Question 14.5: Analyzing the CMOS op-amp TLC1078 The CMOS amplifier in Fig....

We have

\begin{aligned}K_{ p } &=\frac{K_{ x } W}{L}=10 \mu \times \frac{160}{10}=160 \mu A / V ^{2} \quad\left(\text { for all MOSFETs expect } Q _{10}\right) \\K_{ p 10} &=\frac{K_{ x } W}{L}=10 \mu \times \frac{40}{10}=40 \mu A / V ^{2} \quad\left(\text { for } Q _{10}\right)\end{aligned}

For I_{ Q }=40 \mu A \text { and } K_{ p }=160 \mu A / V ^{2} , Eq. (14.69) gives

V_{ GS }=V_{ t }+\sqrt{\frac{I_{ D }}{K_{ p }}}                                     (14.69)

V_{ GS }=V_{ t }+\sqrt{I_{ D } / K_{ p }}=0.5+\sqrt{40 / 160}=1 V

Thus,

V_{ GS 3}=V_{ GS 6}=V_{ GS 11}=V_{ GS 12}=V_{ GS 13}=1 V

Also,

I_{ D 3}=I_{ D 6}=I_{ D 11}=I_{ D 12}=I_{ D 10}=I_{ D 13}=I_{ Q }

The drain current I_{ D 13} \text { of } Q _{13} is given by

I_{ D 13}=K_{ p 13}\left(V_{ GS 13}-V_{ t }\right)^{2}                                     (14.74)

Since V_{ GS 10}=V_{ GS 13}-I_{ Q } R_{7} , the drain current I_{ D 10} \text { of } Q _{10} is given by

I_{ D 10}=K_{ p 10}\left(V_{ GS 10}-V_{ t }\right)^{2}=K_{ p 10}\left(V_{ GS 13}-I_{ Q } R_{7}-V_{ t }\right)^{2}                                     (14.75)

The V_{ GS } \text { values of } Q _{10} \text { and } Q _{13} are different, but their drain currents are equal; that is,

I_{ D 13}=I_{ D 10} \quad \text { and } \quad K_{ p 13}\left(V_{ GS 13}-V_{ t }\right)^{2}=K_{ p 10}\left(V_{ GS 13}-I_{ Q } R_{7}-V_{ t }\right)^{2}

which relates R_{7} \text { to } I_{Q} as follows:

R_{7}=\frac{V_{ GS 13}-V_{ t }}{I_{ Q }}\left[\left(\frac{K_{ p 13}}{K_{ p 10}}\right)^{1 / 2}-1\right]                                         (14.76)

Since K_{ p } is proportional to the ratio W⁄L, Eq. (14.76) can be expressed as

R_{7}=\frac{V_{ GS 13}-V_{ t }}{I_{ Q }}\left[\left(\frac{W_{13}}{W_{10}}\right)^{1 / 2}-1\right]                                           (14.77)

which, for I_{ Q }=40 \mu A , W_{10}=40 \mu m , W_{13}=160 \mu m , V_{ GS 13}=1 V \text {, and } V_{ t }=0.5 V \text {, gives } R_{7}=12.5 k \Omega . Thus, the ratio \sqrt{W_{13}/ W_{10}} sets the value of resistance R_{7} \text { or } I_{ Q } .