Energy within an electric field

[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.