electronics fundamentals circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA chapter 19

Comparators Op-amps can be used to compare the amplitude of one voltage with another. Although general-purpose op-amps can be used as comparators, special op-amps are available to optimize speed and add features. +V An example of a comparison circuit is shown. The input is compared with a reference set by the voltage-divider. Notice that there is no feedback; the op-amp is operated in open-loop, so the output will be in saturation. R1 - Vout Vin + R2

Example 1: Sketch the output of the comparator in relationship to the input; assume the maximum output is ±13 V. Solution: +10 V The threshold is +4.2 V. The output is in positive saturation when Vin > +4.2 V +4.2 V Vin 0 V V = +15 V -10 V R1 +13 V 10 kW - Vout Vin + 0 V R2 3.9 kW -13 V

Example 2: Show the output of the comparator for the last example if the inputs to the op-amp are reversed. Solution: +10 V The threshold is still +4.2 V but now the output is in negative saturation when Vin > +4.2 V. +4.2 V Vin 0 V V = +15 V -10 V R1 +13 V 10 kW + Vout Vin - 0 V R2 3.9 kW -13 V

Summing amplifier There are a number of useful applications for the basic inverting amplifier configuration. One is the summing amplifier that uses two or more inputs and one output. Rf R1 The virtual ground isolates the inputs from each other. Input current from each input is passed to Rf, which develops an output voltage that is proportional to the algebraic sum of the inputs. VIN1 R2 VIN2 - R3 VOUT VIN3 + Rn VINn Virtual ground

Averaging amplifier An averaging amplifier is a variation of the summing amplifier in which all input resistors are equal. The feedback resistor is the reciprocal of the number of inputs times the input resistor value. For example, if there are three input resistors, each with a value of 10 kW, then Rf = 3.3 kW to form an averaging amplifier. R1 Rf VIN1 10 kW 3.3 kW R2 VIN2 - 10 kW VOUT + R3 VIN3 10 kW

Scaling adder A scaling adder is another variation of the summing amplifier in which the input resistors are adjusted to weight inputs differently. The input “weight” is proportional to the current from that input. Larger resistors will allow less current for a given input voltage, so they have less “weight” than smaller resistors. In the case shown, VIN3 is “weighted” 2 times more than VIN2, which is 2 times more than VIN1. R1 Rf VIN1 10 kW 10 kW R2 VIN2 - 5.0 kW VOUT + R3 VIN3 2.5 kW

Example: Solution: Scaling adder What is VOUT for the scaling adder if all inputs are + 1.0 V? Solution: By Ohm’s law, the currents into Rf are I1 = 0.1 mA, I2 = 0.2 mA and I3 = 0.4 mA. R1 Rf Using the superposition theorem, the current in Rf is 0.7 mA. From Ohm’s law, VOUT = VIN1 10 kW 10 kW R2 VIN2 - 7 V 5.0 kW VOUT + R3 VIN3 2.5 kW

Integrators Mathematical integration is basically a summing process. Within certain limitations, an integrator circuit simulates this process. The ideal integrator is essentially a summing amplifier with a capacitor in place of the feedback resistor. In practical circuits, a large value resistor is usually in parallel with the capacitor to prevent the output from drifting into saturation. Rf C R Vin - Vout +

For the ideal integrator, the rate of change of the output is given by Integrators For the ideal integrator, the rate of change of the output is given by The minus sign in the equation is due to the inverting amplifier. If the input is a square wave centered about 0 V, the output is a negative triangular wave (provided saturation is not reached). C Vin 0 V R Vin - Vout 0 V + Vout

Example: A 5 kHz square wave with 10 Vpp is applied to a practical integrator. Show the output waveform voltages. Solution: During the positive input (½ the period), the change in the output is 5.6 V Rf The feedback resistor (Rf) is large compared to R, so has little effect on the shape of the waveform. In a practical circuit, it will cause the output waveform to center on zero as shown on the following slide. 270 kW C 33 nF R Vin - 2.7 kW Vout +

