Impedance

Bifilial line impedance

The impedance (Z) is a measure of opposition that a circuit presents to a current when a voltage is applied. Impedance extends the concept of resistance to alternating current (AC) circuits, and has both magnitude and phase, unlike resistance, which only has magnitude. When a circuit is supplied with direct current (DC), its impedance is equal to the resistance, which can be interpreted as the zero phase angle impedance.

By definition, impedance is the relationship (quotient) between the voltage phasor and the current intensity phasor:

Z=VI{displaystyle Z={frac {V}{I}}}}

Where Z{displaystyle Z} It's impedance, V{displaystyle V} is the tension fasor I{displaystyle I} corresponds to the fasor intensity.

The concept of impedance is especially important if the current varies over time, in which case the magnitudes are described with complex numbers or functions of harmonic analysis. Its module (sometimes inappropriately called impedance) establishes the relationship between the maximum values or the effective values of the voltage and current. The real part of impedance is resistance and its imaginary part is reactance.

The concept of impedance makes it possible to generalize Ohm's law in the study of alternating current (AC) circuits, giving rise to the so-called Ohm's law of alternating current which indicates:

I=VZ{displaystyle I={frac {V}{Z}}}}

The term was coined by Oliver Heaviside in 1886. In general, the solutions for the currents and voltages of a circuit made up of resistors, capacitors, and inductors and without any component of nonlinear behavior, are solutions of differential equations. But, when all voltage and current sources have the same constant frequency and their amplitudes are constant, the steady-state solutions (when all transients have disappeared) are sinusoidal and all voltages and currents have the same frequency as the sources. generators and constant amplitude. The phase, however, will be affected by the imaginary part (reactance) of the impedance.

Mathematical formalism

Definition

Be an electrical or electronic component or a circuit fed by a sinusoidal current I # (ω ω t){displaystyle scriptstyle {I_{circ }cos(omega t)},!}. If the tension between their ends is V # (ω ω t+φ φ ){displaystyle scriptstyle {V_{circ }cos(omega t+varphi)}},}, the impedance the circuit or component is defined as a complex number Z{displaystyle scriptstyle {Z},!}that expressed in polar form has a module equal to the quotient V I {displaystyle scriptstyle {V_{circ } over I_{circ }},}! and an argument that is φ φ {displaystyle scriptstyle {varphi },!}:

日本語Z日本語=V I arg (Z)=φ φ {displaystyle {begin{aligned}{aligned}{V_{circ } over I_{circ }}\arg(Z) fake=varphi end{aligned}},,!}

that is

Z=V I ejφ φ =V I (# φ φ +jwithout φ φ ){displaystyle Z={V_{circ } over I_{circ}}{jvarphi }={V_{circ } over I_{circ }}}}left(cos varphi +jsin varphi right,!}.

Which sometimes, especially in Electrotechnical texts, is also usually written with the format:

Z=V I φ φ {displaystyle Z={V_{circ } over I_{circ }angle varphi ,!}.

As indicated above, impedance is also defined by the quotient between the voltage and current phasors, representing the total opposition (Resistance, Inductive Reactance, Capacitive Reactance) over the current.

Since voltage and currents are sinusoidal, peak values (amplitudes), rms values, peak-to-peak values, or average values can be used. But care must be taken to treat them uniformly and not to mix the types. The result of the calculations will be of the same type as the one used for the voltage or current generators.

Binomial representation

Impedance can be represented in binomial form as the sum of a real part and an imaginary part:

Z=R+jX{displaystyle Z=R+jX,}

R{displaystyle scriptstyle {R}} It's the part. resistive or Real of impedance and X{displaystyle scriptstyle {X}} It's the part. reactivation or imaginary of impedance. Basically there are two kinds or types of reactances:

• Inductive or XL{displaystyle X_{L}}: Due to the existence of inducers.
• capacitive or XC{displaystyle X_{C}}: Due to the existence of capacitors.

Admittance

The admittance is the inverse of the impedance:

And=1Z=andc+jands{displaystyle Y=textstyle {1 over Z}=y_{c}+jy_{s}}}}

Conduct andc{displaystyle scriptstyle {y_{c}}}} is the real part of admission and susceptance ands{displaystyle scriptstyle {y_{s}}}} the imaginary part of the admission.

