Amplifiers x C Chegg Study I Guided So x GO screenshot current windc x Files c S

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Amplifiers

x C Chegg Study I Guided So x GO screenshot current windc x Files c Secure https:// instructure.com/courses/427923/files/folder/Homework?preview 65756916 utah HW-10.pdf O Download 6. You are given the following characteristics for a real amplifier Input offset voltage, Vios 5mV Ria 1M2 Input Resistance, Unity-gain bandwidth, fr 2MHz within 2Volts of power supply Output swing limits, Slew Rate, SR-4 Given the following circuit with the operational amplifier powered at t 12V 24kQ 16 20k2 3 k2 a) Find the ideal gain of the above circuit. b) For small input signals, what is the bandwidth of the circuit c) What is the bandwidth when the circuit is operated to produce the maximum possible peak voltage value? d) For Vin 0.00lsin(2m90kt), what is the ideal value for the peak to peak voltage value at the output? e) For Vin 0.002sin(2T90kt what is the peak to peak voltage value at the output considering the input offset voltage? f How should the circuit above be modified to minimize the effect of the input bias current? Draw the schematic of the modified circuit and state values of added component(s). 7. You are given the following characteristics for a real amplifier Info Close

Explanation / Answer

The input is given to non inverting terminal of OpAmp.

So,Ideal Gain = - (FeedBack Resistor - Rf) / ( Resistor connected to input - R1)

The feedback resistor Rf = 24k || (4k + 20k)
                       = 24k   || 24k  
                       = 24 x 24 / (24 + 24)
                       = 12k ohm
                         
Resistor connected to input - R1 = 3k ohm

(a)

Gain =    - 12k /3k = -4 .       

(b)

Bandwidth Of the circuit = Unit Gain Bandwidth / |Gain|.
                       = 12 Mhz / 4
                       = 3 Mhz
(c) Max Peak to Peak volatage at ouput will be 24 volt.
   Bandwidth = Slew Rate /Max Peak to Peak volatage
   = 4 x 10^3 / (24)   
           = 166.67 Hz

                         
(d) Vm = 0.001 sin(2 x pi x 90 x k x t)
   Vout = -4 x 0.001 sin(2 x pi x 90 x k x t)
   = -0.004 sin(2 x pi x 90 x k x t)
   So, The Peak to Peak Out volatage will be 0.008 volt.

(e)   Vm = 0.002 sin(2 x pi x 90 x k x t)
Now, If we consider Input offset volatage than,
     
   Vout = Vin x (-Rf/R1) + Voffset x (1 + ( Rf/R1 ))
       = 0.002 x sin(2 x pi x 90 x k x t) x (-4) + 0.005 x ( 1 + 4 )
       = 0.025 - 0.008 x sin(2 x pi x 90 x k x t)
      

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