Md. Shahadat Hasan Sohel #0606028 1 Introduction: Transformer is one of the most used electrical equipments of todays world. It is widely used in the power transmission. It has made possible the transmission of power, from where it is generated, to where it will be used, with very little loss. Without transformer, the transmission of power would have been quite impossible. Basic principle: Transformer uses magnetic usk to transfer electrical energy. There are two types of transformers: 1) Step up: Changes alternating low voltage to alternating high voltage 2) Step down: Changes alternating high voltage to alternating low voltage If the number of turns in the primary and secondary windings are ns Vp np and respectively and the voltage across the primary and secondary windings are Vs respectively, then : np Vp = Vs ns and Figure 1: Transformer 2 Theories and Corresponding Calculation: Core Area: If the core area of the transformer is taken as a and va is the corresponding volt-ampere, thena=c √ va Where c is a constant. If a is taken in sq. inch then- √ a=0.16 va Calculation for core area: Required output voltage for my transformer is 18 V and its current capacity in the secondary winding is 1 A. So,Volt-ampere, va = 18×1=18 √ Now, core area, a = 0.16 18 = 0.6788 inch2 Selection of the core: For a shell type transformer, which I planned to make, the core area is equal to the area of the bobbin. But in the market, the available bobbins are of some xed areas. So, I had to choose a bobbin of area which is not less than the calculated core area and close to the calculated value. The most suitable bobbin that I found for this purpose, which fullls the required conditions, had an area of 1inch 2 . Number of turns per volt: Number of turns per volt can be calculated from the following equation- N 108 V = 4F f aB Here, V = Voltage across the winding in volts in rms F = Form factor of the ac waveform f = Frequency of the wave in Hz a = Cross section area of the core in sq. cm N = Number of turns on the considered B = Flux density in maxwells per unit area Calculation for number of turns per volt: Here, F = 1.1 f = 50 Hz a = 6.4516 cm2 B = 9, 600 maxwells per So, cm2 3 N 108 80 V = 4×1.1×50×6.4516×9600 = 11 Voltage in the primary, V p = 220 V Number of turns in the primary, Voltage in the secondary Vs = np = 80 11 ×220 = 1600 18 V Number of turns in the secondary, ns = 80 11 ×18= 130.91 Number of turns given: In the primary winding, calculated turns i.e. 1600 turns are given. But, as to counter leakage eect while loaded, the number of turns in secondary is increased a bit. The turns given in the secondary winding are 140. Selection of wire: The selection of wire is done on the basis of how much current will be drawn. In the secondary winding, maximum current that will be drawn is 1A. So, wire of gauge 21 is used in the secondary, which has a current carrying capacity of 1.0377 A. The maximum current in the primary can be calculated from the following formula: Vs 18 Vp ×Is = 220 × 1 = 0.0818A As a result, the wire used in the secondary is of 34 gauge, which has a current Ip = carrying capacity of 0.0858 A. Determination of Transformer Parameters: Open Circuit Test: The purpose of the open circuit test is to determine the magnetizing reactance (XM )and the equivalnet core-loss resistance (Rf e ) In the open cercuit test, the secondary was kept open and 220 volt ac was applied to the primary. A power-factor meter and an ammeter were connected with the primary coil to measure the angle between applied voltage and current. Then the open circuit voltage (Voc ), between input current and input voltage, primary current, (θoc ) (Ioc ) and the angle were measured. Data from the open-circuit test: The data is given beow: Voc = 220V Ioc = 0.025A θoc = 14◦ The no-load voltage found in the secondary was 19.6. As there is very little leakage at no load, the voltage was a bit higher than as it should be. 4 Calculation of magnetizing branch: From the open circuit