Solution: continued… The results of a computer simulation on Multisim confirm the calculated change (5.6 V) in output voltage (blue line). Rf 270 kW C 33 nF R Vin - 2.7 kW Vout +

Differentiators In mathematics, differentiation is the process of finding the rate of change. An ideal differentiator circuit is shown. It produces an inverted output that is proportional to the rate of change of the input. In practical circuits, a small value resistor is added in series with the input to prevent high frequency ringing. Rf Rin Vin C Vin - Vout +

The output voltage for the ideal differentiator is given by Differentiators The output voltage for the ideal differentiator is given by The minus sign in the equation is due to the inverting amplifier. If the input is a ramp, the output is a negative dc level for the positive slope and a positive dc level for the negative slope. Rf Vin C Vin - Vout + Vout

Example: A 1.0 kHz, 10 Vpp triangular wave is applied to a practical differentiator as shown. Show the output in relationship to the input. Solution: When the input has a positive slope, the output is -5.4 V By symmetry, when the input has a negative slope, the output will be +5.4 V. Rf 2.7 kW Rin C Vin - Vout +5.0 V 120 W 100 nF + Vin 0 V -5.0 V 0 1 ms 2 ms See next slide for waveforms…

Solution: continued… The results of a computer simulation on Multisim confirm the calculated output voltages (±5.4 V). The output voltage is the blue line.

Oscillators The feedback oscillators introduced in Chapter 17 and other types of feedback oscillators can be implemented with op-amps. One type of feedback oscillator is called the Wien-bridge oscillator. This circuit is useful for generating low distortion sine waves. Lead-lag circuit Negative feedback with JFET gain control Rf C1 R1 - Vout JFET bias circuit + D1 Q1 R2 C2 C3 R3 R4

Oscillators The lead-lag circuit in the Wien-bridge oscillator has a maximum response at the resonant frequency given by Vout Vin fr f ⅓Vin The lead-lag circuit response is… + - R3 Rf Q1 C3 D1 R4 Vout C1 This equation is valid when R’s and C’s in the lead-lag circuit are equal. R1 Because the attenuation is ⅓ at fr, the gain of the Wien bridge must set for 3. R2 C2

Example: Solution: Wien-bridge oscillator What is the frequency of the bridge? Solution: Rf C1 10 kW 47 nF The frequency is given by R1 - Vout 6.8 kW + D1 Q1 R2 C2 6.8 kW 47 nF C3 R3 R4 1.0 kW 10 kW 1.0 mF 498 Hz

Triangular-wave oscillator A triangular-wave oscillator can be made from a comparator and an integrator. The integrator produces a ramp due to the constant current charging of the capacitor. When the ramp reaches a trip point, the comparator suddenly switches to opposite level and the ramp changes direction. C Vout (square) - R1 - + Vout (triangle) R2 + Comparator Integrator R3

Square-wave relaxation oscillator The square-wave relaxation oscillator uses a comparator to switch the output based on the charging and discharging of a capacitor. R1 - Vout C + Vout R2 R3

Active filters A filter selects certain frequencies and excludes others. Active filters use op-amps to optimize the frequency response. A 2-pole low-pass filter and its response is shown. The gain for this filter is 1 (0 dB) for f < fc. By reversing the resistors and capacitors in the low-pass circuit, a high-pass filter is created. This filter has a gain of 1 at frequencies where f > fc. Gain (dB) f fc -3 -40 dB/decade Gain (dB) Vout + - R2 R1 C2 C1 Vin fc -3 -40 dB/decade + - R2 R1 C2 C1 Vin * -40dB/decade means a 40dB decrease per 10 fold change in frequency

Voltage regulators Voltage regulators are made from integrated circuits. A basic series IC regulator has four blocks: Control element VIN VOUT Reference voltage Sample circuit Error detector

