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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
electrical_engineering_1:dc_circuit_transients [2023/03/31 14:48] mexleadmin |
electrical_engineering_1:dc_circuit_transients [2023/12/03 16:53] (aktuell) mexleadmin [Exercises] |
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- | ====== 5. DC Circuit Transients (on RC elements) ====== | + | ====== 5 DC Circuit Transients (on RC elements) ====== |
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* **open switch**: If there is a voltage between the two metal parts, charges can also accumulate there. \\ Since the distances are usually large and the air is used as the dielectric, the capacitance of the capacitor formed in this way is very small. | * **open switch**: If there is a voltage between the two metal parts, charges can also accumulate there. \\ Since the distances are usually large and the air is used as the dielectric, the capacitance of the capacitor formed in this way is very small. | ||
- | * **Overhead line**: An overhead line also represents a capacitor against the ground potential of the earth. The charging and discharging by the alternating current leads to the fact that polarizable molecules can align themselves. For example, the water drops near the line are rolled through the field and hum with $100~Hz$ and many times that (harmonics). Peak discharge results in a high-frequency crackle. | + | * **Overhead line**: An overhead line also represents a capacitor against the ground potential of the earth. The charging and discharging by the alternating current leads to the fact that polarizable molecules can align themselves. For example, the water drops near the line are rolled through the field and hum with $100~\rm Hz$ and many times that (harmonics). Peak discharge results in a high-frequency crackle. |
* **Conductor trace**: A trace on a PCB can also be a capacitor against a nearby ground plane. This can be a problem for digital signals (see the charge and discharge curves below). | * **Conductor trace**: A trace on a PCB can also be a capacitor against a nearby ground plane. This can be a problem for digital signals (see the charge and discharge curves below). | ||
* **Human body**: The human body can likewise pick up charge. The charge thus absorbed forms a capacitor with respect to other objects. This can be charged up to some $kV$. This is a particular problem in electrical laboratories, | * **Human body**: The human body can likewise pick up charge. The charge thus absorbed forms a capacitor with respect to other objects. This can be charged up to some $kV$. This is a particular problem in electrical laboratories, | ||
* **Membrane of nerve cells**: Nerve cells also result in a capacitor due to the lipid bilayer (membrane of the nerve cell) and the two cellular fluids with different electrolytes (ions). The nerve cells are surrounded by a thick layer (myelin layer) for faster transmission. This lowers the capacitance and thus increases the successive charging of successive parts of the nerve cell. In diseases such as Creutzfeldt-Jakob or multiple sclerosis, this layer thins out. This leads to the delayed signal transmission which characterizes the disease patterns. | * **Membrane of nerve cells**: Nerve cells also result in a capacitor due to the lipid bilayer (membrane of the nerve cell) and the two cellular fluids with different electrolytes (ions). The nerve cells are surrounded by a thick layer (myelin layer) for faster transmission. This lowers the capacitance and thus increases the successive charging of successive parts of the nerve cell. In diseases such as Creutzfeldt-Jakob or multiple sclerosis, this layer thins out. This leads to the delayed signal transmission which characterizes the disease patterns. | ||
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In the following, the charging process of a capacitor is to be considered in more detail. For this purpose, one has to realize, that during the charging of the capacitor, besides the voltage source $U_{\rm s}$ and the capacitor $C$, there is always a resistance $R$ in the circuit. This is composed of the internal resistance of the (non-ideal) voltage source, the internal resistance of the capacitor, and the parasitic (=interfering) resistance of the line. In practical applications, | In the following, the charging process of a capacitor is to be considered in more detail. For this purpose, one has to realize, that during the charging of the capacitor, besides the voltage source $U_{\rm s}$ and the capacitor $C$, there is always a resistance $R$ in the circuit. This is composed of the internal resistance of the (non-ideal) voltage source, the internal resistance of the capacitor, and the parasitic (=interfering) resistance of the line. In practical applications, | ||
