I am measuring the admittance of a circuit, by switching a load off and on at different loads. The admittance value I got from the measured voltage drop and current flowing through my circuit. The problem is calculation is only good when the circuit is only resistive or having high power factor (0.9 to 1.0). As soon as the inductive impedance value goes up, the admittance measurement shows discrepancies. Can somebody please throw some light on it?
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\$\begingroup\$ what is the circuit? How did you measure? My crystal ball says you need to consider spectrum of admittance vs load and step response affects spectrum. \$\endgroup\$– Tony Stewart EE75Jun 21, 2017 at 19:59
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\$\begingroup\$ Normally averaged over one cycle for repetitive switching with duty cycle, d then Z(f)'= Z(f)/d fundamental or with spectrum harmonic content of asymmetry for even harmonics then Z(f)=FFT of V(t)/I(t) = 1/S(f). A true voltage source is expected such that Rs/Z(f) <<1% or if you expect max power transfer as Glenn suggests use conjugate matching \$\endgroup\$– Tony Stewart EE75Jun 21, 2017 at 20:09
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\$\begingroup\$ Sir, Here the resistor R8 is my load and I am measuring the voltage and current across this one. Anytime I connect a new load (in this case R1 to R7), I switch off and on R8 and note down the voltage and current change, (Change in Current)/(change in voltage) gives me my admittance value. But works only till power factor is 0.9 to 1.0, any drop my formula fails !!! \$\endgroup\$– Kunal SarafJun 22, 2017 at 19:55
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\$\begingroup\$ ΔI/ΔV works in steady state, but during Δt sample interval reactive loads with nonlinear switches with non-ideal components ( R+X(f)) then will be inaccurate unless time constant T is considered. Consider switching a cap voltage instantly: What controls the current? C or ESR actually it is ΔV/(Rs+ESR) for switch with Rs. \$\endgroup\$– Tony Stewart EE75Jun 22, 2017 at 22:04
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\$\begingroup\$ @Kunal Who said (complex) admittance is calculated as change in current over change in voltage? It is calculated as (phasor) current over (phasor) voltage. Am I missing something? By the way, can you share the Simulink model? \$\endgroup\$– alejnavabSep 22, 2021 at 7:31
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1 Answer
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Two givens:
- Admittance is the reciprocal of impedance.
- Due to the maximum power theorem, maximum power in the load is attained when the load is the conjugate match of the output impedance. This means Rsource=Rload and jsource=-jload
Find the load that draws the maximum power, compute its conjugate (since the source impedance is its conjugate), and take the reciprocal of this conjugate to derive the complex admittance of the circuit's output.
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\$\begingroup\$ Sir, I have updated my circuit. Kindly have a look.Here the resistor R8 is my load and I am measuring the voltage and current across this one. Anytime I connect a new load (in this case R1 to R7), I switch off and on R8 and note down the voltage and current change, (Change in Current)/(change in voltage) gives me my admittance value. But works only till power factor is 0.9 to 1.0, any drop my formula fails !!! \$\endgroup\$ Jun 22, 2017 at 19:56
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\$\begingroup\$ Your switched loads are all resistors. Where is the power factor change coming from? \$\endgroup\$ Jun 22, 2017 at 20:02
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\$\begingroup\$ From the Pi Section I have connected at the beginning, near the trasnformer!!!, Any increase in inductance shows bad results about admittance. \$\endgroup\$ Jun 23, 2017 at 21:31
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1\$\begingroup\$ If you have inductance in the source impedance, then you must have a capacitive load with equal (but opposite of course) reactance. This is the conjugate load. \$\endgroup\$ Jun 23, 2017 at 21:35