RC circuit

File:Discharging capacitor.svg

The simplest RC circuit consists of a resistor with resistance {{mvar|R}} and a charged capacitor with capacitance {{mvar|C}} connected to one another in a single loop, without an external voltage source. The capacitor will discharge its stored energy through the resistor. If {{Mvar|V(t)}} is taken to be the voltage of the capacitor's top plate relative to its bottom plate in the figure, then the capacitor current–voltage relation says the current {{Mvar|I(t)}} exiting the capacitor's top plate will equal {{mvar|C}} multiplied by the negative time derivative of {{Mvar|V(t)}}. Kirchhoff's current law says this current is the same current entering the top side of the resistor, which per Ohm's law equals {{Mvar|V(t)/R}}. This yields a linear differential equation:

:$\backslash overbrace\{\backslash Biggl(C\backslash frac\{-\backslash mathrm\{d\}V(t)\}\{\backslash mathrm\{d\}t\}\backslash Biggr)\}^\backslash text\{capacitor\; current\}\; =\; \backslash overbrace\{\backslash Biggl(\backslash frac\{V(t)\}\{R\}\backslash Biggr)\}^\backslash text\{resistor\; current\}\; ,$

which can be rearranged according to the standard form for exponential decay:

:$\backslash frac\{\backslash mathrm\{d\}V(t)\}\{\backslash mathrm\{d\}t\}\; =\; -\backslash frac\{1\}\{RC\}V(t)\; \backslash ,\; .$

This means that the instantaneous rate of voltage decrease at any time is proportional to the voltage at that time. Solving for {{mvar|V(t)}} yields an exponential decay curve that asymptotically approaches 0:

:$V(t)=V\_0\; \backslash cdot\; e^\{-\backslash frac\{t\}\{RC\}\}\; \backslash ,,$

where {{math|V₀}} is the capacitor voltage at time {{math|t {{=}} 0}} and {{mvar|e}} is Euler's number.

The time required for the voltage to fall to {{math|{{sfrac|V₀|e}}}} is called the RC time constant and is given by:Horowitz & Hill, p. 1.13

:$\backslash tau\; =\; RC\; \backslash ,.$

When using the International System of Units, {{mvar|R}} is in ohms and {{mvar|C}} is in farads, so {{mvar|τ}} will be in seconds. At any time {{mvar|N·τ}}, the capacitor's charge or voltage will be {{Sfrac|1|{{Mvar|e}}^N}} of its starting value. So if the capacitor's charge or voltage is said to start at 100%, then 36.8% remains at {{Math|1·τ}}, 13.5% remains at {{Math|2·τ}}, 5% remains at {{Math|3·τ}}, 1.8% remains at {{Math|4·τ}}, and less than 0.7% remains at {{Math|5·τ}} and later.

The half-life ({{math|t{{sub|{{frac|1|2}}}}}}) is the time that it takes for its charge or voltage to be reduced in half:{{Cite web |last=Hanks |first=Ann |last2=Luttermoser |first2=Donald |title=General Physics II Lab (PHYS-2021) Experiment ELEC-5: RC Circuits |url=https://faculty.etsu.edu/lutter/courses/phys2021/ELEC_5_RC_Circuit.pdf}}

:$\backslash tfrac\{1\}\{2\}\; \{=\}\; e^\{-\backslash tfrac\{t\_\{1/2\}\}\{\backslash tau\}\}\; \backslash longrightarrow\; t\_\{1/2\}\; =\; \backslash ln(2)\; \backslash ,\; \backslash tau\; \backslash approx\; \backslash text\{.693\}\; \backslash ,\; \backslash tau\; \backslash ,\; .$

For example, 50% of charge or voltage remains at time {{math|1·t{{sub|{{frac|1|2}}}}}}, then 25% remains at time {{math|2·t{{sub|{{frac|1|2}}}}}}, then 12.5% remains at time {{math|3·t{{sub|{{frac|1|2}}}}}}, and {{Sfrac|1|2^N}} will remain at time {{math|N·t{{sub|{{frac|1|2}}}}}}.

