The use of operational amplifiers is a compulsory course for electricians. Some people have discussed the use of three major steps based on experience and netizens.
First, how to achieve weak signal amplification?
The sensor + op amp + ADC + processor is a typical application circuit for an op amp. In this application, a typical problem is that the current supplied by the sensor is very low. In this case, how to complete the signal amplification? For the amplification of weak signals It is difficult to achieve good results with only a single amplifier, and some special methods and sensor excitation methods must be used, and the use of the synchronous detection circuit structure can obtain very good measurement results. This synchronous detection circuit is similar to a lock-in amplifier structure, including a square wave excitation of a sensor, a current to voltage amplifier, and synchronous demodulation. It should be noted that the current-to-voltage amplifier needs to use an op amp with a very low input bias current. In addition, synchronous SPD requires a dual SPDT analog switch. Another engineer friend suggested that in the selection of op amps, capacitors, resistors and layout, special attention should be paid to selecting high-impedance, low-noise operation and low-noise resistance. Some netizens have also supplemented the solution to such problems, such as: 1) Pay attention to balanced processing in circuit design, try to balance, and effectively suppress interference. These are in National Semiconductor, BB (acquired by TI), ADI and other companies. It can be found in the design manual of the op amp. 2) It is recommended to add a metal shield to cover the weak signal part (open a small mold), and the metal body is connected to the circuit ground, which can greatly improve the anti-interference ability of the circuit. 3) For the nA level of the sensor output, select the op amp with input current pA level. If there is not much demand for speed, the op amp is not expensive. Instrumentation amplifiers are of course the best, and they are more expensive. 4) If a non-instrument op amp is used, the feedback resistor should not be too large, and the M-class is better. Otherwise, the resistance requirements are relatively high. The latter stage is further amplified by 2 steps, and a simple high-pass circuit is added in the middle to suppress 50 Hz interference.
Second, the operational amplifier's bias setting When the dual power supply is connected to a single power supply circuit, engineers will encounter some dilemmas in the setting of the bias voltage. For example, the bias voltage is a good voltage divider. Or is it better to use the reference voltage source? Some netizens recommend using a reference voltage source. The reason is high precision. In addition, it can provide a lower AC bypass. Some netizens recommend using a resistor. The reason is low cost and convenience. In this regard, I believe that when a dual-supply op amp is changed to a single-supply circuit, the best results are obtained if the reference voltage is used. This reference voltage allows the system to be designed for minimal noise and maximum PSRR. However, if the resistor divider method is used, the influence of the power supply ripple on the system must be considered. This usage noise is relatively high and the PSRR is relatively low. 
Third, how to solve the zero drift problem of the operational amplifier? Some netizens pointed out that the general piezoelectric acceleration sensor will be connected to a primary charge amplifier to achieve charge-voltage conversion, but when the sensor is working dynamically, the output voltage of the charge amplifier will not return. Zero phenomenon occurs, how to solve this problem? In this regard, netizens analyzed that there are several possibilities that will lead to zero drift: 1) feedback capacitor ESR characteristics are not good, change with the change of charge; 2) both ends of the feedback capacitor The resistor is not connected, for the operation of the amplifier is stable, the zero drift is reduced, and the resistor is applied across the feedback capacitor to form a DC negative feedback to stabilize the DC operating point of the amplifier; 3) the input impedance of the operational amplifier that may be selected is not high enough, resulting in The charge leaks, causing zero drift. Some netizens also analyzed the causes of zero drift from the perspective of mathematical analysis. It is believed that in addition to making the interference source drift small, the sensor and cable resistance must be large, and the open-loop input impedance of the op amp should be high. The feedback resistance is small, that is, the feedback resistor is used to prevent drift and stabilize the DC operating point. However, if the feedback resistance is too small, it will also affect the lower frequency limit of the amplifier. Therefore, it must be considered comprehensively! For the phenomenon that the output voltage of the charge amplifier does not return to zero, the following methods are generally used to solve: 1) using the technique of the switched capacitor circuit, the offset voltage can be effectively eliminated by using the CDS sampling method; 2) using the synchronous detection circuit structure, Can effectively eliminate the offset voltage.
In fact, more important than these three steps are the four usage details.
