Calculate Turns per Volt- transformer for Amplifier Thank you for watching! Please like and subscribe. Comment if you have any ideas for my next video.
Hello, my friend, thank you very much for this video, but we could not see the transfer accounts sheet because of its distance from the camera. Is it possible to take a picture of those accounts? Thank you
A suggestion for your next video? How about how to wind a 120volt primary power transformer for a stereo tube amp with 8-6L6's and 3-12ax7's per side? Haha, IDK, just thought I'd ask, because that's what I'm trying to figure out the math on how to wind and build.
made a 435 turns primary coil with a 18 SWG copper wire in my DIY transformer. Unfortunately that primary burned out 5 seconds after plugging the transformer to the 220 volts imput source. May I know the reason?
You need to calculate the impedance....to see if you were handling...to much current .... Amp = 230/XL ...It's a thick cable...it can draw many amps...and it will get hot...
However, maybe your problem was the fact that handling a thick cable might have damage the insulation and you created a short circuit. The impedance seems about 1000 mH...if you use the right iron core.....if you use something else...the impedance might drop to dangerous level...less than 100 mH....and you will be pulling....a lot of current..ufffff and it will get hot...very hot...
Howdy. The Transformer equation. E(rms) = 4,44 x N x f x Phi(peak) Where E(rms) = Electro-magnetic force (= - winding voltage) root-mean-square value N = number of winding turns f = frequency Phi(peak) = Peak value of total flux. Unit Wb = Vs. And Phi(peak) = A x B(peak) Where A = Core cross section area. Unit m2. B(peak) = Flux density peak value. Unit T = Wb/m2 = Vs/m2. Regards.
I have a question for you my friend . How to calculate copper wire diametre . And if secondary wire diametre is bigger than primary is it correct , please answer me .
It is derived from 2*pi / sqrt(2) which is actually the RMS Value (that's why you divide a Peak Value with sqrt(2)) and the 2*pi term is from the derivation of the sinusoidal linkage flux. - Let λ = Ν*Φ = Ν * Φmax * sin(2pi*f + φ) - e = - dλ/dt ==> e = - N * 2pi * Φmax * f - Φmax = Bmax * S - also: e_rms = e / sqrt(2) - So.... e_rms = N * 2*pi * Βmax * f * S / sqrt(2) = 4.44 * N * Bmax * f * S ==> To = N / e_rms = 1 / (4.44 * f * Bmax * S) in units [Turns / V]