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For instance, {{nowrap|{{calculator|id=resistance_number|type=number|min=1|size=3|default=1|max=999|step=.1}} {{calculator|id=resistance_unitlog10|type=select|class=cdx-select|mapping={ "teraohms":12, "gigaohms":9, "megaohms":6, "kiloohms":3, "ohms":0, "milliohms":-3 }|style=min-width:1ch|value=6}}}} of resistance with {{nowrap|{{calculator|id=capacitance_number|type=number|min=1|size=3|default=1|max=999|step=.1}} {{calculator|id=capacitance_unitlog10|type=select|class=cdx-select|mapping={ "farads":0, "millifarads":-3, "microfarads":-6, "nanofarads":-9, "picofarads":-12}|style=min-width:1ch|value=-6}}}} of capacitance produces a time constant of approximately {{nowrap|{{calculator

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}}.}} This {{Mvar|τ}} corresponds to a cutoff frequency of approximately {{nowrap|{{calculator

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}}.}} If the capacitor has an initial voltage {{Math|V₀}} of {{nowrap|{{calculator

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The RC circuit's behavior is well-suited to be analyzed in the Laplace domain, which the rest of this article requires a basic understanding of. The Laplace domain is a frequency domain representation using complex frequency {{mvar|s}}, which is (in general) a complex number:

:$s\; =\; \backslash sigma\; +\; j\; \backslash omega\; \backslash ,,$

where

• {{mvar|j}} represents the imaginary unit: {{math|j² {{=}} −1}},
• {{mvar|σ}} is the exponential decay constant, and
• {{mvar|ω}} is the sinusoidal angular frequency.

When evaluating circuit equations in the Laplace domain, time-dependent circuit elements of capacitance and inductance can be treated like resistors with complex-valued impedance instead of real resistance. While the complex impedance {{mvar|Z_R}} of a resistor is simply a real value equal to its resistance {{mvar|R}}, the complex impedance of a capacitor {{mvar|C}} is instead:

:$Z\_C\; =\; \backslash frac\{1\}\{C\; s\}\; .$

File:RC Series Filter (with V&I Labels).svg RC circuit]]

Kirchhoff's current law means that the current in the series circuit is necessarily the same through both elements. Ohm's law says this current is equal to the input voltage $V\_\backslash mathrm\{in\}$ divided by the sum of the complex impedance of the capacitor and resistor:

:$\backslash begin\{align\}$

I(s) &= \frac{V_\mathrm{in}(s) }{R + \frac{1}{Cs}} \\

&= \frac{ Cs }{ 1 + RCs } V_\mathrm{in}(s)\,.

\end{align}

By viewing the circuit as a voltage divider, the voltage across the capacitor is:

:$\backslash begin\{align\}$

V_C(s) &= \frac{\frac{1}{Cs}}{R + \frac{1}{Cs}}V_\mathrm{in}(s) \\

&= \frac{1}{1 + RCs}V_\mathrm{in}(s)

\end{align}

and the voltage across the resistor is:

:$\backslash begin\{align\}$

V_R(s) &= \frac{R}{R + \frac{1}{Cs}}V_\mathrm{in}(s) \\

&= \frac{RCs}{1 + RCs}V_\mathrm{in}(s)\,.

\end{align}

The transfer function from the input voltage to the voltage across the capacitor is

:$H\_C(s)\; =\; \backslash frac\{\; V\_C(s)\; \}\{\; V\_\backslash mathrm\{in\}(s)\; \}\; =\; \backslash frac\{\; 1\; \}\{\; 1\; +\; RCs\; \}\; \backslash ,.$

Similarly, the transfer function from the input to the voltage across the resistor is

:$H\_R(s)\; =\; \backslash frac\{\; V\_R(s)\; \}\{\; V\_\{\backslash rm\; in\}(s)\; \}\; =\; \backslash frac\{\; RCs\; \}\{\; 1\; +\; RCs\; \}\; \backslash ,.$

Both transfer functions have a single pole located at

:$s\; =\; -\backslash frac\{1\}\{RC\}\; \backslash ,.$

In addition, the transfer function for the voltage across the resistor has a zero located at the origin.

