Re :: Pos. Feedback Op Amp

Definitions ::

\[\boxed{\frac1{R}=\frac1{R_1}+\frac1{R_2}=\frac1{R_3}+\frac1{R_4}\\ \frac1{R_{1_2}}{}^{↓}=^{↓}\frac1R-\frac1{R_{2_1}}\\ \frac1{R_{3_4}}{}^{↓}=^{↓}\frac1R-\frac1{R_{4_3}}}\]

\[{}^{↓↓}\ \frac{R_3}{R_1}+\frac{R_3}{R_2}-\frac{R_3}{R_3}=\frac{R_3}{R_4}=\\ =\mathbf{R_3·\frac{R_1+R_2}{R_1·R_2}-1}\]

\[V_B=V_X·\frac{R_1}{R_1+R_2}\\ V_A=V_S+\left({V_X-V_S}\right)·\frac{R_3}{R_3+R_4}\]

\[V_X·\frac{R_1}{R_1+R_2}=V_S+\left({V_X-V_S}\right)·\frac{R_3}{R_3+R_4}\\ V_X·\left({\frac{R_1}{R_1+R_2}-\frac{R_4}{R_3+R_4}}\right)=V_S·\left({1-\frac{R_4}{R_3+R_4}}\right)\]

\[A_V=N=\frac{V_X}{V_S}=\frac{\frac{\cancel{R_3}+R_4\cancel{-R_3}}{\bcancel{R_3+R_4}}}{\frac{\cancel{R_1·R_3}+R_1·R_4\cancel{-R_1·R_3}+R_2·R_3}{\left({R_1+R_2}\right)·\bcancel{\left({R_3+R_4}\right)}}}=\]

\[=\frac{R_1·R_4+R_2·R_4}{R_1·R_4-R_2·R_3}=\frac{1+\frac{R_2}{R_1}}{1-\frac{R_2}{R_1}·\frac{R_3}{R_4}}\]

\[N-N·\frac{R_2}{R_1}·\frac{R_3}{R_4}=1+\frac{R_2}{R_1}\]

\[\left({N-1}\right)·R_1·R_4=R_2·\left({N·R_3+R_4}\right)\]

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\[\frac{R_2}{R_1}=\frac{\left({N-1}\right·R_4)}{N·R_3+R_4}=\frac{N-1}{N·\frac{R_3}{R_4}+1}\]

\[{}^{↑↑}\ \frac{R_2}{R_1}·\left({N·\mathbf{\left({R_3·\frac{R_1+R_2}{R_1·R_2}-1}\right)}+1}\right)=N-1\]

\[\frac{R_2}{R_1}·\left({N·R_3·\frac{R_1+R_2}{R_1·R_2}-\left({N-1}\right)}\right)=N-1\]

\[\frac{R_2}{R_1}=\frac1{\frac N{N-1}·R_3·\frac{R_1+R_2}{R_1·R_2}-1}\]

\[\frac N{N-1}·R_3·\frac{\frac{R_1}{R_2}+1}{R_1}-1=\frac{R_1}{R_2}\]

 \[\frac N{N-1}·R_3·\frac{1+\frac{R_2}{R_1}}{R_1}-\frac{R_2}{R_1}=1\]

 \[\frac N{N-1}·R_3·\frac{\bcancel{\frac{R_2}{R_1}+1}}{R_1}=\bcancel{\frac{R_2}{R_1}+1}\]

 \[\frac N{N-1}=\mathbf{\frac{R_1}{R_3}}\]

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 \[\frac{R_2}{R_1}=\frac{N-1}{N·\frac{R_3}{R_4}+1}=\frac1{\frac N{N-1}·\frac{R_3}{R_4}+\frac1{N-1}}=\]

\[=\frac1{\mathbf{\frac{R_1}{\cancel{R_3}}}·\frac{\cancel{R_3}}{R_4}+\frac1{N-1}}=\frac1{\frac{R_1}{R_4}+\frac1{N-1}}\]

\[\frac{R_2}{R_1}·\left({\frac{R_1}{R_4}+\frac1{N-1}}\right)=1\\ \frac1{R_4}+\frac1{R_1}·\frac1{N-1}=\frac1{R_2}\\ \frac1{R_1}·\frac1{N-1}=\frac1{R_2}-\frac1{R_4}\]

\[N-1=\frac1{R_1·\left({\frac1{R_2}-\frac1{R_4}}\right)}\]

