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Difficulty: HardDiodes

Two standard silicon rectifier diodes are arranged in a dual-input diode OR logic configuration. The anode of Diode 1 is connected to a +5.0 V+5.0\text{ V} DC supply, and the anode of Diode 2 is connected to a +2.0 V+2.0\text{ V} DC supply. Both cathodes connect to a common junction that ties to ground through a 2.2 kΩ2.2\text{ k}\Omega load resistor. Assuming each forward-biased silicon diode exhibits a nominal barrier potential drop of 0.7 V0.7\text{ V}, what is the resulting output voltage measured at the common cathode junction?

  1. A
    1.3 V1.3\text{ V}
  2. 4.3 V4.3\text{ V}Answer
  3. C
    5.0 V5.0\text{ V}
  4. D
    6.3 V6.3\text{ V}

Answer

The voltage measured at the common output cathode junction is 4.3 V4.3\text{ V}.
The output potential is determined by the diode with the highest anode voltage. Diode 1 has +5.0 V+5.0\text{ V} applied to its anode, which forward-biases its PN junction and establishes a common cathode potential of 5.0 V−0.7 V=4.3 V5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}. Because Diode 2 has only +2.0 V+2.0\text{ V} at its anode while its cathode is held at +4.3 V+4.3\text{ V}, it experiences a net reverse bias of −2.3 V-2.3\text{ V} and remains completely non-conductive (OFF).

Step-by-Step Solution

1
Determine which diode enters forward conduction first by comparing anode potentials.
Diode 1 has an anode voltage of +5.0 V+5.0\text{ V}, which is higher than Diode 2's anode voltage of +2.0 V+2.0\text{ V}.
In a common-cathode configuration, the diode with the most positive anode potential forward-biases first and sets the output voltage.
2
Calculate the output node voltage established by the conducting diode.
Vout=5.0 V−0.7 V=4.3 VV_{\text{out}} = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}.
A conducting silicon diode drops approximately 0.7 V0.7\text{ V} across its PN junction.
3
Evaluate the biasing state of Diode 2 with the cathode node established at 4.3 V4.3\text{ V}.
Vanode2−Vcathode=2.0 V−4.3 V=−2.3 VV_{\text{anode2}} - V_{\text{cathode}} = 2.0\text{ V} - 4.3\text{ V} = -2.3\text{ V} (Reverse Biased / OFF).
Because the cathode is more positive than the anode, Diode 2 is reverse-biased, acts as an open circuit, and contributes no current to the load.

Key Concept

Diode conduction states, forward barrier potential drop, and reverse-bias isolation in multi-diode circuits
Estimated Time:1m 30s
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Topics

Diodes

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