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WEBENCH® Tools/TPS61089: Simulation clarification

d_zero
Genius 4775 points
Part Number: TPS61089

Tool/software: WEBENCH® Design Tools

Dear *

we want to use TPS61089 in our new design but have following questions below:

-----------------------------------------------------------------------------------------------------------

Info:

Vin 4.5 to 5V ( typ 5V)

Vout 12V

Iout max = 1.2A Iout typ from 200mA to 600mA, but can be also as low as 80mA in some cases 

desired Fsw =2MHz

------------------------------------------------------------------------------------------------------------

1) i made a simulation in webench Vin=5V, Vout=12V, Iout=1.2A (CCM) (a case)

and get an eff of 90.3% and now i use the slider to set the iout to 0.3A (DCM)and the eff is 88.7% ( this is acceptable)

The webench uses 470nH 15mOhm inductor 

second i made a simulation with  Vin=5V, Vout=12V, Iout=0.3A (b case)

and now the simulation shows eff of 85.1% which is lower , and the inductor is 1.5uH 20mohm

so for me the first solution is better, please can you comment the negative effect of DCM mode?

2) Also in the two simulations the compensation network is different

a) case 0.3A in DCM mode. b) case 0.3A in CCM mode

So when simulating a) case only when the Iout slider is at 1.2A it is showing phase marking, when move the slider to 0.3A then we can't see phase margin?

Please can you comment the impact of the compensation network when the TPS61089 is working in DCM mode?

3)

Also for 470nH we want to use CIGW252010GLR47MNE

can we use also 1uH ( IHLP2020CZER1R0M01) with 2MHz ? Is this better than 470nH?

4)

a) how can we now form calculation after what Iout will the TPS61089 go to DCM mode

b) how can we now form calculation after what Iout will the TPS61089 go to PSM mode

c) is there any connection between PSM and DCM mode?

Best Regards,

David.

  • 0
    TI__Mastermind 40080 points

    Dear Sir,

    To determine when the TPS61089 goes into the DCM, you need to check the inductor ripple. For example, in your application, when Vin=4.5V, Vo=12V, fsw=2MHz, when L=1uH, then peak-peak inductor current is around 1.4A. which means when the input current is around 0.7A, it will enters into the DCM. And 0.7A Iin mean around 0.3A (Vin*Iin*eff/Vo)load current considering efficiency.

    In your design, you need to select the inductor according to the maximum load condition. Why do you care about the DCM and the PFM, is it because of the efficiency?

  • 0 in reply to Helen chen
    Genius 4775 points

    Dear Helen,

    our max load condition can be 1 to 1.2A. If setting the inductor to 470nH while 2MHz the Ipp is around 3A -> In=1.5A -> IDCM below 0.6A.

    But the load current in our case depends how many device do we connect to host unit. So we can connect 4 devices, and most of the time 1 device is connected so the load current will be 0.3A in most cases.

    a) is it normal for TPS61089 tha it will work most of the time in DCM, what are the drawbacks?

    b) in simulation when we put the slider from 1.2A to 0.3A and do the recalculation the phase marking parameter is gone.

    Do we need to care about the phase and gain margin in DCM mode?

    Best Regards,

    D.

  • 0 in reply to d_zero
    TI__Mastermind 40080 points

    Dear Sir,

    It is OK for the TPS61089  to work in the DCM mode or PFM mode.

    In the DCM mode, the phase margin should be OK. And in the PFM mode, the phase margin can't be tested out because of the hysteresis control under PFM mode.

    Are you sure the converter works in the DCM mode?  Could you please attach the waveform?

  • 0 in reply to Helen chen
    Genius 4775 points

    Dear Helen,

    1)we don't have the wave form because we are in PCB layout phase .

    I only have the simulation file in Webench (fsw 2.022MHz)

    if we set the Iout slider to 0.5A or below the Mode says BOOST DCM

    2) regarding Ilim, how much should we set the Ilim regarding to Ilpeak

    current u upper simulation is Rlim 178k = 5.7A is it possible to set to 200k = 5.15A ( simulation will not allow it )?

    Best Regards,

    D.

  • 0 in reply to d_zero
    TI__Mastermind 40080 points

    Dear Sir,

    Suggest you use 1uH inductor in your application.   With 1uH inductor , the peak inductor is around 5.43A, considering the TPS61089's Ilim tolerance, suggest add around 1A margin, which is 6.4A. So the current limit setting resistor should be around 150k ohm.

    One question, why you consider it is unstable in PFM mode?

  • 0 in reply to Helen chen
    Genius 4775 points

    Dear Helen,

    1) i have one question regarding calculated peak inductor current: i calculated with DS formulas with

    Fsw= 2MHzz

    L= 1uH

    max Iout 1.2A

    eff=90%

    Idc=3.2A

    Ipp=1.45A

    ILpeak = 4A -> +20% -> 4.7A -> Rlim =220kOhm

    or am i wrong?

    2) I would rather use CIGW252010GLR47MNE because of space, price and i also already use the 470nH in the design for other DCDC, do oyu see any down side of using the smaller inductor except efficiency ?

    3) Regarding your question why use in unstable mode.

    I'm not considering it i need to cover working area from 300mA to 1.2.A.

    Do you thing we need to take another approach ?

    Best Regards,

    David.

  • 0 in reply to d_zero
    TI__Mastermind 40080 points

    Dear Sir,

    you need to calculate the peak inductor current according to the Vinmin condition. That is Vin=4.5V, then if choose inductor ripple current ratio 0.85, then you can get a 0.47uH inductor.  Iin=Po/0.9/4.5V=3.56A, ILpeak=Iin*1.425=3.56X1.425=5.05A, considering the tolerance, you need to left around 1A margin, that is at least 6A to calculate the Rlim.  So suggest you use a 150k-170k ohm resistor as the current limit setting resistor.

    You can use 470nH inductor, but current rating should be >=6A.