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    Parallel and Series Parallel Configurations

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    sean ballocanag
    This document contains 4 example problems demonstrating parallel and series-parallel diode configurations: 1. A parallel diode configuration with 2 diodes is solved, showing how placing diodes in parallel limits current to a safe level. 2. A series-parallel network is solved for the current I, showing one diode is on and one is off. 3. A series-parallel network with a silicon and germanium diode is solved for the voltage Vo, showing the germanium diode turns on first at 0.3V. 4. A series-parallel network is solved for currents I1, I2, and ID, using voltage loops and diode equations.

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    Parallel and Series Parallel Configurations

    Uploaded by

    sean ballocanag
    484 views3 pages
    This document contains 4 example problems demonstrating parallel and series-parallel diode configurations: 1. A parallel diode configuration with 2 diodes is solved, showing how placing diodes in parallel limits current to a safe level. 2. A series-parallel network is solved for the current I, showing one diode is on and one is off. 3. A series-parallel network with a silicon and germanium diode is solved for the voltage Vo, showing the germanium diode turns on first at 0.3V. 4. A series-parallel network is solved for currents I1, I2, and ID, using voltage loops and diode equations.

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    This document contains 4 example problems demonstrating parallel and series-parallel diode configurations: 1. A parallel diode configuration with 2 diodes is solved, showing how placing diodes in parallel limits current to a safe level. 2. A series-parallel network is solved for the current I, showing one diode is on and one is off. 3. A series-parallel network with a silicon and germanium diode is solved for the voltage Vo, showing the germanium diode turns on first at 0.3V. 4. A series-parallel network is solved for currents I1, I2, and ID, using voltage loops and diode equations.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    484 views3 pages

    Parallel and Series Parallel Configurations

    Uploaded by

    sean ballocanag
    This document contains 4 example problems demonstrating parallel and series-parallel diode configurations: 1. A parallel diode configuration with 2 diodes is solved, showing how placing diodes in parallel limits current to a safe level. 2. A series-parallel network is solved for the current I, showing one diode is on and one is off. 3. A series-parallel network with a silicon and germanium diode is solved for the voltage Vo, showing the germanium diode turns on first at 0.3V. 4. A series-parallel network is solved for currents I1, I2, and ID, using voltage loops and diode equations.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 3

    Lesson 4P

    PARALLEL AND SERIES-PARALLEL CONFIGURATIONS

    Illustrative Problems

    1. Determine Vo, I1, ID1, and ID2 for the parallel diode configuration.

    Solution:
    Since the resulting current direction matches that of the arrow in each symbol and the applied voltage
    is greater than 0.7 V, both diodes are in the “on” state. The voltage across parallel elements is always the
    same and
    Vo = 0.7 V
    The current
    𝑉𝑅 𝐸−𝑉𝐷 10 𝑉−0.7 𝑉
    𝐼 =
    1 = = = 28.18 mA
    𝑅 𝑅 0.33 𝑘𝛺

    This problem demonstrated one reason for placing diodes in parallel. If the current rating of the
    diodes is only 20 mA, a current of 28.18 mA would damage the device if it appeared alone. By placing two
    in parallel, the current is limited to a safe value of 14.09 mA with the same terminal voltage.

    2. Determine the current I for the following network.


    Solution:

    Redrawing the network reveals that the resulting current direction is such as to turn on diode D 1 and
    turn off diode D2. The resulting current I is then

    𝐸1 −𝐸2 −𝑉𝐷 20 𝑉−4𝑉−0.7 𝑉


    I= = ≅ 6.95 𝑚𝐴
    𝑅 2.2 𝑘𝛺

    3. Determine the voltage Vo for the network shown.

    Solution:
    Initially, it would appear that the applied voltage will turn both diodes “on”. However, if both were
    “on”, the 0.7-V drop across the silicon diode would not match the 0.3 V across the germanium diode as
    required by the fact that the voltage across parallel elements must be the same. At the instant during the
    rise that 0.3 V is established across the germanium diode, it will turn “on” and maintain a level of 0.3 V.
    The silicon diode will never have the opportunity to capture its required 0.7 V and therefore remains in its
    open-circuit state as shown. The result
    Vo = 12 V – 0.3 V = 11.7 V

    4. Determine the currents I1, I2, and ID for the network.


    Solution:

    The applied voltage (pressure) is such as to turn both diodes on, as noted by the resulting current
    directions in the network. Note the use of the abbreviated notation for “on” diodes and that the solution is
    obtained through an application of techniques applied to dc series-parallel networks.

    𝑉𝑇2 0.7 𝑉
    𝐼1 = = = 0.212 mA
    𝑅1 3.3 𝑘𝛺

    Applying Kirchhoff’s voltage law around the indicated loop in clockwise direction yields
    −𝑉2 + 𝐸 − 𝑉𝑇1 − 𝑉𝑇2 = 0
    and 𝑉2 = 𝐸 − 𝑉𝑇1 − 𝑉𝑇2 = 20𝑉 − 0.7𝑉 = 18.6 𝑉
    𝑉2 18.6 𝑉
    with 𝐼2 = = = 3.32 mA
    𝑅2 5.6 𝑘𝛺
    At the bottom node (a),
    𝐼𝐷2 + 𝐼1 = 𝐼2
    and 𝐼𝐷2 = 𝐼2 −𝐼1 = 3.32 𝑚𝐴 − 0.212 𝑚𝐴 = 3.108 𝑚𝐴

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