1. Which kind of amplifier operational amplifier basic circuit has inverting amplifier and non-inverting amplifier, how to choose in actual use? If the input and output requirements are reversed, of course, inverting amplifier should be used. If the amplified signal is AC signal, there is no phase. A non-inverting amplifier or an inverting amplifier can be used as required. Which is better? This should be analyzed according to the specific situation.
The advantage of using an inverting amplifier is that the op amp's potential at both inputs is always zero, with or without input signals. There is only a differential signal (or differential mode signal) below the μV level between the two inputs. In addition to the extremely small differential mode signal between the two inputs of the non-inverting input amplifier, there is also a large common mode voltage.
Although the op amp has a large common mode rejection ratio, it will also cause some error due to the common mode voltage. The advantage of the non-inverting amplifier is that the input impedance is extremely high, so the input resistance has little effect on the large and small, and the input impedance Zi of the inverting amplifier is related to the magnitude of the input resistance Ri (the input impedance Zi is equal to the input resistance Ri)
For example, the input impedance requires 100kΩ; if the gain is 300, if an inverting amplifier is used, Ri=100kΩ, Rf=30MΩ. Such a large feedback resistance is difficult for a general-purpose operational amplifier to work properly. In this case, it is more appropriate to use a non-inverting amplifier. In addition, look at the internal resistance of the signal source. Some sensors have large internal resistance. If an amplifier circuit with a small input impedance is used, the measurement accuracy will be affected. In this case, it is more suitable to use a non-inverting amplifier. Here is a circuit that uses both an inverting amplifier and a feedback resistor with a large resistance. As shown in Figure 1, the feedback resistor Rf in this circuit is not directly connected to the output, but is composed of R1 and R2. The midpoint A of the voltage divider. Some analysis of this circuit is now done. This circuit requires an input impedance of 100KΩ and a gain of -500. According to the general inverting amplifier design, Ri = 100 KΩ, Rf = 50MΩ. The partial pressure ratio at point A is R1/(R1+R2)=1/500, and there is R1 "Rf. According to the principle of “virtual short†and “virtual breakâ€, the following formula can be listed: Ii=Vi/100KΩ=If, IfRf=-VA,
Substituting VA=-Vi.
The node current equation can be listed from point A: Ii+If=I2 and Ii=(0-VA)/R1=Vi/100; therefore, I2=(Vi/100Ω)+(Vi/100KΩ)≈Vi/100Ω. This can be found:
V0=VA-I2R2=-Vi-(Vi/100Ω)X49.9kΩ=-500Vi, which satisfies the requirement of Vo/Vi=-500. Excluding If in the calculation will cause some errors, but since Rf>>R1, in this case If is only one thousandth of Ii, this error is not large.
If a non-inverting amplifier is required to make the input and output in phase, the circuit of Figure 2 can be used. Readers can derive their own gains.
Second, the choice of Ri, Rf From the gain formula of the inverting amplifier and the non-inverting amplifier, the gain depends on the ratio of Rf to Ri. And through experiments, within a certain range, Ri and Rf change, as long as the ratio between them does not change, the gain does not change. In the specific design, the resistance value is selected as the servo? In the previous experiment, there was no load. The actual amplifier circuit has a load resistor RL; as shown in Figure 3.
1. The value of the resistor is too small. Design an inverting amplifier. If the gain is -100, can I take Ri=10Ω, Rf=100Ω, Rf=lkΩ? This ratio is correct, but it is not practical. This can be analyzed from two aspects: (1) The current output by the op amp is flowing to the load resistor and Rf, which can be regarded as parallel connection between RL and Rf. Therefore, when Rf is very small (such as 10-100 Ω), the current flowing through Rf is large, and the limited output current capability of the op amp cannot be fully utilized, and even the amplifier is quickly saturated, and the output voltage range of the amplifier becomes very small. , that is, the linear range is very narrow. For example, if Rf is 500Ω, when the output voltage is 10V, 20mA will flow into Rf (100mA if Rf is 100Ω), and the general output current of the op amp is only 10-20mA. In addition, Rf is too small, so that the tube is too expensive, the heat is serious, and the device may be burned out. (2) The input impedance of the inverting amplifier is equal to Ri, so the Ri is very small and its input impedance is small. When the internal resistance of the signal source is large, the signal will not be output.
