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]

3 - Resistor Voltage Divider

$$I_0+I_1+I_2=0\qquad,\qquad\text{! Notice the directions of the currents !}$$
$${\large \cases{I_0=\frac{V_0-V_X}{R_0}\\ {\ }\\ I_1=\frac{V_1-V_X}{R_1}\\ {\ }\\ I_2=\frac{V_2-V_X}{R_2}}}$$
$$\frac{V_0-V_X}{R_0}+\frac{V_1-V_X}{R_1}+\frac{V_2-V_X}{R_2}=0$$
$$\frac{V_0R_1R_2-V_XR_1R_2+V_1R_0R_2-V_XR_0R_2+V_2R_0R_1-V_XR_0R_1}{R_0R_1R_2}=0\qquad|\qquad×R_0R_1R_2$$
$$\frac{V_0R_1R_2+V_1R_0R_2+V_2R_0R_1}{R_1R_2+R_0R_2+R_0R_1}=V_X$$
Note: the $V_X$ is the unmarked voltage point on the schematic !

It is easy to see the calculation scheme easily adjusts for N voltages all connecting to the single $V_X$
$$V_{X_N}=\frac{\displaystyle{\sum_{i=0}^{N-1}V_i\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}{\displaystyle{\sum_{i=0}^{N-1}\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}$$
thus BWD chek ::
for N=2 :
$$V_X=\frac{V_1R_0+V_0R_1}{R_0+R_1}$$
incase of the $V_0=0$ it reduces to a trivial 2 resistor divider formula :
$$V_X=V_1\frac{R_0}{R_0+R_1}$$

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Simulation example/chk. for N=4 :

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[Eop]

INA formula chk

$$E_1+\left({V_X-E_1}\right)·\frac{R_2}{R_6+R_2}=E_2+\left({V_R-E_2}\right)·\frac{R_5}{R_7+R_5}$$
$$E_1+\left({E_2-E_1}\right)·\frac{R_1}{R_1+R_4+R_3}=V_1\qquad\frac{E_1-V_1}{R_1}=\frac{E_1-E_2}{R_{\Sigma3}}$$
$$E_2+\left({E_1-E_2}\right)·\frac{R_4}{R_1+R_4+R_3}=V_2\qquad\frac{V_2-E_2}{R_4}=\frac{E_1-E_2}{R_{\Sigma3}}$$
$$E_1=\frac{R_1}{R_4}\left({V_2-E_2}\right)+V_1\qquad E_2=\frac{R_4}{R_1}\left({V_1-E_1}\right)+V_2$$
$$E_2-E_1+\frac{\left({E_1-E_2}\right)R_4+\left({E_1-E_2}\right)R_1}{R_{\Sigma3}}=V_2-V_1$$

$$1-\frac{R_1+R_4}{R_{\Sigma3}}=\frac{V_2-V_1}{E_2-E_1}=\frac{V_2-V_1}{E_2-\frac{R_1}{R_4}\left({V_2-E_2}\right)-V_1}=\frac{V_2-V_1}{\frac{R_4}{R_1}\left({V_1-E_1}\right)+V_2-E_1}$$
$$E_2\left({1+\frac{R_1}{R_4}}\right)-\left({V_2\frac{R_1}{R_4}+V_1}\right)=\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}$$
  $$E_1\left({1+\frac{R_4}{R_1}}\right)-\left({V_2+V_1\frac{R_4}{R_1}}\right)=-\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}$$
$$E_2=\frac{\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}+V_2\frac{R_1}{R_4}+V_1}{1+\frac{R_1}{R_4}}=\frac{\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}R_4+V_2R_1+V_1R_4}{R_1+R_4}$$
$$E_1=\frac{-\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}+V_2+V_1\frac{R_4}{R_1}}{1+\frac{R_4}{R_1}}=\frac{-\frac{V_2-V_1}{1-\frac{R_1+R_4}{R_{\Sigma3}}}R_1+V_2R_1+V_1R_4}{R_1+R_4}$$
$$.\ .\ .$$
$$V_X\frac{R_2}{R_6+R_2}-V_R\frac{R_5}{R_7+R_5}=E_2\frac{R_7}{R_7+R_5}-E_1\frac{R_6}{R_6+R_2}$$

