Energy within an electric field

[Bhope69199]

I am trying to calculate the energy within an electric field that is generated between two plates by a pulse but am unsure of what voltage value to use. The pulse is a sinc wave.

I am assuming I can still use the equation $E= \frac{1}{2}CV^2$. I know the $V_{rms}$ and $V_{max}$ which is measured on an oscilloscope. The C value is measured by an LCR meter. I am thinking to use $V_{rms}$ as this provides the same work and energy as the equivalent DC value, however I am not sure as it is a sinc wave.

If I also know the frequency can I also calculate how much energy is within this electric field in 1 second by multiplying the Energy in one pulse by how many pulses occur in 1 second.

 

[BvU]

Still don't see no picture ...

Good to see you have not disappeared from this world ...

Sinc is partially negative. The integral is positive ($\pi$). I would use that.

 

[Bhope69199]

Good to see you have not disappeared from this world ...

Thanks, but I don't have much understanding of integration of a sinc function. I'm confused on how the integral takes into consideration the peak voltage and not just the width of the function.

On a small aside. If the pulse wasn't a sinc wave what would I use?

 

[BvU]

I'm confused on how the integral takes into consideration the peak voltage and not just the width of the function.

Is the link clear to you ?
If the pulse is $\operatorname{sinc} x$ the integral is $\pi$
If the pulse is $A\operatorname{sinc} x$ the integral is $A\pi$

If the pulse wasn't a sinc wave what would I use?

$\displaystyle \int i \;dt$

Did you ever study charging a capacitor, e.g. for smoothing rectified AC ?

Still don't see no picture ...

Of the setup. Of the pulse. And are we talking low, high frequency ?

 

[BvU]

I am thinking to use Vrms as this provides the same work and energy as the equivalent DC value,

The rms of a sine is positive, but the average is zero and a sine wouldn't charge a capacitor

 

[Bhope69199]

If the pulse is $A sinc ⁡x$ the integral is $A π$

Ah OK.

Why would I use

$\displaystyle \int i \;dt$

if the pulse wasn't a sinc wave? If I didn't know the current but had the measured voltage curve, can I use:

$\displaystyle \int \frac{V}{R} \;dt$

I understand that AC passes through the capacitor and pulsed DC essentially charges then discharges the capacitor. So using a pulse DC the capacitor will charge until the voltage gets to a maximum then the charge will fall as the pulse drops to zero. What I would like to find out is the energy within the electric field during the pulse.

The frequency is low at 250Hz.

What would you like a picture of? The setup is a pulse source connected to a capacitor and the voltage drop across the capacitor is measured.

 

[BvU]

Why would I use $\displaystyle \int i \;dt$

Because $Q = \int i$

Re

Can I use $\displaystyle \int \frac{V}{R} \;dt$

If you have an $R\ , \$yes. But so far I haven't seen an R in your story or your picture ...

a pulse source

A $\ \operatorname{sinc}$ source ?

 

[Bhope69199]

It was more to check my understanding. There is no R.

A sinc is a pulse isn't it?

Using $A π$ where A is the peak voltage of the sinc wave to find the energy in one pulse, then multiply it by the number of pulses in 1 second gives me more energy than the power supply delivers. Where am I going wrong in calculating the Energy within the electric field?

 

[BvU]

Perhaps in forgetting the time scale?

 

[Bhope69199]

Ah so, it's not Aπ. It's the limits of the size of the sinc wave that I need to integrate between. If the pulse is 150us wide, then integrate between 0 and 150us:

$\int_{0}^{150} {A \, sinc \, x} \, dx$

 

[tech99]

The energy in the capacitor increases as the pulse amplitude increases and decreases as he pulse amplitude decreases. The peak energy occurs at peak voltage. E = 1/2 CV^2. Are you trying to find mean energy, in which case find the mean voltage of ther pulse and apply that.

