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Negative impedance converter
Transfer function derivation
edit
First, assigning Va = voltage at + terminal = Vs
and assigning Vb = voltage at - terminal
and assigning Vo = voltage at output
I
s
+
V
o
−
V
s
R
3
=
0
,
{\displaystyle I_{s}+{\frac {V_{o}-V_{s}}{R_{3}}}=0,}
Kirchoff's current law
V
a
=
V
o
+
I
s
R
3
,
{\displaystyle V_{a}=V_{o}+{I_{s}}{R_{3}},}
rearranging to express
V
a
{\displaystyle V_{a}}
in terms of
V
o
{\displaystyle V_{o}}
and
I
s
{\displaystyle I_{s}}
V
o
=
V
s
−
I
s
R
3
,
{\displaystyle V_{o}=V_{s}-{I_{s}}{R_{3}},}
rearranging to express
V
o
{\displaystyle V_{o}}
in terms of
V
s
{\displaystyle V_{s}}
and
I
s
{\displaystyle I_{s}}
V
b
R
1
+
V
b
−
V
o
R
2
=
0
,
{\displaystyle {\frac {V_{b}}{R_{1}}}+{\frac {V_{b}-V_{o}}{R_{2}}}=0,}
Kirchoff's current law
V
b
=
V
o
R
1
R
1
+
R
2
,
{\displaystyle V_{b}={\frac {{V_{o}}{R_{1}}}{R_{1}+R_{2}}},}
rearranging
V
o
=
(
V
a
−
V
b
)
α
,
{\displaystyle V_{o}=(V_{a}-V_{b})\alpha ,}
where
α
{\displaystyle \alpha }
= the opamp open loop gain
(
V
o
+
I
s
R
3
−
V
o
R
1
R
1
+
R
2
)
α
=
V
o
,
{\displaystyle (V_{o}+{I_{s}}{R_{3}}-{\frac {{V_{o}}{R_{1}}}{R_{1}+R_{2}}})\alpha =V_{o},}
substituting for
V
a
{\displaystyle V_{a}}
and
V
b
{\displaystyle V_{b}}
α
V
o
(
1
−
R
1
R
1
+
R
2
)
+
α
I
s
R
3
=
V
o
,
{\displaystyle \alpha {V_{o}}(1-{\frac {R_{1}}{R_{1}+R_{2}}})+\alpha {I_{s}}{R_{3}}=V_{o},}
rearranging
V
o
(
α
(
1
−
R
1
R
1
+
R
2
)
−
1
)
=
−
α
I
s
R
3
,
{\displaystyle {V_{o}}(\alpha (1-{\frac {R_{1}}{R_{1}+R_{2}}})-1)=-\alpha {I_{s}}{R_{3}},}
rearranging
V
o
(
1
−
R
1
R
1
+
R
2
)
=
−
I
s
R
3
,
{\displaystyle {V_{o}}(1-{\frac {R_{1}}{R_{1}+R_{2}}})=-{I_{s}}{R_{3}},}
cancelling out the
α
{\displaystyle \alpha }
term and simplifying
V
o
=
V
s
−
I
s
R
3
,
{\displaystyle V_{o}=V_{s}-{I_{s}}{R_{3}},}
shown earlier
V
s
(
1
−
R
1
R
1
+
R
2
)
−
I
s
R
3
(
1
−
R
1
R
1
+
R
2
)
+
I
s
R
3
=
0
,
{\displaystyle V_{s}(1-{\frac {R_{1}}{R_{1}+R_{2}}})-{I_{s}}{R_{3}}(1-{\frac {R_{1}}{R_{1}+R_{2}}})+{I_{s}}{R_{3}}=0,}
substituting for
V
o
{\displaystyle V_{o}}
V
s
(
R
2
R
1
+
R
2
)
=
−
I
s
R
3
(
R
1
R
1
+
R
2
)
)
,
{\displaystyle V_{s}({\frac {R_{2}}{R_{1}+R_{2}}})=-{I_{s}}{R_{3}}({\frac {R_{1}}{R_{1}+R_{2}}})),}
V
s
I
s
=
−
R
3
R
1
R
2
{\displaystyle {\frac {V_{s}}{I_{s}}}=-{\frac {{R_{3}}{R_{1}}}{R_{2}}}}