What is a Transformer? How to measure losses & equivalent circuit parameters.

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In its most elementary form, a transformer consists of two wound coils of wire that are inductively coupled to each other. When alternating (AC) current at a given frequency flows in either coil, an alternating voltage of the same frequency is induced in the other coil. The value of this voltage depends on the degree of coupling, and the flux linkages in the two coils. The coil connected to a source of alternating voltage is usually called the primary coil, and the voltage across this coil is the primary voltage. Voltage induced in the secondary coil may be greater than or less than the primary voltage, depending on the ratio of the primary to secondary turns. A transformer is termed a step-up, or a step-down transformer, accordingly. 

If both varying voltage and D.C. voltages are impressed across the primary, only the varying part is delivered to the load. This comes about because the voltage, e, in the secondary is induced in the winding by the core flux, ф, according to the law. 

This law may be stated as follows: The voltage induced in a coil is proportional to the number of turns, and to the time rate of change of magnetic flux in the coil. This rate of change of flux may be large or small. For a given voltage, if the rate of change of flux is small, many turns must be used. Conversely, if a small number of turns is used, a large rate of change of flux is necessary. 

Name Plate Rating (Voltage Ratio): 

The voltage ratio of the transformer is specified as V1 (rated)/V2 (rated). It means that when voltage V1 (rated) is applied to the primary winding, the secondary winding voltage, on full load at a specified power factor, is V2 (rated). The ratio V1 (rated)/V2 (rated) is not exactly equal to NI/N2, because of voltage drops in the primary and secondary. These drops, being small, are neglected, and it is assumed for all practical purposes. 

The rating of the transformer is specified in units of VA/kVA/MVA, depending upon its size. 
kVA (rated) = V (rated) x I (full-load)/1000 

Transformer Losses: 

The transformer has no moving parts, so its efficiency is much higher than that of rotating machines. The various losses in a transformer are enumerated below: 

Core-Loss: These are hysteresis and eddy-current losses, resulting from alternations of magnetic flux in the core. It is emphasized that the core-loss is constant for a transformer operated at constant voltage and frequency, as are all power frequency transformers. 

Copper-loss (I2R-loss): This loss occurs in winding resistances when the transformer carries the load current; varies as the square of the loading expressed as a ratio of the full-load. 
Load (stray) loss: It largely results from leakage fields inducing eddy-currents in the transformer walls and conductors. 

Dielectric-loss: The source of this loss is in the insulating materials; particularly, in solid insulation. 

The major losses are, by far, the first two: Pi, the constant core (iron)-loss, and PCu, the variable copper-loss. It will easily be seen in the following sections that transformer losses, and the parameters of its equivalent circuit, can easily be determined by two simple tests, without actually loading it. 

  

TRANSFORMER TESTING: 

Two major difficulties, which do not warrant the testing of large transformers by direct load test, are: 
(i) A large amount of energy has to be wasted in such a test. 
(ii) It is impossible for a large transformer to arrange a load large enough for direct loading. 

Thus, performance characteristics of a transformer must be computed from a knowledge of its equivalent circuit parameters which, in turn, are determined by conducting simple tests, involving very little power consumption, called non-loading tests. 

In these tests, the power consumption is simply that which is needed to supply the losses incurred. The two non-loading tests are the open-circuit (OC) test, and short-circuit (SC) test. In both these tests voltage, current and power are measured from which the resistance and reactance of the input impedance can be found. Thus, only four parameters can be determined, which correspond to the approximate equivalent circuit of Fig. 1. 

Open Circuit Test: 

The purpose of this test is to determine the shunt branch parameter of the equivalent electrical circuit of transformer. 

  • Connect the Unit under Test (UUT, the transformer) low voltage side to the supply rated AC voltage 
  • Keep the high voltage side open-circuited 
  • Connect the power analyzer or power meter between the AC supply and the UUT. 
  • At no load (open circuit) condition, the no load current (Io) should be small; around 2-6% of the rated current. 
  • Since the Io is small, it’s safe to assume V1 is equal to E1 by neglecting the series impedance. Thus, for all practical purposes, the power input on no-load equals the core (iron) loss. 
  • The circuit parameters can be computed from three readings at different operating points 
    Yo = Gi – jBm 
    Yo = Io/V1 
    Gi = Po/(V1)2 
    Bm = sqrt (Yo^2 – Gi^2) 

Thus, the OC test yields the core–loss and the parameter of the shunt branch of the equivalent circuit. 