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\[N=1+\frac1{R_1·\left({\frac1{R_2}-\frac1{R_4}}\right)}{}^{↑}=^{↑}1+\frac1{R_1·\left({\frac1{R_2}+\frac1{R_3}-\frac1R}\right)}=1+\frac1{R_1·\left({\frac1{R_3}-\frac1{R_1}}\right)}=\]

\[=\boxed{A_V=1+\frac1{\frac{R_1}{R_3}-1}}\]

Reminder :: \(\frac1{R_4}=\frac1{R_2}+\frac1{R_1}-\frac1{R_3} \) or \(\cases{\frac1{R_2}=\frac1R-\frac1{R_1}\\ \frac1{R_4}=\frac1R-\frac1{R_3}}\)

\({Def\ ::\ d=\frac{R_1}{R_3}\\ then\ :\ N-1=\frac1{d-1}\ and\ d=1+\frac1{N-1}}\)

if R is also given the rest can be computed ... as \(R_1=d·R_3\)

in "general" situation the LM308 likes the biasing resistance of the double 7k5 Ω --e.g.-- the R = 3.75 kΩ (for the noisy input the 100 kΩ in parallel with the 5 pF or less may be more suitable/stable however - so ... )

Related Post : More Op-Amp biasing schemes

The LTSpice Example ::

PS! the negative gain can result by the reached formula for the voltage gain -- but in practise it won't work !!!

a slew-rate versus a common mode input impedance ::

Note :: the realistic/conventional Op.-Amp.-s have the impedance value a bit lower than shown in the simulation

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[Eop]

The variety of XOR-s

\[ A\oplus B=\left({A+B}\right)\overline{AB}=^1\]

\[ =^1\overline{\overline{A\overline{AB}}}+\overline{\overline{B\overline{AB}}}=^{21} \]

\[ =^{21}\overline{\overline{A\overline{AB}}\ ·\ \overline{B\overline{AB}}}=^{22} \]

\[ =^1=\left({A+B}\right)\overline{\overline{\overline{A}}\ ·\ \overline{\overline{B}}}=\left({A+B}\right)\overline{\overline{\overline{A}+\overline{B}}}=\left({A+B}\right)\left({\overline{A}+\overline{B}}\right)=^{31} \]

\[ =^{31}\overline{\overline{A+B}}\ ·\ \overline{\overline{\overline{A}+\overline{B}}}=\overline{\overline{A+B}\ +\ \overline{\overline{A}+\overline{B}}}=\overline{\overline{A+B}\ +\ AB}=^{32} \]

\[ =^{31}\cancel{A\overline{A}}+A\overline{B}+B\overline{A}+\cancel{B\overline{B}}=A\overline{B}+B\overline{A}=^{41} \]

\[  =^{41}\overline{\overline{A\overline{B}+B\overline{A}}}=\overline{\overline{A\overline{B}}\ ·\ \overline{B\overline{A}}}=^{42}\overline{\left({A+\overline{B}}\right)\left({\overline{A}+B}\right)}= \]

\[ =\overline{\cancel{A\overline{A}}+AB+\overline{A}\ \overline{B}+\cancel{B\overline{B}}}=\overline{AB+\overline{A}·\overline{B}}=^{32} \]

\[ =^{32}\overline{\overline{\overline{\overline{A+B}+A}}\ ·\ \overline{\overline{\overline{A+B}+B}}}=\overline{\overline{A+B}+A}+\overline{\overline{A+B}+B}=^{33} \]

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note : since the Boolean Arithmetic has no priority in between the conjunction and the disjunction -- the following applies (the ⁽² might be used above ... somewhere) ::

\[ \left({A+B}\right)\left({C+D}\right)=AC+AD+BC+BD\qquad^{(1} \]

\[ A·B+C·D=\left({A+C}\right)·\left({A+D}\right)·\left({B+C}\right)·\left({B+D}\right)=_{"PROOF"}\qquad^{(2} \]

\[ =_{"PROOF"}\left({A+AD+CA+CD}\right)\left({B+BD+CB+CD}\right)= \]

\[ _{=AB+ABD_1+ACB_2+ACD_3+ADB_1+ADCB_4+ACD_3+CAB_2+CABD_4+CAB_2+CAD_3+CDB_5+CD=} \]

\[ =\mathbf{AB}_1+\mathbf{AB}D_2+\mathbf{AB}C_3+A\boxed{CD}+\mathbf{AB}\boxed{CD}_4+B\boxed{CD}+\boxed{CD}= \]

\[ =AB\left({1+D+C+CD}\right)+\left({A+AB+B+1}\right)CD=AB+CD\]

also

\[A+AX=A\left({X+\overline{X}}\right)+AX=AX+A\overline{X}=A=A\left({1+X}\right)=A\left({X+\overline{X}+X}\right) \]

etc. ...

