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[讨论] AC power measurement

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发表于 2016-1-18 22:48:28 | 显示全部楼层 |阅读模式

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AC power measurement uses PWM & PAM by eenuts



Measurement of average power in a 60Hzcircuit (equivalent to VRMS × IRMS × cos(ø)) can be done by sampling theproduct of voltage and current and averaging. This requires four quadrantmultiplication since the instantaneous voltage and current can have oppositepolarities. It can be accomplished by a number of means, including analog todigital (A/D) conversion followed by digital signal processing, or by using arelatively expensive analog multiplier chip followed by analog or digitalprocessing. This Design Idea presents a third approach. It uses inexpensiveop-amps and an analog switch to implement a pulse width/pulse amplitudemodulator (PWM/PAM) used as a four quadrant multiplier. This circuit can beadapted to many different applications.

1.png

The basic concept ofa PWM/PAM multiplier is that the average value of a (non-overlapping) pulsewaveform over a single cycle is the area of the pulse divided by the pulserepetition period. With each rectangular pulse amplitude proportional tovoltage, and width proportional to current, the area of the rectangle isproportional to the product: voltage times current. If the pulse repetitionrate is much higher than the frequency being measured, it can be assumed thatthe voltage and current do not change appreciably during one cycle of thepulse waveform. The PWM/PAM output is followed by a low-pass filter to removethe pulse frequency and its harmonics and recover the desired average.


The circuit uses two transformers: a step-down transformer TR1 to producea low-level voltage signal, and a current transformer TR2 to produce alow-level current signal with full galvanic isolation.

2.png

Figure 1  MeasureAC power using pulse modulation techniques



Op-ampIC1-A acts as a comparator and generates a fairly linear triangular waveform (Figure 2) with a frequency ofabout 6kHz.  This over-sampling (100 times the 60Hz signal frequency, or50 times the theoretical Nyquist rate) is required to achieve an error below 1%of full scale. The triangle wave peak-to-peak amplitude must meet two criteriafor an acceptably linear response. First, it must be greater than or equal tothe AC current maximum peak-to-peak voltage from T2 so that a 100% PWM dutycycle is possible. Second, it must be at least 10% less than ±VCC.

3.png

Figure 2  IC1-B(-) input


Op-ampIC1-B is configured to generate a zero-baseline bipolar square wave (Figure 3) with duty cycledetermined by the instantaneous value of the current signal from T2. CapacitorC2 is required to obtain sharp output voltage transitions. The op-amps shouldbe capable of a slew rate of 10V/μs or more to produce an optimal square PWMsignal. When the instantaneous value of the current signal from T2 is zero, theduty cycle at IC1-B’s output will be 50%. As the current signal increases fromzero to full scale, the duty cycle increases from 50% to 100%. As the currentsignal decreases from zero to negative full scale, the duty cycle decreasesfrom 50% to 0%.


4.png

Figure 3  IC1-Boutput

Op ampIC2-A, switch IC3-A, and the associated resistors implement a four-quadrantmultiplier (reference 1) in the form of an amplifier with a gain of either +1(switch closed) or -1 (switch open). The switch is controlled by the PWM outputfrom IC1-B; the switch is closed when IC1-B’s output is positive and open whenIC1-B’s output is negative. The resulting output from IC2-A (Figure 4) is a replica of the PWMoutput from IC1-B (duty cycle proportional to current), except with amplitudeproportional to voltage.

optimav�urpo8@�9gnal. When the instantaneous value of the current signal from T2 is zero, theduty cycle at IC1-B’s output will be 50%. As the current signal increases fromzero to full scale, the duty cycle increases from 50% to 100%. As the currentsignal decreases from zero to negative full scale, the duty cycle decreasesfrom 50% to 0%.



5.png

Figure 4  IC2-Aoutput

IC2-B,the final stage of the circuit, is a low-pass filter which removes the6kHz PWM frequency and its harmonics as well as the 60Hz modulation. Its outputis proportional to the average of the PWM signal, and thus the average 60Hzpower.





发表于 2016-1-21 17:01:52 | 显示全部楼层
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