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$$IF\ :\ \cases{R_1=R_4=R_A\\ R_3=R_D\\ R_6=R_7=R_F\\ R_2=R_5=R_G}$$
$$E_1=\frac{\cancel{R_A}\left({V_2+V_1}\right)-\frac{V_2-V_1}{1-\frac{2R_A}{2R_A+R_D}}\cancel{R_A}}{2\ \cancel{R_A}}=\frac{V_2+V_1}2-\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}$$
$$E_2=\frac{\cancel{R_A}\left({V_2+V_1}\right)+\frac{V_2-V_1}{1-\frac{2R_A}{2R_A+R_D}}\cancel{R_A}}{2\ \cancel{R_A}}=\frac{V_2+V_1}2+\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}$$
$$.\ .\ .$$

$$\left({V_X-V_R}\right)\frac{R_G}{\cancel{R_F+R_G}}=\left[{\left({\cancel{\frac{V_2+V_1}2}+\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}}\right)-\left({\cancel{\frac{V_2+V_1}2}-\frac{V_2-V_1}2·\frac{2R_A+R_D}{R_D}}\right)}\right]\frac{R_F}{\cancel{R_F+R_G}}$$
$$\boxed{\frac{V_X-V_R}{V_2-V_1}=\frac{R_F}{R_G}·\frac{2R_A+R_D}{R_D}=\frac{R_6}{R_2}·\frac{2R_1+R_3}{R_3}}$$

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dd

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[Eop]

More Op-Amp biasing schemes

$$\frac{V_X-V_R}{V_H-V_L}=\frac{R_F}{R_G}$$

$$IF\ \cases{V_X\ne V_R\\ R\ \rightarrow\ L\ ,\ U\ \rightarrow\ D\\ R_0=R_2\\ R_1=R_3}$$
$$V_L+\left({V_X-V_L}\right)\frac{R_G}{R_F+R_G}=V_H+\left({V_R-V_H}\right)\frac{R_G}{R_F+R_G}$$
$$\left({V_H-V_L}\right)\frac{R_F+R_G}{R_G}=\left({V_X-V_L}\right)-\left({V_R-V_H}\right)=\left({V_X-V_R}\right)+\left({V_H-V_L}\right)$$
$$\frac{V_X-V_R}{V_H-V_L}=\frac{R_F+R_G}{R_G}-1=\frac{R_F}{R_G}$$

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$$IF\ \cases{V_X=V_R\\ R\ \rightarrow\ L\ ,\ U\ \rightarrow\ D\\ R_0\ne R_2\\ R_1\ne R_3}$$
$${}^{see\ the\ optimizations\ below\ :}\ \frac{V_X}{V_S}=\frac{R_BR_F}{R_D\left({R_B-R_F}\right)}$$

$$V_R+\left({V_X-V_R}\right)\frac{R_G}{R_F+R_G}=V_S+\left({V_X-V_S}\right)\frac{R_D}{R_B+R_D}$$
$$\left({V_S-V_R}\right)=\left({V_X-V_R}\right)\frac{R_G}{R_F+R_G}-\left({V_X-V_S}\right)\frac{R_D}{R_B+R_D}$$
$$\boxed{V_R=0\ :\ }\ V_S=V_X\frac{R_G}{R_F+R_G}-\left({V_X-V_S}\right)\frac{R_D}{R_B+R_D}$$
$$V_S\left({1-\frac{R_D}{R_B+R_D}}\right)=V_X\left({\frac{R_G}{R_F+R_G}-\frac{R_D}{R_B+R_D}}\right)$$
$$\frac{V_X}{V_S}=\frac{\frac{R_B}{R_B+R_D}}{\frac{R_G}{R_F+R_G}-\frac{R_D}{R_B+R_D}}=\frac{R_B}{\frac{R_B+R_D}{R_F+R_G}R_G-R_D}=\ ...$$
$$\left({preferably}\right)\ also\ :\ \frac1{R_F}+\frac1{R_G}=\frac1{R_B}+\frac1{R_D}\ ...\frac{R_B+R_D}{R_F+R_G}=\frac{R_BR_D}{R_FR_G}$$
$$...\ =\frac{R_B}{\frac{R_BR_D}{R_F\cancel{R_G}}\cancel{R_G}-R_D}=\frac{R_BR_F}{R_D\left({R_B-R_F}\right)}$$