The unit of admittance, conductance and susceptance is the siemens (symbol S). One siemens is the reciprocal of one ohm.

Graphic representation

Example of fasores

You can represent the voltages of the voltage generators and the voltages between the ends of the components as rotating vectors in a complex plane. The magnitude (length) of the vectors is the voltage module and the angle they make with the real axis is equal to the phase angle with respect to the reference generator. This type of diagram is also called a Fresnel diagram.

With a bit of practice and a modicum of knowledge of geometry, these representations are much more explicit than values or formulas. Of course, those drawings are not, in our time, a graphical method of calculating circuits. They are a way of "seeing" as the stresses add up. Those drawings can facilitate the writing of the final formulas, using the geometric properties. You will find examples of the graphical representation in the examples below.

Calculation of circuits with impedances

The impedance formalism consists of a few rules that allow you to calculate circuits containing resistive, inductive, or capacitive elements in a similar way to calculating direct current resistive circuits. These rules are only valid in the following cases:

• On a permanent basis with unusoidal alternating current. That is to say, that all voltage and current generators are sinusoidal and of the same frequency, and that all transient phenomena (wrong connections and disconnections, sudden insulation failures, etc.) have been mitigated and completely disappeared.
• If all the components are linear. That is, components or circuits in which the amplitude (or effective value) of the current is strictly proportional to the applied tension. Nonlinear components such as diodes, coils with iron cores and others are excluded. Therefore, if the circuit contains inductances or transformers with ferromagnetic nucleus (which are not linear), the results of the calculations can only be approximate and that, provided to respect the working area of the inductances.

Out-of-phase voltage or current generators

If in a circuit there are several voltage or current generators, one of them is chosen as a phase reference generator. If the true voltage of the reference generator is V # (ω ω t){displaystyle scriptstyle {V_{circ }cos(omega t)}For calculation with impedances we will write your tension as V {displaystyle scriptstyle {V_{circ}}}}. If the voltage of another generator has a phase progression α α {displaystyle scriptstyle {alpha }} with respect to the reference generator and its current is I1# (ω ω t+α α ){displaystyle scriptstyle {I_{1}cos(omega t+alpha}}}For calculation with impedances we will write your current as I1ejα α {displaystyle scriptstyle {I_{1}e^{jalpha }}}}}. The argument of the calculated tensions and currents will be the defax of such tensions or currents with respect to the generator taken as reference.

Kirchhoff's Laws

Kirchhoff's laws apply in the same way: "the sum of the currents arriving at a node is zero" and "the sum of all stresses around a mesh is zero". This time, both the currents and the voltages are, in general, complex.

Generalization of Ohm's Law

The voltage between the ends of an impedance is equal to the product of the current times the impedance:

Vz=ZIz{displaystyle V_{z}=ZI_{z},}

Impedance, current and voltage are, in general, complex.

Impedances in series or parallel

Impedances are treated like resistances with Ohm's law. The impedance of several series-connected impedances is equal to their sum:

Series Z=Z1+Z2+ +Zn{displaystyle Z=Z_{1}+Z_{2}+cdots +Z_{n}}}

The impedance of several impedances connected in parallel is equal to the reciprocal of the sum of their reciprocals:

Stop it. Z=11Z1+1Z2+ +1Zn{displaystyle textstyle {Z}=textstyle {1 over scriptstyle {{1 over Z_{1}}}}+{1 over Z_{2}}}+cdots +{1 over Z_{n}}}}}}}}

Interpretation of results

The current result is usually a complex number. That complex number is interpreted as follows:

• The module indicates the value of the voltage or the calculated current. If the values used for generators were the peak values, the result will also be a peak value. If values were effective values, the result will also be an effective value.
• The argument of that complex number gives the disphase with respect to the generator used as a phase reference. If the argument is positive, the calculated voltage or current will be in progress.

Generalization

When all the generators do not have the same frequency or if the signals are not sinusoidal, the impedance formalism cannot be applied directly. The calculation has to be decomposed into several stages in each of which the impedance formalism can be used.

