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Question:

Explain the relationship of the load impedance angle to the displacement coefficient cos(𝜑1) and the power factor λ at the input of the rectifier enter image description here

Not sure exactly what its asking for.

It probably needs the relationship of all 3. The load is an inductor (1H) and a 50 Ohm redistor in series, 𝜑1 is the angle between voltage and current, the power factor λ is \$\frac{P}{S}\$.

So far I though about using \$Pin=Pout\$ \$I_{s1,rms}V_{s,rms}cos(φ1) =\frac{1}{T}\int_{0}^{T}V_d(t)i_d(t)dt \$ then substitute \$i_d = \frac{\sqrt{2}V_s}{|Z_0|}sin(\omega t - \phi)\$ where \$\phi = arctan(\frac{\omega t}{R})\$

Not really sure what I'm looking for and where this is going. I'm making this post not because I am looking for someone to solve this for me but to throw some ideas about what I might be looking for and ways to approach this problem.

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A rectifier circuit represents a nonlinear load on the AC power system. As a result. the AC input current is not perfectly sinusoidal. It contains harmonic components. In such a situation, there are the two ways of defining power factor give two different results. The angular displacement between the voltage and current gives one result, often called the displacement power factor or displacement factor. I believe "displacement coefficient" may be another name (or translation) for displacement factor.

If S is calculated using the total RMS values of the current and voltage, including the harmonic content, a different value of power factor is obtained. That power factor is often called the "true" power factor. I prefer to call it the "total" power factor because there are important differences between that power factor and the power factor for a system that does not have harmonic distortion. If a load without distortion is and a load with distortion are both connected to the same source, the total value of S is not simply the sum of the values of S for the two loads as would be the case for two undistorted loads. When a load has distortion, it is necessary to calculate the total fundamental current and the total harmonic separately and then calculate the total as the square root of the sum of the squares.

The term "load angle" may mean the cos(R/Z) for the load. However that is not a very meaningful concept for Z in a DC circuit. Displacement and total power factors can only be calculated by determining the harmonic of the AC current. I don't believe that can be done directly from cos(R/Z). I don't know exactly how to do that at the moment.

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This kind of circuit is called a PD3 (parallel double, 3 phases -> q=3, so n=6). (unless "writing" errors).

General assumption made is that Current load Ic is assumed constant = \$Ic\$.

It has some formulas to describe it. (for 2 diodes current only allowed)

Mean value of voltage on load : (Vm is peak voltage);

\$ Uco=2 * q /Pi * Vm * sin(Pi/q)\$

Ripple factor : \$Ko = (Vmax-Vmin)/ (2 * Uco) \$ ; or \$ Ko=Pi/(2*n)*(1-cos(Pi/n)/sin(Pi/n)\$

Voltage versus time : \$Uc(t)=Uco*(1+Sum(-2*(-1)^k * cos(k * 2 * Pi * f * t)/(k^2 * n^2 - 1)) \$ (Summation index from k=1 to infinity, in fact all harmonics)

Load current = Ic (to be divided by 3 for each 2 diodes conducting, bridge)

Secondary rms current = \$ Is = Ic / sqrt (q)\$

Secondary factor \$Fs\$ (if transformer used) : \$ Fs = Uco * Ic / (q * Vrms * Is) \$

or \$ Fs = 2 / Pi * sqrt(2) * sin (Pi / q)\$

With a transformer used, \$Fp\$ primary factor \$ Fs = Fp = 3 / Pi = 0.955 \$

If primary is wired delta, Line factor \$FL\$ : \$FL = 0.955 \$, so transformer can be suppressed.

There is also a loss of voltage on load, because of the "internal impedance" of this circuit which is \$ Zin = q/(2*Pi)*(N2 * 2*Pi*f)\$ (N2 turns of secondary transformer -> inductor ), so "loss of voltage" is proportional to Ic.

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