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~~~~~~~~~~~~~~~
simulation example in Falstad
~~~~~~~~~~~~~~~

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[Eop]

Another Diff. op.-Amp. circuit

$$\frac{V_X-V_R}{V_H-V_L}=\frac{R_AR_F}{R_AR_V+R_G\left({R_A+R_V}\right)}$$

$$V_D=\frac{V_CR_{11}R_3+V_XR_0R_3+V_LR_0R_{11}}{R_{11}R_3+R_0R_3+R_0R_{11}}$$
$$V_U=\frac{V_CR_{22}R_5+V_RR_4R_5+V_HR_4R_{22}}{R_{22}R_5+R_4R_5+R_4R_{22}}$$
$$FROM\ :\ V_X-V_R=\left({V_U-V_D}\right)\frac{R_F}{R_G}$$
$$IF\ :\ \cases{R_1=R_2=R_F\\ R_{10}=R_{20}=R_G\\ R_0=R_4=R_A\\ R_3=R_5=R_V}$$
$$\boxed{!\ note\ that\ the\ above\ condition\ makes\ the\ biasing\ invariant\ of\ the\ \mathbf{V_C}}$$
  $$\left({V_X-V_R}\right)\frac{R_G}{R_F}= \frac{\underline{V_C\left({R_F+R_G}\right)R_V}+\boxed{V_RR_AR_V}+V_HR_A\left({R_F+R_G}\right)}{\left({R_F+R_G}\right)R_V+R_AR_V+R_A\left({R_F+R_G}\right)}-\frac{\underline{V_C\left({R_F+R_G}\right)R_V}+\boxed{V_XR_AR_V}+V_LR_A\left({R_F+R_G}\right)}{\left({R_F+R_G}\right)R_V+R_AR_V+R_A\left({R_F+R_G}\right)}$$
$$\left({V_X-V_R}\right)\left({\frac{R_G}{R_F}+\frac{R_AR_V}{R_{\Sigma3}}}\right)=\left({V_H-V_L}\right)\frac{R_A\left({R_F+R_G}\right)}{R_{\Sigma3}}$$
$$\frac{V_X-V_R}{V_H-V_L}=\frac{R_A\left({R_F+R_G}\right)}{\cancel{R_{\Sigma3}}}·\frac{R_F·\cancel{R_{\Sigma3}}}{R_GR_{\Sigma3}+R_AR_VR_F}=$$
$$=\frac{R_A\left({R_F+R_G}\right)\frac{R_F}{R_G}}{\left({R_F+R_G}\right)R_V+R_AR_V+R_A\left({R_F+R_G}\right)+R_AR_V\frac{R_F}{R_G}}=$$
$$=\frac{R_A\cancel{\left({R_F+R_G}\right)}\frac{R_F}{R_G}}{\cancel{\left({R_F+R_G}\right)}\left({R_A+R_V}\right)+R_AR_V\frac{\cancel{\left({R_F+R_G}\right)}}{R_G}}=$$

$$=\frac{R_AR_F}{R_AR_V+R_G\left({R_A+R_V}\right)}$$

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a simulation example ::

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[Eop]

Differential Op-Amp formulas check

the case for :

$$\cases{signal\ :\ U_S=U_1\\ reference\ :\ U_R=U_0\\ gain\ :\ R_G=R_0\\ feedback\ :\ R_F=R_1\\ output\ :\ U_O=U_X}$$

$$U_R=U_\overline{IN}=U_S+\left({U_O-U_S}\right)·\frac{R_G}{R_G+R_F}$$

$$U_S-U_R=-\left({U_O-U_R+U_R-U_S}\right)·\frac{R_G}{R_G+R_F}$$

$$\frac{U_O-U_R}{U_S-U_R}-1=A_V-1=-1-\frac{R_F}{R_G}$$

$$\boxed{A_V=-\frac{R_F}{R_G}}$$

$$\begin{align*}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad  \\ \hline \end{align*}$$

$$\cases{signal\ :\ U_S=U_0\\ reference\ :\ U_R=U_1}$$

$$U_S=U_\overline{IN}=U_R+\left({U_O-U_R}\right)·\frac{R_G}{R_G+R_F}$$

$$U_S-U_R=\left({U_O-U_R}\right)·\frac{R_G}{R_G+R_F}$$

$$\frac{U_O-U_R}{U_S-U_R}=\boxed{A_V=1+\frac{R_F}{R_G}}$$

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$$\frac{U_Y}{U_1}=\frac{R_0+R_1}{R_0}\qquad \frac{U_X-U_0}{U_Y-U_0}=-\frac{R_3}{R_2}\\ {\ }$$