In the case of having linear elements, you can use the superposition theorem: a separate calculation is made for each of the frequencies (replaced in each of the calculations all different voltage generators for a short circuit and all different frequency current generators for an open circuit). Each of the total tensions and currents of the circuit will be the sum of each of the tensions or currents obtained at each of the frequencies. Of course, to make these last sums you must write each of the tensions in the real way, with the dependence of time and the disphase: Vi# (ω ω it+φ φ i){displaystyle scriptstyle {V_{i}cos(omega _{i}t+varphi _{i}}}}}}}}} for tensions and similar formulas for currents.

If the signals are not sinusoidal, but are periodic and continuous, you can decompose the signals into Fourier series and use the superposition theorem to separate the calculation into one calculation for each of the frequencies. The final result will be the sum of the results for each of the frequencies of the series decomposition.

Source of impedances

We are going to try to illustrate the physical meaning of the imaginary part j (where this letter is used instead of i to avoid confusion with the intensity) of the impedances calculating, without using these, the current flowing through a circuit formed by a resistor, an inductor and a capacitor in series.

The circuit is fed with a sinusoidal tension and we have waited long enough for all transient phenomena to disappear (we have a permanent regime). Since the system is linear, the current of the permanent regime will also be sinusoidal and will have the same frequency as that of the original source. The only thing we don't know about the current is its breadth and the gap it can have with respect to the power supply. So, if the power voltage is V=V # (ω ω t){displaystyle scriptstyle {V=V_{circ}cos(omega t)}} the current will be of the form I=I # (ω ω t+φ φ ){displaystyle scriptstyle {I=I_{circ}cos(omega t+varphi)}}Where φ φ {displaystyle scriptstyle {varphi}} It's the disdain we don't know. The equation to be resolved will be:

V # (ω ω t)=VR+VL+VC{displaystyle V_{circ }cos(omega t)=V_{R}+V_{L}+V_{C}}}

where VR{displaystyle scriptstyle {V_{R}}}}, VL{displaystyle scriptstyle {V_{L}}}} and VC{displaystyle scriptstyle {V_{C}}}} are the tensions between the extremities of the resistance, the inductance and the capacitor, respectively.

Applying Ohm's law to the resistance, it results:

VR{displaystyle V_{R},} = RI # (ω ω t+φ φ ){displaystyle RI_{circ }cos(omega t+varphi)}

The definition of inductance tells us that:

VL=d≈ ≈ cdt=dLIdt{displaystyle V_{L}={dPhi c over dt}={dLI over dt}}

If L is constant, it remains:

VL=LdIdt=Ld(I # (ω ω t+φ φ ))dt=− − ω ω LI without (ω ω t+φ φ ){displaystyle V_{L}=Ltextstyle {dI over dt}=Ltextstyle {dleft(I_{circ }cos(omega t+varphi)right) over dt}=-omega LI_{circ }sin(omega t+varphi)}}}.

The definition of capacitance tells us that:

I=dqdt=d(CVC)dt{displaystyle I={{dq} over {dt}}={{{d(CV_{C}}}} over {dt}}}}}

If C is constant:

I=C⋅ ⋅ dVdt{displaystyle I=Ccdot {frac {dV}{dt}}}}

Doing the integral, it can be verified that:

VC=1ω ω CI without (ω ω t+φ φ ){displaystyle V_{C}=textstyle {1 over omega C}I_{circ }sin(omega t+varphi)}.

Thus, the equation to be solved is:

V # (ω ω t)=RI # (ω ω t+φ φ )− − ω ω LI without (ω ω t+φ φ )+1ω ω CI without (ω ω t+φ φ ){displaystyle V_{circ }cos(omega t)=RI_{circ }cos(omega t+varphi)-omega LI_{circ }sin(omega t+varphi)+textstyle {1 over omega C}I_{circ }sin(omega t+varphi

We have to find the values of I {displaystyle scriptstyle {I_{circ}}}} and φ φ {displaystyle scriptstyle {varphi}} to make this equation satisfied for all values t{displaystyle scriptstyle {t}}.