 

[BvU]

Ah so, it's not Aπ. It's the limits of the size of the sinc wave that I need to integrate between. If the pulse is 150us wide, then integrate between 0 and 150us:

$\int_{0}^{150} {A \, sinc \, x} \, dx$

No. You want to scale the time axis

 

[Bhope69199]

Scale the time on the oscilloscope? Or scale the time when calculating the integral? How do I scale the time when calculating the integral and to what?

 

[Bhope69199]

The energy in the capacitor increases as the pulse amplitude increases and decreases as he pulse amplitude decreases. The peak energy occurs at peak voltage. E = 1/2 CV^2. Are you trying to find mean energy, in which case find the mean voltage of ther pulse and apply that.

I'm trying to find the total energy. Essentially the sum of the energy as the pulse amplitude increases and decreases.

 

[Bhope69199]

Scale the time on the oscilloscope? Or scale the time when calculating the integral? How do I scale the time when calculating the integral and to what?

I have the oscilloscope data. I cropped the data so I have one wavelength of the pulse. Are any of the methods below accurate in calculating the total energy:

1. Sum the voltages of every point recorded in the wavelength. Take the voltage total and divide by the number of points. This gives an average voltage per point. Use this to calculate the average energy per point. Multiply this by the total number of points in one wavelength to give the total energy per wavelength. Multiply this by the frequency to get total energy per second. Final value around 0.03J.

2. Calculate the energy per point using the voltage at that point in the whole wavelength. Sum all the energies at each point in the whole wavelength to get the total energy. Divide this by the number of points recorded in the wavelength to get the average energy per wavelength. Multiply this by frequency to get the total energy. Final value around 0.9J.

Each method gives a very different result. Capacitance is 80pF and Freq is 250Hz. Vmax is about 10kV. Pulse width approx 150us.

Are any of the above methods accurate? If not what would be the best method?

 

[tech99]

Sorry to labour my previous point. If you apply a pulse of voltage to a resistor, the energy dissipated increases with the duration of the pulse. But if you apply it to a capacitor the energy put into it does not depend on the length of pulse, only on the peak amplitude.
It's like applying a mechanical pulse to a spring. The energy put into the spring depends only on the peak of the distance, and if you apply the pulse slowly the stored energy is unaltered.

 

[Bhope69199]

Ah OK thanks. That analogy has helped.

So to find the energy I'll use $E = \frac{1}{2}CV_{peak}^2$

 

[BvU]

Scale the time on the oscilloscope? Or scale the time when calculating the integral? How do I scale the time when calculating the integral and to what?

Easiest (if your pulse is really $\operatorname{sinc} x$ ) is to take the first zero point: $\operatorname{sinc} x = 0$ at $x = \pi/2$.

[edit ] that should be $x = \pi$

If that is 10 $\mu$s on your scope, you have the scale factor.
Do you understand what I mean ?
It follows that

So to find the energy I'll use $E = \frac{1}{2}CV_{peak}^2$

Can't work. A wide peak should deliver more charge than a narrow one.

Of course $E$ is proportional to $V_{\text {peak}}$.

 

[Bhope69199]

Easiest (if your pulse is really $\operatorname{sinc} x$ ) is to take the first zero point: $\operatorname{sinc} x = 0$ at $x = \pi/2$. If that is 10 $\mu$s on your scope, you have the scale factor.
Do you understand what I mean ?

Not really. Is there an example I can look up, or have you got a link to further details?

 

[BvU]

Any calculus textbook on integration ?
$$\begin {align*} \int_{-\infty}^\infty \operatorname {sinc} x \, dx &= \pi \Rightarrow \\
\int_{-\infty}^\infty \operatorname {sinc} \left (t\over a\right ) \, dt&= a\pi \end {align*}$$

 

[tech99]

Easiest (if your pulse is really $\operatorname{sinc} x$ ) is to take the first zero point: $\operatorname{sinc} x = 0$ at $x = \pi/2$.

[edit ] that should be $x = \pi$

If that is 10 $\mu$s on your scope, you have the scale factor.
Do you understand what I mean ?
It follows that

Can't work. A wide peak should deliver more charge than a narrow one.

