An accurate current meter isn't cheap for sure, neither is a good current probe for the scope as it appears - $300 and up to $1.5k!
Hello,
When the power plant 1000 miles down the road generates AC electricity voltage and current are in phase. As industry and commerce start switching on large inductive loads aka, large motors, compressors and pumps ELI sticks his hand into the mix, current begins to lag voltage.
Normally we think of watts as volts times amps. When volts and amps are in phase it works out that peak volts occurs at the same time as peak amps so watts are peak volts times peak amps and are real watts.
When ELI is at work it is not Peak volts Times peak amps it is peak volts times peak amps times a percentage and the power utility is not being paid for what they produced. This is what I know as Power Factor.
When I read Merlin I have the sense that some of the details are missing.
Thanks
DT
When the power plant 1000 miles down the road generates AC electricity voltage and current are in phase. As industry and commerce start switching on large inductive loads aka, large motors, compressors and pumps ELI sticks his hand into the mix, current begins to lag voltage.
Normally we think of watts as volts times amps. When volts and amps are in phase it works out that peak volts occurs at the same time as peak amps so watts are peak volts times peak amps and are real watts.
When ELI is at work it is not Peak volts Times peak amps it is peak volts times peak amps times a percentage and the power utility is not being paid for what they produced. This is what I know as Power Factor.
When I read Merlin I have the sense that some of the details are missing.
Thanks
DT
Amplifiers rectifier (linear power supply) have large capacitive reactive factor and this is the reason why we have bad power factor as all we know.
To compensate we have to use inductance in series (cheaper option) or parallel to our input source (load with no reason, but accumulated energy for transfer to capacitive load), this will not be accurate due non linear load (music isn't constant load) and design is not so easy for linear power supply.
For switch mode power supply it is easy because we force current flow with booster and it is called active power factor correction.
With AC current we have apparent power in VA and this is our lead in desing, not true power in W.
So why bother with power factor when we pay for domestic use only real power, not apparent power, despite your current is twice the value You want, payment will be E=P*t [Wh]; t=hr's, or by power P=U*I*cos fi [W].
So if You like to minimise heat loss, or another look minimise heating in Your room then compensate your capacitive load, as all we know what produces heat, losses due high current (P=I2*R) at components.
To compensate we have to use inductance in series (cheaper option) or parallel to our input source (load with no reason, but accumulated energy for transfer to capacitive load), this will not be accurate due non linear load (music isn't constant load) and design is not so easy for linear power supply.
For switch mode power supply it is easy because we force current flow with booster and it is called active power factor correction.
With AC current we have apparent power in VA and this is our lead in desing, not true power in W.
So why bother with power factor when we pay for domestic use only real power, not apparent power, despite your current is twice the value You want, payment will be E=P*t [Wh]; t=hr's, or by power P=U*I*cos fi [W].
So if You like to minimise heat loss, or another look minimise heating in Your room then compensate your capacitive load, as all we know what produces heat, losses due high current (P=I2*R) at components.
I think that I didn't explain myself very well. 
This post is NOT about power companies nor trying to criticize what other forum members build/do/measure, which I find to be highly respectable, worthy of admiration.
The post is about this:
If my design draws 200 mA of HT, I can use, let's say, .3mm wire to wind that winding (following the 3 A / mm^2 ratio)
BUT, IF in reality we are getting a much higher RMS current, than DC current, through that winding (just suppose, power factor is around 0.65), then .3mm is too thin wire. It will get much hotter than we expected.
This post is about sharing thoughts about if we take thin into account or not.
I built a couple of power transformers not considering this. All are built with a large safety margin, but one of them is not, and it definitelty gets warm (much warmer than the rest), even hot now that summer comes.
The next power transformer I'm about to build, will have a PF of 0.7 in mind (which is even optimistic), so I'll use a thicker wire. These are my 2 cents.
This post is NOT about power companies nor trying to criticize what other forum members build/do/measure, which I find to be highly respectable, worthy of admiration.
The post is about this:
If my design draws 200 mA of HT, I can use, let's say, .3mm wire to wind that winding (following the 3 A / mm^2 ratio)
BUT, IF in reality we are getting a much higher RMS current, than DC current, through that winding (just suppose, power factor is around 0.65), then .3mm is too thin wire. It will get much hotter than we expected.
