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- aloulated m O1 the value you Up = A cos wt +1o Calaulate maarli)-min(i) =? Use constant voltage drop mad. Don7 to write units. . - 3.7V forgeta. If you do not go completely around the loop when applying Kirchhoff’s voltage law, then the algebraic sum of the voltages cannot be determined the algebraic sum of the voltages will always be positive the algebraic sum of the voltages will always be negative the algebraic sum is the voltage between the start and finish points b. A p -type semiconductor is a semiconductor doped with impurity atoms whose electron valence is +4 pentavalent impurity atoms trivalent impurity atomsb) 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…
- In the circuit below, we have a Zener Diode circuit with a load, RL. Use the same measurements for source voltage V1 and resistor R3 resistance as in the circuit schematic. V1 -15V R3 2700 D1 1N753A RL Determine the lowest possible value for the load RL, so that the voltage drop across it is equal to the Zener voltage. For your diode, use model Zener 1N753A. This diode has a Zener voltage of 6.2V, a Zener current of 60mA, a Zener impedence of 72, and a test current of 20mA.Diodes Are Connected In Series To Share A Total... 21 Question • Two diodes are connected in series to share a total reverse voltage of VD=D 5 KV. Reverse leakage currents of the diodes are ID1=30 mA and Ip2=35 mA. - Find Vp1 and Vp2 for R,=R2=100 k2 - Find R, and R2 for VD1=VD2 R1 VD1 Hint: Is=L,+IRj= I,2+IR2 VD R2 VD2*Use 4 decimal places. **Note D1 is Germanium, D2 is Silicon, For the circuit below, determine the diode voltages VD1 and VD2, diode currents ID1 and ID2, and voltage across the resistor, VRLIMIT. Use practical model. Assume r'a=150. VBIAS = 15V and RLIMIT = 1.5kohms. VD1 = Blank 1 V; VD2 = Blank 2 V; Ip1 = Blank 3 mA; Ip2 = Blank 4 mA ; VRLIMIT = Blank 5 V Capture3(1).JPG +Vo1- +VRLIMIT +VD2- Io2 VBIAS
- For self bias configuration. Solve for Ib and Ic. ... 4 RC 15002 RB VCC 5 16V 100A/A 80 kQ Re 10002 Ib=Blank 1 HA Ic=Blank 2mAQ2/ Looking at the directions of the four diodes .What is the current through 80 2 resistor in the circuit shown in figure. Assume the diodes to be of silicon and forward resistance of each diode is 1Q. D1 A D3 80 0 +20 V D4 A D2 O A. 225 mA О в. 227 mА О С. 229 mА О D. 230 mA 中 中liting Q9. For the circuit shown in Figure C9, Calculate the following i) Current through the Silicon diode (Isi). ii) Current through the Resistor R2 (1). iii) Current through the diode D4 (IGe4). iv) Voltage at points V3 and V4. Is D1. D2 Si Si oV4 RI 1.3K2 R3 2.7K R2 3K2 IGe4| 15V D5 Si 03 D4 Ge Ge V2 1V Figure C9
- For Fixed Bias Configuration Solve for Ib, Ic and le. Atleast 2 decimal place. Be mindful of units. Assume Ic is not equal to le. ... 4 RC 15002 RB VCC 20V 90A/A 100k2 Ib=Blank 1 HA Ic=Blank 2mA le=Blank 3 mA Blank 1 Add your answer Blank 2 Add your answer Blank 3 Add your answerO * 33% A moodle1.du.edu.om e SIM aY The measured temperature is 173 OF and the reference is 45 °F. The voltage. in mV is Select one: O a. 4.643 O b. 4.097 O e. 3.732 O d. -3.732 Previous page Next pagewe Q2/ what Ms meant by Pulsse Moduletim? why need PM ?