Referring to the figure below, Given Vcc= 20 V, R1 = 500k, R2 = 3k and R3 = 1k. Calculate the lg if beta = 150 +Vcc 82 ww Select one: O a. 29.65 UA O b. 2.95 UA O c. 2.95 mA O d. 29.65 mA
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- ll SuperNe 4G 3:00 PM C @ 6% Question 9 1 points Save A Power supply circuit is delivering0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to I2-Dc 205A 220V omsh 5OHZ O 001 F O 0 02 F O 00167 F O None of the above1) Diode half-wave rectifier a) Consider the half-wave rectified sinewave voltage waveform at the right. What is the DC voltage and RMS voltage in terms of Vpeak? Remember: Vp peak ann T/2 1 Vay = DC = v(0)de avg T Sv(t)dt and V = Hve v² (t)dt TConsider the circuit shown in the figure below with R1=16 ko and R=8 klQ. The Zener diode voltage is Vz = 3.5 V, then the currents Iz and lu respectively, are equal to: Ri of ww RL 20 V O a. 0.594 mA, 0.438 mA Ob. 0.394 mA 0.237 mA Oc 0.794 mA, 0.637 mA O d. 0.994 mA, 0.838 mA
- Consider a n-type silicon sample. If a power supply is connected between two ends of the silicon sample with 5 volts of voltage, 5 milliampers of current flows through the sample. If Na=3x1014 cm-3, N3=0, Dn=10 cm2/sec, Dp=20 cm2/sec, kt/q=D0.025 volts, obtain mobilkity if the sample area is two times its length in magnitude. (q=1.6x10-19 C, T=300 °K, kT=0.025 Ev, n=1010 cm-3) O a. 0.75 O b. 12.2 O c. 10.42 O d. 0.0675 O e. 20.85 O f. 0.45 O g. 8.5 O h. 885b) Use the ALU 74181 in the figure below and the tables (one is enough; second maybe, need to check the new kit) to implement : A + B, A – B, A. B. (MSB) (LSB) 10 (MSB) F3 F2 F1 FO S3 Cn+4 16 s2 20 U14 15 S1 14 so A3 A2 A1 AO B3 B2 B1 B0 M Cn 24 23 22 21 28 27 T26 25 (LSB) (MSB) (LSB)(MSB) (LSB) And given that under M = 1 the circuit performs the following arithmetic and logic functions according to Table 11.1. Input selection S3 Output M-H S2 SI Cn=L. F3 F2 F1 FO A 1 -A В -B A&B AxB 1 A'B 1 Ax(-B) (-A)xB (-A)x(-B) 1 1 ACTIVE HIGH DATA M-L: ARITHMETIC OPERATIONS SELECTION M-H LOGIC s3 s2 s1 so Ino carry) (with carry) F-A PLUS 1 F- LA + B) PLUS 1 F- (A + BI PLUS 1 FUNCTIONS L L F-A F-A+B F-AB F-A H FA+B F-A+T F- MINUS 1 12s COMPL) F-A PLUS AB F- IA + BI PLUS AB L L L L F-0 F-ZERO F- AB F-A PLUS AB PLUS 1 F- (A + BI PLUS AB PLUS 1 L L L H L F-AOB F- AB L L F-A MINUS B MINUS 1 F-A MINUS B F- AB MINUS 1 F- A F-A PLUS AB PLUS1 F-A PLUS B PLUS 1 F- IA + B PLUS AB PLUS 1 L H L F-A PLUS AB…Shown below is a diode circuit. If the input signal Vs = 31 Vp-p and the DC voltage Vdc = 8 V, what is the output voltage (in volts) of the circuit during the negative alteration? No need to show your solution. Just write your numeric answer on the space provided. Round off your answer to 2 decimal places. C1 ... + Vdc 1.0kn Vo Vs Si D1
- For the Zener regulated power supply a shown, the rms value of vI is 15V, the operating frequency is 60Hz, R =100Ω, C =1000μF, the on voltage of diodes D1 and D2 is 0.75V, and the Zener voltage of diode D3 is 15V.(a)What type of rectifier is used in this power supply circuit?(b) What is the dc voltage at V1? (c) What is the dc output voltage VO?(d)What is the magnitude of the ripple voltage at V1? (e) What is the minimum PIV rating for the rectifier diodes? (f) Draw a new version of the circuit that will produce an output voltage of−15 V.Q 2) A single phase half-wave uncontrolled rectifier with RL load is shown With R=1002, La 0. 1H w-377rad/s and V 100V, Determine: a) An expression for the load current i(t). b) The value of the current at t3D16 ms. c) The average load current (1C WLialx197=ア7.9 =121 V, = V, sinfar) 17.2 0.360*Use 4 decimal places. For the circuit below, determine the diode voltages VD1 and Vp2, diode currents Ip1 and Ip2, and voltage across the resistor, VRLIMIT- Use ideal model. Assume r'd (for both diodes)=200. Note D, is Germanium, D2 is Silicon, VBIAS = 14V and RLIMIT = 760 ohms . VD1 = Blank 1 V; VD2 = Blank 2 V; Ip1 = Blank 3 mA; Ip = Blank 4 mA ; VRUMIT = Blank 5 V ... Capture3.JPG +Vp1- +VRLIMIT- +Vp2- Ip2 BIAS
- In the circuit shown below. Let Vm-35 V and i-20 mA and VpQ -0.7 V: iD 3 kQ 0.8 kO Vmcos(@t)V For t Os, the current in the diode equals: Oa. 0 mA Ob. 4 mA Oc. 2 mA Od. 6 mA If t = T/4, then the current in the diode equals: Oa. 6.29 mA Ob. 5.03 mA Oc. 4,03 mA Od. 7.86 mA If t = T/2, then the current in the diode equals: Oa. 15.24 mA Ob. 13.24 mA Oc. 19.24 mA Od. 17.24 mAThe input to a full-wave rectifier is 40 V AC (RMS). The load resistor is 10 ohms. The DC voltage at the output is: O A. 27 V O B. 18 V O C. 36 V O D. None of the other choices are correct O E. 21 VThe input to a full-wave rectifier is 30 V AC (RMS). The load resistor is 10 ohms. The DC power at the output is: O A. 73 W O B. 18 W C. None of the other choices are correct D. 36 W O E. 146 W