The sinusoidal steady state is a special case of complex frequency that considers the input to consist only of pure sinusoids. Hence, the exponential decay component represented by $\backslash sigma$ can be ignored in the complex frequency equation $s\; \{=\}\; \backslash sigma\; \{+\}\; j\; \backslash omega$ when only the steady state is of interest. The simple substitution of $s\; \backslash Rightarrow\; j\; \backslash omega$ into the previous transfer functions will thus provide the sinusoidal gain and phase response of the circuit.

Image:Passive_integrator_circuit_transfer_function.png

The magnitude of the gains across the two components are

:$G\_C\; =\; \backslash big|\; H\_C(j\; \backslash omega)\; \backslash big|\; =\; \backslash left|\backslash frac\{V\_C(j\; \backslash omega)\}\{V\_\backslash mathrm\{in\}(j\; \backslash omega)\}\backslash right|\; =\; \backslash frac\{1\}\{\backslash sqrt\{1\; +\; \backslash left(\backslash omega\; RC\backslash right)^2\}\}$

and

:$G\_R\; =\; \backslash big|\; H\_R(j\; \backslash omega)\; \backslash big|\; =\; \backslash left|\backslash frac\{V\_R(j\; \backslash omega)\}\{V\_\backslash mathrm\{in\}(j\; \backslash omega)\}\backslash right|\; =\; \backslash frac\{\backslash omega\; RC\}\{\backslash sqrt\{1\; +\; \backslash left(\backslash omega\; RC\backslash right)^2\}\}\backslash ,,$

As the frequency becomes very large ({{math|ω → ∞}}), the capacitor acts like a short circuit, so:

:$G\_C\; \backslash to\; 0\; \backslash quad\; \backslash mbox\{and\}\; \backslash quad\; G\_R\; \backslash to\; 1\; \backslash ,.$

As the frequency becomes very small ({{math|ω → 0}}), the capacitor acts like an open circuit, so:

:$G\_C\; \backslash to\; 1\; \backslash quad\; \backslash mbox\{and\}\; \backslash quad\; G\_R\; \backslash to\; 0\; \backslash ,.$

The behavior at these extreme frequencies show that if the output is taken across the capacitor, high frequencies are attenuated and low frequencies are passed, so such a circuit configuration is a low-pass filter. However, if the output is taken across the resistor, then high frequencies are passed and low frequencies are attenuated, so such a configuration is a high-pass filter.

The range of frequencies that the filter passes is called its bandwidth. The frequency at which the filter attenuates the signal to half its unfiltered power is termed its cutoff frequency. This requires that the gain of the circuit be reduced to

:$G\_C\; =\; G\_R\; =\; \backslash frac\{1\}\{\backslash sqrt\; 2\}$.

Solving the above equation yields

:$\backslash omega\_\backslash mathrm\{c\}\; =\; \backslash frac\{1\}\{RC\}\; \backslash quad\; \backslash mbox\{or\}\; \backslash quad\; f\_\backslash mathrm\{c\}\; =\; \backslash frac\{1\}\{2\backslash pi\; RC\}$

which is the frequency that the filter will attenuate to half its original power.

The phase angles are

:$\backslash phi\_C\; =\; \backslash angle\; H\_C(j\; \backslash omega)\; =\; \backslash tan^\{-1\}\backslash left(-\backslash omega\; RC\backslash right)$

and

:$\backslash phi\_R\; =\; \backslash angle\; H\_R(j\; \backslash omega)\; =\; \backslash tan^\{-1\}\backslash left(\backslash frac\{1\}\{\backslash omega\; RC\}\backslash right)\backslash ,.$

As {{math|ω → 0}}:

:$\backslash phi\_C\; \backslash to\; 0\; \backslash quad\; \backslash mbox\{and\}\; \backslash quad\; \backslash phi\_R\; \backslash to\; 90^\{\backslash circ\}\; =\; \backslash frac\{\backslash pi\}\{2\}\backslash mbox\{\; radians\}\backslash ,.$

As {{math|ω → ∞}}:

:$\backslash phi\_C\; \backslash to\; -90^\{\backslash circ\}\; =\; -\backslash frac\{\backslash pi\}\{2\}\backslash mbox\{\; radians\}\; \backslash quad\; \backslash mbox\{and\}\; \backslash quad\; \backslash phi\_R\; \backslash to\; 0\backslash ,.$

While the output signal's phase shift relative to the input depends on frequency, this is generally less interesting than the gain variations. At DC (0 Hz), the capacitor voltage is in phase with the input signal voltage while the resistor voltage leads it by 90°. As frequency increases, the capacitor voltage comes to have a 90° lag relative to the input signal and the resistor voltage comes to be in-phase with the input signal.