To find them, imagine that we feed another identical circuit with another sinusoidal voltage source whose only difference is that it begins with a quarter of a delay period. I mean, the tension will be V=V # (ω ω t− − π π 2)=V without (ω ω t){displaystyle scriptstyle {V=V_{circ }cos(omega t-{pi over} 2)=V_{circ }sin(omega t)}}}}}. In the same way, the solution will also have the same delay and the current will be: I=I # (ω ω t+φ φ − − π π 2)=I without (ω ω t+φ φ ){displaystyle scriptstyle {I=I_{circ }cos(omega t+varphi -{pi over 2})=I_{circ }sin(omega t+varphi)}}}}}. The equation of this second delayed circuit will be:

V without (ω ω t)=RI without (ω ω t+φ φ )+ω ω LI # (ω ω t+φ φ )− − 1ω ω CI # (ω ω t+φ φ ){displaystyle V_{circ }sin(omega t)=RI_{circ }sin(omega t+varphi)+omega LI_{circ }cos(omega t+varphi)-textstyle {1 over omega C}I_{circ }cos(omega t+varphi

There are signs that have changed because the retarded cosine becomes breast, but the retarded breast becomes − − {displaystyle textstyle {mathbf {-}}}Cosme. Now we will add the two equations after we have multiplied the second by j. The idea is to be able to transform expressions of form # x+jwithout x{displaystyle scriptstyle {cos x+jsin x}} in ejx{displaystyle scriptstyle {e^{jx}}}}using the Euler formulas. The result is:

V ejω ω t=RI ej(ω ω t+φ φ )+jω ω LI ej(ω ω t+φ φ )+1jω ω CI ej(ω ω t+φ φ ){displaystyle V_{circ }e^{jomega t}=RI_{circ }e^{jleft(omega t+varphi right)}+jomega LI_{circ}{jleft(omega t+varphi right)} !

Like ejω ω t{displaystyle scriptstyle {e^{jomega t}}} is different from zero, the entire equation can be divided by that factor:

V =RI ejφ φ +jω ω LI ejφ φ +1jω ω CI ejφ φ {displaystyle V_{circ }=RI_{circ }e^{jvarphi }+jomega LI_{circ }e^{jvarphi }+textstyle {1 over jomega C}I_{circ }e^{jvarphi }

it follows:

I ejφ φ =V R+jω ω L+1jω ω C{displaystyle I_{circ }e^{jvarphi }=textstyle {V_{circ} over R+jomega L+scriptstyle {1 over jomega C}}}}}}}

On the left we have the two things we wanted to calculate: the amplitude of the current and its phase shift. The amplitude will be equal to the modulus of the complex number on the right, and the offset will be equal to the argument of the complex number on the right.
And the term on the right is the result of the usual calculation using the impedance formalism in which the impedances of resistors, capacitors, and inductors are treated in the same way as resistors with Ohm's law.
It is worth repeating that when we write:

I=V R+jω ω L+1jω ω C{displaystyle I=textstyle {V_{circ} over R+jomega L+scriptstyle {1 over jomega C}}}}

We admit that the person reading that formula can interpret it and will not believe that the current can be complex or imaginary. The same assumption exists when we find expressions like "feeding tension" Vejω ω t{displaystyle scriptstyle {Ve^{jomega t}}}» or “the current is complete”.

Because the signals are sinusoidal, the factors between rms, maximum, peak-to-peak, or average values are fixed. So, in the impedance formalism, if the input values are peaking, the results will also be peaking. Same for effective or others. But you don't have to mix them.

Examples

Impedance in basic elements

The impedance of an ideal resistor only contains a real component such as:

ZR=R{displaystyle Z_{R}=R}

In this case, the voltage and current are proportional and in phase.

The impedance in an ideal inductor or an ideal capacitor has a purely imaginary component:

The impedance in an inductor increases with frequency;

ZL=jω ω L{displaystyle Z_{L}=jomega L}

The impedance of a capacitor decreases as the frequency increases;

ZC=− − jω ω C{displaystyle Z_{C}={frac {-j}{omega C}}}

A unique generator

A serial inductance and resistance fed by a sinusoidal generator

In the diagram on the right we have a sinusoidal generator V=10# (ω ω t){displaystyle scriptstyle {V=10cos(omega t)}} 10 volts of width and 10 kHz frequency. In series there is a 10 mH inductance and a 1.2 kΩ resistance.

