TPS61175 flyback configuration

Hi

Before answer your questiton, I want to know:
1) Is the +13V and --13V output both isolated from the input source?  Is the +13V and -13V  isolated from each other or they share the same ground?
2) What is your target operation frequerncy ? different frequency will results in different inductance.

BR,Helen

Hi,

1) +13V and -13V are both isolated and they share the same ground.

2) 400kHz

I attach the design schematic.

Thank's

Thanks for the information.

Usually We will let the circuit work at the DCMB to get higher efficiency (because the reverse recovery loss of the output diode is very big under CCM working condition), my calculate will based on this point.

I'll give the calculated result to you tomorrow.

BR, Helen

The control loop in AN SLYT323 is very complex . I sugest to add an extra widing , this will make the loop compensation much easier. Like what in SLVA496 do(Fig 2.).

I not need to isulate the ground in my application.If you look my schematic i take the fedback voltage signal directly. I dont use the optocoupler.
This should simplify the control loop.

I have done it again  the calculations and the suggested values are ok, but I have a doubt about the transformation ratio (1:2).

Transformer turns ratio (K = NS/NP) can be calculated as K=(Vout+Vd)/Vinmin * (1-Dmax)/Dmax)

With my values:

K=(13+0.5)/(4)*(1-0.45)/(0.45)=4.125

The turn ratio is 1:4 and secondary inductance is 16uH with 1uH primary inductance.

Its ok?

Thank's

If the turns ratio is 1:4, then the primary peak current will be higher than 4A. But this device 's peak current limit point is 3.8A, so please take care.

The  He only effects very  high frequency performance, at the cross over frequency of about 10k, it is near to 1, so you needn't take it into account. In the real application, Se>0.75Sn is enough for the stability(datasheet left lots of margin). Sorry, I made a mistake, the pimary inductance can be 8uH and the seconday inductance can be 2uH (I use a wrong current before).

My calculations.

Energy stored in inductance:

E_L=0.5*L_p*I_p,max²

In dicontinuous mode the power transfert to load depends on switching frequency:

P_IN,MAX= E_L*f_SW=0.5*L_p*I_p,max²* f_SW

I_p,MAX depends on primary inductance:

I_p,MAX=(V_IN-V_DROP)/L_p*T_ON

Rearranging the terms:

P_IN,MAX=0.5*(V_IN-V_DROP)²*T_ON²*f_sw/L_p

But T_ON=D_MAX/f_SW and rearranging rhe final formula is:

P_IN,MAX=0.5*(V_IN-V_DROP)²*D_MAX²/f_sw*L_p

With:

P_IN,MAX=13*0.3*1.25=4.87W

V_IN=4V

V_DROP=0.5V

D_MAX=0.45

f_SW=400000Hz

I obtain L_P=0,59uH

It’s correct?

How do you calculus 8uH?

Thank's

I check your calculate, why you set dmax=0.45. In the real application, for the flyback converter, even it is work in the DCM mode, the Dmax can be higher than 0.5.

After first fast tests my prototype works fine!

Thank you Helen!

Dario

Dario,

You're welcome!

This is also a good news for me!

Please click answered if you satisfied with the answer:)

BR,Helen