Of course $E$ is proportional to $V_{\text {peak}}$.

Are we saying then that the width of the pulse has no effect?

 

[BvU]

Are we saying then that the width of the pulse has no effect?

No. I'm bending over backwards to make that clear to the OP, hence this time scale factor !

 

[Bhope69199]

No. I'm bending over backwards to make that clear to the OP, hence this time scale factor !

You are not saying that the width of the pulse has no effect? I'm confused even more now.

$π\frac{A}{B}$ where A is the peak and B is the first zero point.

 

[vanhees71]

You can calculate the response of you AC circuit to any signal by using Fourier or Laplace transform together with the usual harmonic signal. A sinc in the time domain is "box" in the frequency domain, i.e., something $\propto \Theta(-\alpha<\omega<\alpha)$. That's easiest to prove doing it the other way:
$$f(t)=\int_{-\alpha}^{\alpha} \mathrm{d} \omega \frac{1}{2 \pi} \exp(-\mathrm{i} \omega t) = -\frac{1}{2 \pi \mathrm{i} t} [\exp(-\mathrm{i} \alpha t)-\exp(\mathrm{i} \alpha t)]=\frac{\sin(\alpha t)}{\pi t}.$$
Through the inverse you also get the discussed integral:
$$\tilde{f}(\omega)=\int_{\mathbb{R}} \mathrm{d} t \frac{\sin(\alpha t)}{\pi t} \exp(\mathrm{i} \omega t)=\Theta(-\alpha<\omega<\alpha).$$
For $\omega=0$ you get
$$\int_{\mathbb{R}} \mathrm{d} t \frac{\sin(\alpha t)}{t}=\pi.$$

 

[tech99]

No. I'm bending over backwards to make that clear to the OP, hence this time scale factor !

Why does the energy in the capacitor depend on the length of pulse? We have no resistance.

 

[Windadct]

Sorry to come in so late on this...but... E=1/2CV^2 is not really a time dependent equation. Yes - If V is a varying in time then E varies in time.

The E at ANY time depends on V at that moment. So I am totally missing all of this discussion of integration over time.

A Cap (ideal) is a lossless energy storage device, what goes in comes back out. Over the period of any V signal if V=0 then E=0.

The OP was not "what is the Max E?", so it is just : E(t) = 1/2 C V(t)^2 ( The waveform of V is irrelevant)

 

[Bhope69199]

Sorry to come in so late on this...but... E=1/2CV^2 is not really a time dependent equation. Yes - If V is a varying in time then E varies in time.

The E at ANY time depends on V at that moment. So I am totally missing all of this discussion of integration over time.

A Cap (ideal) is a lossless energy storage device, what goes in comes back out. Over the period of any V signal if V=0 then E=0.

The OP was not "what is the Max E?", so it is just : E(t) = 1/2 C V(t)^2 ( The waveform of V is irrelevant)

I am looking to calculate the total Energy. So I assume that if I have a voltage that varies, the total energy is the sum of all energies.

 

[Windadct]

No - there is no "sum"- total energy is the energy at a specific point in time. If the V on a capacitor decreases - then the energy has flowed OUT of the capacitor.

An ideal cap is lossless.

It is like water in a measuring cup - if the height is 1/2 way up indicating 1/2 cup - that is how much you have, pour out some water, you have less water, we do not SUM up the water when we fill and empty the cup.

I find it helpful to consider energy to be a quantity - just like matter. We can not instantly move it from point A to point B ( in a capacitors case this means we can not instantly change the voltage on a cap), we can not create it from "nothing" and we can not destroy it.

 

[BvU]

I have the oscilloscope data. I cropped the data so I have one wavelength of the pulse.

Can you explain what this means ?
What does the pulse look like ? Not like this ? :

If not, then what ? Post a picture of the scope image !

Still don't see no picture ...

Of the setup. Of the pulse. And are we talking low, high frequency ?

