DW EditShow pageOld revisionsBacklinksAdd to bookExport to PDFFold/unfold allBack to top This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong. #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Circuit with multiple diodes: which lamps light up? #@TaskText_HTML@# The following simulation includes multiple diodes and several lamps. A lamp lights brightly when a voltage of approximately \[ \begin{align*} U_{\rm lamp}\geq 5~{\rm V} \end{align*} \] drops across it. Close the switch in the simulation. * Which lamps light up brightly? * Which lamps remain dark? * Explain the result using the idea of diode bypass paths. {{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxG3KQBZsQEBTAWjDACgBzEa4wkTFN14U0UKGwAmQvgPAYZGQYIkMAZgEMArgBsALmwBK4MILDFBeSOHNir1K7ltRoCSf3vuHhPJ-4gVGjr6AO4U3r7Y4WaCkGyhKB4JVknWMWxgeBC0pjY8fNFiuFYQSDBwEGWQ7KFg8vyKcvk2saG41KkUkO0mPi2NHbX5KL1ubeDD-T1+AVp6o1ET2eM+ymqzIdzYOYJLU7EAzsbbk83gIBra+wzpmWE+BbunRWelsFXO5TcQKQVjBQ5wF4fd6VdgZCB-GyRe5PQElYEVN5g26DDo-WHFegIhFxaT1HbCf646EdEl7NhAA noborder}} 1. Determine which lamps light up brightly. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12001~@# <WRAP leftalign> Number the lamps from left to right: \[ \begin{align*} L_1,\;L_2,\;L_3,\;L_4,\;L_5. \end{align*} \] After closing the switch, check the voltage across each lamp in the simulation. A lamp is assumed to light brightly if \[ \begin{align*} U_{\rm lamp}\geq 5~{\rm V}. \end{align*} \] The simulation shows that the outer lamps have a sufficiently large voltage across them, while the inner lamps are bypassed by conducting diodes. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12001~@# The lamps \[ \begin{align*} L_1 \quad \text{and} \quad L_5 \end{align*} \] light up brightly. #@ResultEnd_HTML@# </WRAP> </WRAP> 2. Determine which lamps remain dark. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12002~@# <WRAP leftalign> The inner lamps are connected in parts of the circuit that are bypassed by forward-biased diodes. A forward-biased diode has only a small voltage drop. Therefore, a lamp in parallel with such a diode path receives only a small voltage. If \[ \begin{align*} U_{\rm lamp}<5~{\rm V}, \end{align*} \] the lamp does not light brightly. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12002~@# The lamps \[ \begin{align*} L_2,\;L_3,\;L_4 \end{align*} \] remain dark or almost dark. #@ResultEnd_HTML@# </WRAP> </WRAP> #@TaskEnd_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Circuit with multiple diodes II: current calculation #@TaskText_HTML@# The following simulation includes two diodes and two resistors. Assume a simple constant-voltage diode model: \[ \begin{align*} U_{\rm F}=0.6~{\rm V}. \end{align*} \] The source voltage is \[ \begin{align*} U_0=4.0~{\rm V}. \end{align*} \] The resistors are \[ \begin{align*} R_1=200~\Omega, \qquad R_2=100~\Omega. \end{align*} \] Calculate the currents through * \(D_1\), * \(R_1\), * \(R_2\). {{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZykLsQEBTAWjDACgA3EFFC7iyN17gUeKOIGVxgmAjYAnENjQixywbxnc4CpXj5hRevpvFgdAd2P9B6m1DZW7pnicmQ2AD24IU4c9wE-nRuIAAi7N7Y2EistoSxYCH84SheSmTgGH4UmFk0KQBKaVG4WXTYhIJgGIRSwoWRQmI1dCgILbX1fACqHgAmQgZGdoZifv0MAGYAhgCuADYALmyDoyP6qtwgk7OLK0A noborder}} 1. Calculate the current through \(R_1\). <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12004~@# <WRAP leftalign> The current through \(R_1\) passes through one forward-biased diode. Therefore the voltage across \(R_1\) is \[ \begin{align*} U_{R1} = U_0-U_{\rm F}. \end{align*} \] Insert the values: \[ \begin{align*} U_{R1} &= 4.0~{\rm V}-0.6~{\rm V} \\ &= 3.4~{\rm V}. \end{align*} \] Now apply Ohm's law: \[ \begin{align*} I_{R1} = \frac{U_{R1}}{R_1} = \frac{3.4~{\rm V}}{200~\Omega} = 17~{\rm mA}. \end{align*} \] </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12004~@# \[ \begin{align*} I_{R1}=17~{\rm mA} \end{align*} \] #@ResultEnd_HTML@# </WRAP> </WRAP> 2. Calculate the current through \(R_2\). <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12005~@# <WRAP leftalign> The current through \(R_2\) passes through two forward-biased diodes. Therefore the voltage across \(R_2\) is \[ \begin{align*} U_{R2} = U_0-2U_{\rm F}. \end{align*} \] Insert the values: \[ \begin{align*} U_{R2} &= 4.0~{\rm V}-2\cdot 0.6~{\rm V} \\ &= 2.8~{\rm V}. \end{align*} \] Now apply Ohm's law: \[ \begin{align*} I_{R2} = \frac{U_{R2}}{R_2} = \frac{2.8~{\rm V}}{100~\Omega} = 28~{\rm mA}. \end{align*} \] </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12005~@# \[ \begin{align*} I_{R2}=28~{\rm