Let's calculate the current I{displaystyle scriptstyle {I}} that circulates in the circuit:

I=VZL+ZR=Vjω ω L+R=10j2π π 1040,01+1200{displaystyle I=textstyle {V over Z_{L}+Z_{R}}={V over jomega L+R}={10 over j2pi 10^{4}0,01+1200}}}}}}

=101200+j628,3A{displaystyle textstyle {={10 over 1200+j628,3}}}mathrm {A} }

The application of calculus with complex numbers is necessary if this notation is used.

The application of calculus with complex numbers is necessary if this notation is used.

I=日本語101200+j628,3日本語=7,38mA{displaystyle I=leftSDtextstyle {10 over 1200+j628,3}right.

As the value of the voltage of the generator we take was a peak value (amplitude), the value of the current obtained is also a peak value. The effective current is: Ief=7,382=5,22mA{displaystyle scriptstyle {I_{mathrm {ef} }={7,38 over {sqrt {2}}}}=5,22mathrm {mA} }

The phase of the current is the argument of the complex number

101200+j628,3{displaystyle scriptstyle {10 over 1200+j628,3}}:

arg(101200+j628,3)=− − 0,4823rad=− − 27,63 {displaystyle mathrm {arg} left(textstyle {10 over 1200+j628,3}right)=-0{,}4823,mathrm {rad} =-27{,}63^{circ}}}}}.

The current is in phase delay with respect to the phase of the generator. That is logical, since the circuit is inductive.

Fresnel diagram (or fasor) of inductance and serial resistance. The gray circle only helps the drawing of the right angle between the tension of the resistance and the tension of the inductance.

Only the resistor dissipates power:

PR=12R日本語I日本語2=121200⋅ ⋅ (7,3810− − 3)2=32,7mW{displaystyle P_{R}=textstyle {1 over 2}RleftINDIrightknowledge^{2}=textstyle {1 over 2}1200cdot left(7{,}38,10^{-3}right)^{2}={32,}7,mathrm {mW}}}}}

The fraction 12{displaystyle scriptstyle {1 over 2}} appears because the current value is the peak value.

The tension between the ends of the resistance is VR=IR=(0.00654− − j0.003424)1200=7,84− − j4,109Vpicor{displaystyle scriptstyle {V_{R}=I,R=(0{,}00654-j0{,}003424),1200=7{,}84-j4{,}109,V_{mathrm {pico}}}}}}}}

The effective tension that would be read with a voltmeter would be the module of this tension divided by 2{displaystyle scriptstyle {sqrt {2}}}}: 6,26Vef{displaystyle scriptstyle {6{,}26,V_{mathrm {ef} }}}}}

The tension between the extremities of inductance is
VL=jω ω LI=j628,3(0.00654− − j0.003424)=2,15+j4,109Vpicor{displaystyle scriptstyle {V_{L}=jomega L,I,=j628{,}3,(0{,}00654-j0{,}003424)=2{,}15+j4{,}109,V_{mathrm {pico}}}}}}

The effective voltage read with the voltmeter would also be: 3,28Vef{displaystyle scriptstyle {3,28,V_{mathrm {ef}}}}}

We find that the sum of the two "complete" tensions gives (taking into account the rounds) the tension of the generator. Instead, the sum of the two tensions read with a voltmeter is larger than the generator (7,07Vef{displaystyle scriptstyle {7,07V_{mathrm {ef}}}}}}). This result is typical of the measures made with a voltmeter in circuits in which tensions are not in phase. A voltmeter measures us modules in effective value, which we cannot add directly as we are dealing with fastors with their different orientations.

