Bipolar transistors

What is the aim of a transistor with a Schottky barrier diode connected across its base-collector junction (a Schottky clamp)?

  1. ATo increase the current gain
  2. BTo keep the transistor out of deep saturation, reducing stored charge and speeding up switching
  3. CTo increase the breakdown voltage
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Answer: B. To keep the transistor out of deep saturation, reducing stored charge and speeding up switching

A Schottky diode conducts at a lower voltage than a PN junction, so it bypasses current before the base-collector junction becomes strongly forward biased. This keeps the transistor out of deep saturation and reduces the minority carriers stored in the base, shortening the delay when turning off. The technique was used in Schottky TTL.

  • A. It is not a configuration for changing the current gain.
  • B. Correct. It prevents deep saturation to speed up switching.
  • C. It is not intended to raise the breakdown voltage.
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More Bipolar transistors questions

  1. Which bias condition is correct when an NPN transistor is used for amplification (operated in the active region)?
  2. Which expression gives the DC current gain hFE of a bipolar transistor? (IC: collector current, IB: base current, IE: emitter current)
  3. What is the main reason for making the base of a bipolar transistor thin?
  4. When a bipolar transistor is used as a switch, which operating region corresponds to the on state?
  5. How does the phase of the output voltage of a common-emitter amplifier compare with the input voltage?