Actually I am just missing the design equation to find out the No of turns in primary and secondary winding
http://nptel.ac.in/courses/108105066/PDF/L-23(DP)(PE)%20((EE)NPTEL).pdf
Hello,
The number of turns of the primary is N= (Vin*Dmax)/(Ae*Fs*Bmax). For the secondary you just need to chose the conversion ratio. (Vin = input voltage, Ae = core effective area, Fs = switching frequency, Bmax = max magnetic flux density
Don't forget that flyback converter doesn't have a transformer, it's an inductor coupled, so it' better use a gap in it.
Dear Sir,
- Not very clear what you want to do. Lets assume that the current through the inductor is at all times greater than 0. (the delta-I is smaller than 2*Iavg).
- Lets assume the number of turns of the primary is N1, the number of turns of the secondary is N2.
- Lets assume the primary voltage is V1, the secondary voltage is V2.
- Lets assume there are no losses. If there are losses, you need to use V1', and V2', the primary and secondary voltages corrected for the losses.
- Lets assume the ON time of the primary transistor is T1, the on time of the secondary diode is T2.
- Now we calculate the flux difference during T1, which is equal to the flux difference during T2. (there is a steady-state in the circuit)
delta-flux = V1*T1/N1 = V2*T2/N2
- This gives you the equation for the turns ratio:
N1/N2 = (V1*T1)/(V2*T2)
Cheers,
Henri.
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