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Consider the folded-cascode op amp of Fig.earlier when loaded with a 10pF10-\mathrm{pF} capacitance. What should the bias current II be to obtain a slew rate of at least 10 V/μs\mathrm{V} / \mu \mathrm{s} ? If the input-stage transistors are operated at overdrive voltages of 0.2 V0.2 \mathrm{~V}, what is the unity-gain bandwidth realized? If the two nondominant poles have the same frequency of 25MHz25 \mathrm{MHz}, what is the phase margin obtained? If it is required to have a phase margin of 7575^{\circ}, what must ftf_t be reduced to? By what amount should CLC_L be increased? What is the new value of SRS R ?

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Refer to Figure 12.8 of the textbook and considering :

CL=10 pFSR=10  V/μsV0v=0.2  Vfp=25 MHz\begin{align*} C_{L} &=10 \ \mathrm{pF} \\ S R &=10 \ \mathrm{~V} / \mathrm{ \mu s} \\ V_{0 v} &=0.2 \ \mathrm{~V} \\ f_{p} &=25 \ \mathrm{MHz} \end{align*}

The bias current I is :

SR=ICLI=(SR)(CL)=(10V/μs)(10pF)=104 A=0.1  mA\begin{align*} S R &=\cfrac{I}{C_{L}} \\ I &=(S R)\left(C_{L}\right) \\ &=(10 \mathrm{V} / \mu s)(10 \mathrm{pF}) \\ &=10^{-4} \ \mathrm{A} \\ &=0.1 \ \mathrm{~mA} \end{align*}

The unity-gain band width ftf_{t} is:

ft=GmR0fp=Gm2πCL=IV0v2πCL=(0.1  mA/0.2  V)2π×10 pF=7.957 MHz\begin{align*} f_{t} &=G_{m} R_{0} f_{p} \\ &=\cfrac{G_{m}}{2 \pi C_{L}}\\ &=\cfrac{\cfrac{I}{V_{0v} }}{2 \pi C_{L}}\\ &=\cfrac{(0.1 \ \mathrm{~mA} / 0.2 \ \mathrm{~V})}{2 \pi \times 10 \ \mathrm{pF}} \\ &=7.957 \ \mathrm{MHz} \end{align*}

The phase margin is given by:

 phase margin =90tan1(ftfp)=90tan1(7.957 MHz25 HHz)=9017.655=72.34572.35\begin{align*} \text { phase margin }&=90^{\circ}-\tan ^{-1}\left(\cfrac{f_{t}}{f_{p}}\right)\\ &=90^{\circ}-\tan ^{-1}\left(\cfrac{7.957 \ \mathrm{MHz}}{25 \ \mathrm{HHz}}\right) \\ &=90^{\circ}-17.655^{\circ} \\ &=72.345^{\circ} \\ &\approx 72.35^{\circ} \end{align*}

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