### Design of Single Phase Transformer for Instrument Power Supply

The single phase transformers used in power supplies of various instruments are small transformers which required to deliver few amperes at small voltages having VA ratings from 10 VA to 1000 VA. Such transformers are mainly used to drive the rectifier circuit of the power supply.
The design of such transformers divided as,
1. Core design
2. Winding design
3. Stamping design
1. Core design
The first step in designing such small transformers is to select turns per volt value depending upon VA rating of the transformer. The table 1 gives the values of turns per volt (Te) for various VA ratings.
 Table 1
From the e.m.f. equation,
E = 4.44 Φm f T
...                   T/E = 1/(4.44 Φm f) =  Te                                    ..........(1)
Knowing the frequency of operation which is generally 50 Hz, the value of Φm can be decided using equation (1).
Key Point : Assume maximum value of the flux density in the core Bm as 1 Wb/mm2 which is enough for small transformers.
Ac = Net core area =  Φm/Bm                                ..........(2)
Assuming stacking factor as 0.9,
Agc  = Gross area of core = /0.9                          .............(3)
Key Point : The shell type construction is generally used for small transformers. The core is made up of stampings as the construction of core is laminated. The stampings can be a) E  and I, b) T and U or c) E type as shown in the Fig. 1.
 Fig. 1 Types of stampings
The companies manufacturing the stampings provide the information of the values of A, B, C, D and E for various standard available stampings.
Key Point : For central limb, square section is generally preferred.
Thus for the central limb, the depth and width is made equal.
Wc = Width of central limb =  √Agc         ............(4)
Referring to stampings tables provided by the manufacturers, a standard stamping with Wc nearly equal to the value A is selected.

2. Winding design
The VA ampere rating of such transformers is known. Assume efficiency to be between 80 to 95% the primary current can be obtained as,
Ip = (VA/η Vp ) A                                      .............(5)
where              Vp = Primary voltage
Assume current density in the primary winding conductors to be 2.5A/mm2 which is sufficient for small transformers.
...       δp = Current density in primary conductors = 2.5A/mm          ........(6)
But     δp = Ip /ap    where ap = Area of primary winding conductors
...                       ap = Ip/2.5 mm2                                                  ...........(7)
For small transformers enamelled around conductors are used. The standard tables for the sizes of enamelled round conductors are provided by the manufacturers. Referring the tables, properly sized conductors can be selected.
As turns per volt Te and primary voltage Vis known, the number of turns can be obtained on primary side as,
T1 = Vp Te                                                            .....(8)
The current in the secondary winding is,
I= VA /Vs                                       ............(9)
where                      Vs = Secondary voltage
...                             a= Is s   mm2                                             ..........(10)
where      δs = Current density in secondary conductors in A/mm2
Referring standard table, proper conductors can be selected.
While calculating secondary number of turns, allowance of 5% is selected so as to compensate for the voltage drop in the windings.
T2 = 1.05 Vs Te                                             ............(11)
3. Stamping Design
This includes the decision of window space which is required to accomodate,
1. Primary winding
2. Secondary winding
3. Insulation and former used to support the windings.
The steps to design the stampings are,
Space factor              S = 0.8 (d /d1)3                                                 ............(12)
where                        d = diameter of bar conductor
d1 = diameter of insulated conductor
The values of d1 and can be obtained from the standard tables provided for conductor sizes.
...    Space required for primary winding = T1 ap /S                              ............(13)
and space required for secondary winding = T2 as /S                       ..........(14)
While the space for insulation and former is assumed to 20 to 25% of the space required for both the windings.
Aw = Total window area
= 1.2 (Window area required for both windings)                         .........(15)
Key Point : Standard table can be referred to select the proper stampings which gives window area slightly greater than the value calculated above.

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