Short Circuit Test: 
 

The series parameters of the transformer, as well as its copper losses, could be evaluated using the short circuit test. 

Depending on the transformer voltage and the power source available in your lab, you can short the low or the high side of the transformer for this test. 

Transformer resistances and leakage reactance are very small; the input voltage to circulate the rated load current under short circuit is as low as 5-8% of the rated voltage (in most transformers). 

Due to low input voltage, the exciting current (Io) is only about 0.1-0.5% of the full load current. (Io at the rated voltage is 2-6% of the full load current). Thus, the shunt branch of the equivalent circuit can be ignored, as depicted below. 

  • Raise the input supply voltage gradually from zero until the transformer draws full-load current. 
  • The power analyzer gives the reading as under: 

Voltage = Vsc; Current = Isc; Power input = Psc. 

  • Since they are at low voltages, transformer iron losses can be neglected. Thus, only copper losses of the transformer can be obtained. 

Psc = Pc (copper-loss) 

Z= Vsc/Isc = sqrt (R^2+ X^2) 

R= Psc/(Isc^2) 

X = sqrt (Z^2 – R^2) 

Calculate weight of wire based on transmission voltage and losses

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Hello friends and colleagues, 

If you need to calculate the weight of wire based on transmission voltage, use the math lab model that I created. All what you have to do is: 

  1.  Enter the amount of power you need to transmit (red text “enter power”) 
  2. The allowable percentage power losses (it is 2% in the code below, and you can change that by typing different values) 
  3. The wire length  

 Copy and paste the code below in math lab, and you should be good to go. 

 Start of the code: 

% Calculate weight of wire based on transmission voltage 
clc 
clear 

p = “enter power“; % watts 
res_cu = 1.7e-8; %ohms 
res_al = 2.8e-8; 
rho_cu = 8940; %kg/m^3 
rho_al = 2700; 
rho_ins = 860; %kg/m^3 for ethylene polypropylene rubber 
l = “enter wire distance“; % length, meters 

lossVec = [0.02];% total wire losses (percentage of the power) 

for i = 1:length(lossVec) 
loss = lossVec(i); 
V = 400:100:3000; %line-line voltages, rms 
%V = V*(sqrt(3)); 
% AC 
I = p./(sqrt(3)*V); % RMS current in each line 
Vdrop = loss*V; %voltage drop across the wire for each phase 

thick_ins_AC = .0254*(8.1e-6*(V*sqrt(2)) + .057); %thickness in meters 
thick_ins_DC = .0254*(8.1e-6*V + .057); %thickness in meters 

A_al = res_al * l * I ./ Vdrop; %cross sectional area 
for k = 1:length(V) 
if A_al(k) < 1.33e-6 
A_al(k) = 1.33e-6; 
end 
end 
m_al = A_al*l*rho_al; % total mass for one wire(one phase) 
m_ins_al = (pi*((sqrt(A_al/pi)+thick_ins_AC).^2)-A_al)*l*rho_ins; 
m_al_tot = 3*(m_al + m_ins_al); % model using 3x mass_al +3x mass_ins, 

A_cu = res_cu * l * I ./ Vdrop; 
m_cu = 1*A_cu*l*rho_cu; 
m_ins_cu = (pi*((sqrt(A_cu/pi)+thick_ins_AC).^2)-A_cu)*l*rho_ins; 
m_cu_tot = 3*(m_cu + m_ins_cu); 

% DC 
Idc = p./V; %DC voltage 

Adc_cu = res_cu * (2*l) * Idc./ Vdrop; %cross sectional area, for one wires 
mdc_cu = Adc_cu*l*rho_cu; % total mass for one wires 
Rdc_cu = res_cu*l./Adc_cu; % resistance per wire 
mdc_ins_cu = (pi*((sqrt(Adc_cu/pi)+thick_ins_DC).^2)-Adc_cu)*l*rho_ins; 
mdc_cu_tot = 2*(mdc_cu + mdc_ins_cu); 

Adc_al = res_al * (2*l) * Idc./ Vdrop; %cross sectional area, for two wires 
for n = 1:length(V) 
if Adc_al(n) < 1.33e-6 %assumes that limit for aluminum is AWG 6 
Adc_al(n) = 1.33e-6; 
end 
end 
mdc_al = 1*Adc_al*l*rho_al; % total mass for one wire 
Rdc_al = res_al*l./Adc_al; 
mdc_ins_al = (pi*((sqrt(Adc_al/pi)+thick_ins_DC).^2)-Adc_al)*l*rho_ins; 
mdc_al_tot = 2*(mdc_al + mdc_ins_al); 