This post is about sharing thoughts about if we take thin into account or not.
I built a couple of power transformers not considering this. All are built with a large safety margin, but one of them is not, and it definitelty gets warm (much warmer than the rest), even hot now that summer comes.
The next power transformer I'm about to build, will have a PF of 0.7 in mind (which is even optimistic), so I'll use a thicker wire. These are my 2 cents.
I recommend you think of PF as having two separate and independent factors - PF due to cos phi of fundamental waveform, and PF due to distortion of fundamental. One can't be compensated by adding in the other (eg. inductance parallel to input source).
@Elerion
RMS current will be the same as calculated, but peak current will be much higher in short pulses, what I mean, if You put high capacitance after rectifier you'll have the case of high current in short pulse, but RMS value will be the same, design with care your rectifier, that is all what I mean not to explain utility bills.
Stress is on the diode (rectifier) and capacitor (ripple current limitation).
and finally current THD produced by this rectifier is pretty high.
RMS current will be the same as calculated, but peak current will be much higher in short pulses, what I mean, if You put high capacitance after rectifier you'll have the case of high current in short pulse, but RMS value will be the same, design with care your rectifier, that is all what I mean not to explain utility bills.
Stress is on the diode (rectifier) and capacitor (ripple current limitation).
and finally current THD produced by this rectifier is pretty high.
@Elerion...
when i design my power traffos, i take an assumption of 0.6 power factor,
reason being, so i can size my traffos even bigger, some commercial winders will use 0.8 or 0.9 and realise savings...
but i am not a commercial winder and bottomline was never a concern for me...
this way my traffos are somewhat bigger than off the shelf traffos...
but more than power factor, the quality of the iron laminates matters a lot too,
M6 is of course best for available EI's but expensive, so RM18/H18 are the next best thing for me,both at 0.35mm thick, has a lower heat and magnetising currents than
the H50 that is commonly used and is 0.5mm thick...
realize too that the assumed 0.6 power factor is not the same as the power factor obtained in actual use...with actual amps...
when i design my power traffos, i take an assumption of 0.6 power factor,
reason being, so i can size my traffos even bigger, some commercial winders will use 0.8 or 0.9 and realise savings...
but i am not a commercial winder and bottomline was never a concern for me...
this way my traffos are somewhat bigger than off the shelf traffos...
but more than power factor, the quality of the iron laminates matters a lot too,
M6 is of course best for available EI's but expensive, so RM18/H18 are the next best thing for me,both at 0.35mm thick, has a lower heat and magnetising currents than
the H50 that is commonly used and is 0.5mm thick...
realize too that the assumed 0.6 power factor is not the same as the power factor obtained in actual use...with actual amps...
No. There is very little phase shift, as the peak current almost coincides with peak voltage. You are confusing two quite different causes of poor power factor: phase shift and harmonics. Hence simply adding an inductor, in series or parallel, will not help.pitbul said:
It is unclear exactly what you mean by "RMS value will be the same", but it certainly is not the case that the RMS current into a rectifier+capacitor is equal to the DC current out. That is precisely the point of this thread.
It has already been made clear in this thread that people who know what they are doing take account of this, and people who do not know what they are doing do not take account of this. It is unclear to me what else there is to discuss.Elerion said:
btw, in choosing wire sizes, i use the ampacity of 500 circular mils per ampere as initial estimates,
this can still go down or up depending on available space in the winding window...
the ARRL wire table states that 300 to 700 circular mills can be applied...
this is what i go by...
you may want to look at this: RECTIFIER TRANSFORMER CALCULATION
or this: http://www.hammondmfg.com/pdf/5c007.pdf
i can do it another way, if you know your load power, then traffo secondary volt amperes are determinable...
this can still go down or up depending on available space in the winding window...
the ARRL wire table states that 300 to 700 circular mills can be applied...
this is what i go by...
you may want to look at this: RECTIFIER TRANSFORMER CALCULATION
or this: http://www.hammondmfg.com/pdf/5c007.pdf
i can do it another way, if you know your load power, then traffo secondary volt amperes are determinable...
@DF96
I didn't say that adding inductor will solve compensation, my conclusion was
"this will not be accurate due non linear load (music isn't constant load) and design is not so easy for linear power supply".
