Home
Class 12
PHYSICS
If voltage gain in CE configuration is 2...

If voltage gain in CE configuration is 24. If `R_(e)=10 k, `R_(c)` = 10. B = 100 and input resistance R =2.5kohm, find the output voltage for an input voltage of 1 mV.

Text Solution

AI Generated Solution

To solve the problem step by step, we will use the given data and the formula for voltage gain in a common emitter (CE) configuration. ### Step 1: Understand the Given Data We are provided with the following information: - Voltage Gain (AV) = 24 - Input Voltage (Vin) = 1 mV = 1 × 10^(-3) V - Other parameters (RE, RC, B) are given but are not needed for this calculation. ...
Promotional Banner

Topper's Solved these Questions

  • SEMICONDUCTOR DEVICES

    AAKASH SERIES|Exercise Problems (LEVEL-II)|9 Videos

  • SEMICONDUCTOR DEVICES

    AAKASH SERIES|Exercise EXAMPLES|56 Videos

  • SEMICONDUCTOR DEVICES

    AAKASH SERIES|Exercise Exercise (very Short answer questions)|36 Videos

  • RAY OPTICS

    AAKASH SERIES|Exercise PROBLEMS ( LEVEL-II)|60 Videos

  • UNITS AND MEASUREMENT

    AAKASH SERIES|Exercise PRACTICE EXERCISE|45 Videos

Similar Questions

Explore conceptually related problems

A transistor has a current amplification factor (current gain) of 50. In a common emitter amplifier circuit, the collector resistance is chosen as 5Omega and the input resistance is 1Omega . The output voltage if input voltage is 0.01 V is

In a common emitter amplifier, using output reisistance of 5000 ohm and input resistance fo 2000 ohm, if the peak value of input signal voltage is 10 m V and beta=50 , then the peak value of output voltage is

In common emitter amplifier, the current gain is 62. The collector resistance and input resistance are 5 k Omega an 500 Omega respectively. If the input voltage is 0.01V, the output voltage is

An N-P-N transistor is being used in CE mode for which the current transfer ratio is alpha=(25)/(26) . The input resistance is 1000Omega and amplitude of A.C. input voltage is 10 mV. The amplitude of the amplified output collector current is

Power gain for N-P-N transistor is 10^(@) , input resistance 100Omega and output resistance 10000Omega . Find out current gain.

A n-p-n transisitor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Omega is connected in the collector circuit and the voltage drop across it is 0.8 V . If the current amplification factor is 0.96 and the input resistance of the circuits is 192 Omega , the voltage gain and the power gain of the amplifier will respectively be

A transistor connected in common emitter configuration has input resistance R_("in")=2 K Omega and load resistance of 5 K Omega . If beta=60 and an input signal 12 mV is applied , calculate the resistance gain, voltage gain and power gain.

The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10, 20 and 30 , respectively and the input signal is 1mV peak value, then what is the output signal voltage (peak value) (i) if dc supply voltage is 10V ? (ii) if dc supply voltage is 5V?

The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10, 20 and 30 , respectively and the input signal is 1mV peak value, then what is the output signal voltage (peak value) (i) if dc supply voltage is 10V ? (ii) if dc supply voltage is 5V?

A common emitter transistor amplifier has a current gain of 50 . If the load resistance is 4k Omega , and input resistance is 500 Omega , the voltage gain of amplifier is.

AAKASH SERIES-SEMICONDUCTOR DEVICES-Problems (LEVEL-I)
  1. For the circuit shown in Fig. find 1) the output voltage 2) the vo...

    Text Solution

    |

  2. The applied input AC power to a half wave rectifier is 190W. The DC ou...

    Text Solution

    |

  3. A p-n diode is used in a half wave rectifier with a load resistance of...

    Text Solution

    |

  4. A full wave rectifier uses two diodes with a load resistance of 100 O...

    Text Solution

    |

  5. A full-wave rectifier is used to convert 50 Hz A.C into D.C, then the ...

    Text Solution

    |

  6. In an N-P-N transistor operating in the active region, the collector c...

    Text Solution

    |

  7. In a common base configuration, with a base current of 0.005 mA, the e...

    Text Solution

    |

  8. For a transistor 'alpha' = 0.98 and emitter current I(E) = 2.5 mA. Cal...

    Text Solution

    |

  9. In common base configuration of a transistor, a change of 200 mV in em...

    Text Solution

    |

  10. In a transistor the emitter current is 1.01 times as large as the coll...

    Text Solution

    |

  11. In a transistor circuit the base current changes from 30 muA to 90muA....

    Text Solution

    |

  12. The constant a of a transistor is 0.9. What would be the change in col...

    Text Solution

    |

  13. For a transistor, the current gain of common-base configuration is 0.8...

    Text Solution

    |

  14. The current gain of a transistor in common emitter circuit is 49. Calc...

    Text Solution

    |

  15. A change of 0.5 mA in the emitter current of a transistor produces a c...

    Text Solution

    |

  16. A change of 200mV in base-emitter voltage causes a change of 100μA in ...

    Text Solution

    |

  17. For a single transistor amplifier, the collector load is R(L) =22kOmeg...

    Text Solution

    |

  18. AP-N-P transistor is used in common-emitter mode in an amplifier circu...

    Text Solution

    |

  19. The input resistance of a common-emitter amplifier is 600 Omega and lo...

    Text Solution

    |

  20. If voltage gain in CE configuration is 24. If R(e)=10 k, R(c) = 10. B ...

    Text Solution

    |