Power Rating Calculation
Power Rating Calculation
(OP)
A single phase transformer is feeding a half-wave rctifier with a resistive load across it.
What is the real value of the power cosumed in the resistor load( dc + ripples ),is it to :
- Multiply transformer's secondary rms values of the voltage and current- assume 100% efficiency.
-Or to mutiply the the intatanious values of the voltage and current at the load terminals and then thake the average over one cycle period- as the wattmeter does?
What is the real value of the power cosumed in the resistor load( dc + ripples ),is it to :
- Multiply transformer's secondary rms values of the voltage and current- assume 100% efficiency.
-Or to mutiply the the intatanious values of the voltage and current at the load terminals and then thake the average over one cycle period- as the wattmeter does?






RE: Power Rating Calculation
Better is to take a TrueRMS meter and measure it.
Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com
RE: Power Rating Calculation
(Esin x /R) and (Esin x) ,multiply them together and ,make the integral over one period cycle and devide over the period cycle and that would be the wattmeter reading( the mean value of the product of current and voltage)
Now go to the secondary winding, multiply the rms value of the current and voltage ( transformer rating )and you will find it larger than the wattmeter reading though we assumed 100% efficiency.
Why that difference ?
Which one is the true power consumed in the resistor ?
RE: Power Rating Calculation
RE: Power Rating Calculation
You are asking why the DC load power and the power on the low voltage side of the transformer are different? And specifically why the transformer side is higher?
If that is a the case. The power at the transformer has got to be higher because their are substantial heating losses at the rectifier. Also if you have a filter cap you have losses due the ESR of the capacitor. So - out with the 100%!
If this isn't what you are asking you must further clarify. Draw a picture perhaps.
Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com
RE: Power Rating Calculation
single phase transformer , its secondery is connected in series with a single diode and resistive load( simple half-wave rectifier)
The DC power Wdc = Idc x Vdc -- average values of the rectified voltage and current.
If you calculate the transformer's secondary VA rating using the rms values of the voltage and current there you will find it = 3.49Wdc
Now the wattmeter at the resistive load will caculate the power as the mean value of the product of the instanious current and voltage over one period - If you do this calculation mathematically you will find it =2.46Wdc
Why that difference between 3.49Wdc and 2.46Wdc ?
2.46
RE: Power Rating Calculation
If you are measuring at one point and calculating for another point, that's different.
RE: Power Rating Calculation
one at the load and the other at the transformer secondary.
The two calculations give different results.
No measurments are taken- Just mathematical calculations and neglect any losses.
RE: Power Rating Calculation
I think even on simple issues it is very difficult to convey specifics of engineering problems in pure text.
That being said, if I understand your situation, the answer is that one calculation is based on the transformer rating and the other is based upon the power consumed by the load.
You wrote: "If you calculate the transformer's secondary VA rating using the rms values of the voltage and current there you will find it = 3.49Wdc"
The XFMRs VA rating is the amount of apparent power that it can deliver without being overloaded. The XFMR does not always deliver this amount of power, but rather delivers the power demanded by the load. Loads that demand less than this amount of power can be served by the XFMR, loads that demand more will overload it.
It sounds like you're serving 2.46W worth of load with 3.49W worth of XFMR. No problem.
Another possible interpretation of your question would be that you know the circuit draws 3.49W (as calculated at the XFMR secondary), but you only calculate 2.46 watts of that power going to the resistor.
If this is the case, then you have ommitted the fact that the diode dissipates power as well. Diodes have characteristic voltage drops when conducting (Forward biased). 0.7V for Si based or 0.3V for Ge based diodes. I devise that has a voltage drop and is passing current is consuming or dissipating power.
I think one or the other of these explanations accounts for your discrepancy.
Regards.
RE: Power Rating Calculation
Thanks for the detailed explaination.
I still have one point is not clear.If the load is pure resistive,that means that the current and voltage are in phase and the power consumed by the load is real power,so if we calculate the power at the load should equal the power deliverd by the secondary winding-assume in all our calculations ideal diode and ideal transformer.
Is the product Vrms.Irms at the transformer's secondary does not the represent the real power delivered by it for that resistive load?
RE: Power Rating Calculation
RE: Power Rating Calculation
RE: Power Rating Calculation
Are you saying that only the fundamental component of the current will cause the resistor to heat up?
RE: Power Rating Calculation
When calculating the power from the AC voltage and the fourier coefficients of the AC current, only the instantaneous values of fundamental current times the voltage produces a non-zero result. That explains why voltamperes calculated from the AC RMS values is higher than the watts.
All of the current contributes to the power in the resistor, but I think that you need to go to the fourier components to make sense of the voltamperes vs. watts.
RE: Power Rating Calculation
Seems like either a homework problem or a troll.
RE: Power Rating Calculation
If you take the diode out of the circuit and imagine a 10V RMS AC source feeding a 2 Ohm resistor, a current of 5A RMS flows and dissipates 50W in the resistor. If you plug the diode back in the current will be halved to 2.5A RMS because you only have half a cycle, and the power dissipated in the resistor will be 12.5 W.