test data: Poc = Voc × Ioc × cosθoc = 220 × .025 × cos14 Now, P = Voc × If e ⇒ If e = VPoc = 5.33665 220 = 0.02426 A = 5.3366 W Again, 2 2 Ioc = If2e + IM q 2 − I2 ⇒ IM = Ioc fe √ 2 ⇒ IM = 0.025 − 0.024262 ⇒ IM = 0.006A = 6mA Then, Rf e = VIfoce 220 ⇒ Rf e = 0.025 ⇒ Rf e = 8800Ω And, oc XM = VIM 220 ⇒ XM = 0.006 ⇒ XM = 36666.67Ω Short Circuit Test: The purpose of the open circuit test is to determine the equivalent resistance (Req ) ,equivalent leakage reactance (Xeq ) and equivalen impedance (Zeq ). In the short circuit test, the secondary was shorted. An ammeter was connected in series with the secondary coil. 220 volt ac was applied to the primary. A power-factor meter and an ammeter were connected with the primary coil to measure the angle between applied voltage and current. Rated current (1A) was allowed to ow through the secondary and primary current (Isc ), primary voltage (Vsc ), and the angle between voltage and current (θsc ) were measured. Data from the open-circuit test: The data is given beow: Isc = 0.1A Vsc = 19.5V θsc = 3◦ Calculation of equivalent circuit: From the open circuit test data: Psc = Vsc × Isc × cosθsc = 220 × .1 × cos3= Now, sc Zeq,HS = VIsc ⇒ Zeq,HS = 19.5 0.1 1.9473 W 5 ⇒ Zeq,HS = 195Ω And, 2 Psc = Isc Req,HS ⇒ Req,HS = PI 2sc sc ⇒ Req,HS = 1.9473 0.12 ⇒ Req,HS = 194.73Ω Again, 2 2 2 Zeq,HS = ReqHS + XeqHS q 2 2 ⇒ Xeq,HS = ZeqHS − ReqHS √ ⇒ Xeq,HS = 1952 − 194.732 ⇒ Xeq,HS = 10.258Ω Eciency Test: For eciency test, 220 Volt was applied in the primary and a load was connected that draws a current of 1A through it. For this load, primary current and powerfactor angle was measured. The data is given below: Vp = 220V Ip = 0.12A voltage, Vs = 18.2V current, Is = 1A Primary voltage, Primary current, Secondary Secondary Angle between primary voltage and primarey current, θ = 14◦ Calculation of eciency: From the data, Pin = Vp Ip cosθ = 220 × 0.11 × cos14◦ = 23.481W att Output power, pout = Vs Is = 18.2 × 1 = 18.2W att P 18.2 Eciency, η = out × 100% = Pin 23.481 × 100% = 77.51% Input power, Equivalent Circuit: From the parameters calculated from short circuit and open circuit tests, we can draw the equivalent circuit as: 6 Figure 2: Equivalent circuit 5V DC Output: The output of the transformer is applied to the input of a bridge rectier consisting of four diodes (1N4007). Then the rectied voltage is supplied to a 7805 IC through a capacitor(1000µF ). And from the output of the IC, through another capacitor(1000µF ), 5V constant dc output is taken. The circuit is as following: 7 Figure 3: 5V DC Output Circuit Discussion: Eciency: The eciency of the transformer is about 78%. The probable reasons behind this are1) Presence of core losses such as: Eddy current loss and hysteresis loss 2) As the exact B-H curve of the core was not available, the maximum B that was taken in designing the transformer, may not be at the highest point of the linear region of the B-H curve. Output Voltage: The output voltage required was 18 for my transformer. I designed the transformer to give 18 V at rated load which draws 1A current. That was made by taking in consideration the leakage eect. To counter the leakage eect, few 8 additional turns were given in the secondary winding as mentioned in the calculation part. Accordingly, at loaded conditon the secondary voltage is 18.2. But at no load, the leakage eect is suciently low. So, the output voltage becomes around 20 (19.8 to be exact). On the day of submission, the input voltage was also higher (230V instead of 220). And, the output was checked at no load condition. So, the output voltage got even higher (around 21V). Comment: The transformers that are available in the market, which give similar voltage, found to have eciencies between 65% to 70%. So, I think, it is quite a success to have a transformer with eciency around 80%. And, as it was my rst practical project, I enjoyed a lot, doing every bit of it. It was really great to do something in practical and acquire rst hand experience.