All,
Belows a link to a webpage I did on the subject. Please excuse the ads as that
was the only place I had to put this at the time. The current drawn by the
transformer is from the drops in the power supply plus the load. In a class AB
amp, it's only 50% efficient so the current will be X2 what is needed to
produce the wanted output power. Another thing is voltage sag under load. My
experience has shown me that it is always more than 10% drop. I always use
12-13% but have seen as high as 15% in some Heathkit amps using smaller
transformers on the edge so to speak. The extra power needed by the transformer
itself is from it's efficiency which is ruled by watts per pound loss in the
iron, and I^2R losses in the wire. The higher the current rating on the wire in
circular mils per ampere, the higher the efficiency. Also, a better grade of
iron in the core gives better efficiency. The lower you go in M numbers like
M-2 to M-50 is the efficiency drop in watts per pound loss. M-2 givin
g the highest efficiency, but being the most costly.
Formulas;
http://amateur-transformer.125mb.com/pf.htm
Best,
Will
*********** REPLY SEPARATOR ***********
On 2/10/06 at 7:35 AM Steve Thompson wrote:
>KD7QAE wrote:
>> For the major types of filters the secondary RMS current is related to
>> the DC load current as:
>>
>> Inductor input: 0.7
>> FWCT capacitor input: 1.0 to 1.2
>> FWB capacitor input: 1.6 to 1.8
>>
>> Data from O.H. Schrade: "Analysis of Rectifier Operation"
>I agree these are the ranges of figures that 'drop out' most often - but
>I'd caution against automatically assuming that they cover every
>instance. In the last case, if you have 1.9 instead of an assumed 1.7,
>there's 25% more heat.
>
>In the absence of using PSU Designer or somesuch to predict values,
>0.7xIdc for choke input and 1.7xIdc for capacitor input are also useful
>estimates of the rms ripple current in the smoothing caps.
>
>Steve
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