$$\boxed{U_X=}\ U_0\left({\frac{R_2+R_3}{R_2}}\right)-\frac{R_3}{R_2}U_1\left({\frac{R_0+R_1}{R_0}}\right)\ {^?= {}_?}\ \mathbf{...}\ =\left({U_0-U_1}\right)\frac{R_2+R_3}{R_2}\ \boxed{=\Delta U_{IN}\left({1+\mathbf{M}}\right)}\\ {\ }$$
$$?\qquad \frac{R_2+R_3}{R_2}=\frac{R_3R_0+R_3R_1}{R_0R_0}\ {^?= {}_?}\ ...\ =\frac{R_3\frac{R_0}{R_1}+R_3}{\frac{R_0}{R_1}R_2}=\frac{R_3\frac{R_3}{R_2}+R_3}{\frac{R_3}{R_2}R_2}=\frac{R_2+R_3}{R_2}\\ {\ }$$
$$R_0=\frac{R_3\left({R_0+R_1}\right)}{R_3+R_2}\\ {\ }$$
$$\cases{\underline{R_0R_3}+R_0R_2=\underline{R_0R_3}+R_1R_3\\ {\ }\\ \boxed{R_0R_2=R_1R_3}}\qquad \frac{R_3}{R_2}=\frac{sR_0}{sR_1}=\mathbf{M}\qquad \mathbf{...} \uparrow$$

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some tests :

MAX input impedance test :

Max. frequency TEST :

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[Eop]

Initialization fix for an old simple square root algorithm

. . . it's multiple proven that in general situation the iteration formula :
$$\sqrt{A}\leftarrow a_{n+1}=\frac{\frac{A}{a_n}+a_n}2$$
gives the fastest convergence , initialization :
$$a_0=A\rightarrow a_1=a_{1_1}=\frac{A+1}2$$

however i just explored a different approach with iterative formula :
$$\sqrt{A}\leftarrow a_{n+1}=\frac{2Aa_n}{A+a_n^2}=\frac2{\frac1{a_n}+\frac{a_n}A}=\frac{2A}{\frac A{a_n}+a_n}=\frac A{\frac{\frac{A}{a_n}+a_n}2}$$
which has an initialization :
$$a_0=1\rightarrow a_1=a_{1_2}=\frac{2A}{A+1}=1+\frac{A-1}{A+1}$$

the "Old" initialization gives quite large positive error ... while my "Latest" initialization gives quite large negative error . . . however the both averaged gives more reasonable positive error , where the :
$$a_1=a_{1_3}=\frac{a_{1_1}+a_{1_2}}2=\frac{A+1}4+\frac{A}{A+1}$$

while for each $a_n$ where $n>1$ the first iteration formula on this page applies

...

yet a lesser negative error is got by combining the two first initializations as :
$$a_1=a_{1_4}=\frac{2a_{1_1}a_{1_2}}{a_{1_1}+a_{1_2}}=\frac1{\frac{A+1}{4A}+\frac1{A+1}}=\frac{A}{a_{1_3}}$$

...

+ yet a significantly lesser positive error is got by combining the two last initializations as :
$$a_1=a_{1_5}=\frac{a_{1_3}+a_{1_4}}2=\frac{a_{1_3}+\frac{A}{a_{1_3}}}2$$

the last formula is actually the value of $a_2$ for $a_{1_3}$ . . . so , . . .

. . . anyway it give us quite accurate estimation formula for the near A=1 , as :

$$\sqrt{A}≈a=\frac{\frac{A+1}4+\frac A{A+1}+\frac A{\frac{A+1}4+\frac A{A+1}}}2$$

the error ≤ about 5% - from $\frac1{20}...20$

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[Eop]

Right Triangle

$α+ß=\frac{π}2=90°\ ,\ R=\frac{C}2$

$r=\sqrt{{\left({\frac{C}2}\right)}^2+\frac{A·B}2}-\frac{C}2←\cases{A=r+x\\ B=r+y\\ C=x+y}$

$\frac{a}h=\frac{h}b=\frac{A}B=tan\ ß=\frac1{tan\ α}$

$a·b=h^2$

$a+b=C$

$A^2+B^2=C^2={\left({a+b}\right)}^2=a^2+2·a·b+b^2$

