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The CMOS inverter in Figure 16.21 is biased at
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Microelectronics: Circuit Analysis and Design
- Using the sine PWM method with the full bridge inverter below, it is desired to generate a voltage of 50 Hz on the serial RL load. A voltage of 120 V DC is applied to the input of the inverter circuit. Amplitude modulation rate ma -0.9 and frequency modulation rate mf -19. The resistance of the series RL load is 15 OHM and the coil inductance is 40 mH. a) What is the power drawn by the load resistor?b) What is the total harmonic distortion value (THD) of the load current?arrow_forwardYou are asked to design a BJT inverter, as shown in Figure 2.b. The input voltage is equal to either zero (logic level 0) or 5 V (logic level 1), and Vcc = 5 V. The transistor has VBe = 0.6 V when it is on and Vce (sat) = 0.2 V. i) Calculate the value of Rc to have Iç = 3 mA when the transistor is on. ii) Calculate the maximum value of Rp to make sure the transistor is in saturation when it is on, assuming Bmin = 80 in the active mode. +Vc Re Rg Figure 2.b.arrow_forwardFor an inverter with VIL = 0.46 V, VIH = 0.77 V, VOL= 0.08 V, and VOH = 1.2 V, find out the maximum value for the noise voltage in presence of which the inverter can work properly. Ans: 0.38 V.arrow_forward
- The total rms output voltage of a single-phase bridge inverter operating from a nominal 200V DC supply is to be kept constant at 100V. If the supply voltage varies from 180 to 200V. Calculate : 1. 2. The range of conduction pulse widths (2d). For nominal input voltage find rms value of fundamental output voltage and rms value of 3rd harmonic in the output.arrow_forwardThe load voltage waveform of a single phase full bridge inverter supplied from 300V DC voltage source is shown in below figure. Output frequency is 50HZ and the load consists of series R-L components. The load values are 2N and 0.04H, respectively. (Conduction interval for half period is 120°) 150 100 50 T -50 3 -100 -150 .002 .004 .006 .008 .014 ,02 .01 Time (sec) .012 .016 .018 a. Find the RMS values of the load voltage and the load current b. Calculate and draw the voltage across the load inductance c. Calculate and draw the source current and find its average valuearrow_forward- The single-phase half-bridge inverter has a resistive load of R = 2.40 and the DC input voltage is 48V. Determine: 1- the rms output voltage at the fundamental frequency 2- fundamental displacement factor. 3- fundamental power, drown from each source . 4- fundamental load power. 5- total power drown from each source . 6- total load power. %3Darrow_forward
- Problem 3 Consider a CMOS static inverter. Please be careful with units. The NMOS transistor has the following characteristics. The channel length is 350 nanometers. For NMOS: tox=10nm, Vm=0.25V, un=560cm²/vsec, W=700nm, L=350nm For PMOS: tox=10nm, Vtp=-0.35V, up=240cm²/vsec, W=700nm, L=350nm Compute the resistance R₁ for NMOS if the power supply voltage (Vad) is 1.25 volts.arrow_forwardA full bridge inverter with RLC load having the following values: R-7.5 Ohms, L=12.5 mH. C-22 uF. The switching frequency is 500 Hz and the DC input voltage is 180V. The RMS load current is 6.275A. The average and RMS currents of each transistor are equal to (Consider up to the fifth harmonics in calculation): Select one: Oa. 3.14A, 3.63A b. 0.82A, 4.437A O c. 0.82A, 3.63A O d. 3.14A, 4.437Aarrow_forwardThe load voltage waveform of a single phase full bridge inverter supplied from 300V DC voltage source is shown in below figure. Output frequency is 50HZ and the load consists of series R-L components. The load values are 22 and 0.04H, respectively. (Conduction interval for half period is 120°) 150 100 50 T -50 3 -100 -150 .002 .004 .006 .008 .01 .012 .014 .016 .018 .02 Time (sec) a. Calculate and draw the load current for the first two-period interval b. Calculate and draw the load current for the steady-state conditionarrow_forward
- A single phase bridge inverter has an RLC load with R= 20 ohms, L= 32 mH and C= 0.115 mF. The inverter frequency is fo= 60 Hz and DC input voltage is Vs 110 V. If the peak magnitudes of the output current and its fundamental are equal to 6.3A and 6.14A, then the power absorbed by the load Po and the fundamental power Po1 are equal to: Select one: O a. 397.56W and 377W b. 795.12W and 754W O c. 562.23W and 533.16W Od. None of thesearrow_forwardA single phase half bridge inverter is operating from a 25 V DC source and supplies power to resistive load. What is the distortion factor of fundamental output voltage?arrow_forwardIn a half wave bridge inverter circuit, the power delivered to the load by each source is given by a. VsIs b. 4VsIs c. 2VsIs d. VsIs/2arrow_forward
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