%figure 
%plot(V,m_al,V,m_ins_al,V,m_al_tot) 
%legend(‘Aluminum’,’Insulation’,’Total’) 
%title(‘Weight Composition for AC Aluminum Cables’) 
%xlabel(‘Line-line Voltage [V]’) 
%ylabel(‘Conductor weight [kg]’) 
%grid on 

%figure 
%plot(V,m_cu,V,m_ins_cu,V,m_cu_tot) 
%legend(‘Copper’,’Insulation’,’Total’) 
%title(‘Weight Composition for AC Copper Cables’) 
%xlabel(‘Line-line Voltage [V]’) 
%ylabel(‘Conductor weight [kg]’) 
%grid on 

%figure 
%plot(V,mdc_cu,V,mdc_ins_cu,V,mdc_cu_tot) 
%legend(‘Copper’,’Insulation’,’Total’) 
%title(‘Weight Composition for DC Copper Cables’) 
%xlabel(‘Line-line Voltage [V]’) 
%ylabel(‘Conductor weight [kg]’) 
%grid on 

%figure 
%plot(V,mdc_al,V,mdc_ins_al,V,mdc_al_tot) 
%legend(‘Aluminum’,’Insulation’,’Total’) 
%title(‘Weight Composition for DC Aluminum Cables’) 
%xlabel(‘Line-line Voltage [V]’) 
%ylabel(‘Conductor weight [kg]’) 
%grid on 

figure 
%semilogy(V,m_cu_tot,V,m_al_tot,V,mdc_cu_tot,V,mdc_al_tot) 
plot(V,m_cu_tot,V,m_al_tot,V,mdc_cu_tot,V,mdc_al_tot) 
legend(‘AC Cu’,’AC Al’,’DC Cu’,’DC Al’) 
title(‘Weight Comparison with 2% losses’) 
xlabel(‘Line-line Voltage [V]’) 
ylabel(‘Conductor weight [kg]’) 
grid on 
end 

Example to calculate short circuit current for circuit breaker

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The purpose of this write up is to help the reader estimate the sExample, before  you hire  the PE to do the  official  calculation. 

All electrical systems are susceptible to short circuits.  The short circuit currents can produce considerable thermal and mechanical stresses in electrical distribution equipment. Therefore, it’s important to protect personnel and equipment by calculating short-circuit currents during system upgrade and design. Since these calculations are life-safety related, they’re mandated by 110.9 of the NEC. 

In the example shown in Fig. 1, our goal is to find the irrupt rating for the two breakers (1600A and 250A), and also find the withstand rating of the two panels (main and sub-panel).

First, we will calculate the full load current for the 1MVA transformer. 
I F.L = P  / (1.73 * V L-L ); where P is the transformer power rating in VA, and V L-L is the line to line RMS voltage at the secondary side of the transformer. 
I F.L = 1,000,000/ 1.73*480 =  1,202 A; the I F.L is the full load current of the transformer. 
Now, to find the short circuit rating of point 1 (i.e. if we bolt all the three wires of the secondary side of the transformer, and how much current we will see at that point):  
I S.C. for 3 Ph = I F.L /Z%; where I.S.C. is the short circuit current, and Z% is the transformer impedance; which usually can be obtained from the local utility company. 
I S.C. 3 ph for point 1 = 1202 /0.05 = 24,506 A or 24.5 KA (this is the available short circuit current at point 1). 
The short circuit current at point 2 will be lower than point 1, due to the line impedance (line impedance for the cables will impose the current). 
The calculation below will show the short circuit current at point 2: 
From the drawing; cable length = L = 50 ft. 
F = (1.73 * L* I S.C. 3 ph) / (C *n*V L-L); F for the three-phase fault       ——– [1] 
F = (2 * L* I S.C. L-L) / (C *n*V L-L); F for line to line fault            ——– [2] 
F = (2 * L* I S.C. L-N) / (C *n*V L-N); F for Line to Neutral fault      ——–[3] 
In our case, we will use equation #1 for this example. 
C = the conductor factor; the value can be obtained from the table (1) by locating the conductor’s size in the proper configuration 
n = the number of parallel runs 
L = length in ft.
Table (1): These values are equal to one over the impedance per foot, based upon resistance and reactance, found in IEEE std. 241 

F = (1.73 *50 *24,506) / (28,303*5*480) = 0.031206 

M = 1/(1+F) = 0.969738; where M is the multiplier 

I S.C. 3 ph for point 2 = 24506* M = 24506*0.969738 = 23,764 A or 23.7 KA (this is the available short circuit current at point 2) 

It’s worth mentioning that the further the main panel is from the transformer, the lower the short circuit current, but you will have a bigger voltage drop, due to the larger line impedance. 

