In message <6CEB3A36096CD2119FEE00A0C9410EC2228017@excalibur.swindon.msl
.mitel.com>, Peter Chadwick <Peter_Chadwick@mitel.com> writes
>
>A good way of testing a transformer for suitability is to start by measuring
>the secondary resistance as Rich says. Then measure the primary resistance,
>and convert that to an equivalent series secondary resistance: sum it with
>the measured secondary resistance. You can now use the rectifier resistance
>and the load resistance to come up with the ratio of RL/RS, the load to
>series resistance ratio. From the graphs in the books, you figure the output
>voltage for various values of RL for the capacitor. You can use standard
>formulas to figure the RMS current in the xfmr secondary.
Be careful which standard formulas you use. In a cap. input supply, the
short duration current spikes give a heating effect which can be much
higher than a continuous current value calculated from the dc current. I
checked some designs using PSpice and found that the *real* rms (heating
effect) current was 1.7-2.2 x the value calculated from 1.414*Idc - the
power dissipated in the windings was 3-5x higher than might have been
expected.
The RSGB little data book has a graph which gives a multiplying factor
related to Rs, Rl and C - I've not seen this elsewhere.
Don't forget that if you voltage double, the current doubles so rms
current goes up by 4.
>Now use a Variac
>on the primary, an AC ammeter across the secondary, and wind up the primary
>volts until you reach the calculated rms secondary current. Leave it to
>cook, while keeping a eye on the xfmr to see how much it heats up. If it
>doesn't get too hot (smells too much!) and is OK after say, an hour (for
>RTTY or contesting), remove the supply and measure the secondary ohms again.
>Apply the formula for the increase of resistance with temperature, and you
>have the winding temp of the xfmr.
You should also leave the transformer on full primary volts with the
secondary o/c for a while to check the temperature rise from iron
losses, and add this to the copper losses measured as above. Typically
30-50% of total dissipation is in the iron.
Steve
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