Dif. Pos.-fbk. op.-Amp.

see also : Pos. fbk. Op. Amp. - (spice) noise

 

$$V_a=V_S+\left({V_O-V_S}\right)\frac{R_0}{R_0+R_2}$$

$$V_b=V_R+\left({V_O-V_R}\right)\frac{R_1}{R_1+R_3}$$

$$\left({V_O-V_R}\right)\frac{R_1}{R_1+R_3}= \left({V_S-V_R}\right)+\left({V_O-V_S}\right)\frac{R_0}{R_0+R_2}$$

$$\frac{V_O-V_R}{V_S-V_R}= \frac{R_1+R_3}{R_1}\left({1+\frac{V_O-V_S}{V_S-V_R}·\frac{R_0}{R_0+R_2}}\right)$$

$$A_V·\frac{R_1}{R_1+R_3}=1+\frac{V_O-V_R+V_R-V_S}{V_S-V_R}·\frac{R_0}{R_0+R_2}=1+\left({A_V-1}\right)\frac{R_0}{R_0+R_2}$$

$$A_V\left({\frac{R_1}{R_1+R_3}-\frac{R_0}{R_0+R_2}}\right)=1-\frac{R_0}{R_0+R_2}$$

$$A_V=\frac{R_2\left({R_0+R_2}\right)\left({R_1+R_3}\right)}{\left({R_0+R_2}\right)\left[{R_1\left({R_0+R_2}\right)-R_0\left({R_1+R_3}\right)}\right]}=$$

$$=\frac{R_2\left({R_0+R_2}\right)\left({R_1+R_3}\right)}{R_2\left({R_0+R_2}\right)\left({R_1+R_3}\right)\left[{\frac{R_1}{R_2}·\frac{R_0+R_2}{R_1+R_3}-\frac{R_0}{R_2}}\right]}=$$

$$=\left[{Def.:\ R\ ,\ \ \frac1{R_0}+\frac1{R_2}=\frac1{R_1}+\frac1{R_3}=\frac1{R}}\right]=$$

$$=\frac1{\frac{R_1R_0R_2}{R_2R_1R_3}-\frac{R_0}{R_2}}=\frac1{R_0\left({\frac1{R_3}-\frac1{R_2}}\right)}=\frac1{R_0\left({\frac1{R_0}-\frac1{R_1}}\right)}=\frac1{1-\frac{R_0}{R_1}}={A_V}^+$$

---

What the above means - is that in case of the shown configuration - the non-common mode signal is extracted from common mode one , amplified ... and added back to the common mode one . . . shortly put :

$$V_O=V_R+A_V·\left({V_S-V_R}\right)$$

PS! : It is also possible to show - as for the above positive voltage gain derivation - that when we swap V_S and V_R then the negative/inverting voltage gain becomes $A_V=\frac1{1-\frac{R_1}{R_0}}$ (see below) . . .

assuming the relation $R_2=R_0·\left({A_V-1}\right)$ for the positive gain and the relation $R_2=-A_V·R_0$ for the inverting gain -- the following applies :

the neg. gain case :

$${A_V}^+=1-{A_V}^-=1-\frac1{1-\frac{R_1}{R_0}}=\frac{1-\frac{R_1}{R_0}-1}{1-\frac{R_1}{R_0}}=\frac{-\frac{R_1}{R_0}}{1-\frac{R_1}{R_0}}=\frac1{1-\frac{R_0}{R_1}}$$

e.g. $\quad\left|{{A_V}^-}\right|={A_V}^+-1\quad$ ← that
for the same resistor values or for the same R₀ : R₁ ratio

about formulas :

parameter	non-inverting	inverting
$A_V\quad$	user set $\displaystyle{\frac1{1-\frac{R_0}{R_1}}}$	user set $\displaystyle{\frac1{1-\frac{R_1}{R_0}}}$
$R_0$	user set	user set
$R_2$	$R_0·\left({A_V-1}\right)$	$-A_V·R_0$
$\frac1R$	$\displaystyle{\frac1{R_0}+\frac1{R_2}=\frac1{R_1}+\frac1{R_3}}$
$R_1$	$\displaystyle{\frac{R_0}{1-\frac1{A_V}}}$	$R_0\left({1-\frac1{A_V}}\right)$
$R_3$	$\displaystyle{\frac1{\frac1R-\frac1{R_1}}}$
$R_3$	$\displaystyle{R_1\left({\frac1{1-{\left({1-\frac1{A_V}}\right)}^2}-1}\right)}$	$\displaystyle{\frac{R_1}{{\left({1-\frac1{A_V}}\right)}^2-1}}$

---

$$V_b=V_S+\left({V_O-V_S}\right)\frac{R_1}{R_1+R_3}$$

$$V_a=V_R+\left({V_O-V_R}\right)\frac{R_0}{R_0+R_2}$$

$$\left({V_O-V_R}\right)\frac{R_0}{R_0+R_2}= \left({V_S-V_R}\right)+\left({V_O-V_S}\right)\frac{R_1}{R_1+R_3}$$

$$\frac{V_O-V_R}{V_S-V_R}·\frac{R_0}{R_0+R_2}= 1+\frac{V_O-V_R+V_R-V_S}{V_S-V_R}·\frac{R_1}{R_1+R_3}$$

$$A_V·\frac{R_0}{R_0+R_2}=1+\left({A_V-1}\right)·\frac{R_1}{R_1+R_3}$$

$$A_V·\left({\frac{R_0}{R_0+R_2}-\frac{R_1}{R_1+R_3}}\right)=1-\frac{R_1}{R_1+R_3}=\frac{R_3}{R_1+R_3}$$

$$A_V=\frac{R_3\left({R_1+R_3}\right)\left({R_0+R_2}\right)}{\left({R_1+R_3}\right)\left[{R_0\left({R_1+R_3}\right)-R_1\left({R_0+R_2}\right)}\right]}=\frac{\mathbb{Z}}{\mathbb{Z}\left[{\frac{R_0}{R_3}·\frac{R_1+R_3}{R_0+R_2}-\frac{R_1}{R_3}}\right]}=$$

$$=\frac1{\frac{R_0R_1R_3}{R_0R_2R_3}-\frac{R_1}{R_3}}=\frac1{R_1\left({\frac1{R_2}-\frac1{R_3}}\right)}= \frac1{\frac{R_1}{R_1}-\frac{R_1}{R_0}}=\frac1{1-\frac{R_1}{R_0}}={A_V}^-$$

---

 [Eop]