mA} \end{align*} \] #@ResultEnd_HTML@# </WRAP> </WRAP> 3. Calculate the current through \(D_1\). <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12006~@# <WRAP leftalign> The diode \(D_1\) supplies both current paths. Therefore, by Kirchhoff's current law, \[ \begin{align*} I_{D1} = I_{R1}+I_{R2}. \end{align*} \] Insert the values: \[ \begin{align*} I_{D1} &= 17~{\rm mA}+28~{\rm mA} \\ &= 45~{\rm mA}. \end{align*} \] </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12006~@# \[ \begin{align*} I_{D1}=45~{\rm mA} \end{align*} \] #@ResultEnd_HTML@# </WRAP> </WRAP> #@TaskEnd_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Circuit with multiple diodes III: switch-dependent currents #@TaskText_HTML@# The following simulation includes two diodes and a switch. Assume a simple constant-voltage diode model: \[ \begin{align*} U_{\rm F}=0.7~{\rm V}. \end{align*} \] The source voltage is \[ \begin{align*} U_0=5.0~{\rm V}. \end{align*} \] The resistor is \[ \begin{align*} R_1=1.0~{\rm k}\Omega. \end{align*} \] Calculate the currents through * \(R_1\), * \(D_1\), * \(D_2\), depending on the switch state \(S\). {{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWKswDZ0A4BMYDMAWfAdiIE59UFcRVIQl9qEBTAWjDACgA3ELLfH3x1+gsFgxQpw+lLowEnAE5C64ybkhiJUsPE4B3EIyyqQuVJIHzD5y2dFnInACZ3J69w-AA5fGHwMTgAPcwxqMCJBfwiSYyEQABEuUKwENUhSPgw1cXiBEAAlFL4dSOo0jyJUfMEAVWc3E3AdZus+X39AkONicCjjOMiiWqSsTgBnL09mz3kQADMAQwAbCeZOXCI6TW0NCxaPOVtHT3a521wDzwsPHWdQ3HJwTOM9cDzBAoBlTiA noborder}} 1. Calculate the currents for open switch \(S\). <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12007~@# <WRAP leftalign> With the switch open, only \(D_1\) is connected to the resistor path. The conducting diode clamps the node voltage to approximately \[ \begin{align*} U_{\rm node}\approx U_{\rm F}=0.7~{\rm V}. \end{align*} \] The resistor current is therefore \[ \begin{align*} I_{R1} &= \frac{U_0-U_{\rm F}}{R_1} \\ &= \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} \\ &= 4.3~{\rm mA}. \end{align*} \] Since only \(D_1\) conducts, \[ \begin{align*} I_{D1}=I_{R1}, \qquad I_{D2}=0. \end{align*} \] </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12007~@# For open switch: \[ \begin{align*} I_{R1}&=4.3~{\rm mA}, \\ I_{D1}&=4.3~{\rm mA}, \\ I_{D2}&=0. \end{align*} \] #@ResultEnd_HTML@# </WRAP> </WRAP> 2. Calculate the currents for closed switch \(S\). <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12008~@# <WRAP leftalign> With the switch closed, \(D_1\) and \(D_2\) are connected in parallel. The resistor current is still determined by the source voltage, the forward diode voltage, and \(R_1\): \[ \begin{align*} I_{R1} &= \frac{U_0-U_{\rm F}}{R_1} \\ &= \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} \\ &= 4.3~{\rm mA}. \end{align*} \] Kirchhoff's current law gives \[ \begin{align*} I_{D1}+I_{D2}=I_{R1}. \end{align*} \] With the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined. If both real diodes are approximately identical, the current splits approximately equally: \[ \begin{align*} I_{D1} \approx I_{D2} \approx \frac{4.3~{\rm mA}}{2} = 2.15~{\rm mA}. \end{align*} \] </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12008~@# For closed switch: \[ \begin{align*} I_{R1}=4.3~{\rm mA} \end{align*} \] and \[ \begin{align*} I_{D1}+I_{D2}=4.3~{\rm mA}. \end{align*} \] For approximately identical real diodes: \[ \begin{align*} I_{D1} \approx I_{D2} \approx 2.15~{\rm mA}. \end{align*} \] #@ResultEnd_HTML@# </WRAP> </WRAP> 3. Explain why the current sharing is not unique in the simple model. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~12009~@# <WRAP leftalign> The constant-voltage diode model assumes that each conducting diode has exactly the same voltage drop: \[ \begin{align*} U_{\rm D}=U_{\rm F}. \end{align*} \] For two parallel diodes, this condition is true for many possible current distributions. Therefore, the model only determines the sum \[ \begin{align*} I_{D1}+I_{D2}, \end{align*} \] not the individual diode currents. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~12009~@# The constant-voltage diode model determines only \[ \begin{align*} I_{D1}+I_{D2}=4.3~{\rm mA}. \end{align*} \] It does not uniquely determine \(I_{D1}\) and \(I_{D2}\) separately. <callout type="warning" icon="true"> Parallel diodes are sensitive to small differences in real diode characteristics. Current sharing should not be assumed to be perfect without checking the design. </callout> #@ResultEnd_HTML@# </WRAP> </WRAP> #@TaskEnd_HTML@# CKG Edit