Behaviour of a capacitor+resistor (frequency domain)

Question

why a capacitor behaves as an open circuit when \$\omega<\omega_p\$ and as a short circuit when \$\omega>\omega_p\$ (where \$\omega_p=1/RC\$ is a pole)? My question is generic.

Thank you for your time.

Answer 1

why a capacitor behaves as an open circuit when \$\omega<\omega_p\$ and as a short circuit when \$\omega>\omega_p\$ (where \$\omega_p=1/RC\$ is a pole)?

It doesn't.

Q=CV and differentiating to obtain current we get I = C dv/dt. Basically the current that flows is proportional to the rate of change of voltage. It's as simple as that.

Answer 2

To understand the complex impedance of the capacitor, we have to think about what happens in the capacitor if we apply a sinusoidal voltage \$ u_c = U * sin(\omega t) \$. The current will be, by definition:

$$ i_c = C * \frac{du_c}{dt} $$

now we can derive from \$u_c\$:

$$ \frac{du_c}{dt} = U * \omega * cos(\omega t) $$

The impedance is the quotient of voltage and current, so if we put in everything we have:

$$ Z = \frac{u_c}{i_c} = \frac{U * sin(\omega t)}{C * U * \omega * cos(\omega t)} = \frac{1}{\omega C} * \frac{sin(\omega t)}{cos(\omega t)} $$

We can interpret this as the impedance consisting of a 90° phase shift (cos to sin) and an amplitude limiting factor \$\frac{1}{\omega C} \$. This will often be written in exponential form as \$ Z_C = \frac{1}{j \omega C} \$, where \$ \frac{1}{j}\$ is the 90° phase shift.

From the simple formula for Z, we can observe the extreme values at \$ \omega \to 0 \$ and \$ \omega \to +\infty \$

$$ \lim_{\omega \to +\infty} \frac{1}{\omega C} = 0 $$

As you can see, the impedance at an infinite \$\omega\$ is 0, which is a short circuit. It also works the other around if we set \$\omega\$ to 0:

$$ \lim_{\omega \to 0} \frac{1}{\omega C} = +\infty$$ .

An infinite impedance means an open circuit. We can remember this by thinking of how capacitors are often used to block DC (\$\omega = 0\$) in circuits, while passing higher frequencies without much attenuation.

Everything in between 0 and \$\infty\$ will present a capacitive load.

Answer 3

^{simulate this circuit – Schematic created using CircuitLab}

$$ V_o = V_i(\frac{Z_C}{Z_C + R}) \Rightarrow H(\omega) = \frac{1/j\omega C}{R+(1/j\omega C)}\\ \Rightarrow H(\omega)=\frac{1}{1+j\omega RC} $$ Thus \$ \omega_p = 1/RC \$.

Also \$ \Rightarrow |H(\omega)| = \frac{1}{\sqrt{1+(\omega RC)^2}} \$

As \$\omega >> \omega_p \$ (not \$ \omega>\omega _p\$), \$(1+(\omega RC)^2) \rightarrow \infty \$ thus \$ |H(\omega)| \rightarrow 0\$ ........ (1)

Likewise, as \$\omega << \omega_p \$ (not \$ \omega<\omega _p\$), \$\omega RC \rightarrow 0\$ thus \$ |H(\omega)| \rightarrow 1\$ ........ (2)

From (1), if output (\$V_o\$) goes to 0 then \$|X_C|\$ should be 0 (i.e. short circuit) because R does not change with frequency.

From (2), if output (\$V_o\$) follows input (\$V_i\$) then \$|X_C|\$ should be infinity (i.e. open) because zero current flows due to total "infinite" resistance.

P.S. Sorry for bad English.