AC vs. DC - who can explain this?

[sorinsm]

Who can explain this?
A DC power supply vs. AC power supply according to the attached picture having the same load, the same current and voltage on it, but different power distributed on resistor.
Practically I configured the electric circuit and measured its parameters, both for DC and AC with almost the same results as well as into the simulation (I used NI MultiSIM v.11.0.1).
Is anybody able to prove that I’m wrong somewhere?
And what’s wrong here?!

I'm waiting your comments by sending mails to: <email.sorin@yahoo.com>
A lot of thanks in advance !

 

[ADWSystems]

I'm not entirely sure what is going on either. The intersting thing to note is the Voltage and Current measured (as displayed) equate to the lower power reading of 2.041kW (8.843*230.814=2.041kW). Hopefully someone has a full explanation.

 

[JimB]

A lot will depend on how the meters are calibrated.
In the first case, the supply will be a distorted rectified sinewave, the capacitor is too small to give smooth DC.
In the second case, the supply to the resistor is a sinewave.

If the meters were responding to the RMS value and calibrated to display RMS, then I would expect the power to be VxI in both cases.

However, if the meters are responding to the average value and calibrated to display RMS, then there will be an error.

JimB

 

[ADWSystems]

Jim, according to the OP, the simulation results are similiar to the bench results. What gets me is the DC measurement of power does not equal the DC current times the DC voltage. But after considering the remainder of your statement, the small filter capacitor (versus the load amerage) may be upsetting the algorithms for calculate the current and/or the power. I don't normally deal with 8A draws, so at first look the cap size looked fine. A quick search revealed the ripple voltage = current / (ripple frequency =120Hz * filter cap). In this case 8.84A/(120Hz*.000220F)=334V.

To Sorinsm: can you retry with the simulation and test with a filter capacitor of atleast 22,000uF (x100 the original). This should reduce the ripple voltage to 3.34V. Even 47,000uF or larger would be even better.

 

[JimB]

That circuit does not produce smooth DC.
The capacitor is too smal.

The DC is half wave recitfied AC, which although unidirectional, the average value (which is what the meters probably respond to) is not the same as the RMS value.

JimB

 

[ADWSystems]

That circuit does not produce smooth DC.
The capacitor is too smal.

The DC is half wave recitfied AC, which although unidirectional, the average value (which is what the meters probably respond to) is not the same as the RMS value.

JimB

So the average value of the half wave rectified signal for measuring the DC is different than the rms value of the AC sine wave for measuring the AC signal. In short, the DC meters are upset and the readings given are not trustworthy. The simulation and circuit need to be updated to produce a constant voltage in the DC side.

 

[sorinsm]

Yes ADWSystems, you're right. The higher capacitor value will produce a low ripple wave, or a smooth voltage output as well I did at the beginning of the experiment and the outputs were accordingly. I'm sure that the DC circuit is not efficient but this isn't the problem (see the attached picture).

I'm interested what's wrong with the measurements which were done & the simulation parameters. Anyhow I would like to understand why different power output on both circuits....or how to calculate the power on the resistor from DC circuit?!
About my experiment :
- I used an Yokogawa 230 as powermeter and multimeter type Meterman 37 as voltmeter / ammeter. These were set up according to current / voltage type of the power supply (ac/dc) before each measurement.

 

[ADWSystems]

Neither power supply is DC. Your trace in the post shows that you don't have a DC signal. What I believe is wrong is the use of the DC mode to measure a AC voltage and current. The results are unpredictable because of the mismatch.

 

[BrownOut]

I'm sure that the DC circuit is not efficient but this isn't the problem (see the attached picture).

Nobody said efficiency was the problem. They said the measurement is incorrect because of the significant ripple.

 

[MrAl]

Who can explain this?
A DC power supply vs. AC power supply according to the attached picture having the same load, the same current and voltage on it, but different power distributed on resistor.
Practically I configured the electric circuit and measured its parameters, both for DC and AC with almost the same results as well as into the simulation (I used NI MultiSIM v.11.0.1).
Is anybody able to prove that I’m wrong somewhere?
And what’s wrong here?!

I'm waiting your comments by sending mails to: <email.sorin@yahoo.com>
A lot of thanks in advance !

Hi there,

Think about this for a moment. You've got an energy storage element in one circuit but not in the other. The peak voltage gets partially stored in the cap in one circuit, but not in the other. Which one should show more power dissipated in the load?
Dont increase the value of the cap, instead, remove it entirely and test again.

DC meters usually average the input, which is not the same as measuring power. It doesnt matter if the average voltage is the same, sometimes it doesnt even matter if the rms reading is the same, because power is defined as the voltage times the current averaged over time. Without doing an exact analysis, here is the difference:

PowerFirstCircuit=avg(v(t)*i(t)) with the constraints avg(v(t))=K1 and avg(i(t))=K2,
PowerSecondCircuit=avg(v(t))*avg(i(t)) with the constraints avg(v(t))=K1 and avg(i(t))=K2,

So we see in the first case we first multiply, then take the average, but in the second case we first take two averages and then later multiply. We have two constraints but they are really kind of arbitrary and not as important as they may appear to be at first. The important point is that if we calculate power differently in each case we will get a different answer in each case.

You would do yourself a favor to look into other cases like these two also, especially pulse circuits. Pulse circuits show very clearly the differences like this.

We can do an exact analysis if you prefer to show the difference.