$A^2=a^2+h^2=a^2+a·b=a·\left({a+b}\right)=a·C$

$B^2=b^2+h^2=b^2+b·a=b·\left({b+a}\right)=b·C$

$a=\frac{A^2}C$

$b=\frac{B^2}C$

$h=\frac{A·B}C=\sqrt{a·b}$

 $S=\frac{A·B}2=\frac{\sqrt{a·C·b·C}}2=\frac{h·C}2=\frac{\frac{A·B}C·C}2$

 $P=A+B+C=A+B+\sqrt{A^2+B^2}=\sqrt{a·C}+\sqrt{b·C}+\sqrt{C·C}=\sqrt{C}·\left({\sqrt{a}+\sqrt{b}+\sqrt{a+b}}\right)$

$b=h^2\ ,\ a=1\ →\ A^2=C=h^2+1\ ,\ B^2=b·A^2=b·C=h^2·\left({h^2+1}\right)\ ,\ R=\frac{C}2$

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[Eop]

About the Quadratic Equation

Derivation of the solutions ::

$x^2+px+q=0=\left({x-u}\right)\left({x-v}\right)=x^2-\left({u+v}\right)x+uv$

$p=2a$

$x^2+2ax+a^2-a^2+q=0$

$x+a=±\sqrt{a^2-q}$

$a=\frac p2$

$x=-\frac p2±\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_1=-\frac p2-\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_2=-\frac p2+\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_2-x_1=2\sqrt{{\left({\frac p2}\right)}^2-q}$

$x_2+x_1=-p$

$x_2=-p-x_1$

$x_1=-p-x_2$

$x+p+\frac qx=0$

$\displaystyle{x=-p-\frac qx\quad \rightarrow \quad x_n=x_n+x_\overline{n}-\frac q{x_n}}$

$\displaystyle{x_n=\frac q{x_\overline{n}}\ \equiv\ x_nx_\overline{n}=q\ |\ about:\ q=uv}$

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Example ::

$\mathbf{p}$	$\mathbf{q}$	$\mathbf{x_n}$	$\mathbf{x_1}\\ -p-x_2\\ \displaystyle{\frac q{x_\overline{n}}}$	$\mathbf{x_2}\\ -p-x_1\\ \displaystyle{\frac q{x_\overline{n}}}$
$-1$	$-1$	$-\frac{-1}2±\sqrt{{\left({\frac{-1}2}\right)}^2-\left({-1}\right)}=$ $\frac{1±\sqrt{5}}2=$ $=\cases{-\frac{\sqrt{5}-1}2\\ \frac{\sqrt{5}+1}2}$ $=\cases{-\frac{5-1}{2\left({\sqrt{5}+1}\right)}\\ \frac{5-1}{2\left({\sqrt{5}-1}\right)}}$ $=\cases{\frac{-2}{\sqrt{5}+1}\\ \frac2{\sqrt{5}-1}}$	$\frac{-2}{\sqrt{5}+1}\mathbf{=}$ $\mathbf{=}-\left({-1}\right)-\frac{2}{\sqrt{5}-1}=$ $=1-\frac{\sqrt{5}+1}2=$ $=-\frac{-2+\sqrt{5}+1}2=$ $=-\frac{\sqrt{5}-1}2=$ $=-\frac2{\sqrt{5}+1}$ $\mathbf{=}\frac{-1}{\left({\frac2{\sqrt{5}-1}}\right)}=$ $=-\frac{\sqrt{5}-1}2=$ $=\frac{-2}{\sqrt{5}+1}$	$\frac2{\sqrt{5}-1}\mathbf{=}$ $\mathbf{=}-\left({-1}\right)-\frac{-2}{\sqrt{5}+1}=$ $=1+\frac{\sqrt{5}-1}2=$ $=\frac{2+\sqrt{5}-1}2=$ $=\frac{\sqrt{5}+1}2=$ $=\frac2{\sqrt{5}-1}$ $\mathbf{=}\frac{-1}{\left({\frac{-2}{\sqrt{5}+1}}\right)}=$ $=\frac{\sqrt{5}+1}2=$ $=\frac2{\sqrt{5}-1}$

Dynamic Formulas for the Capacitor and for the Inductor

Charging the capacitor from the CV(=Vs) src. ::

The voltage drop on the series resistor $$V_R=V_S-V_C$$

$$t_{0_{↑}^{↑}→V_C}=-R·C·ln\left({1-\frac{V_C}{V_S}}\right)=R·C·ln\frac{V_S}{V_R}$$

$$t_{V_{1_C}{}_{↑}^{↑}→V_{2_C}}=R·C·ln\frac{V_S-V_{1_C}}{V_S-V_{2_C}}=R·C·ln\frac{V_{1_R}}{V_{2_R}}$$

Dis-charging the capacitor through a fixed value resistor ::

