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Cell 1The basic cell has a structure as shown below:AnalysisWe have:$$i_{out}=i_1i_2$$ $$i_1=g_m(Δvv_A)$$ $$i_2=g_m(Δvv_B)$$ Putting these in the 1st equation we have: $$i_{out}=g_m(v_Bv_A)....(1)$$ Also $[g_m(Δvv_B)+g_m(Δvv_A)]R_2=v_A$ . From this we can solve for $v_A$ and we have:$$v_A=(2Δvv_B){g_mR_2}/{1+g_mR_2}...(2)$$ Also $v_Bv_A=g_m(Δvv_B)R_1$ . From this and using the $v_A$ value from (2) we have:$$v_B=Δv{g_mR_1+{2g_mR_2}/{1+g_mR_2}}/{1+g_mR_1+{g_mR_2}/{1+g_mR_2}}...(3)$$ Substituting $v_B$ from (3) and $v_A$ from (2) into (1) we have:$$i_{out}=g_m[Δv{g_mR_1+{2g_mR_2}/{1+g_mR_2}}/{1+g_mR_1+{g_mR_2}/{1+g_mR_2}}(1+{g_mR_2}/{1+g_mR_2}){2Δvg_mR_2}/{1+g_mR_2}]$$ Dividing by $Δv$ and simplifying we have:$$G_M=i_{out}/{Δv}=1/{1/{g_m^2R_1}+1/g_m(1+2R_2/R_1)+R_2}$$ $$G_M={g_m^2R_1}/{1+g_mR_1+2g_mR_2+g_m^2R_1R_2}$$ By substituting $g_m={\ln n}/R_1$ and taking approximations we have:$$G_M≈1/{1/g_m(1+2R_2/R_1)+R_2}=1/{R_1/{\ln n}(1+2R_2/R_1)+R_2}≈1/{2R_2/{\ln n}+R_2}=1/R_2[{\ln n}/{2+\ln n}]$$ Useful numbersNormally the n factor is set to 8 since that allows good layout. With n = 8 we have$$G_M=0.51/R_2≈1/{2R_2}$$ Even if n=4 we have $G_M=0.4/R_2$ which is still close with about 25% error. So the number $1/{2R_2}$ is good to keep handyCell 2This alternate topology has a structure as shown below:AnalysisBoth topologies follow the same analysis:$$i_{out}=i_1i_2$$ $$i_1=g_m(Δvi_1R_2)={g_mΔv}/{1+g_mR_2}$$ $$i_2=g_m(Δvi_2(R_1+R_2))={g_mΔv}/{1+g_m(R_1+R_2)}$$ Therefore we have: $$i_{out}={g_mΔv}/{1+g_mR_2}{g_mΔv}/{1+g_m(R_1+R_2)}$$ $$⇒i_{out}/{Δv}=G_M={g_m^2R_1}/{(1+g_m(R_1+R_2))(1+g_mR_2)}$$ If $g_mR_2≫1$ then we can approximate $G_M$ as:$$G_M≈R_1/{R_2(R_1+R_2)}$$ Useful NumbersRemember that the bandgap voltage is about 1.2V and Vbe is about 0.6V so in this case we have:$${V_T\ln n}/R_1R_2≈0.6V$$ So for n=8 we have $R_2=11.1R_1$ , putting this in the $G_M$ equation we have$$G_M≈1/{12.1R_2}$$ Comparing the Transconductances$$({g_m^2R_1}/{1+g_mR_1+2g_mR_2+g_m^2R_1R_2})_I◌({g_m^2R_1}/{1+g_mR_1+2g_mR_2+g_m^2R_1R_2+g_m^2R_2^2})_{II}$$ It is clearly seen than the $G_{MI}>G_{MII}$ because of the extra term $g_m^2R_2^2$ in the denominator. Hence the topology 1 always provides better accuracy by making the effects of mismatch and offset for the same bias currents lower.From the numbers we also see that for n=8 the 1st topology provides 6 times more $G_M$ .References    


Copyright 2018 Milind Gupta 