I am using a single phase to three phase converter to drive a induction motor. I found that the single phase input power to the converter is smaller than the three phase output power from the converter.
Can you tell me why?

 

You have either made an error or a remarkable discovery. If the later, patent it quickly.

Fred Townsend

 

I have just recently read a book relating to 3-phase alternating current, so I think I can help you.

The actual single phase current is only equal to the root-mean-square of the maximum current level. When this is combined with the root-mean-square of the maximum voltage, your total power will not be very high, especially if your power factor is low.
When using three phase power, the power is used much more efficiently than single phase because there is a higher constant potential differnce between the combination of the three phases. This increases the actual voltage and current.

I think this is correct.

Adam Clark

 

from [email protected]
I can tell you that either you have just invented something that will make you an overnight billionaire, or you have made a measurement error. Which is more likely?
Measuring AC power with some precision for either single phase or three phase(and using ordinary volt and current meters) is not easy.
In a DC circuit, volts times amps gives you power in Watts.
In an AC circuit, volts times amps gives you volt-amps. This number is always larger than the AC power in Watts unless it is a purely resistive circuit (no inductance or capacitance).
This is due to the phase angle between the voltage and the current. You need a way to determine E*I*cosine theta on both the single phase side and the three phase side.
There are meters that can do this, but not on my budget. If you have access to an oscilloscope, you could determine the phase angle on both the single and three phase sides.
for the single phase side :
power = E*I* cosine of phase angle
for the three phase side :
power = E*I*1.732 * cosine of phase angle
(where 1.732 is square root of 3)
Whether you are using a rotary or solid state converter, there will be energy losses in the converter. Power out will always be smaller than power in.
And if by some incredible good fortune, your converter really does "amplify power", I would like to be the first investor in this device. Oil from the Middle East will no longer be necessary. Maybe someone should tell the president.

 

What are your input and output measurements?

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}

 

Because you haven't stated what the values were, nor how they were obtained! By power measuing instruments? Or by calculation of measured
paameters, i.e., phase-phase volts, phase-neutral volts, line-amps, power factor?

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}

 

By power if you mean active power (kW) what you have found is normal.

But if you measure apperant power (kVA), then input should be higher than the output (reduced by the efficency of the converter which should be about 0,95).

The reason is simple: Power factor.

At the input, power factor is almost unity, however at the output, power factor is that of the motor, which is not higher than 0,9 usually/depending on the rating of the motor.

The magic is coming from the DC link capacitors, which are making compensation.

 

No, but I wanna buy some of those :^)
I think you are having measurement difficulties.

Regards

cww

 

By "power"-if you mean volts, some drives are set up so that 110V single phase input runs at 220V-3 phase output.

 

That is normal to a converter if your only using an ammeter particularly if its a variable speed drive with 1phase input and 3phase output. But it may not be true in terms of power(kw).Im quite sure your using an ammeter as your gauge which are sometimes mistaken as a power meter.Power can only be transformed from one form to another and can never be increased from the source except if you sacrifice something(e.g. speed). Most of the time power produced is lesser than the power used due to losses.