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Methods evaluate power electronics' efficiency

15 May 2015  | Liping Zheng

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Validating system efficiency of a power-electronics circuit is crucial in evaluating the overall system performance, design optimization, and sizing of cooling systems. Figure 1 shows the conventional method of performing efficiency measurement. The power-electronics system operates at the rated output-power level, and, by measuring the input power and output power, you can calculate the system's efficiency using the equation η=(POUT/PIN)×100%, where POUT is output power and PIN is input power. In other words, the measured input power is equal to the output power plus the power loss of the system.

Figure 1: In a conventional method of performing efficiency measurement, the power-electronics system operates at the rated output-power level.

However, measuring the efficiency of a high-power system that delivers power to loads such as motors, generators, or industrial-computer equipment requires a source that delivers the rated power. The infrastructue therefore should comprise a suitably rated source and an equivalent load that can support the rating of the power-electronics system you are evaluating. These requirements can drive up the facility's infrastructure cost; for one-time design-validation measurements, this cost is difficult to justify.

Figure 2: This method eliminates the test load by shorting the output-load terminals.

This design idea describes alternative methods of measuring the efficiency of a high-power power-electronics system that simplifies the test-infrastructure requirement by eliminating the test load and using a source that must support only the loss of the power-electronics system. Figure 2 shows the proposed method, which eliminates the test load by shorting the output/load terminals. The system's control algorithm maintains the required input- and output-current amplitude and frequency by developing circulating reactive power. IGBTs (insulated-gate bipolar transistors) and magnetic components dominate the system's losses, which are functions of the amplitude and frequency of the input and output currents. The loss is also less sensitive to the power-factor and PWM (pulse-width-modulation) index.

To know the required input and output current, you must estimate the system's power factor, the motor's back EMF (electromotive force), and the systems's source voltage. This example uses a field-oriented control for both source- and load-side inverters, resulting in the following equations:

where IROUT is the required output current, which comprises real current, IROUT_RE, and reactive current, IROUT_IM; IRIN is the required input current, which comprises the real current, IIN_RE, and the reactive current, IIN_IM; PRIN is the required input power; POUT is the output power at the test condition; VBEMF is the motor's back EMF; VGRID is the grid voltage; and ηE is the estimated efficiency of the circuit.

By maintaining the input current to be IRIN and the output current to be IROUT, the measured input real power will be close to the power loss, PLOSS, at the actual output-power level, POUT. Therefore, you can calculate the efficiency as follows: η=(POUT)/(POUT+PLOSS)×100%.

If the measured efficiency, which you calculate using this equation, does not quite match the estimated efficiency, ηE, update the second equation using the measured efficiency, η, and repeat the measurement until they are close. Calnetix has used this method to evaluate the efficiency of a 125-kW power-electronics system, compared the results with the conventional measurements, and found them to be closely matching.

Figure 3: This method uses two identical power-electronics systems. The second system offsets the input reactive current that the test system creates.

Most high-power power-electronics systems have high efficiency, which means that the real current is much less than the reactive current. To reduce the required current from the grid, you can use the method in figure 3, which uses another identical system to offset the input reactive current that the test system creates. By providing a path for circulating reactive power, the utility sources the lost power only, not the total power. In figure 3, the input current of the second power-electronics circuit is IRIN= IRIN_RE+jIRIN_IM. By setting the first circuit to have an input current of IRIN1≈IRIN_RE− jIRIN_IM,the power from the source is only ISOURCE=IRIN1+IRIN≈IRIN_RE+IRIN_RE+ j(IRIN_IM−IRIN_IM)=2IRIN_RE. The circuit uses the input current from the source only to overcome the power losses of the two circuits, thereby eliminating the need for a high-power infrastructure.

About the author
Liping Zheng is with Calnetix in Yorba Linda, California, United States.

This article is a Design Idea selected for re-publication by the editors. It was first published on July 29, 2010 in

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