The gain and phase expressions together may be combined into these phasor expressions representing the output:

:

Image:Passive_integrator_circuit_impulse_response_1.png

The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse or Dirac delta function.

The impulse response for the capacitor voltage is

:$h\_C(t)\; =\; \backslash frac\{1\}\{RC\}\; e^\{-\backslash frac\{t\}\{RC\}\}\; u(t)\; =\; \backslash frac\{1\}\{\backslash tau\}\; e^\{-\backslash frac\{t\}\{\backslash tau\}\}\; u(t)\backslash ,,$

where {{math|u(t)}} is the Heaviside step function and {{math|τ {{=}} RC}} is the time constant.

Similarly, the impulse response for the resistor voltage is

:$h\_R(t)\; =\; \backslash delta\; (t)\; -\; \backslash frac\{1\}\{RC\}\; e^\{-\backslash frac\{t\}\{RC\}\}\; u(t)\; =\; \backslash delta\; (t)\; -\; \backslash frac\{1\}\{\backslash tau\}\; e^\{-\backslash frac\{t\}\{\backslash tau\}\}\; u(t)\backslash ,,$

where {{math|δ(t)}} is the Dirac delta function

:This section relies on knowledge of the Laplace transform.

The most straightforward way to derive the time domain behaviour is to use the Laplace transforms of the expressions for {{mvar|V_C}} and {{mvar|V_R}} given above. Assuming a step input (i.e. {{math|V_in {{=}} 0}} before {{math|t {{=}} 0}} and then {{math|V_in {{=}} V₁}} afterwards):

:$\backslash begin\{align\}$

V_\mathrm{in}(s) &= V_1\cdot\frac{1}{s} \\

V_C(s) &= V_1\cdot\frac{1}{1 + sRC}\cdot\frac{1}{s} \\

V_R(s) &= V_1\cdot\frac{sRC}{1 + sRC}\cdot\frac{1}{s} \,.

\end{align}

Image:Series RC capacitor voltage.svg

Image:Series RC resistor voltage.svg

Partial fractions expansions and the inverse Laplace transform yield:

:

These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is charging; for discharging, the equations are vice versa. These equations can be rewritten in terms of charge and current using the relationships {{math|C {{=}} {{sfrac|Q|V}}}} and {{math|V {{=}} IR}} (see Ohm's law).

Thus, the voltage across the capacitor tends towards {{mvar|V₁}} as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged.

The product {{math|RC}} is both the time for {{mvar|V_C}} and {{mvar|V_R}} to reach within {{math|{{sfrac|1|e}}}} of their final value. In other words, {{math|RC}} is the time it takes for the voltage across the capacitor to rise to {{math|V₁·(1 − {{sfrac|1|e}})}} or for the voltage across the resistor to fall to {{math|V₁·({{sfrac|1|e}})}}. This RC time constant is labeled using the letter tau ({{math|τ}}).

The rate of change is a fractional {{math|1 − {{sfrac|1|e}}}} per {{mvar|τ}}. Thus, in going from {{math|t {{=}} Nτ}} to {{math|t {{=}} (N + 1)τ}}, the voltage will have moved about 63.2% of the way from its level at {{math|t {{=}} Nτ}} toward its final value. So the capacitor will be charged to about 63.2% after {{mvar|τ}}, and is often considered fully charged (>99.3%) after about {{math|5τ}}. When the voltage source is replaced with a short circuit, with the capacitor fully charged, the voltage across the capacitor drops exponentially with {{mvar|t}} from {{mvar|V}} towards 0. The capacitor will be discharged to about 36.8% after {{mvar|τ}}, and is often considered fully discharged (<0.7%) after about {{math|5τ}}. Note that the current, {{mvar|I}}, in the circuit behaves as the voltage across the resistor does, via Ohm's Law.

