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would an engineer use this equation?

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dl09

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i found this equation on the internet 2 × √ ( i/n × π × a) for calculating the gauge of both the primary wire and secondary wire of a flyback converter. i is effective current, and n is number of turns, a is current density, and the article said pick 4 amps per millimeter squared to 8 amps per millimeter squared. would an engineer actually use this equation?
 

A real engineer skilled in the art would consider this
a very rough guide and from there, move on to more
analysis and characterization. The Ridley power supply
group on F*c*book currently has a lot of discussion
about such things - measuring and modeling and
theory and some experience based anecdotes.

Your current density tolerable depends a lot on the
xfmr construction, and application environment
(temperature called out, and the thermal path to it).
 

so that means yes he would?
 

ONLY as a preliminary estimate, yes.
The temperature rise will have to be measured with worst case conditions.
 

so an engineer would use this equation. then he would have to measure temperature rise. i think i understand this.
 

R = rho. L / Area is what an engineer would use, watts dissipated = R . i^2 ( for rms current in the wire )

then there is an added heating effect from a close wound coil of wire, i.e. internal hot spotting

kind regards - an engineer.
 

so an engineer would use this equation R = rho. L / Area to calculate temperature rise?
 

temp rise is a function of the watts dissipated in the Tx and the surface area and shape of the Tx and any cooling air flow - there is no simple formula for this only experience ...
 

what is the Tx?
 

Here, it is "transformer". But in other discussions it
could as well be transistor or transmitter, depending
on context.
 

i am thinking i will use the equation 4 amps per millimeter squared equals amps through copper wire over cross section of copper wire to calculate gauge. i will select the input current and solve for cross-section of copper wire, and then use a wire gauge table to convert from millimeter squared to gauge. if i use this equation, will the flyback converter still work?
 

if i use this equation, will the flyback converter still work?

You are asking a lot of questions expecting to have definite answers, and this is not how things works, and particularly these kind of open ended questions are not suited to ask in any forum for electronics. Engineers, technicians, hobbists (whoever) do not do their jobs based solely on formulas and equations, but rather in a set of knowledge gathered along years to determine what else should be considered besides the variables and relationship on the thitherto given equations. People do not have the whole scenario of your problem, so they cannot answer Yes or No as you are expecting with the incomplete description of the issue. For example, depending on the error margin allowed, one could answer Yes, the above formula suffice for you, go ahead...
 

i am trying to learn how to design and build a flyback converter. should i start with the equation i used in post 11 and the other equations i found on the internet? i just trying to learn how to do this.
 

Hi,

i am trying to learn how to design and build a flyback converter. should i start with the equation i used in post 11 and the other equations i found on the internet? i just trying to learn how to do this.
It's not only formulas, there are design rules, you need to make decisions...

There are a lot of online courses, design notes, tutorials on how to design flyback converters.
Look for some reliable ones, from big semiconductor manufacturers or universities.
They are good and free.

They tell you all you need to know.

There also are free simulators.

Klaus
 

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