Two outdated generators

Condensor and serial resistance between two disfased senoidal generators

On the right circuit, a capacitor 1μ μ F{displaystyle scriptstyle {1,mu F}} and resistance 3kΩ Ω {displaystyle scriptstyle {3,kOmega}} in series, they are connected between two sinusoidal generators. We take as generators two phases of the three phase supply. The left generator will be our reference generator V1=2302# (314t){displaystyle scriptstyle {V_{1}=230{sqrt {2}}cos(314,t}}}}. The right generator is in progress phase 2π π /3{displaystyle scriptstyle {2pi /3}. I mean, V2=2302# (314t+2π π 3){displaystyle scriptstyle {V_{2}=230{sqrt {2}}cos(314,t+{2pi over 3})}}}. With the formalism of impedances, the left generator will be V1=230Vef{displaystyle scriptstyle {V_{1}=230,V_{mathrm {ef}}}}}} and the right V2=230ej2π π 3Vef{displaystyle scriptstyle {V_{2}=230,e^{j{2pi over 3}}{,V_{mathrm {ef}}}}}}}}. Let's start by calculating the tension difference between the two generators:

V12=230(1− − ej2π π 3)=230(1− − # (2π π 3)− − jwithout (2π π 3)){displaystyle V_{12}=230,left(1-e^{j{2pi over 3}}}right)=230left(1-cos left(textstyle {2pi over 3}right)-jsin left(textstyle {2pi over 3}right)}}

=230(1,5− − j0.866)=345− − j199,19Vef=398,37e− − j0.55774{displaystyle =230,(1{,}5-j0{,}866)=345-j199{},19,V_{mathrm {ef}=398{,}37e^{-j0{,}}}}

The module of this tension is 398,37Vef{displaystyle scriptstyle {398{}37V_{mathrm {ef}} } and is delayed of 0.5574 radians (30°) with respect to the baseline voltage.

Fresnel diagram corresponding to the second example. The first circle guides the tensions of the two generators. The second for the right angle between the tension of the capacitor and that of the resistance.

The current flowing is:

I=V12R+1jω ω C=398,37e− − j0.55363000− − j3185=398,37e− − j0.55364375,41e− − j0.8153=0.0910ej0.2917{displaystyle I=textstyle {V_{12} over R+scriptstyle {1 over jomega C}={398{,}37,e^{-j0{,}{5236} over {3000-j31}{398{,}37,e^{-j0{,}{5236-j}{0}{over}{3⁄4}{,}}}{3}}}}{3}{, {3}{3}}}}{3}}{3}{, {1⁄4}}}}}}}{3}}{, {3}}}}}{3}{3}}{3}{, {3}}{3}{,}}{3}}{3}{, {3}}}}}}}{

Since the voltage values used for the generators were rms values, the calculated current also comes as rms value: 91 mA in 16.71° phase advance with respect to the reference voltage.

The tension between the ends of the resistance is VR=RI=3000⋅ ⋅ 0.0910ej0.2917=273ej0.2917Vef{displaystyle scriptstyle {V_{R}=R,I=3000cdot 0{,09}10,e^{j0{,}2917}=273,e^{j0{,}2917}V_{mathrm {ef}}}}}}}}

The tension between the ends of the capacitor is:
VC=ZCI=− − j3185⋅ ⋅ 0.0910ej0.2917=3185e− − jπ π 20.0910ej0.2917=289,83e− − j1,2791Vef{displaystyle scriptstyle {V_{C}=Z_{C},I=-j3185cdot 0{,0910,e^{j0{,}2917}=3185,e^{-j{pi over 2}0{,}0910,e^{j0{,}2917hr=289{,}{,}{.

The voltage between the ends of the capacitor lags 73.3° with respect to the reference voltage. As in the previous example, the sum of the modules of the voltages (those that would be measured with a voltmeter) of the resistor and of the capacitor (563 V) is greater than the total voltage applied (398 V).

The voltage at point A of the circuit will be:

VA=V1− − VC=230− − 289,83e− − j1,2791=230− − (83,35− − j277,6){displaystyle V_{A}=V_{1}-V_{C}=230-289{,83},e^{-j1{,}2791}=230-(83,35-j277{,}6)}}

=146.65+j277,6=314ej1,085Vef{displaystyle =146.65+j277{,}6=314,e^{j1{,}08},V_{mathrm {ef}}}}}}

The voltage at point A is greater than that at each generator.