I have been trying to understand the maths of it and failing to a large amount, but looking at the wiki page again it says that
'a represents the amplitude of the function' and gives and example. ( can not copy here do to the format! ).
Remembering back many years and from what I can remember sin (90) is 1 and sin (270) being -1 so an amplitude of 8 would give a max of +8 and min of -8 so actually 16 volts peak to peak as I tend to keep using.
So you are correct Jeffrey there is a factor of 2 difference. If I continue to use peak to peak values then the rms secondary voltage must be 51.3 /(2√3) = 14.81 Volts. So apologies, with this correction every thing starts to look do-able.
Primary voltage would be 14.81 x 2.96 is 43.84 Volts rms.
Feeding this back into the formula :-
Number of primary turns is 43.84/(4 × 1.15 × 12.5 × 1.46 × 6.71 ×10⁻⁴) = 778 turns (17.75 turns per volt).
The 100mA secondary peak to peak current would then be 29 mA rms and primary current 29/2.96 near enough to 10 mA add a bit for losses say 11 to 12 mA.
Secondary turns would be 778/2.96 = 263 turns stick on a bit for losses 263 x 1.05 take me to 276 turns.
So with these current figures and turns ratios I am thinking I could get a double winding on the primary section of the transformer to make it push pull. I have just ordered some small reels of 36 awg and 33 awg wire to have a go.
Last question for now is, if I find I can have more turns per volt on the primary, so for example if I use 20 turns per volt rather than the 17.75 is that better to do. I guess there is an increase in resistance of the winding but could it be beneficial to have more turns per volt?
I feel a lot happier and can probably go to sleep now.
Adrian
'a represents the amplitude of the function' and gives and example. ( can not copy here do to the format! ).
Remembering back many years and from what I can remember sin (90) is 1 and sin (270) being -1 so an amplitude of 8 would give a max of +8 and min of -8 so actually 16 volts peak to peak as I tend to keep using.
So you are correct Jeffrey there is a factor of 2 difference. If I continue to use peak to peak values then the rms secondary voltage must be 51.3 /(2√3) = 14.81 Volts. So apologies, with this correction every thing starts to look do-able.
Primary voltage would be 14.81 x 2.96 is 43.84 Volts rms.
Feeding this back into the formula :-
Number of primary turns is 43.84/(4 × 1.15 × 12.5 × 1.46 × 6.71 ×10⁻⁴) = 778 turns (17.75 turns per volt).
The 100mA secondary peak to peak current would then be 29 mA rms and primary current 29/2.96 near enough to 10 mA add a bit for losses say 11 to 12 mA.
Secondary turns would be 778/2.96 = 263 turns stick on a bit for losses 263 x 1.05 take me to 276 turns.
So with these current figures and turns ratios I am thinking I could get a double winding on the primary section of the transformer to make it push pull. I have just ordered some small reels of 36 awg and 33 awg wire to have a go.
Last question for now is, if I find I can have more turns per volt on the primary, so for example if I use 20 turns per volt rather than the 17.75 is that better to do. I guess there is an increase in resistance of the winding but could it be beneficial to have more turns per volt?
I feel a lot happier and can probably go to sleep now.
Adrian
Learning as I go!
Youtube EF91 Valve
Youtube EF91 Valve