True: Current pulse will be in phase with voltage peak, but as load will increase (when ripple increase) phase shift will be visible.
DC RMS current will be defined by load, not by transformer, can be higher than transformer output capability due energy storage in capacitance.
I didn't say that adding inductor will solve compensation, my conclusion was
"this will not be accurate due non linear load (music isn't constant load) and design is not so easy for linear power supply".
True: Current pulse will be in phase with voltage peak, but as load will increase (when ripple increase) phase shift will be visible.
DC RMS current will be defined by load, not by transformer, can be higher than transformer output capability due energy storage in capacitance.
Nothing to do with music not being a constant load (whatever that means). It is because a capacitive input PSU is not a constant load, but a very peaky one.
An inductance will not compensate, even approximately, because it is solving the wrong problem.
The DC load cannot exceed the transformer capability, except for brief periods.
An inductance will not compensate, even approximately, because it is solving the wrong problem.
The DC load cannot exceed the transformer capability, except for brief periods.
this is very true, and if the overload is gross, can lead to core saturation....
power factor for transformer and dc is compensated by the power company at the distribution center for around 0.9
The only problem is when you operate a factory with a lot of electrical motors and inductive loads. If you draw a huge electricity the power company force the installation of PF reducing capacitor bank.
Transformer manufacturers specify a certain power output VA which is reliable, you don't need to think about PF.
The only problem is when you operate a factory with a lot of electrical motors and inductive loads. If you draw a huge electricity the power company force the installation of PF reducing capacitor bank.
Transformer manufacturers specify a certain power output VA which is reliable, you don't need to think about PF.
i get Elerion now, his concern was mostly about how to go about making his traffo, and he has a good advice in this thread....
Once again swept away with the wave of a hand.
Hello,
I have the sense that some of the details are missing.
This topic is not so much about Power Factor as inefficiency, heat and derating the VA capacity of our little transformers. Power Factor as such does not matter.
The transformers we are speaking of are all less than a KW.
Go into a large office building, all the step down transformers and conductors for lighting and receptacles are K-Factor derated to compensate for all the connected switching power supplies. Lots of spikes and pulses on those transformers that look nothing like cosine waves. There is no conversation about PF only K Factor transformers with huge deratings to continue operating with increased heat caused by Harmonics caused by the switching power supplies in the connected computer equipment and lighting loads.
DF96 is correct, adding an inductor will not help.
Thanks DT
FYI http://www.xitrontech.com/assets/002/5787.pdf
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really nothing to be bothered much about...no need to overthink this imho...
but why is pf important when designing traffos....?
say i wanted a 1000va secondary, so assuming a 0.6 pf, primary va then becomes 1667 va...
so that the required core area then becomes, 7.3 sq inches, based on chapter 5 of RDH4,http://www.audiofaidate.org/it/materiale/Radiotron/chaptr05.pdf so that if 1 3/4 inch core was used, stack needed is 4.2 inches...
now assume a power factor of 0.8. the core area now becomes, 6.3 sq inches, see that there is savings in weight and cost of a traffo.....
here we only need a stack of just 3.6 inches using the same `1 3/4 inch core...
so here we see that power factor is vital in dimensioning
the traffo core..
btw, that 5.58 constant used in RDH4, is 5.2 on some japanese traffo i have seen, and Patrick Turner i believe uses a more conservative 4.4 of i am not mistaken..
but why is pf important when designing traffos....?
say i wanted a 1000va secondary, so assuming a 0.6 pf, primary va then becomes 1667 va...
so that the required core area then becomes, 7.3 sq inches, based on chapter 5 of RDH4,http://www.audiofaidate.org/it/materiale/Radiotron/chaptr05.pdf so that if 1 3/4 inch core was used, stack needed is 4.2 inches...
now assume a power factor of 0.8. the core area now becomes, 6.3 sq inches, see that there is savings in weight and cost of a traffo.....
here we only need a stack of just 3.6 inches using the same `1 3/4 inch core...
so here we see that power factor is vital in dimensioning
the traffo core..
btw, that 5.58 constant used in RDH4, is 5.2 on some japanese traffo i have seen, and Patrick Turner i believe uses a more conservative 4.4 of i am not mistaken..
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