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Exercises: | Exercises: | ||
- | - Become familiar with how the capacitor current $i_C$ and capacitor voltage $u_C$ depend on the given capacitance $C$ and resistance $R$. \\ To do this, use for $R=\{ 10~\Omega, 100~\Omega, 1~k\Omega\}$ and $C=\{ 1~\mu F, 10 ~\mu F\}$. How fast does the capacitor voltage $u_C$ increase in each case n? | + | - Become familiar with how the capacitor current $i_C$ and capacitor voltage $u_C$ depend on the given capacitance $C$ and resistance $R$. \\ To do this, use for $R=\{ 10~\Omega, 100~\Omega, 1~k\Omega\}$ and $C=\{ 1~\rm µF, 10 ~ µF\}$. How fast does the capacitor voltage $u_C$ increase in each case n? |
- Which quantity ($i_C$ or $u_C$) is continuous here? Why must this one be continuous? Why must the other quantity be discontinuous? | - Which quantity ($i_C$ or $u_C$) is continuous here? Why must this one be continuous? Why must the other quantity be discontinuous? | ||
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* Ideally, the capacitor is not fully charged to the specified voltage $U_{\rm s}$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_{\rm s}$ | * Ideally, the capacitor is not fully charged to the specified voltage $U_{\rm s}$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_{\rm s}$ | ||
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The process is now to be summarized in detail in formulas. Linear components are used in the circuit, i.e. the component values for the resistor $R$ and the capacitance $C$ are independent of the current or the voltage. Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities: | The process is now to be summarized in detail in formulas. Linear components are used in the circuit, i.e. the component values for the resistor $R$ and the capacitance $C$ are independent of the current or the voltage. Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities: | ||
\begin{align*} | \begin{align*} | ||
- | R = {{u_R(t)}\over{i_R(t)}} = {{{\rm d}u_R}\over{{\rm d}i_R}} = const. \\ | + | R = {{u_R(t)}\over{i_R(t)}} = {{{\rm d}u_R}\over{{\rm d}i_R}} = {\rm const.} \\ |
- | C = {{q(t)} | + | C = {{q(t)} |
\end{align*} | \end{align*} | ||
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Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https:// | Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https:// | ||
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\end{align*} | \end{align*} | ||
- | Thus, for a fully discharged capacitor ($U_{\rm s}=0~{\rm V}$), the energy stored when charging to voltage $U_{\rm s}$ is $\delta W_C={{1}\over{2}} C \cdot U_{\rm s}^2$. | + | Thus, for a fully discharged capacitor ($U_{\rm s}=0~{\rm V}$), the energy stored when charging to voltage $U_{\rm s}$ is $\Delta W_C={{1}\over{2}} C \cdot U_{\rm s}^2$. |
=== Energy Consideration on the Resistor === | === Energy Consideration on the Resistor === | ||
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- | {{drawio> | + | {{drawio> |
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- | <panel type=" | + | # |
- | {{youtube> | + | The following circuit shows a charging/ |
- | </ | + | The values of the components shall be the following: |
+ | * $R_1 = 1.0 \rm k\Omega$ | ||
+ | * $R_2 = 2.0 \rm k\Omega$ | ||
+ | * $R_3 = 3.0 \rm k\Omega$ | ||
+ | * $C = 1 \rm \mu F$ | ||
+ | * $S_1$ and $S_2$ are opened in the beginning (open-circuit) | ||
- | <panel type=" | + | {{drawio>electrical_engineering_1: |
- | {{youtube> | + | 1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\ |
- | </ | + | 1.1 What is the value of the time constant $\tau_1$? |
- | <panel type=" | + | # |
- | {{youtube> | + | The time constant $\tau$ is generally given as: $\tau= R\cdot C$. \\ |
+ | Now, we try to determine which $R$ and $C$ must be used here. \\ | ||
+ | To find this out, we have to look at the circuit when $S_1$ gets closed. | ||
- | </WRAP></WRAP></panel> | + | {{drawio> |
+ | |||
