A signal occurring in a television set is illustrated in Figure E.46. Write a mathematical description of it. -10 x(t) Signal in Television t (us) 60 -10
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- A clamper circuit has 20 Vp-p. 100Hz square wave input voltage. The circuit consists of silicon diode IN4001 and 3V battery as shown in Figure 1 C. 0.1 µF D R Vi(t) 50 k2 Vo(t) 3 V Figure 1 a) Find the output voltage for all input voltages values. b) Sketch the output waveform, Vo(t).Please answer all subpart in short please Asap for like this Please all subpart in short please..I need the answer as soon as possible
- hello there can u help for this questionQ4) Attempt to answer ONE branch only: a) Regarding ripple vector parameter, What are the differences between full wave and half wave rectifiers? b) From the circuit shown below which shows combining a positive clipper with a negative dipper Determine and draw the output voltage waveform? Note that Vp = 0.7V 1.h Vout 15 Vek 3 He V11) AC power in a load can be controlled by using a. Two SCR's in parallel opposition b. Two SCR's in series c. Three SCR's in series d. Four SCR's in series 2) The advantage of using free - wheeling diode in half controlled bridge converter is that a. There is always a path for the ac current independent of the ac line b. There is always a path for the dc current independent of the ac line c. There is always a path for the dc current dependent of the ac line d. There is always a path for the ac current dependent of the ac line 3) Silicon controlled rectifier can be turned on a. By applying a gate pulse and turned off only when current becomes zero b. And turned off by applying gate pulse c. By applying a gate pulse and turned off by removing the gate pulse d. By making current non zero and turned off by making current zero
- What is CEMF?(c) A 50 Hz sinusoidal AC voltage source of peak value V, = 340 V delivers power to a resistive load of 50 n. (1) Calculate the average power delivered. (i) The AC source is now connected to the load via a full wave diode bridge. The turn-on voltage of the diodes is 0.3 V. Calculate the peak voltage across the load and the average voltage across the load. (ii) Explain how a smoothing capacitor can be used to hold up the voltage across the load. Compare the efficiency of the rectifier with and without the smoothing capacitor. (iv) Define "regulation" as applied to the output voltage of a rectifier and explain why an LR filtered rectifier bridge provides better regulation of the load voltage than an RC filtered rectifier bridge.Determine 1) the maximum load current. 2) the mean load current .3)the rms alternating load current. 4) the DC power supplied to the load. 5) the input power to the anode circuit. 6) the rectification efficiency. 7) the percentage of regulation. 8) the ripple current. R=10°0 Vi R=10°n AC Figure (2-72) the circuit of problem (2-6). (2-7) A full-wave bridge rectifier with a 120 Vrms sinusoidal input has a load resistance of Ikn. Determine:- 1) The de voltage available at the load. 2) Find the maximum current through each diode during conduction. 3) What is the required power rating of each diode. (2-8) A zener diode has a de power dissipation of Iw and a zener voltage rating of 27v. What is the value of I, max for the device? (2-9) Calculate and draw the output voltage for the circuit shown in Figure (2-74). +100v Vo 2.2kd4 2.2kQ -100v Figure (2-74) the circuit of problem (2-10).
- ۹:۰۱۱ • ۶۹۳}{ نقطة واحدة نقطة واحدة نقطة واحدة 36. Z 4.5G+ KOREK TELECOM. the rms value of half wave rectifier symmetrical square wave current of 2Ais 1.414 1 0.707 1.73 a sinusoidal current has a maximum value of 10A.if the signal is half rectifier its rms value will be 10A 7.07A 5A 14.14A Ripple factor of Half Wave Rectifier isThe waveform displayed on an oscilloscope is as shown in Figure The 'time/cm' switch is set to 10 ms/cm, and the 'volts/cm' switch is set to 50 V/cm. Determine the (i) amplitude of waveform Q, (ii) peak to peak value of waveform P, (iii) frequency of waveform P and (iv) phase angle difference between P and Q in Degrees. QFacts: Facts about Clampers A clamper is an electronic circuit that fixes either the positive or the negative peak excursions of a signal to a defined value by shifting its DC value. The clamper does not restrict the peak-to-peak excursion of the signal, it moves the whole signal up or down so as to place the peaks at the reference level. A diode clamp (a simple, common type) consists of a diode, which conducts electric current in only one direction and prevents the signal from exceeding the reference value; and a capacitor, which provides a DC offset from the stored charge. The capacitor forms a time constant with the resistor load, which determines the range of frequencies over which the clamper will be effective. Question: A clamping circuit has to have an independent source, a diode, a resistor, and a capacitor. To keep a constant voltage on the capacitor over the period of the input, the RC time constant must be large. A design rule of thumb is to make the RC time constant at…

