Time-domain considerations RC circuit
capacitor voltage step-response.
resistor voltage step-response.
partial fractions expansions , inverse laplace transform yield:
v
c
(
t
)
=
v
(
1
−
e
−
t
r
c
)
v
r
(
t
)
=
v
e
−
t
r
c
.
{\displaystyle {\begin{aligned}v_{c}(t)&=v\left(1-e^{-{\frac {t}{rc}}}\right)\\v_{r}(t)&=ve^{-{\frac {t}{rc}}}\,.\end{aligned}}}
these equations calculating voltage across capacitor , resistor respectively while capacitor charging; discharging, equations vice versa. these equations can rewritten in terms of charge , current using relationships c = q/v , v = ir (see ohm s law).
thus, voltage across capacitor tends towards v time passes, while voltage across resistor tends towards 0, shown in figures. in keeping intuitive point capacitor charging supply voltage time passes, , charged.
these equations show series rc circuit has time constant, denoted τ = rc being time takes voltage across component either rise (across capacitor) or fall (across resistor) within 1/e of final value. is, τ time takes vc reach v(1 − 1/e) , vr reach v(1/e).
the rate of change fractional 1 − 1/e per τ. thus, in going t = nτ t = (n + 1)τ, voltage have moved 63.2% of way level @ t = nτ toward final value. capacitor charged 63.2% after τ, , charged (99.3%) after 5τ. when voltage source replaced short-circuit, capacitor charged, voltage across capacitor drops exponentially t v towards 0. capacitor discharged 36.8% after τ, , discharged (0.7%) after 5τ. note current, i, in circuit behaves voltage across resistor does, via ohm s law.
these results may derived solving differential equations describing circuit:
v
i
n
−
v
c
r
=
c
d
v
c
d
t
v
r
=
v
i
n
−
v
c
.
{\displaystyle {\begin{aligned}{\frac {v_{\mathrm {in} }-v_{c}}{r}}&=c{\frac {dv_{c}}{dt}}\\v_{r}&=v_{\mathrm {in} }-v_{c}\,.\end{aligned}}}
the first equation solved using integrating factor , second follows easily; solutions same obtained via laplace transforms.
integrator
consider output across capacitor @ high frequency, i.e.
ω
≫
1
r
c
.
{\displaystyle \omega \gg {\frac {1}{rc}}\,.}
this means capacitor has insufficient time charge , voltage small. input voltage approximately equals voltage across resistor. see this, consider expression
i
{\displaystyle i}
given above:
i
=
v
i
n
r
+
1
j
ω
c
,
{\displaystyle i={\frac {v_{\mathrm {in} }}{r+{\frac {1}{j\omega c}}}}\,,}
but note frequency condition described means that
ω
c
≫
1
r
,
{\displaystyle \omega c\gg {\frac {1}{r}}\,,}
so
i
≈
v
i
n
r
{\displaystyle i\approx {\frac {v_{\mathrm {in} }}{r}}}
which ohm s law.
now,
v
c
=
1
c
∫
0
t
i
d
t
,
{\displaystyle v_{c}={\frac {1}{c}}\int _{0}^{t}i\,dt\,,}
so
v
c
≈
1
r
c
∫
0
t
v
i
n
d
t
,
{\displaystyle v_{c}\approx {\frac {1}{rc}}\int _{0}^{t}v_{\mathrm {in} }\,dt\,,}
which integrator across capacitor.
differentiator
consider output across resistor @ low frequency i.e.,
ω
≪
1
r
c
.
{\displaystyle \omega \ll {\frac {1}{rc}}\,.}
this means capacitor has time charge until voltage equal source s voltage. considering expression again, when
r
≪
1
ω
c
,
{\displaystyle r\ll {\frac {1}{\omega c}}\,,}
so
i
≈
v
i
n
1
j
ω
c
v
i
n
≈
i
j
ω
c
=
v
c
.
{\displaystyle {\begin{aligned}i&\approx {\frac {v_{\mathrm {in} }}{\frac {1}{j\omega c}}}\\v_{\mathrm {in} }&\approx {\frac {i}{j\omega c}}=v_{c}\,.\end{aligned}}}
now,
v
r
=
i
r
=
c
d
v
c
d
t
r
v
r
≈
r
c
d
v
i
n
d
t
,
{\displaystyle {\begin{aligned}v_{r}&=ir=c{\frac {dv_{c}}{dt}}r\\v_{r}&\approx rc{\frac {dv_{in}}{dt}}\,,\end{aligned}}}
which differentiator across resistor.
more accurate integration , differentiation can achieved placing resistors , capacitors appropriate on input , feedback loop of operational amplifiers (see operational amplifier integrator , operational amplifier differentiator).
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