Do yourself a favor, post a schematic !

The setup is a pulse source connected to a capacitor and the voltage drop across the capacitor is measured.

Each method gives a very different result. Capacitance is 80pF and Freq is 250Hz. Vmax is about 10kV. Pulse width approx 150us.

Very interesting. 10 kV capacitor ! Perhaps you also know the ouput impedance of the pulse generating device ? And of the oscilloscope input ? With a 10 kV vertical scale ?

After all, 80 pF discharges over 1 M$\Omega$ in 80 $\mu$s !
$\$

 

[Bhope69199]

No - there is no "sum"- total energy is the energy at a specific point in time. If the V on a capacitor decreases - then the energy has flowed OUT of the capacitor.

An ideal cap is lossless.

It is like water in a measuring cup - if the height is 1/2 way up indicating 1/2 cup - that is how much you have, pour out some water, you have less water, we do not SUM up the water when we fill and empty the cup.

I find it helpful to consider energy to be a quantity - just like matter. We can not instantly move it from point A to point B ( in a capacitors case this means we can not instantly change the voltage on a cap), we can not create it from "nothing" and we can not destroy it.

OK. So leading on from this how would I calculate the energy required to charge that capacitor over that pulse?

If the energy within the electric field is at a specific point in time, the energy the source provides is CV^2, but do I have to add up all the specific point in times that the energy of the capacitor is known (i.e over the width of the pulse? or half a pulse as the decrease is energy going back into the source)?

 

[Bhope69199]

Can you explain what this means ?
What does the pulse look like ? Not like this ? :

If not, then what ? Post a picture of the scope image !

The pulse is repeating, with a set amount of time at zero voltage, so I took one wavelength. Yes the pulse is very similar to your image but not exact.

What would the scope image and schematic provide you? When you ask for frequency are you asking for pulse frequency or some other frequency, can you be more explicit?

Oscilloscope 1MΩ impedance, Scope 900MΩ impedance. Not sure of the impedance of the pulse generating device but the resistance is approx. 9kΩ.

 

[vanhees71]

If I understood it right then you just have a capacitor connected to a voltage source which provides a sinc signal. In the usual quasistationary approximation the voltage at the capacitor in this case is just the voltage of the source, and the energy in the electric field within the capacitor is at all times $E_C=C V^2(t)/2$.

 

[Bhope69199]

If I understood it right then you just have a capacitor connected to a voltage source which provides a sinc signal. In the usual quasistationary approximation the voltage at the capacitor in this case is just the voltage of the source, and the energy in the electric field within the capacitor is at all times $E_C=C V^2(t)/2$.

Yes your understanding is correct.

So how would I calculate the energy provided by the power source?
Would it be $CV^2(t)$? Or would I need to find the sum of all $CV^2(t)$ over the whole pulse?

 

[vanhees71]

I don't understand the question. The energy in the capacitor at any instant of time is $C V^2(t)/2$.

Of course this scenario is unrealistic, because in reality you have a finite resistance and there's a Ohmic loss, which you can calculate by integrating the corresponding power (energy per unit time) dissipated in the resistor $P(t)=R i^2(t)$:
$$E_{\text{diss}}=\int_{-\infty}^{\infty} \mathrm{d} t R i^2(t).$$
You can calcuate $i(t)$ by solving the differential equation of this more realistic circuit (most conveniently using a Fourier or Laplace transform).

 

[Bhope69199]

OK the energy in the capacitor at any instant is $CV^2(t)/2$.

How do I find the energy supplied to the capacitor at any time, and the total energy supplied to the capacitor, if I don't know the finite resistance you mention.

 

[BvU]

I have the oscilloscope data. I cropped the data so I have one wavelength of the pulse.

Can you explain what this means ?

Did I miss the answer ?

Yes the pulse is very similar to your image but not exact.

Your reluctance to show the image is almost commendable.