$$t_{V_{2_C}{}_{↓}^{↓}→V_{1_C}}=-R·C·ln\frac{V_{1_C}}{V_{2_C}}=R·C·ln\frac{V_{2_C}}{V_{1_C}}$$

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"Charging" the inductor from the CV(=Vs) src. ::

The peak current $$I_{MAX}=\frac{V_S}R$$

$$t_{0_{↑}^{↑}→I_L}=-\frac LR·ln\left({1-\frac{I_L}{I_{MAX}}}\right)$$

$$t_{I_{1_L}{}_{↑}^{↑}→I_{2_L}}=\frac LR·ln\frac{I_{MAX}-I_{1_L}}{I_{MAX}-I_{2_L}}$$

"Dis-charging" the inductor through a fixed value resistor ::

$$t_{I_{2_L}{}_{↓}^{↓}→I_{1_L}}=\frac LR·ln\frac{I_{2_L}}{I_{1_L}}$$

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about how to get linear I U ramps on the inductor and on the capacitor : simulation

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[Eop]

Diluting the solution

If there is solution with the concentration $x=\frac b{b+a}$ and we need to get the solution with concentration $y=\frac b{b+c}$ then from $x\left({b+a}\right)=b=y\left({b+c}\right)$ we get the ratio of masses $\displaystyle{\frac xy=\frac{b+c}{b+a}=\frac{m_{FINAL}}{m_{INITIAL}}}$ . . . so - if the $\displaystyle{m_{FINAL}}$ is given we need the $\displaystyle{m_{INITIAL}=\frac yx·m_{FINAL}}$ of the initial solution , the amount of water needed to be added is $c-a=\mathbf{d}$ , from $\displaystyle{\frac xy=\frac{m_F}{m_I}=1+\frac d{m_I}=\frac1{1-\frac d{m_F}}}$ we see that $\displaystyle{d=\left({\frac xy-1}\right)·m_I=\left({1-\frac yx}\right)·m_F}$ . . . checking
$\displaystyle{d=\left({\frac xy-1}\right)·m_I=\left({\frac xy-1}\right)·\frac yx·m_F}$ and
$\displaystyle{d=\left({1-\frac yx}\right)·m_F=\left({1-\frac yx}\right)·\frac xy·m_I}$

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PS! -- the below DEMO works and "in reverse" for the initial percentage being less than the final ... the result is given as a negative amount of the water to be "added" --e.g.-- describing the amount of water to be (evaporated/)extracted from the solution ... Now updated! to match the possible concentrating option ...

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▼ ▼ ▼
► ► ► ◄ ◄ ◄
▲ ▲ ▲

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. . . a Q.C. (in MSO Excel) for some non-intuitive cases ::

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Physics

Linear	Circular
Time $t\ \left({s}\right)$	Time $t\ \left({s}\right)$
Mass $m\ \left({kg}\right)$	Moment of inertia $I=\sum{m_i·r_i}\ \left({kg·m^2}\right)$
Distance $s\ \left({m}\right)$	Angle $\overrightarrow{φ}\ \left({rad=1}\right)$
Velocity $v\ \left({\frac ms}\right)$	Angular velocity $\overrightarrow{ω}=\frac{∂φ}{∂t}=\frac vR\ \left({\frac {rad}s=\frac1s}\right)$
Acceleration $a\ \left({\frac m{s^2}}\right)$	Angular_acceleration $\overrightarrow{ɛ}=\frac{∂ω}{∂t}=\frac{a_{\tau}}R\ \left({\frac{rad}{s^2}=\frac1{s^2}}\right)$
