Increasing efficiency and reducing ripple & noise in [email protected]?

[Bob J.]

Hi all,

It is designed output, [email protected], that efficiency and ripple & noise of
travel adapter are around 70% and under 100mV at input volts
100-240VAC. However, it is not now like below.

1. Efficiency
1.1 63%: Input Volts(100VAC), Output [email protected], Transformer primary
turns 96
1.2 67%: Input Volts(100VAC), Output [email protected], Transformer primary
turns 136

2. Ripple and Noise
2.1 260mV: Input Volts(100VAC), Output [email protected], Transformer primary
turns 96
2.2 230mV: Input Volts(100VAC), Output [email protected], Transformer primary
turns 136

3. Switching frequency is 60KHz

You can see schematic about it

http://my.dreamwiz.com/jyeonsuk/qna/q_schem_for_rn_n_0601.pdf

Please give me any suggestion to improve and reduce them.

Thanks,

 

[Fritz Schlunder]

Hi all,

It is designed output, [email protected], that efficiency and ripple & noise of
travel adapter are around 70% and under 100mV at input volts
100-240VAC. However, it is not now like below.

1. Efficiency
1.1 63%: Input Volts(100VAC), Output [email protected], Transformer primary
turns 96
1.2 67%: Input Volts(100VAC), Output [email protected], Transformer primary
turns 136

2. Ripple and Noise
2.1 260mV: Input Volts(100VAC), Output [email protected], Transformer primary
turns 96
2.2 230mV: Input Volts(100VAC), Output [email protected], Transformer primary
turns 136

3. Switching frequency is 60KHz

You can see schematic about it

http://my.dreamwiz.com/jyeonsuk/qna/q_schem_for_rn_n_0601.pdf

Please give me any suggestion to improve and reduce them.

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%.

 

[Bob J.]

Thanks for your answer.

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.

Impedance is 0.14Ω, ripple current is 450mA, size is 8x11.5mm and
longevity is 3,000h at 105℃. I will try it changing capacitors.

Thanks,

Bob.

 

[Pooh Bear]

Thanks for your answer.


Impedance is 0.14Ω,

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

ripple current is 450mA,

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

size is 8x11.5mm and
longevity is 3,000h at 105℃. I will try it changing capacitors.

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

Graham

 

[Bob J.]

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

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

Thanks,

Bob.

 

[Terry Given]

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:

50% duty cycle, 1A dc gives 4A peak and 1.63Arms, with 1.29Arms cap current


HTH

Cheers
Tery

 

[Bob J.]

I will follow you.

Thanks a lot Terry,

Bob.