The calculation below will show the short circuit current at point 3 (this is similar to the point 2 calculation): 

F = (1.73 * L* I S.C. 3 ph at point 2 ) / (C *n*V L-L) 

F = (1.73 *100*23,764) / (16673*1*480) = 0.51370 

M = 1/(1+F) = 1/1.51370 = 0.6606 

I S.C 3 ph for point 3 = 23,764 *0.6606 = 15,699 A or 15.7 KA 

I hope you find this example and the short circuit calculation useful. Please let me know if you have any questions. 

80AMPS All In One ESC for drones

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DRF is pleased to announce the release of the prototype design of our high current Brushless Motor Drive. Capable of delivering a peak current of 80 Amps, it offers a wide operating voltage range of 30 to 100 Volts, and it can safely operate at temperatures up to 50°C. The unit is designed for use as an Electronic Speed Control (ESC) for drones, as well as for battery-powered tools; such as drills, grinders, and cutters. 

 DRF’s proprietary all-in-one board includes: 

  • A micro controller 
  • A power stage 
  • A gate driver 
  • Temperature monitoring for the switching MOSFETs 
  • Protection against voltage transients 
  • A Controller Area Network (CAN Bus) 
  • Onboard protection against under-voltage, over-current, and over-temperature 

 Handling 80 Amps on a Single-Board Device: 

Designing an 80 Amp peak current with SMD switching devices was very challenging; particularly, because of the size and weight of the heat sinks normally required for cooling the switching of the MOSFETs. To achieve this goal, each of the PCB’s eight layers included two ounces of copper to spread the heat away from the MOSFET junction. Then, we added more than 500 ten mil (0.254 mm) thermal micro vias, in order to reduce thermal resistance of the system. In this manner, we caused the removal of excess heat from the die, through the PCB, to the ambient environment.

Great for UAV’s! 

The 80 Amp peak current, and the superior efficiency that our Brushless Motor Drive provides, makes it possible for UAVs to stay in the air longer, in order to endure harsh headwinds, and all the while carry larger payloads. 

Who We Are:

We are an innovative design and development team, specializing in power electronic circuitry and power conversion products. We primarily provide product design/electrical engineering services and have worked with cutting edge companies like yours in the past.

 Our services include: 

  • Analog/Digital/Power circuit design 
  • Linear/Switching regulator design 
  • Electro-mechanical product design and program management 
  • Power Management  
  • Manufacturing support 
  • Pre-compliance review and test support (certification and test management for IEC and UL)  
  • Due Diligence and Private Equity Advisory Services 
  • CAD design services, schematics capture, and board layout 

To learn more about our services, customers feedback, and product success stories please visit us at: http://www.drfengineeringservices.com 

Design Guide To Creating Your Next Inductor

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Welcome to the December edition of DRF’s newsletter!  This month, Saad Thabit presents a guide for designing a custom inductor, using C Cut-Cores. 

Do you have a new inductor design in mind, but don’t know what to use? 

Look no further! Amorphous metal and nanocrystalline are the best available options for your designs requiring the highest efficiency, and here’s why: 

  • High permeability leads to increased inductance and reduces winding turns, resulting in reduced I2R losses. 
  • High saturation induction will reduce size of the inductor. 
  • High frequency will range from 50Hz up to 100KHz. 
  • High operating temperatures can reach up to 120℃. 
  • Low coercivity increases the efficiency, and reduces hysteresis loss.   
  • Low core loss reduces energy consumed, and minimizes the temperature rise.  

Filter inductor designs, the values of the inductance, inductor current, operating frequency, ripple current and power losses will be determined by the following application:  

V_ind = 2*π*f*L*I_Peak 

Where: 

  • f = Inductor operating fundamental frequency 
  • L = Inductance value 
  • I Peak = Inductor peak current 

How to calculate the cross-sectional area product Ap: 

 Ap = [V ind*I peak*10^4] / [Kf*Ku*B peak*f*J] 

Where:  

  • Ap units are in cm4 
  • Kf =4.44 for sine waveforms 
  • Ku is the core window utilization fill factor 
  • B peak is the flux density in Tesla 
  • f is the operating frequency in Hz 
  • J is the current density in Amp / cm2  

The Ap product of a C-type cut core is the product of the available window area (Wa) of the core in square centimeters (cm2), multiplied by the effective cross-sectional area (Ac) in square centimeters (cm2), which may be stated as: 

A p = W a x Ac [cm4] 

Figure 1 (shown above) shows the outline form of a C-core type inductor; typical of those shown in the catalogs of suppliers. 

From this, it can be seen that [Wa] is the BC product, and [Ac] is the AD product. 

The AC inductor must support the applied voltage V ind. The number of turns is calculated from Faraday’s Law, shown below: 

N = [V ind*10^4] / [L*Kf*B peak*f*Ac] [Turns] 

Now, we can calculate the air gap with the equation below: 

Lg = [0.4*π*N^2*Ac*10^-8]/ L 

Where: 

  • L is the inductance, measured in Henry 
  • AC is measured in cm2 
  • Lg is the gap in cm 

Keep in mind, the fringing flux will decrease the total reluctance of the magnetic path, therefore increasing the inductance, by a factor F; all of which can be found in the formula below: 

F = [1+(Lg/sqrtAc) *(Ln(2*C/Lg))]  

Now that our major inductor parameters have been identified, we need to determine the inductor wire size. 

 Calculate inductor bare wire area Awire: 
  Awire =   I peak/J [cm2]  

Select the wire from the wire manufacturer table. For reference, you can use the  table supplied in this link.    

Now, we have determined the wire size. If we also know the mean turn length, as well as the number of turns, then we can determine the total length of the wire. From there, we can calculate the DC winding resistance, using the formula: 

R dc = MLT*N*resistance per unit length [Ohm] 

Where: 

Resistance per unit length as read from the wire tables for the selected wire size. 

 The losses in an ac inductor are made up of three components: 

  1. Copper loss, Pcu
    2. Iron loss, Pfe
    3. Gap loss, Pg  
  • Copper Losses, Pcu 

Pcu = (IL)^2 *R dc [Watts] 

  • Iron losses, Pfe 

For amorphous metal, use the following equation: 

Watt/Kilogram = 6.5 * f(KHz)^1.51*Bac(T)^1.74 

For Nano-crystalline use the following equation: 

Watt/Kilogram=1.8 * f(KHz)^1.53*Bac(T)^1.52

Where:

Bac is the flux density, and can be calculated using the formula below: 

 Bac = [ L*di] / [2*N*A] [Tesla]  

Variables: 

 L inductance in Henry 
 Ac core cross section area in m² 

  • Gap loss, Pg 

Pg = Ki* a*Lg*f*Bac^2 [watt]

Where:  

  • a is the core strip width in cm (see Figure 1 above) 
     
  • f is frequency in Hertz 
     
  • Lg is the gap width in cm 
     
  • Ki is the Gap loss Coefficient 

 To calculate total indicator losses, use the following equation: 

P total = Pcu + Pfe + Pg [Watts] 

From here, you can calculate the inductor temperature rise, after calculating the inductor surface area. 

As the resistivity of the copper winding increases with temperature, the winding loss increases with temperature as well. In the magnetic materials, the core loss increases with the increasing temperature, above approximately 100-degrees C. The value of the saturation flux density becomes smaller with increases in the temperature. 
Winding and core loss causes the temperature increase; therefore, the loss must be kept below some maximum value. In practice, the maximum temperature is usually limited to 100-125 degrees C by several considerations, which include the reliability of the insulation on the copper winding, and inductor insulation system. 

For Additional assistance in inductor designs, contact DRF Engineering Services with your custom inductor design request. 

Who We Are: 

We are an innovative design and development team, specializing in power electronic circuitry and power conversion products. We primarily provide product design/electrical engineering services, and have worked with cutting edge companies like yours in the past.  

Our services include, but are not limited to the following:  

  • Analog/Digital/Power circuit design 
  • Linear/Switching regulator design 
  • Electro-mechanical product design and program management 
  • Power Management  
  • Manufacturing support 
  • Pre-compliance review and test support (certification and test management for IEC and UL) 
  • Due Diligence and Private Equity Advisory Services 
  • CAD design services, schematics capture, and board layout