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6th April 2011, 12:51 AM  #1 
diyAudio Member
Join Date: Jan 2008

Single Ended Output Transformer
Hi Guys,
I want to build EL34/6L6 SE guitar power amp, but I also want to wound OT. According to Valveheart, full calculation goes like this(for my example): Power P= 10W(3.6K EL346L6) Low freq response fd=82Hz Primary Impedance Ra=3600 Ohm Secondary Impedance Rs=4/8 Ohm Flux density B  5000Gauss According to Valveheart core dimension is determinated by the eq: Q=sqrt( (Pout*G*10^6) / (fmin*B*Is))  where:  Pout is the output power  G is coefficient of weight between copper and iron /G=1.5  1.0/  f min is the lowest desirable frequency  Is is the density of current in the wire  Ampere/sq. millimeter (1.5  2.5 A/sq. mm).  B is Magnetic Field Intensity in Gs I don't get this equation cause every time I calculate it I think it's wrong. Here's mine example: Q=sqrt( (10*1.5*10^6) / (82*5000*1.5))=4.9cm^2 I'm confused because of the lower equation, cause for the same input I get: Q=20*sqrt(Pout/fmin) Q=20*sqrt(10/82)=6.98cm^2 I wonder how they got 20?Can someone explain this to me I will take 6.25cm^2 core for the rest of an example: Primary turns: n1=10^8/(4.44*B*fmin*Q)=10^8/(4.44*5000Gauss*82Hz*6.25cm^2)=8.78 U1=sqrt(Ra*Pout)=sqrt(3600*10)=190VAC Np= n1*U1=1667 turns  1665 (5layers  111turns, primary is divided in 3 parts, so (5*111)*3=1665 Secondary turns: Ns4=1667/(sqrt(3600/4))=56 turns Ns8=1667/(sqrt(3600/8))=79 turns Primary current:Ip=Ia+Ia' Ia=SQRT(Pout/Ra))*1000=SQRT(10/3600))*1000 = 52.70mA Ia'=73mA(this is bias for tube) Ip= Ia+Ia'=125.70mA Secondary currents: Is4=SQRT(Pout/4)=1.58A Is8=SQRT(Pout/8)=1.12A Wire diameter: wp=SQRT(Ip*0,001/3)*1,13=SQRT(125.70*0,001/3)*1,13=0.23mm  0.3m wire is chosen ws=SQRT(Is4*0,001/3)*1,13=SQRT(1.58*0,001/3)*1,13 = 0.82mm  0.7m wire is chosen Can someone who is experienced in winding OT's can confirm this, before i start wounding the transformer. Thanks 
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