These results may also be derived by solving the differential equations describing the circuit:

:

The first equation is solved by using an integrating factor and the second follows easily; the solutions are exactly the same as those obtained via Laplace transforms.

Consider the output across the capacitor at high frequency, i.e.

:$\backslash omega\; \backslash gg\; \backslash frac\{1\}\{RC\}\backslash ,.$

This means that the capacitor has insufficient time to charge up and so its voltage is very small. Thus the input voltage approximately equals the voltage across the resistor. To see this, consider the expression for $I$ given above:

:$I\; =\; \backslash frac\{V\_\backslash mathrm\{in\}\}\{R+\backslash frac\{1\}\{j\backslash omega\; C\}\}\backslash ,,$

but note that the frequency condition described means that

:$\backslash omega\; C\; \backslash gg\; \backslash frac\{1\}\{R\}\backslash ,,$

so

:$I\; \backslash approx\; \backslash frac\{V\_\backslash mathrm\{in\}\}\{R\}$

which is just Ohm's Law.

Now,

:$V\_C\; =\; \backslash frac\{1\}\{C\}\backslash int\_\{0\}^\{t\}I\backslash ,dt\backslash ,,$

so

:$V\_C\; \backslash approx\; \backslash frac\{1\}\{RC\}\backslash int\_\{0\}^\{t\}V\_\backslash mathrm\{in\}\backslash ,dt\backslash ,.$

Therefore, the voltage across the capacitor acts approximately like an integrator of the input voltage for high frequencies.

Consider the output across the resistor at low frequency i.e.,

:$\backslash omega\; \backslash ll\; \backslash frac\{1\}\{RC\}\backslash ,.$

This means that the capacitor has time to charge up until its voltage is almost equal to the source's voltage. Considering the expression for {{mvar|I}} again, when

:$R\; \backslash ll\; \backslash frac\{1\}\{\backslash omega\; C\}\backslash ,,$

so

:

Now,

:

Therefore, the voltage across the resistor acts approximately like a differentiator of the input voltage for low frequencies.

Integration and differentiation can also be achieved by placing resistors and capacitors as appropriate on the input and feedback loop of operational amplifiers (see operational amplifier integrator and operational amplifier differentiator).

File:RC Parallel Filter (with I Labels).svg RC circuit]]

The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltage {{math|V_out}} is equal to the input voltage {{math|V_in}} — as a result, this circuit acts as a filter on a current input instead of a voltage input.

With complex impedances:

:

This shows that the capacitor current is 90° out of phase with the resistor (and source) current. Alternatively, the governing differential equations may be used:

:

When fed by a current source, the transfer function of a parallel RC circuit is:

:$\backslash frac\{V\_\backslash mathrm\{out\}\}\{I\_\backslash mathrm\{in\}\}\; =\; \backslash frac\{R\}\{1+sRC\}\backslash ,.$

It is sometimes required to synthesise an RC circuit from a given rational function in s. For synthesis to be possible in passive elements, the function must be a positive-real function. To synthesise as an RC circuit, all the critical frequencies (poles and zeroes) must be on the negative real axis and alternate between poles and zeroes with an equal number of each. Further, the critical frequency nearest the origin must be a pole, assuming the rational function represents an impedance rather than an admittance.

The synthesis can be achieved with a modification of the Foster synthesis or Cauer synthesis used to synthesise LC circuits. In the case of Cauer synthesis, a ladder network of resistors and capacitors will result.Bakshi & Bakshi, pp. 3-30–3-37

• RC time constant
• RL circuit
• LC circuit
• RLC circuit
• Electrical network
• List of electronics topics
• {{Slink|Johnson–Nyquist noise|Thermal noise on capacitors}} – gives the derivation of kTC noise caused by a resistor with a capacitor to be:

:: $$

V_\text{rms} = \sqrt{ k_\text{B} T \over C }.

• Step response

{{reflist}}

• Bakshi, U.A.; Bakshi, A.V., Circuit Analysis - II, Technical Publications, 2009 {{ISBN|9788184315974}}.
• Horowitz, Paul; Hill, Winfield, The Art of Electronics (3rd edition), Cambridge University Press, 2015 {{ISBN|0521809266}}.

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Category:Analog circuits

Category:Electronic filter topology