Chapter 19: Problem 2567
The behavior of Ge as semi-conductor is due to width of: (A) Conduction band being large (B) Forbidden band being large (C) Conduction band being small (D) Forbidden band being small and narrow
Chapter 19: Problem 2567
The behavior of Ge as semi-conductor is due to width of: (A) Conduction band being large (B) Forbidden band being large (C) Conduction band being small (D) Forbidden band being small and narrow
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Get started for freeA light emitting diode has a voltage drop of \(2 \mathrm{~V}\) across it when \(10 \mathrm{~mA}\) current is passed. If this \(\mathrm{LED}\) is to be operated with \(6 \mathrm{~V}\) battery the value of limiting resistor would be (A) \(400 \Omega\) (B) \(4000 \Omega\) (C) \(40 \mathrm{k} \Omega\) (D) \(300 \Omega\)
In a N-P-N transistor circuit, the emitter, collector and base current are respectively \(\mathrm{I}_{E}, I_{C}\) and \(I_{B} .\) The relation between them is (A) \(\mathrm{I}_{\mathrm{C}}<\mathrm{I}_{\mathrm{E}}<\mathrm{I}_{\mathrm{B}}\) (B) \(\mathrm{I}_{\mathrm{B}}<\mathrm{I}_{\mathrm{C}}<\mathrm{I}_{\mathrm{E}}\) (C) \(\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{C}}<\mathrm{I}_{\mathrm{E}}\) (D) \(\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{C}}>\mathrm{I}_{\mathrm{E}}\)
By adding impurity in intrinsic semiconductor \(\mathrm{P}\) type semiconductor is made. charge of these P type semiconductor is (A) trivalent, neutral (B) pentavalent, neutral (C) pentavalent, positive (D) trivalent, negative
For a transistor, in a common base configuration the alternating current gain \(\alpha\) is given by: (A) $\left[\Delta \mathrm{I}_{\mathrm{C}} / \Delta \mathrm{I}_{\mathrm{B}}\right]_{(\mathrm{V}) \mathrm{C}=\mathrm{const}}$ (B) $\left[\Delta \mathrm{I}_{\mathrm{B}} / \Delta \mathrm{I}_{\mathrm{C}}\right]_{(\mathrm{V}) \mathrm{C}=\mathrm{const}}$ (C) $\left[\Delta \mathrm{I}_{\mathrm{C}} / \Delta \mathrm{I}_{\mathrm{E}}\right]_{(\mathrm{V}) \mathrm{C}=\text { const }}$ (D) $\left[\Delta \mathrm{I}_{\mathrm{E}} / \Delta \mathrm{I}_{\mathrm{C}}\right]_{(\mathrm{V}) \mathrm{C}=\text { const }}$
In an P.N.P transistor circuit, the collector current is \(10 \mathrm{~mA}\). If \(90 \%\) of the electrons emitted reach the collector: (A) \(\mathrm{I}_{\mathrm{E}}=9 \mathrm{~m} \mathrm{~A}\) (B) \(\mathrm{I}_{\mathrm{E}}=10 \mathrm{~mA}\) (C) \(\mathrm{I}_{\mathrm{B}}=1 \mathrm{~mA}\) (D) \(\mathrm{I}_{\mathrm{B}}=-1 \mathrm{~mA}\)
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