Analog electronics, Basic concepts, Projects

Star and delta configurations

In addition to parallel and series, impedances also can make star and delta connections. Also are shown the transformations.

Star and delta connections

star and delta connections
On the left, resistors linked in delta and on the right, star. Source: electrical4u.

Some sources label delta connection as pi and star as T.

Star-delta transformation (Y-Δ)

transformations
Source: EletroWorld.

Equations to transform a star configuration to delta.

Ra=R1R2+R1R3+R2R3R1Ra=\frac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}

Rb=R1R2+R1R3+R2R3R2Rb=\frac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{2}}

Rc=R1R2+R1R3+R2R3R3Rc=\frac{R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{3}}

Delta-star transformation (Δ-Y)

Equations for (Δ-Y) transformation.

R1=RbRcRa+Rb+RcR_{1}=\frac{RbRc}{Ra+Rb+Rc}

R2=RaRcRa+Rb+RcR_{2}=\frac{RaRc}{Ra+Rb+Rc}

R3=RaRbRa+Rb+RcR_{3}=\frac{RaRb}{Ra+Rb+Rc}

If all resistors are equal, the equations become much simpler. Considering RYR_{Y} as resistor value in star and RΔR_{\Delta } of resistor in delta.

RY=RΔ3R_{Y}=\frac{R_{\Delta }}{3}

RΔ=3RYR_{\Delta }=3\cdot R_{Y}

Transformation with capacitors and coils

What if instead of resistors, we have capacitors and coil?

Demonstrating the equation of (Δ-Y) transformation for capacitors.

1c1=1C2C31C1+1C2+1C3\frac{1}{c_{1}}=\frac{\frac{1}{C_{2}C_{3}}}{\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}}

c1=C2C3+C3C1+C1C2C1c_{1}=\frac{C_{2}C_{3}+C_{3}C_{1}+C_{1}C_{2}}{C_{1}}

c2=C2C3+C3C1+C1C2C2c_{2}=\frac{C_{2}C_{3}+C_{3}C_{1}+C_{1}C_{2}}{C_{2}}

c3=C2C3+C3C1+C1C2C3c_{3}=\frac{C_{2}C_{3}+C_{3}C_{1}+C_{1}C_{2}}{C_{3}}

Equations for inverse transformation (Y-Δ).

1C1=1c1c2+1c1c3+1c2c31c1\frac{1}{C_{1}}=\frac{\frac{1}{c_{1}c_{2}}+\frac{1}{c_{1}c_{3}}+\frac{1}{c_{2}c_{3}}}{\frac{1}{c_{1}}}

C1=c2c3c1+c2+c3C_{1}=\frac{c_{2}c_{3}}{c_{1}+c_{2}+c_{3}}

C2=c1c3c1+c2+c3C_{2}=\frac{c_{1}c_{3}}{c_{1}+c_{2}+c_{3}}

C3=c1c2c1+c2+c3C_{3}=\frac{c_{1}c_{2}}{c_{1}+c_{2}+c_{3}}

And for inductors, the calculations are similar to resistors.

configurations with coils
Source: toppr.

Equivalent values from Y to Δ.

LRS=LRLS+LSLT+LRLTLTL_{RS}=\frac{L_{R}L_{S}+L_{S}L_{T}+L_{R}L_{T}}{L_{T}}

LRT=LRLS+LSLT+LRLTLSL_{RT}=\frac{L_{R}L_{S}+L_{S}L_{T}+L_{R}L_{T}}{L_{S}}

LST=LRLS+LSLT+LRLTLRL_{ST}=\frac{L_{R}L_{S}+L_{S}L_{T}+L_{R}L_{T}}{L_{R}}

Equivalent values from Δ to Y.

LR=LRSLRTLRS+LRT+LSTL_{R}=\frac{L_{RS}L_{RT}}{L_{RS}+L_{RT}+L_{ST}}

LS=LRSLSTLRS+LRT+LSTL_{S}=\frac{L_{RS}L_{ST}}{L_{RS}+L_{RT}+L_{ST}}

LT=LSTLRTLRS+LRT+LSTL_{T}=\frac{L_{ST}L_{RT}}{L_{RS}+L_{RT}+L_{ST}}

And for impedances.

impedances
Source: Wikimedia.

