with that.

#### shortbus

Joined Sep 30, 2009
8,951
Reason I asked is it's for an ongoing project, an EDM, electrical discharge machine. In an EDM the ideal wave form is for the voltage to lead the current. But the voltage also needs to be "measured consistent" pulse of energy discharged through the working gap of an electrode. Usually from a capacitor, but in a cap the current leads the voltage, just the opposite of an inductor. But the inductor gives the correct wave form needed. So, this is why I thought discharging a cap through an inductor then into the gap would work.

#### Ele1

Joined Dec 3, 2014
17
In theory, yes. But it's not as simple as that.

What you have to do is throw the switch to connect the source, inductor, and capacitor in series. Then at just the right moment when the inductor current reaches a maximum, you have to throw the switch to put the inductor and the capacitor in series (or parallel -- same thing with only two components) and then, just as the current dies in the inductor, open the switch to prevent the capacitor from discharging back through the inductor. When you throw the switch to remove the source you will have an inductive kick that will cause arcing across the switch contacts unless you snub them out with a diode or capacitor across the contacts and there will be losses associated with that.
I get a good view of what is happening.

#### Ele1

Joined Dec 3, 2014
17
If you connect an ideal inductor in series with an ideal capacitor and apply a step voltage, V to the inductor, then the inductor will store energy during the capacitor charge and transfer it to the capacitor (resonant charging).
If you terminate the charging at the point the inductor current drops to zero then all the energy is transferred to the capacitor and the final capacitor voltage is twice the source voltage, giving a final energy on the capacitor of 1/2 C(2V)^2 = 2CV^2, with no energy loss in the transfer.
Terminating the charging at the point the inductor current drops to zero means the capacitor does not release current back into the inductor.

2CV^2 is twice QV from the power supply. Should an increase in the final energy on the capacitor be (1/2QV < final energy ≤ QV)?

#### Bernard

Joined Aug 7, 2008
5,788
The title of this post tickled a faint memory from 1957 when I built a capacitive discharge ignition from " Radio & Electronics" ? which used a charging choke to boost the cap. voltage from around 200 V to 300V. It worked for some time untill something died. Do not think it was the same as a boost converter of SMPS.

#### The Electrician

Joined Oct 9, 2007
2,866
Question:
If an inductor is placed in series with the capacitor then would (some of the) energy
lost to resistance be instead stored in the inductor and further charge the capacitor
passed (1/2)QV as the capacitor charging current decreased?
Detailed discussion can be found here:

Using an inductor can eliminate the problem of half the energy being lost in resistance when charging a capacitor. A switch in series to disconnect at the right time, when the inductor current is zero, is key. A diode can do this.

Also see here:

Last edited:

#### Ele1

Joined Dec 3, 2014
17
Using an inductor can eliminate the problem of half the energy being lost in resistance when charging a capacitor. A switch in series to disconnect at the right time, when the inductor current is zero, is key. A diode can do this.

Also see here:

I like your post, your explanation is clear. The inductor can transform Q into V, but QV remains the same and is equal to the power source QV (ideally). The capacitor is at a higher voltage, but did the energy loss still occur? If the total QV from the power source were determined, would the capacitor discharge more than 1/2(QV)?

I reexamined the article:

Where did half of the capacitor charging energy go?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4

The article states,

"Part of the intuitive part that goes into setting up the integral is that getting the first element of charge dq onto the capacitor plates takes much less work because most of the battery voltage is dropping across the resistance R and only a tiny energy dU = dqV is stored on the capacitor."

The article relates that less energy is required to place charge on a capacitor at the beginning of charging process. This leads to the conclusion that more energy is required to place charge on a capacitor near the end of the charging process. At the beginning, more energy is available to be lost to resistance.

If you use a constant voltage source to charge the capacitor, then most of the energy is lost to resistance at the beginning of the charging process than at the end. So, if the voltage is varied to be very low at the beginning of the charging process and gradually increasing to some value then the capacitor should be charged more efficiently.

An inductor should be able to do this by intercepting charging current to the capacitor. As the inductor charges, current is absorbed and voltage is opposed from the power source. A lower current and voltage is released gradually to the capacitor as the inductor charges. The capacitor more efficiently charges during the beginning.

As the inductor finishes charging, the energy stored in the inductor can than be released (ideally) into the capacitor toward the end of the capacitor charging process, where more energy is required to place charge on the capacitor. So, both charging and discharging the inductor helps make the capacitor charging more efficient.
------------------------

But, I have problem with the concept that less energy is required to place charge on a capacitor at the beginning of charging process. The article shows the charging process as a polynomial integral for a graph of (voltage vs. time). But, uses an intergration of (voltage vs. charge).

I think a graph of (voltage vs. charge) for a capacitor would be more linear with constant slope. This is reflected in the result to the intergration of (voltage vs. charge), 1/2(QV). The the energy to place charge on the capacitor is the same for nearer the beginning or more toward the end.

Inside the capacitor, I think, the first charges have little resistance going on the plate (low voltage) and have more energy to push off charges on the opposite plate (high current). As more charges are added to the plate, more energy is required to place the charges on the plate (higher voltage), and less energy to push charges off the opposite plate (lower current).

Voltage and current change during the charging, but the energy stored is restricted to (1/2)QV for each charge placed on the capacitor. Something specific about the capacitor is limiting the energy stored. I disagree with the article (well, for now). Perhaps, charges placed on a capacitor have to come in with current and are in motion, but then come to rest on the plates coupled to opposite charges, resulting in an unavoidable, but constant energy loss.