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ECE 3410 – Homework 4 Problem 1. A non-inverting op amp configuration is shown below. The op amp has the following characteristics: • DC open-loop gain A0 = 40dB. • Open-loop 3dB cut-off frequency fc = 1MHz. • Slew rate SR = 2V/ µs. • Rail voltages at ±5V. • R1 = 1kΩ and R2 = 1kΩ, so the ideal closed-loop gain is 2V/ V. R2 R1 − vOUT + vIN − + (A) Calculate the unity-gain frequency (ft ) of this amplifier circuit. Give your answer in Hz. Solution ft = fc A0 = (1MHz) (100V/ V) = 100MHz. (B) Calculate the 3dB cut-off frequency for the closed-loop circuit. Solution ft G 100MHz = 2 = 50MHz. fc = Utah State University 1 ECE 3410 – Homework 4 Solution (cont.) (C) Sketch the magnitude response (i.e. Bode plot) for this op amp, showing both the open-loop and closed-loop responses. Identify the unity-gain frequency. Solution 50 Magnitude [dB] 40 open loop 30 20 10 closed-loop ft 0 −10 102 103 104 105 106 Frequency [Hz] 107 108 109 (D) Calculate the full-power bandwidth (FPBW) for this op amp. Solution SR 2πVR 2 × 106 V/ s = 2π × 5V = 63.662kHz FPBW ≤ (E) If the circuit amplifies an input signal described by the equation vIN = VIN sin (2πf t) , Utah State University 2 ECE 3410 – Homework 4 answer these subquestions: i. If f = 1MHz, what is the maximum possible amplitude VIN with no slewing? Solution SR 2πf = 0.31831V (zero-to-peak) VIN ≤ ii. If VIN = 1V, what is the maximum possible f with no slewing? Solution SR 2πVIN = 318.31kHz f≤ (F) If vIN is a square wave, is it possible to avoid slewing in the circuit’s output? Solution No. By definition, square waves change value instantaneously after each halfperiod. The op amp cannot change values instantly. (G) Carefully study the axes and waveforms shown below. The waveforms are shown for the ideal behavior (with no slewing). If slewing is present, draw the resulting waveforms. You may assume that vOUT = 0 at time t = 0. Utah State University 3 ECE 3410 – Homework 4 4 Output [V] 2 0 −2 −4 0 0.5 1 1.5 Time [µs] 2 2.5 3 0 0.5 1 1.5 Time [µs] 2 2.5 3 4 Output [V] 2 0 −2 −4 Utah State University 4 ECE 3410 – Homework 4 Problem 2. An inverting Miller integrator circuit is shown below. The component values are R1 = 10kΩ R2 = 10kΩ C = 10µF C R2 R1 vIN − vOUT + (A) When operating at DC, the capacitor C can be viewed as an open circuit (i.e. as though it isn’t there). What is the DC gain of this circuit? Solution GDC = −1V/ V (B) Using Laplace-transform analysis, show that the steady-state transfer function is given by R2 1 H (s) = − . R1 1 + sR2 C You may assume that the op amp is ideal. Solution Gi = − Z2 (s) Z1 (s) R2 = − 1+sR2 C2 R1 1 R2 = − R1 1 + sR2 C2 Utah State University 5 ECE 3410 – Homework 4 (C) Since an integrator has transfer function of the form K/s, this circuit can only be considered an integrator for high-frequency signals. At low frequencies (close to DC), it is a unity-gain buffer. What is the cut-off frequency that separates the AC integrating mode from the DC buffering mode? Solution 1 2πR2 C2 = 1.5915Hz fc = (D) The magnitude response of a linear system is p |H (f )| = H (j2πf ) H (−j2πf ). If the circuit has an input sine wave with amplitude 100mV at frequency f = 10Hz, predict the amplitude of the output waveform. Solution Based on the formula, we may solve (100mV) |H (f )| = (100mV) R2 R1 s 1 1 + (2πf R2 C)2 ! = 15.718mV (E) Suppose the op amp has an input bias current Ibias = 1µA and an offset voltage of Vofs = 10mV. What is the resulting DC offset that appears at the circuit’s output? (Remember that the offset voltage is modeled as a DC input applied to the op amp’s non-inverting terminal). Solution The DC offset is found by applying the usual formulas. The capacitance C has no effect, since bias current and offset voltage apply only at DC. R2 VOUT = Vofs 1 + + Ibias R2 R1 = 30mV Utah State University 6 ECE 3410 – Homework 4 Solution (cont.) Note that you can explore the behavior of this and other op amp configurations by using simple SPICE simulation models. Here is an example: ∗ Integrator simulation ∗ Input s o u r c e ∗ ( i n p u t i s node 1 ) : VIN 1 0 SIN ( 0 100m 1 0 ) ∗ RC c o n n e c t i o n s : R1 1 2 10 k R2 2 3 10 k C 2 3 10u ∗ Non−i d e a l c h a r a c t e r i s t i c s : IB 2 0 DC 1u VOS 4 0 DC 10m ∗ Op amp model ( j u s t a v o l t a g e −c o n t r o l l e d v o l t a g e −s o u r c e ) ∗ ( output i s node 4 ) E1 3 0 2 4 1 e6 ∗ SIMULATION COMMANDS: . t r a n 1m 1 . end After running this simulation in SPICE, you can plot V(3) to look at the output waveform. It should be very close to what we calculated. It is also interesting to study the integrator’s behavior with a square-wave input: ∗ I n t e g r a t o r s i m u l a t i o n with square −wave i n p u t ∗ Input s o u r c e ∗ ( i n p u t i s node 1 ) : VIN 1 0 PULSE( 1 0 0m −100m 0 1m 1m 0 . 0 5 0 . 1 ) ∗ RC c o n n e c t i o n s : R1 1 2 10 k R2 2 3 10 k C 2 3 10u ∗ Non−i d e a l c h a r a c t e r i s t i c s : IB 2 0 DC 1u VOS 4 0 DC 10m Utah State University 7 ECE 3410 – Homework 4 Solution (cont.) ∗ Op amp model ( j u s t a v o l t a g e −c o n t r o l l e d v o l t a g e −s o u r c e ) ∗ ( output i s node 4 ) E1 3 0 2 4 1 e6 ∗ SIMULATION COMMANDS: . t r a n 1m 1 . end Utah State University 8