54153. Ac signals are given to both inverting and non-inverting terminals of an op-amp. When will the output maximum
54154. An ideal op-amp has zero slew rate.
54155. In a bridge rectifier circuit the rms value of input ac voltage is 10 V. The PIV across each diode is
54156. In a half wave diode rectifier circuit the current flows in the load circuit for
54157. In a push pull circuit
54158. A transistor with a = 0.9 and ICBO = 10 μA is biased so that IBQ = 90 μA. Then IEQ will be
54159. In figure, transistor βdc = 100 and LED voltage when it is conducting is 2 V. Then the base current which saturates the transistor is
54160. In figure the dc emitter current of each transistor is about
54161. The input voltage for starting oscillations in an oscillator is caused by
54162. The current flowing in a certain P-N junction at room temperature is 2 x 10-7 Amp. When a large reverse biased voltage is applied. Calculate the current flowing when 0.1 volts is applied.
54163. In following figure find VDSQ by assuming gate current is negligible for the p-channel JFET. (if IDQ = - 6 mA, RS = 0, VDD = -18 V, RD = 2 kΩ, IDSS = - 10 mA, IPO = - 3 V)
54164. Which of the following power amplifiers has highest efficiency?
54165. A full wave rectifier using centre tapped transformer and a bridge rectifier use similar diodes and have equal no load output voltage. Under equal load conditions
54166. Assume that op-amp in figure is ideal. If input Vi is triangular, the output V0 will be
54167. In an amplifier with a gain of - 1000 and feedback factor β = - 0.1, the change in gain is 20% due to temperature. The change in gain for feedback amplifier will be
54168. The officer .........out five minutes ago.?
54169. A transistor has two p-n junctions. The batteries should be connected such that
54170. The value of parameter re used in re transistor model
54171. Assertion (A): In a full wave rectifier output the lowest ac component is at twice the input frequencyReason (R): In a full wave rectifier, the output waveform consists of half sinusoids
54172. An op-amp has zero gain for common mode inputs. Then CMRR =
54173. Assertion (A): Negative feedback reduces the bandwidth of an amplifier Reason (R): Negative feedback stabilizes the gain of an amplifier
54174. In an amplifier the voltage gain is the ratio of
54175. Provide suitable prepositions: Switzerland is ......... Germany, France, Australia and Italy.?
54176. Find resistance RB to bring transistor to threshold of saturation VCB = 0, VBE = 0.7 V, a = 0.96
54177. In figure ID = 4 mA. Then VGS =
54178. A typical value of gm for a FET is about 25 μs.
54179. The dissipation at the collector is in the quiescent state and increases with excitation in the case of a
54180. In following circuit, RB will be if VCC = 10 V, VS = 2 V, RC = 5 kΩ, RS = 90 kΩ, β = 50, ICE = 0, VCEsat = 0.1 V
54181. The open loop gain of an ideal op-amp is
54182. In following figure, what will be R1 and R2 for maximum symmetrical swing if VCSat ≈ 0. Given that RE = 200 Ω, RC = 400 Ω, VCC = 20 V, β = 99
54183. The ideal characteristics of a stabilizer is
54184. The ripple factor of a half wave rectifier is more than 2.
54185. Assertion (A): The circuit in figure produces repetitive narrow pulses when input is fed with sine or triangular waveform having peak value more than + VReason (R): The high gain op-amp produces voltages at two levels. If input is more than + V, the output is + 15 V otherwise the output is - 15 V.
54186. In a CE amplifier drives a low load resistance directly the result will be
54187. Provide suitable prepositions. My name has been excluded ......... the list .?
54188. For high frequencies a capacitor like
54189. A non-inverting op-amp summer is shown in figure, the output voltage V0 is
54190. In following figure find VGG by assuming gate current is negligible for the p-channel JFET. (if IDQ = - 6 mA, RS = 0, VDD = -18 V, RD = 2 kΩ, IDSS = - 10 mA, IPO = - 3 V)
54191. A difference amplifier using op-amp has closed loop gain = 50. If input is 2 V to each of inverting and non-inverting terminals, output is 5 mA. Then CMRR =
54192. The coupling capacitor in amplifier circuits
54193. In figure if C is replaced by short circuit then RB will be
54194. A full wave rectifier circuit using centre tapped transformer and a bridge rectifier are fed at 100 V, 50 Hz. The frequencies of outputs in these two rectifiers are
54195. When the ac base voltage in a CE amplifier circuit is too high, the ac emitter current is
54196. When load is coupled to class A amplifier through transformer, efficiency decreases.
54197. The gain of an FET amplifier can be changed by changing
54198. The open loop gain of an actual op-amp is about 100.
54199. In the circuit figure LED will be on when v1 is
54200. A differential amplifier consists of two CB amplifiers.