# Power Balance of a Simple RC Circuit

Discussion in 'Off-Topic' started by Omnibus, Mar 8, 2011.

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1. ### Omnibus Thread Starter New Member

Mar 8, 2011
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Hi All,

Recently I've come upon a result which, while inherent in the theory of electricity, has not been noticed so far. Lately, I found a great way to demonstrate it using PSpice which eliminates the need for expensive equipment and/or tedious calculations.

Create a schematic in PSpice (cf. the attachment) of an RC circuit powered by an 800kHz pulse generator, voltage amplitude of 1.17V and a voltage offset of -3.356V. The value of R is 9.9244Ohms and the value of C is 115pF. Run the simulation from 1.25ms to 2.5ms at 1.25ns increment (that's one period). Once the simulation completes transfer the data into an Excel spreadsheet (by going to Edit->Select All->Copy).

Then calculate the input power Ein as the average of the instantaneous products Ii*Vi of the instantaneous current Ii and voltage Vi values. Do the same for the output power Pout where you would average over the instantaneous values of the product Ii*Ii*R. Then form the quotient Pot/Pin and you will see an amazing thing. Not only Pout differs from Pin but Pin has negative value. Negative value of Pin means that all the power is returned to the power source. Thus, not only is Joule heat produced in the dissipative element, active R, but, in addition, energy is returned to the source. A similar effect can be observed with an LRC circuit with properly chosen values of the elements. In the LRC circuit the Pout/Pin > 1 effect can be achieved even in absence of voltage offset.

One warning. Don't use the power analysis, part of SPice because it is based on procedures involving many more arithmetic operations than the procedure I described which leads to inevitable amassing of errors due to the essence of the digital machine we're using for this calculation. These errors will obscure the above effect.

The procedure I described is the most transparent and straightforward possible procedure of processing exact data such as the list of Ii's and Vi's produced by the simulation. You may want to note that the product Ii*Vi is the value of the instantaneous slope of the energy-time curve. The sum of all the Ii*Vi products, divided by the number of points n will give the average power Pin over the entire period. Therefore, we don't need integration or any other procedure, especially containing approximations, to determine what the average power during the period is. The same applies for Pout, when averaging over all Ii*Ii*R values within that period.

Those interested in discussing this phenomenon in more detail or for some other reason may also sign up at actascientiae.org/v. That board is also LaTeX enhanced.

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2. ### CDRIVE Senior Member

Jul 1, 2008
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What does this statement mean?

3. ### Omnibus Thread Starter New Member

Mar 8, 2011
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This means that is a newly discovered effect, inherent in the existing theory. The effect doesn't come out of any new expressions of current, voltage or how power is calculated.

4. ### wayneh Expert

Sep 9, 2010
12,394
3,246
You must have missed a day of beginning calculus. What you've described IS integration, and includes the error of n being anything less than infinity.

5. ### Omnibus Thread Starter New Member

Mar 8, 2011
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Agreed. That's the correct way of integrating power over time, not the way it's usually done (in Pspice as well).

As for the error, how do you explain that mere changing the offset changes the Pout/Pin so dramatically, the quotient Pout/Pin being unity, as expected, only for zero voltage offset?

6. ### Omnibus Thread Starter New Member

Mar 8, 2011
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So, I guess, what we agree on now is that in our case
$(\frac {1} {n} \sum\limits_{i = 1}^n I_i V_i)( t_n - t_1)$
is the correct way of integrating $I_i V_i$ over time, rather than the usual

$\sum\limits_{i = 1}^{n-1} \left( \frac {W_{i+1}-W_{i}}{2} (t_{i+1}-t_{i}) \right)$
where $W_{i} = I_i V_i$ and $t_i$ is the the time at instant i.

P.S. Note, writing just $\frac {1} {n} \sum\limits_{i = 1}^n I_i V_i$ is not the integral, as you inferred.

7. ### wayneh Expert

Sep 9, 2010
12,394
3,246
Those are only integrals at the limit where n approaches infinity. Otherwise they are approximations without the error term estimate.

I'd need to think more about the question you've posed. But what's the point? This stuff has all been worked out long ago and by every EE student since. If you're identifying a quirk or shortcoming in specific software, that's fine but otherwise I don't think you've made it clear to folks here what you're trying to communicate. Imagine explaining it to your grandmother. (no offense to the other forum members! )

8. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Congratulations!!!! You have proven perpetual motion.

That is an interestingly empty web site too...

Last edited: Mar 9, 2011
9. ### Omnibus Thread Starter New Member

Mar 8, 2011
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First, the website you say is empty has been set up by a friend to have a discussion on these matters.

The site which you're citing doesn't contain arguments and just seems to be a place for expressing some general opinions. Correct me if I'm wrong.

On the other hand, I'm providing clear arguments for what I'm claiming. I would be most interested to see counter arguments to my arguments. I guess that's the usual procedure during a scientific discourse. If you can provide such counter arguments not only will I agree this thread is off topic but I would ask you to remove it from the forum. I hope you understand my best intentions.

Last edited: Mar 9, 2011
10. ### Omnibus Thread Starter New Member

Mar 8, 2011
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Let me try to explain it. Aside from quirks in software (which is not the point I'm making; software is just fine) what I'm showing is that even for $n \rightarrow \infty$ the quotient $\frac {P_{out}} {P_{in}$ must be a function of the voltage offset $C$ and only when $C = 0$ will we have $\frac {P_{out}} {P_{in}} = 1$. So, what I'm trying to communicate here is that this fact, having far-reaching consequences, has not been noticed so far. Indeed, let's observe the expression for Pin for a given set of $I_i$ and $V_i$ (notice carefully that for that same set of $I_i$ and $V_i$ Pout is always the same):

$P_{in} = \frac {1} {n} \sum\limits_{i=1}^{n} (C + V_i) I_i = \frac {1} {n} \sum\limits_{i=1}^{n} C I_i + \frac {1} {n} \sum\limits_{i=1}^{n} V_i I_i$

It is obvious from the above that because of the first term on the right of the second equality the value of Pin will be a function of C for a given set of $I_i$ and $V_i$ even when $n \rightarrow \infty$. Pin will equal Pout only when $C = 0$. This we can see in concrete terms by using PSpice although the fact of varying $\frac {P_{out}} {P_{in}}$ is evident from the above purely algebraic considerations. So far, we have known that $\frac {P_{out}} {P_{in}}$ must be unity under all circumstances, offset or no offset of voltage. This understanding should be corrected.

11. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Obviously, the other link did not convince you. These "proofs" that you can get something for nothing have been offered to us for some time. Try this link, and the links within it - http://forum.allaboutcircuits.com/showthread.php?t=28067

We have built up a certain immunity to the claims, and a certain weariness towards the discussions thereof. It can't happen in the real world.