226. In a CE amplifier, the output voltage is equal to product of
227. In figure, VG =
228. A diode can have a very high forward resistance.
229. In figure the voltage drop across diode is 0.7 V. Then the voltage across diode during negative half cycle is
230. In figure vi = 10 mV dc maximum. The maximum possible dc output offset voltage is
231. A transistor with a = 0.9 and ICBO = 10 μA is biased so that IBQ = 90 μA. Then ICQ will be
232. Which of the following is called the Gate way to the Pacific?
233. The current through R1 is(If β = 99, VBE = 0.74 V)
234. Which of the following can be used as a buffer amplifier?
235. The feedback technique employed in the following circuit is
236. 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
237. In figure the approximate voltages of
238. 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
239. An increase in ambient temperature means that maximum power rating of transistor
240. Two CE stages are coupled through a capacitor. To calculate the quiescent base current of the two transistors, the capacitor is treated as
241. If it is desired to have low output impedance in an amplifier circuit then we should use
242. Generally the gain of a transistor amplifier falls at high frequencies due to the
243. Which one of the following is connect expression of id for figure?
244. A voltage tripler circuit and voltage quadrupler circuit use identical components. Then
245. 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
246. In a power amplifier the collector current flows for 270° of the input cycle. The operations is
247. 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
248. In figure the current through resistor R
249. In a negative feedback amplifier A = 100, β = 0.04 and Vs = 50 mV, then feedback will be
250. 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.