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the simulation ::

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[EoP]

Swapping Display Memory on Desk Calculator

v.1
\[\begin{array}{llll}
math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\
M=M+D & [\ M+\ ] &B&A+B\\
D=-D & [\ ±\ ] &-B&A+B\\
D=D+M & [\ +\ ]\ [\ MR\ ]\ [\ =\ ] &-B+A+B&A+B\\
M=M-D & [\ M-\ ] &A&A+B-A\\
M ↔ D &done&A&B
\end{array}
\]

v.2
\[\begin{array}{llll}
math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\
D=-D & [\ ±\ ] &-B&A\\
D=D+M & [\ +\ ]\ [\ MR\ ]\ [\ =\ ] &-B+A&A\\
M=M-D & [\ M-\ ] &A-B&A-A+B\\
D="0,=" & [\ 0\ ]\ [\ =\ ] &0(+A)&B\\
M ↔ D &done&A&B
\end{array}
\]

v.3
\[\begin{array}{llll}
math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\
D=D-M & [\ -\ ]\ [\ MR\ ]\ [\ =\ ] &B-A&A\\
M=M+D & [\ M+\ ] &B-A&A+B-A\\
D=D-M & [\ -\ ]\ [\ MR\ ]\ [\ =\ ] &-A+B-B&B\\
D=-D & [\ ×\ ]\ [\ 1\ ]\ [\ -\ ]\ [\ =\ ]\ \left({\ _{[\ =\ ]}\ }\right)^{\ (1} &(-A)·1-((-A)·1)&B\\
M ↔ D &done&A&B
\end{array}\\
_{NOTE^{\ (1}\ :\ DEPENDING\ ON\ THE\ SPECIFIC\ CALCULATOR\ THE\ ALTERNATE\ SEQUENCE\\ FOR\ THE\ ARITHMETIC\ NEGATION\ [\ ±\ ]\ MAY\ REQUIRE\ DOUBLE\ [\ =\ ]\ PRESS} 
\]

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[Eop]

Resistor Bridge

Simulation / Schematic :

\[U_R=\frac{U_LR_2R_4+U_UR_0R_4+U_DR_0R_2}{R_2R_4+R_0R_4+R_0R_2}=
\frac{U_L\left({R_1R_3+R_0R_3+R_0R_1}\right)-U_UR_0R_3-U_DR_0R_1}{R_1R_3}\]
\[U_L=\frac{U_RR_1R_3+U_UR_0R_3+U_DR_0R_1}{R_1R_3+R_0R_3+R_0R_1}=
\frac{U_R\left({R_2R_4+R_0R_4+R_0R_2}\right)-U_UR_0R_4-U_DR_0R_2}{R_2R_4}\]
\(U_RR_1R_3R_2R_4+U_UR_0R_3R_2R_4+U_DR_0R_1R_2R_4=U_R\sum_E^3\sum_O^3-U_UR_0R_4\sum_O^3-U_DR_0R_2\sum_O^3\)
\[U_R=R_0·\frac{\left({U_UR_3+U_DR_1}\right)R_2R_4+\left({U_UR_4+U_DR_2}\right)\sum_O^3}{\sum_E^3\sum_O^3-R_1R_2R_3R_4}\]
\(U_LR_1R_3R_2R_4+U_UR_0R_4R_1R_3+U_DR_0R_2R_1R_3=U_L\sum_O^3\sum_E^3-U_UR_0R_3\sum_E^3-U_DR_0R_1\sum_E^3\)
\[U_L=R_0·\frac{\left({U_UR_4+U_DR_2}\right)R_1R_3+\left({U_UR_3+U_DR_1}\right)\sum_E^3}{\sum_E^3\sum_O^3-R_1R_2R_3R_4}\]

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[Eop]