Impulse $p=m·v=F·dt\ \left({N·s}\right)$	Angular momentum $N=I·\overrightarrow{ω}=\overrightarrow{R}×\overrightarrow{p}\ \left({\frac{kg·m^2}s}\right)$
Force $F=\frac{∂p}{∂t}=m·a\ \left({N}\right)$	Torque $M=I·\overrightarrow{ɛ}=\overrightarrow{R}×\overrightarrow{F}\ \left({N·m}\right)$
Energy $E=\frac{m·v^2}2=m·a·s\ \left({J}\right)$	Energy $W=\frac{I·ω^2}2=I·ɛ·φ\ \left({J}\right)$
Work $A=F·∂s·Cos\ φ\ \left({J}\right)$	Work $A=M·∂φ\ \left({J}\right)$
Power $P=\frac{∂A}{∂t}\ \left({W}\right)$	Power $P=\frac{∂W}{∂t}\ \left({W}\right)$
$d\left({m·\overrightarrow{v}}\right)=\overrightarrow{F}·dt$	$d\left({I·\overrightarrow{ω}}\right)=\overrightarrow{M}·dt$

p+ , p– , x(T) , f(x)

Form the $x^2-xT-1=0$ where $T\in\mathbb{R}^{+}\ ;\ T≠0$
$x-\frac1x=T\ →\ \frac xT=\frac{xT+1}{xT}=\frac{x+\frac1T}x$
Def.: $x≡p^+$ and $\frac1x≡p^-$
$p^+=\frac{\sqrt{T^2+4}+T}2\ →\ p^-=\frac1{p^+}=\frac2{\sqrt{T^2+4}+T}=\frac{\sqrt{T^2+4}-T}2$

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Def.: $x\left({T}\right)=T+p^-\left({\frac1x}\right)=T+\frac{\sqrt{T^2+4}-T}2=\frac{\sqrt{T^2+4}+T}2$ Def.: $f\left({x}\right)≡f\left({x_T}\right)≡x\left({T}\right)$

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The properties of the $f\left({x}\right)$ :

1. $f\left({-x}\right)=\frac1{f\left({x}\right)}=f\left({x}\right)-x$
   $\frac1{f\left({-x}\right)}=f\left({x}\right)=f\left({-x}\right)+x$
2. $f\left({-x}\right)+x=\frac1{f\left({x}\right)-x}$
3. $x·f\left({x}\right)-x·f\left({-x}\right)=x^2$
   $x·f\left({-x}\right)-x·f\left({x}\right)=-x^2$

NULL

testing it out . . .
embedded super and subscript

$\displaystyle{e^{e^{e^{e^e}}}}\\\displaystyle{a_{1_{2_{3_4}}}}$

over-/under-line :
$$\begin{array}{|c|c|c|} \hline \overline{A}+\overline{B}=\overline{AB} & \overline{A}\ \overline{B}=\overline{A+B} & A\overline{B}+\overline{A}B=A⊕B\\ \hline {R\\ _{NORMAL}} & {\mathbf{R}\\ _{BOLD}} & {\mathbb{R}\\ _{OUTLINE}}\\ \hline \underline{UNDERLINE} & \overrightarrow{VECTOR} & {A·\cancel{B}\\ _{CANCELING\ ^{1)}\ AN\ OPERAND}} \\ \hline \end{array}$$ $$\begin{array}{|c|c|} \hline switch&size\\ \hline \backslash tiny&{\tiny text\ size}\\ \hline \backslash scriptsize&{\scriptsize text\ size}\\ \hline \backslash small&{\small text\ size}\\ \hline \backslash normalsize&{\normalsize text\ size}\\ \hline \backslash large&{\large text\ size}\\ \hline \backslash Large&{\Large text\ size}\\ \hline \backslash LARGE&{\LARGE text\ size}\\ \hline \backslash huge&{\huge text\ size}\\ \hline \backslash Huge&{\Huge text\ size}\\ \hline \end{array}$$ $$\begin{array}{|c|c|} \hline switch&color\\ \hline \backslash color\left\{yellow\right\}&{\color{yellow}{colored\ text}}\\ \hline \end{array}$$ about ¹⁾ MathJax TeX and LaTeX Support — MathJax 2.7 documentation - TeX and LaTeX extensions