Ripple voltage

CannaBruh

Well-Known Member
I guess I don't understand how a capacitor is using real power if not storing it and returning it as in apparent?
 

CannaBruh

Well-Known Member
A larger concern would be that current limiting/voltage dropping cap failing short. A transformer gives you isolation, but then why not drop $10 on a cheap driver?
 

nfhiggs

Well-Known Member
I guess I don't understand how a capacitor is using real power if not storing it and returning it as in apparent?
Same way an inductor burns power through inductive reactance - to the current flow, ohms are ohms whether its reactance or resistance. Again this is NOT the same thing as power factor, which is entirely the result of the current/voltage phase shift of a reactive load. That phase shift causes additional heating in the transformer supplying the mains.
 

CannaBruh

Well-Known Member
These are reactive loads though, the only thing dissipating heat in the inductor is the winding resistance of the conductor.
 

nfhiggs

Well-Known Member
These are reactive loads though, the only thing dissipating heat in the inductor is the winding resistance of the conductor.
That's simply not the case. Total heat dissipated by an inductor is (current) x (resistance + reactance).

Edit: correction, its current times resistance squared.
 
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CannaBruh

Well-Known Member
can you explain further, this is completely counter-intuitive to the notion that reactive components do not dissipate power.

as far as I know reactive power is shuttled back and forth, it does no work, and any losses are from non-ideal components, not in the reactance
 

CannaBruh

Well-Known Member
you're talking about losses in non-ideal components; Rac and core losses, eddy current and parasitic capacitance, foil and dielectric losses etc, this is not the same as saying that capacitive or inductive reactance dissipate power in the form of heat, because they don't
 

nfhiggs

Well-Known Member
can you explain further, this is completely counter-intuitive to the notion that reactive components do not dissipate power.

as far as I know reactive power is shuttled back and forth, it does no work, and any losses are from non-ideal components, not in the reactance
You're thinking of the interaction of a capacitor and an inductor in an oscillator circuit - If you measure the AC current between them it will be higher than the input current - because there is a certain amount of current that moves back and forth between them. Again this is different than the current passing through a series filter. Total power across the device is ALWAYS going to be current times TOTAL impedance squared. Total impedance is resistance plus reactance.

You are essentially correct in saying "reactive power" does not do any WORK - that is, it does not help turn a motor or produce light, but it most certainly does produce waste heat. It's not imaginary, nor is it "returned to the grid" in any way.
 

nfhiggs

Well-Known Member
you're talking about losses in non-ideal components; Rac and core losses, eddy current and parasitic capacitance, foil and dielectric losses etc, this is not the same as saying that capacitive or inductive reactance dissipate power in the form of heat, because they don't
Nope. Reactance is most certainly real - it is impedance to AC current just like resistance, and will ALWAYS produce waste heat in the component.
 

CannaBruh

Well-Known Member
Didn't say reactance isn't real, I'm saying that reactance does not dissipate in the form of true power like a resistor or resistive load

when ac voltage is applied to a cap, it stores energy and returns energy, no net loss via heat (with exceptions to considerations mentioned earlier)
same as when you apply a voltage to an inductor, field builds field collapses,

This is not the same as if we applied the same voltage across a resistor, where we would have say 10VAC / 1k ohms = 10mA = 100mW(true power)

same applied to some capacitor

10VAC to a capacitor at a frequency which brings Xc = 1k

the capacitor is not dissipating 100mW of true power


....this is not the same as the tank current in a resonant circuit
 

nfhiggs

Well-Known Member
Didn't say reactance isn't real, I'm saying that reactance does not dissipate in the form of true power like a resistor or resistive load

when ac voltage is applied to a cap, it stores energy and returns energy, no net loss via heat (with exceptions to considerations mentioned earlier)
same as when you apply a voltage to an inductor, field builds field collapses,

This is not the same as if we applied the same voltage across a resistor, where we would have say 10VAC / 1k ohms = 10mA = 100mW(true power)

same applied to some capacitor

10VAC to a capacitor at a frequency which brings Xc = 1k

the capacitor is not dissipating 100mW of true power


....this is not the same as the tank current in a resonant circuit
But it actually IS the same - that's why reactance is measured in ohms. Ask any electronics engineer if he has to account for reactive heat loss when designing AC circuits.

