6 V 10 V 3Ω 4Ω ww (4) i4 22 is 6Ω 5Ω 6Ω 8Ω i3. 9 V 12 V
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- Vcc = +12 V %3D a. Determine the following values for the fixed-bias configuration in the figure to the right. Rc IBo and ICO 2.2 k2 C2 RB 240 k2 ac output VCEQ, VB, and Vc 10 uF C ac input VCE B = 50 VBC 10 µFexplain how the circuit works 16V RA 10K 15V = v2 INDICADOR LOOICO Vin R1 10K RB LM741 100 led 15V V3 O So IS O O S2 CONVERTIDOR DIGITAL A ANALOGO S3 9999OUTPUT VOLTAGE (V) 2.0 1.8 1.6 1.0 0.8 0.6 0.2 0 <-50 a. TMP35 b. TMP36 c. TMP37 +Vs = 3V -25 0 50 25 TEMPERATURE (°C) 75 Figure 6. Output Voltage vs. Temperature 100 b 125 00337-007 Using the appropriate curve, develop a linear equation for the output temperature as a function of voltage.
- +5V VDO EN Xtal 8ה 7 +5V Freq1 HGND | GND х Mhz Sensor0 crystal Freq26 5 Vss Out 04 PRE Xtal 1k Output HA7210IP 4 CLK CLR SN74HC7491 GND 0.01 μ GND +5V 4 +5V +5V +5V OE O/P3 0. 1µF | 47µF GND SG351P |2 x MHz GND GND Consider the Quartz Crystal Microbalance Sensor shown above, with crystal resonant frequency, x = 3.3 Mhz. You are given the following values: Active crystal area = 1.6 cm², and the change in frequency = - 10 Hz Calculate the change in mass, Am in nano gram, up to 2 decimal places. Answer:If Is = 3x1016A and VT= 26mV, Then Vout is: R 1 kQ 3 V Vout 750 mV Select one: O a. None of these O b. 1.31 V O c.2.66 V O d. 1.99 V e. 1.65 VWhich of these is Inverse Square Law 21 O a. E= 1/d O b. E= 1/d O C.E= 1/d? O d. E= 1/d
- Q6// Determine Ve, VB, VE, Ic, Ig and Ig for the circuit shown in figure below (B=100 & a= 0.9) Vcc = +22 V Rc 10 Kohm 39 Kohm RB C2 • Vo C1 1µF Vin • Q1 1µFDetermine the Thevenin open-circuit voltage in the figure below as seen from terminals a-b. j4 2 V. ell o a -j2 Q 5/0° A 0.2V, 4Ω O 7.3524-72.9° V 73.52472.9° V O 7.352279.2° V O 7.352272.9° V O 735.2472.9° mV wwIf Is = 1x1016A and V= 26MV, Then Vout is: RL1 k2 3 V 750 mV Q1 Select one: a. None of these O b. 1.31 V O c. 1.65 V d. 2.66 V e. 1.99 V 11:20 PM EN 4/15/2021
- n8 The value of cos(3 – 4i) is swered t of 2.50 O a. 5e-0.9273 i estion O b. -27.0349 – 3.8512 i O c. 5e 0.9273 i O d. -27.0349 + 3.8512 iA step down converter is operated with a duty cycle of k 0.75. The input voltage is V, 20V. and the load is L=5mH, R=152 The minimum inductor current is I 0.1A. The maximum inductor current is %3D %3D Select one: O a. 0.9 A O b. 1.9 A O C. None of these O d. 1.1 A O e.2AR= 10e R=22 Ro 24 Re Ha ZR= 102 %3D loor 邱201 Is - V - 6. 7 Is' 8. I7- 9. I lo - l RyonAL - 2. エTormL 15 VG 3. Ragalo011- 14 V7 4. R345678901 5 R12345C7891011- lo. Ill- K Vio ノ NOTE: PLEASE MAKE YOUR ANr IN ORDER. 1. Vo B. 2 Rq 3 Va - 4 Vo - 了 Ii Vo 3A R3 GNEN : R-20; Re =42; Ra GR; R4=88