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Transcript

OpAmps Circuits Tutorial Question 1 www.electronics.dit.ie Q1- Solution www.electronics.dit.ie Q1 - Solution www.electronics.dit.ie Question 2 www.electronics.dit.ie Q2 - Solution www.electronics.dit.ie Q2 - Solution www.electronics.dit.ie Q2 - Solution www.electronics.dit.ie Q2 - Solution www.electronics.dit.ie Question 3 www.electronics.dit.ie Q3 - Solution www.electronics.dit.ie Q3 - Solution www.electronics.dit.ie Q3 - Solution www.electronics.dit.ie Q3 - Solution www.electronics.dit.ie Q3 - Solution www.electronics.dit.ie Question 4 www.electronics.dit.ie Q4 - Solution www.electronics.dit.ie Q4 - Solution www.electronics.dit.ie Q4 - Solution www.electronics.dit.ie Q4 - Solution www.electronics.dit.ie Question 5 An op-amp-based inverting integrator is measured at 100Hz to have an voltage gain of -100V/V. At what frequency is its gain reduced to -1V/V? What is the integrator time constant? Solution: The gain is given by: G (ω) = 1 ωRC that is G (200π) = 1 = 100 200πRC τ = RC = 1 20000π therefore G (ω) = www.electronics.dit.ie 1 20000π = = 1V / V ωRC 2πf f = 10000 Hz Question 5A An op-amp-based inverting integrator is measured at 100Hz to have an voltage gain of -100V/V. At what frequency is its gain reduced to -1V/V? What is the integrator time constant? Solution: For integrator, the gain decays 20dB/decades. That is, when frequency increase by a factor, the gain decreases by the same factor. Therefore, when the gain decrease from 100V/V by factor of 1/100 to -1V/V, the frequency should increase by 100 times. That is, at 10000Hz the gain will reduced to -1V/V. www.electronics.dit.ie Question 6 A differentiator uses an ideal op-amp, a 10K resistor, and a 0.01uf capacitor. What is the frequency fo at which its input and output sine wave signals have equal magnitude? What is the output signal for for a 1-V p-p sine wave input with frequency equal to 10fo ? Solution: The transmission function of differentiator is given by: G (ω) = − for G (ω0 ) = 0.0001ω0 = 1 when f=10fo www.electronics.dit.ie R = − jωRC = − jω ×10000 × 0.01×10 −6 = − j 0.0001ω 1 j ωC ω0 = 10000 f 0 = 10000 / 2π G (10 × 10000) = 0.0001 × 100000 = 10 Question 7 7. A weighted summer circuit using an ideal op-amp has three inputs using 100K resistors and a feedback resistor of 50K. A signal v1 is connected to two of the inputs, while a signal v2 is connected to the third. Express v0 in terms of v1 and v2. If v1=3V, v2=-3V, what is v0? Solution: Rf Rf Rf v3 v2 + v1 + v0 = − R3 R2 R1 50 50 50 = − v1 + v2 + v1 100 100 100 1 = − v1 + v2 = −(3 − 1.5) = −1.5V 2 www.electronics.dit.ie Question 8 1 Design an op-amp circuit to provide an output v0 = − 3v1 + v2 2 Choose relatively low values of resistors but ones for which the input current (for each source) does not exceed 0.1mA for 2-V input signals. Solution: The input resistors can be determined as: Rf Rf 1 = 3 and = ⇒ R2 = 6 R1 R1 R2 2 R1 = www.electronics.dit.ie v1 2V ≥ = 20 KΩ i1 0.1mA R2 ≥ 2V = 120 KΩ, 0.1mA R f = 60 KΩ Question 9 9 For the difference amplifier use superposition to find v0 in terms of the input voltages v1 and v2: v1 = 10 sin( 2π × 60t ) − 0.1sin( 2π × 1000t ),Volts v2 = 10 sin( 2π × 60t ) + 0.1sin( 2π × 1000t ), Volts Solution: Disabling v1, the circuit is a non-inverting amplifier, therefore v02 = (1 + www.electronics.dit.ie 10 R )v2 = 11v2 R Question 9 (cont.) 9 Solution: Disabling v2, the circuit is an inverting amplifier, therefore v01 = − 10 R v1 = −10v2 R Therefore the total output should be: v0 = v01 + v02 = 11v1 − 10v2 v0 = sin( 2π × 60t ) − 2.1sin( 2π × 1000t ), Volts www.electronics.dit.ie