2. The value of the resistor is too large. When Ri and Rf are too large, it may cause large current drift interference. If Rf takes 10MΩ, the 100nA working bias current IiB will form a 1V voltage drop affecting output on Rf. If IiB changes slightly, it will cause serious drift of the output signal and form a circuit sensitive to external interference. If you touch the 10MΩ resistor with your finger, the output is a topped waveform with noise mixed. In addition, due to the factor of distributed capacitance, when the operating frequency is high, the frequency characteristics will deteriorate. Generally, the resistance is selected between 1kΩ and 1MΩ, and the selection between 10 and 100KΩ is more common. The second question is how to choose the accuracy level of the resistor. In applications where the magnification is not critical, such as the preamplifier circuit of the audio circuit, a 5% precision resistor can be used. Amplifiers with high accuracy requirements can use 1% or 0.5% high precision resistors. Metal film resistors are generally used. In amateur conditions, you can use the digital multimeter ohm file to select the resistor, and live it for better results. It should be pointed out here that the resistance of the resistor will also change with temperature, it will affect the temperature drift of the amplifier, so in the amplifier with higher accuracy requirements, the resistor with smaller temperature coefficient should be selected, generally high precision. The resistance has a temperature coefficient of 100 ppm/. Below C (ppm means one percent). Finally, mention the power of the resistor. Since the current flowing through the Ri, Rf, and Rp resistors is small, a metal film resistor of 1/8 W-l/16 W is generally available, and 1/4 W is also available.
3. Can I replace Ri, Rf with a semi-adjustable resistor? Since the resistance of the resistor has a certain error, the set gain can be different from the actual gain, and the precision resistor is expensive and has a large size. Can the semi-adjustable resistor be used instead of Ri and Rf, To meet the gain requirements, as shown in Figure 4 (a), (b), (c): In principle, all three options are possible, but there are actually some problems. First of all, the contact of the semi-adjustable resistor is not reliable, and its resistance may change when it is vibrated or impacted. The resistance layer of a general semi-adjustable resistor is a carbon film, which has a large temperature coefficient and other temperature coefficient of resistance. The number mismatch is prone to temperature drift, so it is not suitable for high-precision amplifiers. Comparing with Figures 4(a) and 4(b), the (b) scheme is better, and the change of Rf has less influence on Rp. Scheme (c) Although the gain adjustment range is extremely large, the actual adjustment range should not be too large. It is better to use precision multi-turn potentiometer in gain adjustment, which is characterized by convenient adjustment, accuracy and small temperature coefficient.
Third, the application of the buffer Figure 5 (a) is a familiar phase amplifier circuit. If Ri = ∞ and Rf = 0 in the circuit, the circuit becomes what looks like Figure 5(b), which is the buffer circuit. It can be known from the non-inverting amplifier gain formula 1+Rf/Ri that if Rf=0 and Ri=∞, the gain is 1. Similarly, the principle of “virtual short†is used to analyze: the voltage at the inverting input is equal to the voltage at the non-inverting terminal, and the inverting terminal is directly connected to the output terminal, that is, V0=Vi. Therefore, the circuit with a gain of 1 is very simple. What are the peripheral components, what is the use? First, do an experiment according to Figure 6. Here is the soil 9V power supply, measured by digital meter, from the experimental results can be seen: at no load (ie not connected to 0.5k In the case of load resistance), the input voltage is equal to the output voltage in the range of -7.06V to +8.93V, and is in phase, and the voltage following range is quite large. When the load current is small, there is no difference between load and no load. If it is greater than 14mA, the output voltage is smaller than the input voltage, that is, the amplifier is saturated in advance. For the relationship between the output current and the output voltage, interested readers can use the circuit shown in Figure 7 to conduct the measurement experiment. It features extremely high input impedance, very low output impedance, and high output current. Therefore, it can be used as a buffer isolation stage "insertion" circuit. For example, some oscillator circuits can be directly connected to the load. The output characteristics of the oscillator are even stopped or interrupted, and the buffer is inserted between the oscillator and the load to work stably. For example, some sensors with high internal resistance, such as piezoelectric sensors, have very weak output signals and internal signals. The resistance is extremely high. If a general amplifier is directly connected to the sensor, a large measurement error will occur. At this time, a buffer is inserted between the sensor and the amplifier to perform impedance transformation.
Nantong Boxin Electronic Technology Co., Ltd. , https://www.ntbosen.com