Quote:
Originally Posted by ChristianFletcher
Primary Resistance P1 - P2 292.5 Ohm
...
Secondary S1 - S2 372 Ohm
primary Voltage 210V Secondary Voltage 171V unloaded
primary Voltage 210V secondary Voltage 159V @ 20mA load
Impedance Voltage 14.9V
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OK. So, playing with figures:
Assuming the off-load primary current is small (so primary resistance loss is negligible) we have:
Turns ratio = 171V / 210V = 57/70.
Secondary load = 20mA so primary on-load current = 20mA x 57/70 = 16.29mA.
Primary voltage drop = 16.29mA x 292.5 ohms = 4.76V giving 'effective' primary voltage 210V - 4.76V = 205.24V.
And primary power loss = 0.01629² x 292.5 W = 0.078W.
Secondary voltage then, allowing for primary drop = 205.24V x 57/70 = 167.1V.
Secondary volt drop = 20mA x 372 ohms = 7.44V.
So actual load voltage is 167.1 V - 7.44V = 159.66V which is pretty close to your measured figure.
Secondary power loss = 0.02² x 372 W = 0.1488W.
Total winding losses thus come to 0.078 + 0.1488 W = 0.227W. This is somewhat lower than your measured 0.3W, though same order of magnitude.
One immediate comment is that with secondary loss about double the primary, a more efficient transformer could have been made with slightly smaller gauge primary wire and slightly larger gauge secondary (to fit in the same space). The losses would have been nearer equal then. However without getting into the designer's mind (or knowing the constraints he/she was under), there could have been a good reason for it being as it is.