Increasing efficiency and reducing ripple & noise in 5.0V@1.5A?

What is the ESR of your capacitors C09 and C10? 220uF each is rather minimal for a flyback topology and a 1.5A output current level, but more important than the size of the capacitors is the ripple current rating (for longevity and reliability) and ESR.

Improving the output ripple voltage performance is probably a simple matter of using much lower ESR capacitors for C09 and C10. Lower ESR output capacitors may also improve efficiency slightly, and if you are already achieving 67%, that isn't that far off the target 70%.

Thanks for your answer.

or

Impedance is 0.14=CE=A9, ripple current is 450mA, size is 8x11.5mm and longevity is 3,000h at 105=E2=84=83. I will try it changing capacitors.

Thanks,

Bob.

That sounds rather high. I assume that's ohms there.

For a 1.5A supply ? You're kidding me !

I think you need to. Make sure it's designed for high frequency use too.

Graham

I want to know what it is exactly or how it is calculated generally.

Thanks,

Bob.

Its easy:

- get a pen

- draw a right-angle triangle

- write out the equation for the area of the triangle

- draw a rectangle

- write out the equation for the area of the rectangle

- equate the two areas

thats the bulk of the maths.

The height of the rectangle is your DC Input current, Idc

Idc = Pin/Vdc pretty much by definition

The width of the rectangle is one switching period, T = 1/Fswitching

The area of the rectangle is the charge taken from the input during one cycle,

Arectangle = Qin = Iin_dc*T

The width of the triangle is the on-time of the switch, Ton = D*T

The height of the triangle is your peak primary current

The area of the triangle is the charge stored in the transformer core during the on time which is also, for a DCM flyback, the charge delivered to the load when the switch is off, which is equal to the total charge supplied by the input during one switching period (Ton + Toff).

When you equate the areas you get 0.5*Ip_peak*D*T = Idc*T

which you can solve for Ip_peak = Iin_dc*2/D

which is the peak primary current.

The same argument works for the output current:

Iout_dc*T = 0.5*Isec_peak*(1-D)*T

Isec_peak = peak secondary current

(1-D)*T = Toff = time during which energy delivered to secondary

giving Isec_peak = Iout_dc*2/(1-D)

for D = 0.5, Ip_peak = 4*Iin_dc and Is_peak = 4*Iout_dc

I cant be bothered going through the derivation cos its trivial, but the rms of these triangle waveforms is Ipeak*sqrt(D/3) so:

Ip_rms = Ip_peak*sqrt(D/3) = 2*Iin_dc/sqrt(3*D)

Is_rms = Is_peak*sqrt((1-D)/3) = 2*Iout_dc/sqrt(3*(1-D))

the capacitor ripple current is the difference between the DC and the winding current, so the RMS cap current is:

Ip_cap_rms = sqrt(Ip_rms^2 - Iin_dc^2) = Iin_dc*sqrt(4*/(3*D) - 1)

Is_cap_rms = sqrt(Is_rms^2 - Iout_dc^2) = Iout_dc*sqrt(4*/(3*(1-D)) - 1)

example:

HTH

Cheers Tery

I will follow you.

Thanks a lot Terry,

Bob.

Terry Given wrote: