Electronics > Beginners
nobody talking about switching PS wasting power on input filter caps?
Ice-Tea:
--- Quote from: radiolistener on August 24, 2019, 04:45:52 pm ---Just imagine DC battery with + and - terminals. And capactior with A and B terminals.
Now connect DC battery to capacitor: + to A and - to B. The capacitor will be charged.
Now connect DC battery to capacitor with reversed polarity (this is how AC current works): - to A and + to B. The capacitor first will be discharged through DC battery and then charged to reversed polarity.
--- End quote ---
One more try. Look at the diagram. What you are describing is 2 phases out of 4. There where you have a positive voltage and you're charging the cap (first 90° section, postive voltage positive current) and there where you have a negative voltage and charging the cap (to a negative voltage) (third 90° section, negative current and negative voltage). This does not take into account to the two other sections that have postive voltage and negative current and the part where the voltage is negative and the current positive.
james_s:
There is no wasted energy. Rather than making an ass of yourself by insisting there is against several people who know what they're talking about, go try it yourself. Find a small portable backup generator and connect some large capacitors across it. You can monitor the engine load by listening to the sound it makes.
Another method to prove there is no wasted energy is take a bunch of capacitors, wire them in parallel and place them inside an insulated styrofoam cooler with a temperature sensor and connect it to the mains. Measure the current draw and calculate the apparent power, then record the temperature rise over a period of say 30 minutes.
Now find an incandescent lamp that will draw roughly the same current as the capacitor bank and place it in the same box, repeating the experiment in place of the capacitor bank. You will find that the lamp results in a dramatic temperature rise while the capacitors do not and this can only mean one thing, the lamp is consuming energy and the capacitors are not. Energy cannot be created or destroyed, for energy to be wasted something has to heat up.
Zero999:
--- Quote from: radiolistener on August 24, 2019, 04:45:52 pm ---
--- Quote from: Ice-Tea on August 24, 2019, 03:52:39 pm ---I have no idea how to explain this to you. However, you will not find anyone on this board or on another tech-savvy forum that agrees with you. Unless it's called "Free Energy For All" or something. This should tell you something. And perhaps it should bring you to read up on the subject.
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There is no "Free Energy For All" at all, there is only wasted energy.
--- Quote from: Ice-Tea on August 24, 2019, 04:09:49 pm ---No. Look at the diagram I posted earlier.
--- End quote ---
Just imagine DC battery with + and - terminals. And capactior with A and B terminals.
Now connect DC battery to capacitor: + to A and - to B. The capacitor will be charged.
Now connect DC battery to capacitor with reversed polarity (this is how AC current works): - to A and + to B. The capacitor first will be discharged through DC battery and then charged to reversed polarity.
Let's assume that capacitor charge takes energy E from battery. So, during experiment you spent 3*E energy (charge+discharge+charge) from DC battery. But capacitor stores just 1*E energy. So, you're lose 2*E energy.
And note, this energy is NOT consumed by capacitor. It is consumed by DC battery. This is wasted energy of DC battery.
The same things happens with AC generator with connected capacitor on the output.
On rise edge of sine wave capacitor will be charged from AC generator. On falling edge of sine wave capacitor will be discharged to AC generator and then charged again in reverse polarity. It happens in a loop.
As you can see, on falling edge capacitor will be discharged to AC generator.
It means that current from capacitor (which flows through AC generator) will make resistance to mechanical rotation. And you will spend more gasoline for AC generator rotation. You're just wasted that energy.
This is something like use gasoline to move car from point A to point B and then move it back to point A. Gasoline wasted and the car is staying at the point where it was before :)
--- End quote ---
Your battery analogy is flawed because it's a square wave, which will theoretically cause infinite current to flow when the polarity is reversed. In reality the power will be dissipated in the internal resistance of the battery and capacitor. The mains is sinusoidal, so you're not comparing like with like.
Look at what happens when the power drawn from V1 is plotted in LTSpice. It starts at zero, gradually goes more negative (in this case a negative number is power being taken from V1)), reaches a peak at 45º, then falls to zero again and climbs to positive, as the energy stored in the capacitor is returned to V1. Note, that average power taken from V1 is zero, over the entire waveform. No one is claiming perpetual motion.
In real life there will be some resistance which will dissipate power and cause the generator to work harder, but it won't be as extreme as with a square wave. The mains also typically has many inductive loads, such as motors and lightly loaded transformers, so a little bit of capacitance will help to bring the power factor closer to unity.
soldar:
--- Quote from: james_s on August 24, 2019, 04:44:37 pm --- Eventually perhaps it will sink in and you will understand what we are trying to say. Until then just accept that a reactive load does not cause the generator to draw more mechanical energy from the engine, turbine or whatever is driving it, beyond the small percentage that becomes heat in the windings and wiring. The fact that you do not grasp how this can happen does not make it any less true. It is easily demonstrated though if you can find a portable AC genset and some capacitors or inductors.
--- End quote ---
You know, it just occurred to me that what he is saying is, in a way, the same as the free energy crowd. They say they can make energy appear out of nowhere and radiolistener is saying energy disappears which is just as impossible. Energy can be stored but it cannot be made to disappear. Energy can be converted to heat and that is precisely the question of the OP: why do capacitors that are apparently consuming power do not get hot? And the answer is because they do not consume power, they store power and then return it. If they "wasted' it then it would be converted to heat.
radiolistener:
--- Quote from: Ice-Tea on August 24, 2019, 05:07:42 pm ---One more try. Look at the diagram.
--- End quote ---
Ok, let's look at diagram.
We have 311 Vpk 50 Hz AC source loaded with 31.83 uF capacitor (100 Ohm at 50 Hz).
So, let's look at diagram of power consumption for AC source:
green diagram = voltage on AC source
red diagram = power consumption on AC source
As you can see, AC source first feeds capacitor with generated power (negative power consumption), and then AC source consume power from capacitor (positive power consumption).
What happens with power consumed by AC source?
I provided you with answer - this power will be used against AC generator. In simple words this power will be used to stop rotation of AC generator rotor. In order to continue AC generating you're needs to apply more mechanical power to AC generator.
This is my explanation. So, what is your explanation, what is happens with power consumed by AC source? :)
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