Find Poles and Zero intuitively of LDO

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  • Опубликовано: 16 янв 2025

Комментарии • 10

  • @SAhellenLily
    @SAhellenLily 4 месяца назад +1

    Nice explanation!Thank you brother again🙏

  • @krishnasreenivas5295
    @krishnasreenivas5295 5 месяцев назад +1

    thank you brother, nice explanation

  • @JgNt3981
    @JgNt3981 8 месяцев назад

    I'm not able to understand sir. How can I understand working of transistors and mosfets ? Any reference videos sir? Thank you,

  • @yaojiexu9914
    @yaojiexu9914 7 месяцев назад

    Hello Amit, thank you for the analysis. I have one more comment about the wpe, wpe = 1/CL*(RFBtot||RL||rdspass), say in higher frequency, the Cc would short the gate and drain of MP, the pass devices, result in E node equivalent R = 1/gmpass. So in this case, wpe = gmpass/CL. This is usually be used as 2 stage amplifier 2nd pole analysis.

  • @sumanchowdhury2724
    @sumanchowdhury2724 3 месяца назад

    Hi... Isnt for point A .. Req= 1/gm10?

    • @妥群
      @妥群 3 месяца назад

      yes

    • @wesley-p2w
      @wesley-p2w 2 месяца назад

      Yes, you are right

  • @j0mell0
    @j0mell0 7 месяцев назад

    Thanks! Keep making more "intuition" videos. I feel like we gain valuable knowledge and skills when it comes to intuition, rather than crunching math.
    It would be nice to continue and explain how the transfer function would like long along with the phase shift. Primarily, It would be nice to see what a RHP or LHP affect the phase margin.

  • @sayanbaidya9724
    @sayanbaidya9724 7 месяцев назад

    Nice analysis !!

  • @SAhellenLily
    @SAhellenLily 4 месяца назад

    Thank you Sir 😊 Simple Conpect
    Very interesting Ldo with feedback Types of the v_v,serial _Parell (voltage Amp ,the B=r2/r1+r2 is also +,Av no feedfack is + ) With Cc to achieve the circuit compensation ❤
    for Rz the only Zero in the Ldo circuit
    vo/vi
    =1-gmpassRz-gmpass(1/SCc)=
    Vo=0 and then 1=gmpassRz+gmpass(1/SCc)..(1)
    1=gmpass(Rz+1/SCc)..(1)
    (1/gmpass)-Rz=1/SCc..(1)
    1/S=(Cc/gmpass)-CcRz..(1)
    So Wz=1/[(1/gmpass)-Rz)]Cc....Answer🤔
    Of all the pole contributions
    in this circuit, there are only two pole contributions. One pole is the capacitance effect of the first pole vo1, and the other pole is vo2. The improve poles are too close. Add Cc to make the effect difference between the two poles.🤔