I'm still not sure what the issue is here, but the confusion may be arising because the RMS value of a sine wave is not the same as the average value. RMS=0.707(Peak)Average= 0.637 (Peak)
Regards
Marmite
RE: Power Rating Calculation
5 amps rms on the first half cycle added to 0 amps rms on the second->
sqrt((5^2+0^2)/2half cycles)= 5/sqrt(2)=3.54 A rms on the full cycle.
Do the same for voltage and you get 10/sqrt(2) volts rms. Multiply them together and you get 25W average power dissipated. Now see jghrist's answer above.
I agree pf is not applicable (or unity) downstream from the diode where the load is resistive and linear. Upstream, the load looks non-linear and there will be less than unity pf due to distortion. Voltage is present during the second half cycle, but it does not contribute to power without the current. pf=25/(3.54*10)=.707
RE: Power Rating Calculation
You get the same result with a SCR controller at part output voltage uf you measure voltage before an after and use that in a power calculation.
Keith hit it exactly in the first post. You would need a RMS power meter that does the rms calculation on instantaneous power measurements to measure the same power before and after. Many power meters don't do this, they calculate rms voltage and current and power factor and then do the math. I've seen a Fluke 43B ($4k meter or so??) display different power on the input and output of a SCR controller.
RE: Power Rating Calculation
Regards
Marmite
RE: Power Rating Calculation
Sure you can. If you have true rms reading meters, and if the other assumptions (linear purely resistive load below a lossless diode) do not introduce too much error, just use the distortion pf I derived above in your calculation.
RE: Power Rating Calculation
The fundamental component of half wave rectified sine wave current is 0.5·I·sin(2·π·f·t) where I is the peak current. The rms value would be Irms = 0.5·I/√2. The fundamental component of the voltage is V·sin(2·π·f·t). The rms value would be Vrms = V/√2. The power is the product Vrms·Irms or 0.5·V/√2·I/√2 = 0.25·V·I. Both the fundamental current and voltage have a zero phase angle so the displacement power factor is unity.
Using the example of 10 volts rms and a 2 ohm resistor, the peak voltage is 10·√2. The peak current is 10·√2/2. The power is 0.25·10·√2·10·√2/2 = 25W.
This is a very complicated method for this simple problem, where it is intuitively obvious that the power would be 1/2 that of unrectified current, but it would be useful in other cases.
RE: Power Rating Calculation
so. at the transformer's secondary side the power delivered to the circuit will be Irms.Vrmsx(.707)
Now what will be the exact tranformer's rating:
Is it the product of Irms.Vrms
Or the real power it delivers Irms.Vrmsx(.707)
RE: Power Rating Calculation
It is easy to back calculate the power factor when you do know the power already but it's not so easy to calculate the power facture to figure out the power. As i said, I've seen fairly expensive power meters wrongly measure power on the line side of SCR circuit.
I don't fully agree with the power factor arguement anyways. If you looked at the voltage and current on a scope you'd see an in-phase current for 1/2 of the voltage cycle and no current for the other 1/2 of the voltage cycle. Where is the phase shift? I could also argue that the only current that flows goes through a resistor. I'm in agreement with Marmite in that an ideal resistor does not have a power factor so power factor isn't applicable.
RE: Power Rating Calculation
The power factor is not a displacement power factor that arises from a phase shift, but a distortion power factor that arises from the current not being sinusoidal over the complete cycle. With unity displacement power factor and a lower distortion power factor, the total power factor (sometimes called the "true" power factor is less than 1. It is not the resistor that causes the reduced power factor it is the diode.
zacky:
Transformers are rated in VA, not watts. The distorted waveform causes transformer heating is excess of what would be caused if the waveform was not distorted. At a minimum, the transformer needs to be sized for Vrms X Irms. You also need to consider that the transformer might be saturated due to the unidirectional secondary current. It would be a lot better to use a full wave rectifier.
RE: Power Rating Calculation
In your 25 Jan post, you said Irms was 5 for the first half cycle and Vrms was 10/?2. Shouldn't Vrms be 10 to get 5 amps rms through the 2 ohm resistor?
According to the McGraw-Hill Science and Technology Dictionary at http://www.answers.com/topic/power-factor-1, Active power is 25W. Irms = 5/?2 amps over the whole cycle. Vrms = 10. Apparent power is 10·5/?2 = 50/?2 = 35.4 VA. Power factor is 25/(50/?2) = 0.707.
The transformer would have to be rated for the apparent power of 35.4 VA.
RE: Power Rating Calculation
Transformers for non-linear loads also have a k factor rating which deals with the extra heating. I agree that full wave rectification would be better.
RE: Power Rating Calculation
>>A single phase transformer is feeding a half-wave rctifier with a resistive load across it.
It can be transformer or just ac line. We assume a perfect AC input, an ideal rectifier and resistive load.
>> What is the real value of the power cosumed in the resistor load
It is one half what it would be without the diode. Without the diode it would be V^2/R and with the rectifier it is V^2/R/2. Rather obvious when you think about it.
>> Multiply transformer's secondary rms values of the voltage and current- assume 100% efficiency.
This is also correct. Basically the same thing.
>> Or to mutiply the the intatanious values of the voltage and current at the load terminals and then thake the average over one cycle period- as the wattmeter does?
This should also give the same result. Are you assuming they would give different results? Am I missing something?
RE: Power Rating Calculation
You're not missing anything. There are multiple complicated ways to arrive at the same intuitively obvious answer.