Electronics > Beginners
nobody talking about switching PS wasting power on input filter caps?
james_s:
--- Quote from: stefon on August 23, 2019, 07:27:08 pm ---> A capacitor or an inductance across the line consume reactive power. This means they take power, store it and return it to the network. This means a power factor of less than unity.
I don't get it, it is AC, so current flows thru cap, so it is wasting power, because it is shorting L to N.
> A rectifier bridge followed by a capacitor does not consume any reactive power. At all. None. It keeps everything it gets and returns nothing. And yet has a bad power factor.
i don't care about caps after bridge.
--- End quote ---
No, you don't get it. It's not shorting anything and it's not wasting power.
Current flows yes, but no power is consumed. Ok in reality nothing is ever 100% but for the sake of simplicity we can call it none. A reactive load draws energy during part of the cycle and dumps it back into the line on another part of the cycle. It is power that is just sort of bouncing around, it is passing through the wires but then getting returned to the source without being consumed.
soldar:
--- Quote from: stefon on August 23, 2019, 07:27:08 pm ---I don't get it, it is AC, so current flows thru cap, so it is wasting power, because it is shorting L to N.
--- End quote ---
Current flows through capacitor. Correct. Charging it and storing energy in the capacitor. This energy is returned back to the grid, it is not lost.
Think of an elevator counterweight. As it is raised it consumes (and stores) energy. This energy is returned to the system the next time the elevator is raised and the weight falls.
radiolistener:
--- Quote from: soldar on August 24, 2019, 07:03:07 am ---Current flows through capacitor. Correct. Charging it and storing energy in the capacitor. This energy is returned back to the grid, it is not lost.
--- End quote ---
yes, but this "returned back" energy creates impedance change for power source. So, a power source (power plant) needs to spend more energy in order to fight with this "returned back" energy. Because this "returned back" energy will works against power source. Isn't it?
Also, there is additional power loss in the capacitor and power loss in the wire, due to higher current. All this energy is lost and cannot be consumed. Yes, not all this loss happens inside capacitor, but this energy is wasted due to capacitor :)
Ice-Tea:
--- Quote from: radiolistener on August 24, 2019, 11:02:36 am ---yes, but this "returned back" energy creates impedance change for power source. So, a power source (power plant) needs to spend more energy in order to fight with this "returned back" energy. Because this "returned back" energy will works against power source. Isn't it?
--- End quote ---
No, there is no net effect. The generator needs to make an extra effort to charge the cap, that effort is returned a little bit after that.
radiolistener:
--- Quote from: Ice-Tea on August 24, 2019, 11:30:24 am ---No, there is no net effect. The generator needs to make an extra effort to charge the cap, that effort is returned a little bit after that.
--- End quote ---
"little bit after" amplitude in the mains grid will be changed, because this is AC. So, returned energy will work against power source. This energy will works to distort sine waveform. For power source it will looks the same as additional load. So, electricity generator will needs to spend more power in order to keep the same voltage on the grid.
The difference with usual load is that this energy will be returned back and wasted on AC electricity generator side instead of capactior. But anyway it will be wasted.
First generator needs to make an extra effort to charge the cap, and then generator needs to make an extra effort to discharge the cap. This is worse than just waste that energy on dummy load. If you spend that energy on dummy load, there is no need for generator to spend extra effort to discharge the cap.
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