Скачать или смотреть Gate Questions on BJT Circuits | 2015-2024 | GATE PYQ | GateBusters ECE | NerdyBug

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In this video, we tackle BJT Circuit Problems for GATE, diving deep into essential concepts to help you ace the exam. From BJT circuit analysis to BJT amplifier analysis, we solve and explain various GATE questions on BJT circuits step by step.

This video covers all GATE problems solved from 2015 to 2024, ensuring a comprehensive preparation experience. Whether you're preparing for Analog Circuits for GATE ECE or Analog Circuits for GATE EEE, this video is tailored to your syllabus with clear and concise explanations. Strengthen your grasp of GATE analog circuits and build confidence for the exam.

Watch now to master these critical concepts for GATE 2025 or GATE 2026 preparation! Don’t forget to like, share, and subscribe for more valuable content.

🔥Full series:    • Analog Circuits 1 | NerdyBug  
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📅 Chapters:
00:00:00 Introduction
00:00:13 Problem 1 [GATE 2020] : For the BJT in the amplifier shown below, VBE = 0.7 V, kT/q=26 mV. Assume the BJT output resistance [ro] is very high and the base current is negligible. The capacitors are also assumed to be short circuited att signal frequencies. The input vi is direct coupled. The low frequency gain vo/vi of the amplifier is

00:05:11 Problem 2 [GATE 2017] : For the DC analysis of the common emitter amplifier shown, neglect the base current and assume that the emitter and collector current are equal. Given that VT=25 mV, VBE = 0.7 V, and the BJT output ro is practically infinite. Under these conditions, the midband voltage gain magnitude, av=vo/vi is

00:09:46 Problem 3 [GATE 2017] : In the figure shown, the npn transistor acts as a switch. For the input vin(t) as shown in the figure, the transistor switches between the cut off and saturation regions of operation, when T is large. Assume collector to emitter voltage at saturation, VCE(sat)=2.0 V and base to emitter voltage VBE=0.7 V. The minimum value of the common base current gain (α) of the transistor for the switching should be

00:13:35 Problem 4 [GATE 2017] : In the circuit shown, transistors Q1 and Q2 are biased at a collector current of 2.6 mA. Assuming that transistor current gains are sufficiently large to assume collector current equal to emitter current and thermal voltage of 26 mV, the magnitude of voltage gain Vo/Vs in the mid band frequency range is [upto second decimal place]

00:19:35 Diode Connected Transistor

00:23:54 Problem 5 [GATE 2017] : The miller effect in the context of a common emitter amplifier explains

00:25:11 Problem 6 [GATE 2017] : Consider the circuit shown in the figure. Assume base to emitter voltage VBE = 0.8 V and common base current gain (α) of the transistor is unity. The value of collector to emitter voltage [in volts] is

00:28:56 Problem 7 [GATE 2016] : Resistor R1 in the circuit below has been adjusted so that I1=1 mA. The bipolar transistors Q1 and Q2 are perfectly matched and have very high current gain, so their base currents are negligible. The supply voltage Vcc is 6 V. The thermal voltage kT/q is 26 mV. The value of R2 (in Ω) for which I2=100 µA is

00:32:13 Problem 8 [GATE 2016] : Which one of the following statements is correct about an ac-coupled common-emitter amplifier operating in the mid-band region?

00:33:55 Problem 9 [GATE 2016] : Consider the circuit shown in the figure. Assuming VBE1=VBE2=0.7 V. value of the dc voltage Vc2 (in volt) is

00:37:38 Problem 10 [GATE 2015] : In the ac equivalent circuit shown, the two BJTs are biased in active region and have identical parameters with β greater than 1. The open circuit small signal voltage gain is approximately

00:38:43 Problem 11 [GATE 2015] : In the circuit shown, I1 = 80 mA and I2 = 4 mA. Transistors T1, and T2 are identical. Assume that the thermal voltage VT is 26 mV at 27 °C. At 50 °C, the value of the voltage V12 = V1-V2 [in mV] is

00:42:01 Problem 12 [GATE 2015] : In the circuit shown in the figure, the BJT has a current gain (b) of 50. For an emitter base voltage VEB = 600mV , the emitter collector voltage VEC [in volts] is

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