Voltage regulators A series regulator can use a comparator to compare the output voltage with a reference voltage. The series transistor drops more or less voltage to keep the output constant. Q1 VIN VOUT When VOUT exceeds the reference voltage, the inverting input exceeds the non-inverting input voltage, causing the op-amp to turn off transistor Q1, and the output voltage to dip. Therefore, voltage at the inverting input is forced to be the same as the reference voltage by feedback action When VOUT goes below the reference voltage, the reverse happens, causing the op-amp to turn on transistor Q1, and the output voltage to pull the voltage up. R1 + R2 - R3

Example: Solution: Voltage regulators What is the output voltage of the series regulator? Solution: VIN = Q1 VOUT +24 V R1 4.7 kW + R2 - 6.8 kW 15.6 V 5.1 V R3 3.3 kW

Voltage regulators A shunt regulator also has four blocks; it controls the current in the parallel control element. A series resistor drops more or less voltage to keep the output constant. R1 VIN VOUT Reference voltage Error detector Control element Sample circuit

Voltage regulators Shunt regulators are not as efficient as series regulators, but have the advantage of short circuit protection. R1 VIN VOUT R2 Error detector Can you identify each element in this circuit? - Q1 R3 + Control element Sample circuit Reference voltage R4

Selected Key Terms Summing amplifier Averaging amplifier Scaling adder An amplifier with several inputs that produces an output voltage proportional to the algebraic sum of the inputs. An amplifier with several inputs that produces an output voltage that is the mathematical average of the input voltages. A special type of summing amplifier with weighed inputs.

Selected Key Terms Integrator Differentiator Active filter Series regulator A circuit that produces an inverted output that approaches the mathematical integral of the input. A circuit that produces an inverted output that approaches the mathematical derivative of the input, which is the rate of change. A frequency selective circuit consisting of active devices such as transistors or op-amps combined with reactive (RC) circuits. A type of voltage regulator with the control element in series between the input and output.

Quiz 1. When an op-amp is configured as a comparator, the gain is equal to a. 0 b. 1 c. a ratio of two resistors. d. the open-loop gain.

Quiz 2. The approximate voltage at the inverting input of the op-amp shown is equal to the average of the input voltages. the sum of the input voltages. 0 V R1 Rf VIN1 10 kW 3.3 kW R2 VIN2 - VOUT 10 kW + R3 VIN3 10 kW

Quiz 3. For the scaling adder shown, the input with the greatest weight is R1 Rf VIN1 VIN1 VIN2 VIN3 they are all equal. 10 kW 10 kW R2 VIN2 - 5.0 kW VOUT + R3 VIN3 2.5 kW

Quiz 4. In a practical integrator, the purpose of the feedback resistor (Rf) is to a. limit the gain. b. prevent drift. c. prevent oscillations. d. all of the above. Rf C R Vin - Vout +

Quiz 5. Assume the top waveform represents the input to a differentiator circuit. Which represents the expected output? Vin a. b. c. d.

Quiz 6. The lead-lag network in a Wien bridge with equal value R’s and C’s attenuates the signal by a factor of a. 2 b. 3 c. 5 d. 10

Quiz 7. A Wien-bridge is used to produce a. sine waves. b. square waves. c. triangle waves. d. all of the above.

Quiz 8. For the circuit shown, the two outputs (in red) produce a. sine and square waves. b. triangle and square waves. c. sine and triangle waves. d. sawtooth and triangle waves. C Vout - R1 - + Vout R2 + Comparator Integrator R3

Quiz 9. The purpose of the op-amp in the series regulator is a. to sample the output. b. to establish a reference. c. as a control element. d. error detection. Q1 VIN VOUT R1 + R2 - R3

Quiz 10. An advantage of a shunt regulator is a. short circuit protection. b. efficiency. c. no need for a reference voltage. d. all of the above.

Quiz Answers: 1. d 2. c 3. c 4. b 5. c 6. b 7. a 8. b 9. d 10. a