+ | We see that for the time constant, we need to use $R=R_1 + R_2$. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | \tau_1 &= R\cdot C \\ | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | 1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values! | ||
+ | |||
+ | # | ||
+ | |||
+ | To get a general formula, we again take a look at the circuit, but this time with the voltage arrows. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | We see, that: $U_1 = u_C + u_{R2}$ and there is only one current in the loop: $i = i_C = i_{R2}$\\ | ||
+ | The current is generally given with the exponential function: $i_c = {{U}\over{R}}\cdot e^{-t/\tau}$, with $R$ given here as $R = R_1 + R_2$. | ||
+ | Therefore, $u_{R2}$ can be written as: | ||
+ | |||
+ | \begin{align*} | ||
+ | u_{R2} &= R_2 \cdot i_{R2} \\ | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
+ | 2. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$. | ||
+ | At this point, the switch $S_1$ will be opened. \\ Calculate $t_1$! | ||
+ | |||
+ | # | ||
+ | |||
+ | We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/ | ||
+ | Therefore, we get $t_1$ by: | ||
+ | |||
+ | \begin{align*} | ||
+ | u_C = 4/5 \cdot U_1 & | ||
+ | 4/5 & | ||
+ | e^{-t/ | ||
+ | | ||
+ | t &= -\tau \cdot \rm ln (1/5) \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | t & | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
+ | 3. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$. The values of the voltage sources are now: $U_1 = 5.0 ~\rm V$ and $U_2 = 10 ~\rm V$. | ||
+ | |||
+ | 3.1 What is the new time constant $\tau_2$? | ||
+ | |||
+ | # | ||
+ | |||
+ | Again, the time constant $\tau$ is given as: $\tau= R\cdot C$. \\ | ||
+ | Again, we try to determine which $R$ and $C$ must be used here. \\ | ||
+ | To find this out, we have to look at the circuit when $S_1$ is open and $S_2$ is closed. | ||
+ | |||
+ | {{drawio>electrical_engineering_1: | ||
+ | |||
+ | We see that for the time constant, we now need to use $R=R_3 + R_2$. | ||
+ | |||
+ | \begin{align*} | ||
+ | \tau_2 &= R\cdot C \\ | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | \tau_2 &= 5~\rm ms \\ | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
+ | 3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. | ||
+ | |||
+ | # | ||
+ | |||
+ | To calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$, we first have to find out the value of $u_{R2}(t_2 = 10 ~\rm ms)$, when $S_2$ just got closed. \\ | ||
+ | * Starting from $t_2 = 10 ~\rm ms$, the voltage source $U_2$ charges up the capacitor $C$ further. | ||
+ | * Before at $t_1$, when $S_1$ got opened, the value of $u_c$ was: $u_c(t_1) = 4/5 \cdot U_1 = 4 ~\rm V$. | ||
+ | * This is also true for $t_2$, since between $t_1$ and $t_2$ the charge on $C$ does not change: $u_c(t_2) = 4 ~\rm V$. | ||
+ | * In the first moment after closing $S_2$ at $t_2$, the voltage drop on $R_3 + R_2$ is: $U_{R3+R2} = U_2 - u_c(t_2) = 6 ~\rm V$. | ||
+ | * So the voltage divider of $R_3 + R_2$ lead to $ \boldsymbol{u_{R2}(t_2 = 10 ~\rm ms)} = {{R_2}\over{R_3 + R2}} \cdot U_{R3+R2} = {{2 {~\rm k\Omega}}\over{3 {~\rm k\Omega} + 2 {~\rm k\Omega} }} \cdot 6 ~\rm V = \boldsymbol{2.4 ~\rm V} $ | ||
+ | |||
+ | We see that the voltage on $R_2$ has to decrease from $2.4 ~\rm V $ to $1/10 \cdot U_2 = 1 ~\rm V$. \\ | ||
+ | To calculate this, there are multiple ways. In the following, one shall be retraced: | ||
+ | * We know, that the current $i_C = i_{R2}$ subsides exponentially: | ||
+ | * So we can rearrange the task to focus on the change in current instead of the voltage. | ||
+ | * The exponential decay is true regardless of where it starts. | ||
+ | |||
+ | So from ${{i_{R2}(t)}\over{I_{R2~ 0}}} = {\rm e}^{-t/ | ||
+ | \begin{align*} | ||
+ | {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} & | ||
+ | -{{t_3 - t_2}\over{\tau_2}} | ||
+ | | ||
+ | | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | t_3 &= 14.4~\rm ms \\ | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
+ | 3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | |||
+ | # | ||
+ | {{drawio> | ||
+ | # | ||
+ | |||
+ | # | ||
+ | |||
+ | {{page> | ||
+ | |||
+ | # | ||
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- What are the energies and the total energy? \\ How is this understandable with the previous total energy? | - What are the energies and the total energy? \\ How is this understandable with the previous total energy? | ||
- | </ | + | # |
- | {{page> | ||