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1 ECE 3336 Introduction to Circuits & Electronics MORE on Operational Amplifiers Spring 2015, TUE&TH 5:30-7:00 pm Dr. Wanda Wosik Set #14.

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Presentation on theme: "1 ECE 3336 Introduction to Circuits & Electronics MORE on Operational Amplifiers Spring 2015, TUE&TH 5:30-7:00 pm Dr. Wanda Wosik Set #14."— Presentation transcript:

1 1 ECE 3336 Introduction to Circuits & Electronics MORE on Operational Amplifiers Spring 2015, TUE&TH 5:30-7:00 pm Dr. Wanda Wosik Set #14

2 2 Basics of Operational Amplifiers Noninverting Case We will focus on operational amplifiers, specifically on Ideal Operational Amplifiers, definitions and requirements for their ideal operation in noninverting configuration Negative Feedback that allows for op-amp to be controlled by external elements

3 3 Solving Op Amp Circuits As for inverting configuration we will have two assumptions for the analysis and design. We will again treat the op amps as ideal circuits. We will again call these assumptions golden rules. i in =0A The first assumption: i - = i + = 0. results from large resistances at the inputs. Currents do not flow into the op-amp. The second assumption v + ≈v - deals with the output that makes the input voltages equal v + ≈v -. This is realized by introducing negative feedback loop, which spans the output and the inverting input. negative feedback loop

4 4 A Note on the Second Assumption The second golden rule v - = v + results in the virtual short, or the summing- point constraint. The constrain refers to the input voltages, which become the same v - = v + if there is the negative feedback and the open loop gain A v(OL) is large. Without negative feedback, even a small input voltage will cause saturation of the output either at V + or V -. That depends on the sign of v in. Inverting Input Noninverting Input + dc V supply Output NO NEGATIVE FEEDBACK yet This is open loop configuration Negative dc power supply

5 5 Op Amp Circuits with the Negative Feedback Loop Golden Rules 1) i - = i + = 0. 2) v - = v +. Virtual short Negative feedback Negative feedback adds a portion of the output signal to the inverting input. Since the signs of these voltages are opposite, the negative feedback acts as if the signal applied to the input decreases. The net result is that the output voltage can be controlled by the external elements and does not saturate. For ideal op-amps we will apply two golden rules to solve circuits ideal

6 6 Op Amp in the Non-inverting Configuration An op amp operates in the noninverting configuration when the input voltage is applied to the noninverting terminal. R F is the feedback resistor R s is the source resistor There is a negative feedback thanks to R F Negative feedback gives the virtual short: v - =v +. Since v + =V s also v - =V s. The op-amp does not draw currents i in =0A These comments are identical as for the inverting configuration ideal A v(OL) ≈∞

7 7 Solving op-amp in the Non-Inverting Configuration Closed Loop As earlier, to find v out we have to find v RF. To find v RF we have to know current i F which can be calculated from i s. The current i s is given by the voltage v - =V s and R s. ideal v + =v S A v(OL) ≈∞ 0A Since we have golden rules (i in =0, v + =v - ) v + =v S Closed loop voltage GAIN:

8 8 Significance of the Closed Loop Gain The negative feedback loop, combined with ideal properties of the op- amp (high open loop gain 10 5 -10 7 and large input resistance) ensures that the gain does not depend on the op amp the gain is the determined by a ratio of two resistors connected to the op-amp. No phase change ideal

9 9 Voltage Follower Important application of the noninverting configuration is obtained when there is no resistance in the negative feedback loop. ideal VSVS VSVS So, the voltage at the input is the same as the voltage at the output v out =v S. Do we gain anything here by doing that? We do! We have a very large input resistance of this circuit: Such op-amps do not show loading effects (i.e. voltage drop due to low resistance connected to an output of a circuit). They work as voltage follower but they also act as impedance buffers. R F =0Ω Golden rules apply: v + =v - and i in =0A

10 10 The Differential Amplifier This is a combination of inverting and noninverting configuration. As earlier we have negative feedback and the op-amp is ideal. i 2 =-i 1 v - =v + i in =0 Group and arrange: Rearrange

11 11 Instrumentation Amplifier (IA) i R1 v2v2 v1v1 i in =0 Now use the results from differential op-amp with v out1 and v out2 replacing original v 1 and v 2 v out2 v out1 Advantages: Very high input resistance Very high common-mode-rejection-ratio CMMR (goal: CMMR   for perfectly matched resistors. That results in v out ≈0V for v 1 =v 2 ) IA are made as integrated circuits

12 12 Integrator Now we add the op-amp and we get an integrator. It also constitutes a part of an analog computer The Golden Rules are used for the op-amp Virtual short The integrating circuit was used earlier Now we integrate both sides and we have the integrator

13 13 Differentiator Now we add the op-amp and we get a differentiator. It also constitutes a part of an analog computer. The Golden Rules are used for the op-amp The differentiating circuit was used earlier

14 14 Active Filters The concept of frequency dependence of the signals seen in the filters (remember that we had |H(j  )| max =1 for those filters) is here combined with the signal amplification. We will use here the negative feedback configuration We will also use impedances instead of resistors We still have the same golden rules: no input currents (high R in ) virtual short

15 15 Active Low-Pass Filter 0V So the cutoff frequency is also the 3dB frequency (as before) -3dB Phase is just like for the simple filter The voltage gain A LP is calculated using Golden Rules Amplification Cutoff frequency Amplification

16 16 Active High-Pass Filter Phase is just like for the simple high pass filter The voltage gain calculated using Golden Rules Negative feedback Inverting configuration cutoff 3dB frequency Phase: Amplification

17 17 Op-Amp as a Level Shifter A useful circuit to adjust DC voltage level = to remove the DC offset from the signal invertingnoninverting 220kΩ 10kΩ We want this to be equal 0V That gives V ref =1.714V We can design such precision voltage sources using R p Potentiometer Power supply Use the superposition principle (one source at a time)

18 18 Active Band-Pass Filter The voltage gain A BP is again calculated using Golden Rules Magnitude of A BP Negative feedback Inverting configuration  1 is the unity gain frequency @1@1 Relations between the frequencies Three characteristic frequencies

19 19 Characteristic Frequencies in the Band-Pass Filters Cancel off  0 -3dB So  LP and  HP are 3dB frequencies while  1 is the unity gain frequency The voltage gain has 3 characteristic frequencies:  1,  LP and  HP Gain around  LP Gain around  HP =0=0=0=0

20 20 Bode Plots for the Active Band-Pass Filter 11  LP  HP  LP  HP 11 We can plot the magnitude of the voltage gain as a function of frequency Relations between the frequencies The phase is like for simple bandpass filters  LP  HP Linear scale dB scale 45° -45°

21 21 Limitations of the Op-Amps Saturation of the voltage at the output occurs at about ±V s. Small signals at the input are required

22 22 Limitations of the Op-Amps Frequency Response Limits refer to the voltage gain of the open loop and closed loop configuration Open loop gain decreases very quickly with frequency The voltage gain decreases in the closed loop configuration but the cutoff frequency increases The gain-bandwidth product is constant K

23 23 Limitations of the Op-Amps Slew rate limitation of op-amp means that the op-amp output voltage does not respond with the same slope as the input signal Increasing frequency means faster changing or steeper slopes at the zero crossing Slew rate is limited by the frequency and amplitude product As the result of limited slew there is a distortion of the output signal.

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