So on the scope you almost see

Re-read post #20: $x = \pi\$ corresponds to t = 75 $\mu$s, so $a = 75 \mu s /\pi\$ $$

10 \,\text{kV} \ \int_{-\infty}^\infty \operatorname {sinc} \left (t\over a\right ) \, dt= 0.75\,{\text Vs} $$

What would you like a picture of? The setup is a pulse source connected to a capacitor and the voltage drop across the capacitor is measured.

Your reluctance to show a schematic of the setup is also almost commendable.
Fortunately you dedicate a few words to it, so we can try to erverse engineer what we are talking and confusing each other about since Thursday:

Oscilloscope 1MΩ impedance, Scope 900MΩ impedance. Not sure of the impedance of the pulse generating device but the resistance is approx. 9kΩ.

So what is it ? 1MΩ or 900MΩ ? I hope the latter...
Now let's try to reproduce our excavations and draw the circuit:

which, according to your description, might be equivalent to

which means the capacitor discharges over the 9 k$\Omega$ in about 70 ns. [edit: 0.7 $\mu$s]

I can't for the life of me understand how the pulse generator can come up with such a weird pulse shape, but other than that it looks like we have a bad case of impedance mismatching with ringing as a consequence.

Without further context in more detail it's hard to say anything sensible about this.

$\$

 

[Bhope69199]

Yes your interpretation is correct! Apologies I meant probe impedance is 900MΩ (scope impedance is 1MΩ).

What further context would you need?

From the posts I believe I have found out the energy within the capacitor at a specific time during the pulse. Now what I would like to understand is how to calculate the energy delivered by the pulse generator over 1 second.

 

[BvU]

What further context would you need?

Brand name and type number of the pulse generator, so I can look up the specs
And what's on the scope if you remove the 80 pF capacitor (if the probe can stand the pulse ...)

From the posts I believe I have found out the energy within the capacitor at a specific time during the pulse.

Freedom of religion, but I have hard time believing it...

Now what I would like to understand is how to calculate the energy delivered by the pulse generator over 1 second.

Basically zero: whatever is pumped into the capacitor is almost immediately dissipated again by the 9 k$\Omega$ .

But you could do $(\int E(t)\,dt )/(\int \,dt )\$
IF (big IF) indeed the voltage over the capacitor is $V(t) = \operatorname {sinc} {\pi t \over 75 \mu{\text s}}$ then $E= {1\over 2} \int CV^2\, dt$ (over 1 pulse) times 4 pulses/s

$\$

 

[Bhope69199]

Great, thanks.

Brand name and type number of the pulse generator, so I can look up the specs
And what's on the scope if you remove the 80 pF capacitor (if the probe can stand the pulse ...)
$\$

It is a bespoke made generator so no specs. Open circuit the pulse is the same shape but about 20% larger in both amplitude and width.

Basically zero: whatever is pumped into the capacitor is almost immediately dissipated again by the 9 k$\Omega$ .
$\$

OK but even if it is dissipated into the 9kΩ there would have been a draw from the power supply surely?. So how do we calculate that energy coming from the power supply?

But you could do $(\int E(t)\,dt )/(\int \,dt )\$
IF (big IF) indeed the voltage over the capacitor is $V(t) = \operatorname {sinc} {\pi t \over 75 \mu{\text s}}$ then $E= {1\over 2} \int CV^2\, dt$ (over 1 pulse) times 4 pulses/s

$\$

Isn't it times 250 pulses in a second? 250Hz, is a pulse every 4ms isn't it?

 

[BvU]

Of course. I mixed up the 250 and the 4. So: yes.

Open circuit the pulse is the same shape but about 20% larger in both amplitude and width.

Most peculiar. I know a capacitor causes lag in a signal. But I never expect lead ...
Or are the peaks shifted in time ?

 

[Bhope69199]

Of course. I mixed up the 250 and the 4. So: yes.Most peculiar. I know a capacitor causes lag in a signal. But I never expect lead ...
Or are the peaks shifted in time ?

The peaks are actually shifted in time.