Conversion from Y to Δ.

Zac=ZaZc+ZaZb+ZbZcZbZ_{ac}=\frac{Z_{a}Z_{c}+Z_{a}Z_{b}+Z_{b}Z_{c}}{Z_{b}}

Zab=ZaZc+ZaZb+ZbZcZcZ_{ab}=\frac{Z_{a}Z_{c}+Z_{a}Z_{b}+Z_{b}Z_{c}}{Z_{c}}

Zbc=ZaZc+ZaZb+ZbZcZaZ_{bc}=\frac{Z_{a}Z_{c}+Z_{a}Z_{b}+Z_{b}Z_{c}}{Z_{a}}

Conversion from Δ to Y.

Za=ZabZacZab+Zac+ZbcZ_{a}=\frac{Z_{ab}Z_{ac}}{Z_{ab}+Z_{ac}+Z_{bc}}

Zb=ZabZbcZab+Zac+ZbcZ_{b}=\frac{Z_{ab}Z_{bc}}{Z_{ab}+Z_{ac}+Z_{bc}}

Zc=ZacZbcZab+Zac+ZbcZ_{c}=\frac{Z_{ac}Z_{bc}}{Z_{ab}+Z_{ac}+Z_{bc}}

3 problem examples

Let’s find the value of I current in this circuit.

Problem 1 with star and delta

Converting internal star in delta.

RΔ=3RYR_{\Delta }=3\cdot R_{Y}

RΔ=36=18ΩR_{\Delta }=3\cdot 6=18 \Omega

Problem 1 with 2 deltas

3 resistor pairs are in parallel. Just simplify the circuit to find I current.

Simplified problem 1

Rt=9189+18=6ΩRt=\frac{9\cdot 18}{9+18}=6\Omega

I=426=8AI=\frac{42}{6}=8A

How to calculate I current in this circuit?

Replacing R1, R2 and R3 with (Δ-Y) tranformation.

Ra=4,7k1,1k4,7k+1,1k+6,8k=5,17k12,6=0,41kΩRa=\frac{4,7k\cdot 1,1k}{4,7k+1,1k+6,8k}=\frac{5,17k}{12,6}=0,41k\Omega

Rb=4,7k6,8k4,7k+1,1k+6,8k=31,96k12,6=2,53kΩRb=\frac{4,7k\cdot 6,8k}{4,7k+1,1k+6,8k}=\frac{31,96k}{12,6}=2,53k\Omega

Rc=1,1k6,8k4,7k+1,1k+6,8k=7,48k12,6=0,59kΩRc=\frac{1,1k\cdot 6,8k}{4,7k+1,1k+6,8k}=\frac{7,48k}{12,6}=0,59k\Omega

With simplified circuit, becomes easier to calculate current.

Problem 2 2
Problem-2-3-1

Rt=9,33k7,39k16,72k=4,12kΩRt=\frac{9,33k\cdot 7,39k}{16,72k}=4,12k\Omega

I=80,41k+4,12k=1,76mAI=\frac{8}{0,41k+4,12k}=1,76mA

How to find total resistance of this resistor association?

Problem 3 of star and delta

Putting this cube in a shape that allows to visualize a group of resistors for conversion.

Problem-3-2 about star and delta

RY=RΔ3=93=3ΩR_{Y}=\frac{R_{\Delta }}{3}=\frac{9}{3}=3\Omega

Converting delta on the left in star.

Ra=Rb=R1R2R1+R2+R7=91233=3,27ΩR_{a}=R_{b}=\frac{R1\cdot R2}{R1+R2+R7}=\frac{9\cdot 12}{33}=3,27\Omega

Rc=12129+12+12=4,36ΩR_{c}=\frac{12\cdot 12}{9+12+12}=4,36\Omega

R7+R8=12,27ΩR7+R8=12,27\Omega

R2+R6=7,36ΩR2+R6=7,36\Omega

12,277,3612,27+7,36=4,6Ω\frac{12,27\cdot 7,36}{12,27+7,36}=4,6\Omega

3,27+4,6=7,87Ω3,27+4,6=7,87\Omega

Finally, the total resistance is:

Rt=97,879+7,87=4,19ΩRt=\frac{9\cdot 7,87}{9+7,87}=4,19\Omega

About Pedro Ney Stroski

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