Your understanding of electronic theory is simply not entirely correct. A capacitor charges and discharges, as AC current passes through it. They most certainly DO heat up when AC is passed through them, and that heating is defined above in the equation I gave you - Current times total impedance squared.
 

CannaBruh

Well-Known Member
AC current doesn't actually pass through the dielectric though, electrons leave one plate as they build on another ....

The "reactive heat loss" you're speaking of I believe would relate again to the power triangle, and accounting for I^2*R losses no?

Power is not dissipated in reactive loads.
 

nfhiggs

Well-Known Member
AC current doesn't actually pass through the dielectric though, electrons leave one plate as they build on another ....

The "reactive heat loss" you're speaking of I believe would relate again to the power triangle, and accounting for I^2*R losses no?

Power is not dissipated in reactive loads.
AC passes through a cap VIA the charging and discharging of the plates. Just as it passes from primary to secondary via the magnetic field interaction.

Reactive power losses are calculated via the power triangle. But that power is most certainly "lost" in the form of heat. How else would it be lost? It cannot just vanish into thin air.

While it is true that a PURE reactance does not dissipate power - for all practical purposes, such a device does not exist. A pure reactance would be a capacitor with infinite resistance or an inductor with no resistance.
 

CannaBruh

Well-Known Member
Electrons only flow through the leads to the source & load during the charging/discharging cycles, they do not flow through the dielectric otherwise DC would pass in a neutral state.

That power is most certainly NOT lost it is stored n the form of flux or charge and then returned, the only energy lost as related to the reactive power is the current component in the form of I^2*R losses and losses in non-ideal components.

That input capacitor is not sitting there cooking off some 25+ Watts
 

mauricem00

Well-Known Member
these type of transformer less power supplies are in common use and the only power loss in the capacitor is from parasitic elements.coils and capacitors store energy and return it to the circuit . perhaps you should have your instructor explain this to you.using rectified AC directly into cobs with no current limiter is dangerous. would advise against it
 

nfhiggs

Well-Known Member
AC passes through a cap VIA the charging and discharging of the plates. Just as it passes from primary to secondary via the magnetic field interaction.

Reactive power losses are calculated via the power triangle. But that power is most certainly "lost" in the form of heat. How else would it be lost? It cannot just vanish into thin air.

While it is true that a PURE reactance does not dissipate power - for all practical purposes, such a device does not exist. A pure reactance would be a capacitor with infinite resistance or an inductor with no resistance.
OK. I will have to stand corrected here. After thinking about it, and referring to my old textbooks (which did not really address this) I did some research and I think you are more correct in saying reactance itself does not dissipate heat, though a device may experience more heating than its internal resistance accounts for, due to the current/voltage phase shift - the "power triangle" you referred to, and that total heat dissipated is not additive (resistance + reactance), but is dependent on the phase vectors of the voltage and current.

My apologies!
 

iHearAll

Well-Known Member
From experience, I have seen caps blow in a current blocking mode iif they couldnt handle enough current pass. They certainly can get hot. For instance if you used a 1mF cap the reactance would be pretty low and allow more current to pass to the rectifier and the cobs. Like, I had a situation where I wanted to run some cheap Chinese cobs at 3a 100w range. I found the farad required to complete this task and also found the capacitance available as an e12 value for fairly cheap. Well, they heated up in each corcuot used them in an failed. Now, I understands this could be due an over load of power supplied but it really performs like itbos passing current in AC and therefore putting of heat like any other component. So, what I had found was that separating the current between a few caps worked really well and not to exceed a half ampere per cap.

If I were to go on he transformer route I would have to just ccs to be more cost effective. Transformers are great but pricey to ship around.
 
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CannaBruh

Well-Known Member
If your capacitor was heating up it likely wasn't rated for the use you were putting in under.

Are these film caps or EL caps we're talking?
 

iHearAll

Well-Known Member
If your capacitor was heating up it likely wasn't rated for the use you were putting in under.

Are these film caps or EL caps we're talking?
They were film, I made an assumption that since the voltage across the cap was only going to see 30v or so, then they could be rated for a low voltage like 120vac but what got me was the power. So, I gave them a whirl six hey were 3$ a piece. I managed to run a crop under them but they eventually leaked dielectric and became an open circuit. Lesson learned to use a parallel bank of caps.
 
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