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TPS40210: FB voltage at Negative output

K.Hamamoto
Guru 12155 points
Part Number: TPS40210
Other Parts Discussed in Thread: PMP9624

Hi

Our customer are going to consider the configuration of the negative output from the negative input.　（Vin=-15V,Vo=-6.8V, Io=1.5A)

So, we found some reference design at TI site. At the one of design,we see the negative input-output configuration with TPS40210 such as PMP9624.

http://www.ti.com/jp/lit/df/tidrf59/tidrf59.pdf

I have some question regrading this configuration.

1. At this configuration, is TP3 is GND and TP4 is -3.3V ?

2. I understand FB voltage need to invert but how can we calculate the FB resister which is R102 and R2 and R7?

(Also EN signal need to invert.)

3.How can we calculate the inductor value and Switch current?

Could you give us if there is any other good references to help?

Best Regards,

Koji Hamamoto

  • 0
    TI__Guru* 78805 points

    Hello Hamamoto-san,

    Thank you for considering the TPS40210.  To answer your questions in the same order:

    1. Yes, TP3 is system GND, and TP4 is negative output rail.

    2. R102 is just a dummy load which I believe the designer used to cope with no load condition.  All Vout is applied across R2  Note that Q2 and Q3 Vbe's cancel each other.  Then, 

    ie = Vout /R2  (emitter current)

    ic = ~ie  (collector current)

    Vfb = ic x R7

    Use these equations to select R7 and R2 in order to set Vout at a desire value.

    3. The circuit is basically a buck.  Use the buck converter's equations to calculate the inductor  and switch current. 

    Best Regards,

    Youhao Xi, Applications Engineering

  • 0 in reply to Youhao Xi
    Guru 12155 points

    Hi Youhao-san,

    Thank you for your support.

    Can I ask regarding Q3 at previous post?

    As you mentioned, we can calculate these components as buck converter. So, here is my understanding.

    is the calculation correct? (in case of TIDRF59)

      Inductor ; if the ripple is 30% of Iout, L(min) = D*(Vi-(Vi-Vo))*/(dIL*Fsw) = (-3.3/-11.7)*(-15-(-11.7)/(Iout*0.3*250kHz) = 6uH.

      SW current : peak current is Iout+(dIL/2)= 2.3A

    Best Regards,

    Koji Hamamoto

  • 0 in reply to K.Hamamoto
    TI__Guru* 78805 points

    Hi Hamamoto-san,

    Its close but not accurate.  Just treat the negative voltage as positive voltage.  For instance, a -15Vin to -3.3Vout buck converter is equivalent to +15Vin to +3.3Vout.  

    Then,

    L(min) = Vo * (1-Dmin) / (dIL *Fsw)

    Where Dmax = Vo /Vinmax

    TFor your case:  -15Vinmax to -3.3Vout:

    Dmin= -3.3/-15 = 0.22

    L(min) = 3.3 * (1-0.22) /(1.5A *0.3 *250KHz) = 22.88 uH.

    You may also consider to choose the ripple current to be 40%, which will leads to L(min)=17.16 uH.

    Hope this clarifies.

    Thanks,

    Youhao

    . 

  • 0 in reply to Youhao Xi
    Guru 12155 points

    Hi Youhao,

    Thank you. 　I understood. But I have one question. (It's might be typo)

    >L(min) = Vo * (1-Dmin) / (dIL *Fsw)

    >Where Dmax = Vo /Vinmax

    Do you mean Duty is Dmax, not Dmin at above equation?

    Best Regards,

    Koji Hamamoto