AD633 Multiplier Problem

Hi,

I am designing a power meter which using the multiplier LM331 to multiply the Voltage and Current signal. The voltage and current signal will be go through the sample and hold circuit and the output of the sample and hold will be go into the LM331 multiplier. Power, P=VIcos(angle) and the LM331 having a transfer function of output W = [(X1-X2)(Y1-Y2)/10V]+Z. The problem is for example V=3V and I=2A, Output should be 6W(ignoring the cos(angle)). Due to the transfer function of LM331 i will never get the output 6W that i want. So anyone there can help me out? =)

Replies

  • shreyasm89
    shreyasm89
    I don't think you can't ignore the power factor because that will denote the phase shift between vlt. & current. If you could explain how you got the transfer probably we could help you out. As ther are 2 inputs(X1 & X2) is 'Z" the o/p impedance of LM331?
  • syee10
    syee10
    The transfer function W=[(X1-X2)(Y1-Y2)/10V]+Z is from the datasheet of AD633. Z is the additional input allows the user to add 2 or more multiplier output. What i mean is the Instantaneous Voltage and Instantaneous Current. The result W of the multiplier should give me 3W if we assume the V=3V and I=2cos60=1A? But with the Transfer function in the multiplier it will never give me that answer? I had attached the datasheet of AD633 for your reference. Thanks =)
  • shreyasm89
    shreyasm89
    What are x1 & X2 as well as Y1 & Y2?
  • syee10
    syee10
    X1X2 and Y1Y2 are differential Input. For example if i want to multiply 3X2 it will be X1=3 and Y1=2. X2 and Y2 will be grounded as 0?
  • shreyasm89
    shreyasm89
    I think judging by the transfer function your power meter is givung o/p in terms of dB? Am I right?

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