67851. If an amplifier with gain of - 100 and feedback of β = - 0.1 has a gain change of 20% due to temperature the change in gain of the feedback amplifier will be
67852. In figure the cut in voltage of diode is 0.7 V and its bulk resistance is 20 Ω. The peak value of circuit current of during positive and negative half cycles are
67853. A voltage gain of a CE amplifier connected in collector to base configuration is 50. Collector to base resistor RF = 1 kΩ, RC = 4Ω, the O/P resistance R0 =?
67854. A voltage multiplier circuit, using diodes and capacitors is suitable for
67855. In a CE amplifier the ac emitter resistance
67856. Heat sink results in
67857. Who was the first to perform the surgery of human heart transplant?
67858. In a CE amplifier, the output voltage is equal to product of
67859. In figure, VG =
67860. A diode can have a very high forward resistance.
67861. In figure the voltage drop across diode is 0.7 V. Then the voltage across diode during negative half cycle is
67862. In figure vi = 10 mV dc maximum. The maximum possible dc output offset voltage is
67863. A transistor with a = 0.9 and ICBO = 10 μA is biased so that IBQ = 90 μA. Then ICQ will be
67864. The current through R1 is(If β = 99, VBE = 0.74 V)
67865. Which of the following can be used as a buffer amplifier?
67866. The feedback technique employed in the following circuit is
67867. A second order active bandpass filter can be obtained by cascading LP second order filter having higher cut off frequency fOH with a second order HP filter having lower cut off frequency fOL provided
67868. In figure the approximate voltages of
67869. The open loop gain of an amplifier is 50 but likely to decrease by 20% due to various factors. If negative feedback with β = 0.1 is used, the change in gain will be about
67870. An increase in ambient temperature means that maximum power rating of transistor
67871. Two CE stages are coupled through a capacitor. To calculate the quiescent base current of the two transistors, the capacitor is treated as
67872. If it is desired to have low output impedance in an amplifier circuit then we should use
67873. Generally the gain of a transistor amplifier falls at high frequencies due to the
67874. Which one of the following is connect expression of id for figure?
67875. A voltage tripler circuit and voltage quadrupler circuit use identical components. Then
67876. A Ge diode operated at a junction temperature of 27°C. For a forward current of 10 mA, VD is found to be 0.3 V. If VD = 0.4 V then forward current will be
67877. In a power amplifier the collector current flows for 270° of the input cycle. The operations is
67878. An amplifier has input impedance of 4 kΩ and output impedance of 80 kΩ. It is used in negative feedback circuit with 10% feedback. If open loop gain is 90, the closed loop input and output impedances are
67879. In figure the current through resistor R
67880. In a negative feedback amplifier A = 100, β = 0.04 and Vs = 50 mV, then feedback will be
67881. For an npn transistor connected as shown in the figure VBE = 0.7 volts. Given that reverse saturation current of the junction at room temperature 300K is 10-13A, the emitter current.
67882. The dc output voltage of the circuit is
67883. A voltage doubler circuit is fed by a voltage Vm sin ωt. The output voltage will be nearly 2 Vm only if
67884. The purpose of connecting a coupling capacitor in the output circuit of an amplifier is
67885. Which component is allowed to pass through it by a choke filter?
67886. In the diode circuit shown in figure the diodes are ideal. The impedance seen by source is
67887. In an amplifier the stray capacitances assume impedance at low frequencies
67888. In CE amplifier the base current is very high.
67889. In the graphical analysis of CE amplifier circuit, the upper end of load line is called
67890. A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltage
67891. In the graphical analysis of an amplifier circuit, the slope of dc load line depends on
67892. In figure we need an ac ground. The proper value of C is
67893. In figure V0 =
67894. In figure, ID = 4 mA. Then VS =
67895. The circuit shown is
67896. Consider 49 cascaded amplifiers having individual rise time as 2 n sec. 3 n sec. ... 50 n sec. The input waveform rise time is 1 n sec. Then the output signal rise time is given time by (Assume output signal rise time is measured within 10 percent range of the final output signal.)
67897. The zener diode in the rectangular circuit shown in the figure has a zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The maximum load current drawn from this circuit ensuring proper functioning over the input voltage range between 20 and 30 volts, is
67898. In figure, ID = 4 mA. Then VDS =
67899. In figure, V0 =
67900. In the circuit of figure diode will conduct
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