54251. Regulation of the DC power supply of 12 V, 100 mA is the effective resistance of power supply is 20 Ω
54252. “The Old Testament” tells of the ......... the Jews made with God.?
54253. In the circuit of figure v1 = 8 V and v2 = 0. Which diode will conduct (Assume ideal diodes)?
54254. Assertion (A): An op-amp has high voltage gain, high input impedance and low output impedanceReason (R): Negative feedback increases output impedance
54255. A transistor has a maximum power dissipation of 350 mW at an ambient temperature of 25°C. If derating factor is 2 mW/°C, the maximum power dissipation for 40°C ambient temperature is
54256. The output voltage waveform of a CE amplifier is fed to a dc coupled CRO. The trace on the screen will be
54257. In an RC phase shift oscillator, the total phase shift of the three RC lead networks is
54258. In the circuit of figure β = 100 and quiescent value of base current is 20 μA. The quiescent value of collector
current is
54259. Values of VT at 20°C for an ideal P-N diode
54260. Assertion (A): In an op-amp the voltage gain and band width can be adjusted as per requirementReason (R): Large value capacitor can also be fabricated on a chip
54261. In figure which diode will conduct and what will be the value of V0?
54262. For the amplifier circuit of figure, the h parameters of transistor are hib = 25 Ω, hfb = 0.999, hob = 10-6 Ω. The voltage gain is
54263. In figure, VCC = + 30 V, R1 = 200 kΩ and R2 = 100 kΩ. If VBE = 0.7 V, the voltage a cross RE =
54264. A negative feedback can be of
54265. In a class C power amplifier the input frequency of ac signal is 1 MHz. If tank circuit has C = 1000 pF, the value of L =
54266. The Vo of the op-amp circuit shown in the given is
54267. In calculating output impedance of an amplifier the source is replaced by an open circuit.
54268. For transistor 2 N 338 the manufacturer specifies Pmax = 100 mW at 250°C free air temperature and maximum junction temperature of 125°C. Its thermal resistance is
54269. In figure base current is 10 μA and βdc = 100. Then VE =
54270. It has been found that in a rectifier circuit with RC filter one RC section reduces ripple by 15%. Two RC sections are used in cascade the reduction in ripple would be
54271. An op-amp has
54272. In the figure, assume the op-amp is to be ideal. The output Vo if the circuit is
54273. As the ratio Rf/RL increases the efficiency of a rectifier increases.
54274. Assuming VCE sat = 0.2 V and β = 50, the minimum base current (IB) required to drive the transistor in the given figure to saturation is
54275. A half wave diode rectifier uses a diode having forward resistance of 50 ohms. The load resistance is also 50 ohms. Then the voltage regulation is
54276. A Hartley oscillator is used for
54277. In Class C operation the collector current looks like
54278. An op-amp integrating circuit uses
54279. Don’t interfere ......... other people’s affairs .?
54280. A CE amplifier has re = 6 Ω, β = 100 and a = 0.98. The input impedance is
54281. A full wave diode rectifier using centre tapped transformer has diodes with forward resistance of 50 ohms each. The load resistance of 50 ohms each. The load resistance is also 50 ohms. The voltage regulation is
54282. Leakage current approximately doubles for every 10°C increase in the temperature of a transistor has ICBO = 400 nA at 25°C, then its leakage current at 80°C
54283. A CB amplifier has IE = 3.5 mA, a = 0.98 and RL = 600 Ω. The input impedance
54284. An amplifier has open loop gain of 100, input impedance 1 kΩ and output impedance 100 Ω. If negative feedback with β = 0.99 is used, the new input and output impedances are
54285. In a BJT amplifier the power gain from base to collector is 4000. The power gain in dB is
54286. In figure, secondary winding has 40 turns. For maximum power transformer to 2 ohm resistance the number of turns in primary is
54287. In a CE amplifier circuit the dc voltage between emitter and ground
54288. A buffer amplifier should have
54289. If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then its common mode rejection ratio is
54290. An npn transistor has a Beta cutoff frequency fβ of 1 MHz, and a common emitter short circuit low frequency current gain β0 of 200. It unity gain frequency fT and the alpha cut off frequency fa 2 respectively are
54291. In figure if the transistor is cut off, the collector voltage is equal to
54292. If input frequency is 50 Hz, the frequency of output wave in a full wave diode rectifier circuit is
54293. A class A transformer coupled power amplifier is to deliver 10 W output. The power rating of transistor should not be less than
54294. A CB amplifier has a = 0.98 and RL = 600 Ω. If IE = 3.5 mA, the current gain is
54295. funity of an op-amp is about 1 kHz.
54296. Assertion (A): In an RC coupled amplifier the junction capacitances can be neglected at low frequencies but not at high frequenciesReason (R): Capacitive reactance is inversely proportional to frequency.
54297. In figure, D1 turns on when
54298. In a full wave rectifier circuit using centre tapped transformer, the peak voltage across half of the secondary winding is 30 V. Then PIV is
54299. In the circuit of figure both diodes are ideal. If v1 = 10 V and v2 = 10 V which diode will conduct?