15. For the network of Fig. 9.80: a. Determine re. b. Find Avmid c. Calculate Z;. Vo/Vị- %3D
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- (a) Define bandwidth of an amplifier and provide explanation using a diagram. (b) A circuit has no lower cut-off frequency but has an upper cut-off frequency of 100 MHz. What is its bandwidth?You are given an audio amplifier. The input to the amplifier is a time varying voltage and the output of the amplifier is also a time varying voltage. Outline a practical procedure that can be used to determine the continuous time frequency response of the amplifier for frequencies ranging from -20kHz to 20kHz. You need to mention the equipment that you will use while working in a standard laboratory and how you will use it.for the following given circuita) find Avs mid = Vs/ Vi.b) define the values fLS, fLC and fLE.c) determine low cutting frequencies. d) bode curves using the cutting frequencies you find defined in stylish edraw asymptotes.h) draw the low frequency response of the amplifier using the results you find in stylish F.
- For The Table 1-Find the cutoff frequency by changing the frequency until Vo becomes 0.707Vi. 2-Graph the gain and the phase against frequency.A trans-impedance amplifier (TIA) has a signal gain of -1500 V/A up to certain operation frequency. (b). You are given a photodiode D1, a capacitor C₁, a resistor R₁ and an operational amplifier which operates at a supply voltage of 5 V. Draw a clearly labelled circuit diagram showing how you can construct a TIA from the given components for detecting and converting weak optical signals into electrical signals.true or false The unity - gain frequency determination depends on the frequency bandwidth of the amplifier and the value of the mid-range gain.
- Design a non-inverting amplifier that has a gain of 12. Assume value for Rf then find Ri.We examined the common source amplifier shown in the figure in the 5th experiment. The selection criterion of the input capacitance is Xcin = 0.1Rin: Calculate the required input capacitance value, Cin , if an input signal with a frequency of 2 kHz is applied. Şekitde gösterilen ortak source yükseltecini 5. deneyde incelemiştik Giriş kapasitesinin seçim kriteri Xcin - 0.1Rin dir. 2 kHzlik giriç sinyali için gerektt tapasite, Cm değerini hesaplayınız. Circuit parameters / Devre Parametreleri R1 = 53 k2, R, = 17 k2 R1 RD Cout Vout D VDD M2 Cin G Vin R2 RS cs O a. 74.19 nF O b. 86.55 nF O c. 49.46 nF O d. 61.82 nF O e. 43.28 nFIn transistor amplifier design, a by-pass capacitor is connected across the emitter resistor, RE, to effectively short out the emitter resistor at signal frequencies. This design improves the gain of the transistor for the desired ac signals. The circuit as shown in (a) is a common-emitter amplifier. The shaded portion is a low-frequency model of the transistor in use. In this problem, the task is to design a proper by-pass capacitor so that there is a pole at s = -300 rad/s. Reduce the circuit to its Thévenin equivalent as shown in (b) and then select the proper capacitor. Rs = 10 K0, R₂ = 2 KQ, RE = 3.3 kQ, R₂ = 1 k0, and 6 = 70. ww Rs Vs(1) www RE iB(1) +Vy(s) (a) ZT(S) M vc(1) Big(1) + RL CE
- Choose the correct answer 1. Common emitter configuration of BJT is a a) Current Amplifier b) Voltage Amplifier c) Both a and b d) None of the above 2. Common base configuration of BJT is a a) Current Amplifier b) Voltage Amplifier c) Both a and b d) None of the above 3. Function generators and AC source in Multisim are a) Similar in nature b) Function generators can generate more waveforms while AC source is a square wave c) AC source is just a 220V, 50Hz wave d) Function generate can generate more waveforms while AC source is a asine wave 4. Common collector configuration of BJT is a a) Current Amplifier b) Voltage Amplifier c) Both a and b d) None of the above 5. How can you control the output voltage of a common emitter amplifier BJT a) By RB b) By RE c) By VCC d) By coupling capacitorCan we obtain a plot of XL against frequency f experimentally?a. Determine the low cutoff frequency for the overall cct. b. Determine the High cutoff frequency for the overall cct c. Sketch approximately the low frequency, and high frequency response for the amplifier overall VI 0.82 ks2 # 0.47 μF 14 V 68 ΚΩ 10 ΚΩ 1.2 ΚΩ Cwi Cwo • 5.6 ΚΩ FIG 9.80 = 5pF Chc= 12 pF = 8 pF Che= 40 pF Cce = 8 pF 0.47 μF HH B = 120 20 μF 13.3 ΚΩ