LM25118-Q1: Questions

Hi Kura, Thank you for considering the LM35118. The best way to get all your questions answered clearly in a short time is having a conf call with the customer. Please help us to connect with the customer directly. Please email to youhao.xi@ti.com so we can set up the conf call.

Thanks,
Youhao Xi, System and Applications Engineering, APP-BMC-BCS, TI

Hi Kura-san,

1. regarding the errors in your attached calculator spreadsheet, you should select component values close to the suggested. These error occurs becuase your selection leads to hit the peak current limit in normal load and hence it cannot operate properly to regulator the output voltage. See comment to yoru question 4 below.
2. Nothing special ans it just follows the generic selection guide for rectifier diode: forware current > Io (max); peak current > didoe peak curent; losses is basically Io x VF, and by Rth you shoudl evluate the temperature rise not exceedign Tj of the diode.
3. the output current still satisfy the basic relationship among the boost switch current, duty cycle, and inductor ramp current, until the outut voltage drops below Vin in excessive overloading like output soft short or even short circuit.
4. Please refer to the datasheet's design guideline in selecting the current sense resistor and ramp capacitor. Their signals are superimposed to compare with the peak current limit threshold. The way you changed the selecton value leads to trip the peak current limit.
5. A larger Rs will trip the peak current limit.
6. Samll inductor may work but the ripple current increases, and it may run early into discontinuous mode if you stay with the same frequency. Please change the inductance to a smaller value in the calculator and it should still genterate a reasonable design.
7. The frequency tolerance will be slightly wider than @RT=29.11k. but tighter than @RT=9.525k.
8. What to you mean by this question? When current limit is hit for >256 cycle, teh UVLO pin will be pull low by the built in hiccup timer. Other than that it does not affect other functions. Note that UVLO must stay >1.25V to keep the controller in operation.

Best Regards,
Youhao Xi, System and Applications Engineering, APP-BMC-BCS, TI

Hi Kura-san

Q1.Inductor Current Limit Calculation(P28)
Please let us know relation ship M value(10%-30%) and actual current limit value.

A1. M is the margin above the calculated peak inductor current that the current limit is being set. For example if M is set to b 10% and the peak current is calculated to be 10A the target current limit will be 10A+(10A*1.1)=11A . If it were selected to be 30% the current limit would be 10A+(10A*1.3)=13A. From this the current sense resistor can be calculated. The actual current limit is based on the current sense resistor and the RAMP capacitor that is selected.


Q2.Please let us know how to decrease inductor current limit value.

A2. Current limit can be reduced by reducing the value of M and recalculating the sense resistor and RAMP capacitor values until a desirable current limit is achieved.


Q3.Please let us know how to calucurate output curent value when Current limit happen at Buck-boost operation.
They used LM25118 in previous model ,they checked IL(ave):8.2A,Iout:2.6A when current limit happen.

A3. As Youhao mentioned above the relationship between output current a peak current limit. Peak current limit is known so the average inductor current can be derived and then average input current. From input current the output current can be derived. This can be derived using standard equations for inductor ripple current, duty cycle and efficiency.


Q4.When they don't use UVLO function,can they leave UVLO pin open ?

A4. I recommend pulling the signal high externally if if is not needed. IT is also possible to connect the UVLO pin to the VIN signal through a resistor. Then the UVLO pin should be clamped using a Zener diode that is under the ABS max rating for the pin.


Q5. Please let us know how to caluculate Fsw variability.
Can they consider only RT variability? (equation1 @P15 datasheet).

A5. Looking under the Oscillator (RT PIN) section of the Electrical Characteristics table it can be seen that for a nominal switching frequency of 200kHz the frequency deviation is -11%/+12% (worst case). For a nominal switching frequency of 515kHz the deviation is -13%/+12% (worst case) . Since the switching frequency selected for this application is 350kHz it can be assumed that the switching frequency deviation (worst case) will be some where between the above values.


Q6.If you can check their circuit and calculator ,we can send both to you by separate mail.

Please send it to the email address that Youhao mentions above.

Thanks,

Garrett