# Transformer Rating Selection

For selecting a transformer, I stumbled across the following equations for obtaining VA and IAC (citation here).

I am dealing with a full wave bridge rectifier with a capacitive load. The wiki explains to use these two equations:

$$IAC = 1.8 \times IDC$$

$$VA = 1.4 \times (watts + (2 \times IDC))$$

I am curious how the equations are derived for both IAC and VA. Also, would watts be the power from the rectifier? For my power supply design, I need my max IDC to be 3 and VDC to be 15.

These kinds of rules of thumb are always dangerous to use since a bunch of assumptions went into them that are not stated. At worst they can be downright misleading. My advice is to stay away from such "rules". Instead understand what is going on. Then you can derive your own relationships for any specific condition you want, and you'll know what premises they depend on and when they are valid.

The first equation:

Iac = 1.8 Idc

is apparently intended as a quick guide as to the current rating of a transformer feeding a full wave bridge, given the DC current drawn from that bridge. I certainly would never use this in a design. It's basically trying to say that the AC current rating needs to be higher than the ultimate DC current. There is definitely truth in that, since the DC voltage will be roughly the AC peak to peak voltage. Therefore from just a power balance alone, the AC current has to be higher since its RMS voltage is lower than the DC voltage.

But, where exactly did the 1.8 come from? What assumptions about conduction angle into the bridge are embodied into it. What about DC voltage droop? Impedance of the transformer, etc? This equation is useless without knowing what criteria were used to derive it.

Again, when you are faced with sizing a transformer, look at your particular situation and do your own analisys. There is no substitute for knowing what you're doing and then applying that to design decisions.

The second equation:

VA = 1.4 (watts + (2 Idc))

It clearly total garbage from a units check alone. You can't add power and current and expect anything sensible as a result.

• 1.8 factor assume conduction angle 40~55 degr (ki=0.1~0.3), and Vd=0.8V for a normal loaded transformer. Dec 10, 2014 at 19:49
• @GRT: Exactly. I wasn't really asking, but trying to point out that such sugar coated "rules" are useless without knowing the assumptions the sugar coating is hiding. Without knowing the assumptions, you have no way to tell if the rule applies to any particular situation. Put another way, such rules are basically useless. At best they allow the incompetent to produce a workable design - sometimes. Dec 10, 2014 at 20:53
• ...agree with you Dec 10, 2014 at 21:18
• I understand conductance angle is the fraction of the wave for an element to conduct, but what is a good assumption for a typical transformer. Also, when is it important to consider Core loss or Magnetizing reactance ? I would like to know some major properties to consider for picking a transformer. thanks! Dec 11, 2014 at 2:22

For full wave rectification with a bridge rectifier and capacitive filter (unregulated), you need a secondary voltage of:

where: Vdc the required voltage of your PS (i.e 15V)

VDF Voltage drop across the diode at full load (i.e 0.8V)

Vrip Maximum ripple voltage across the capacitor filtar (i.e 10%)

Vprim.nom Is your mains power nominal voltage (i.e 230V)

Vprim.min Is the low line voltage (i.e 190V)

0.9 is the rectifier efficiency

For the rated current keep the factor 1.8 times the DC current as per my comment above . Finally a 1,44 times the rated VA it is a safe choice.

• Out of curiosity, Why do you multiply by the ratio of Vprim.nom to Vprim.min for the sum of voltages? Dec 11, 2014 at 2:08
• To include the mains variations Dec 11, 2014 at 5:42