Virtual short "" virtual broken "two axes, deal with 11 operational amplifiers details

The circuit composed of computing amplifiers is varied and confusing. It is the focus of learning analog circuits. When analyzing its working principles, if the core is not grasped, it is often very big. For this reason, I have collected the application of the world's operational amplifier circuit. I hope that you will be engaged in circuit board repair.
All the books and courses on analog electronic technology show that when introducing the operational amplifier circuit, it is nothing more than giving a qualitative review of the circuit. For example, this is a same-direction amplifier, then, the relationship between the output and the input is deduced. Then Vo = (1 + Rf) VI is obtained, which is a reverse amplifier, and Vo =-RF * VI ...... Finally, students often come to the impression that they can remember the formula! If we change the circuit a little bit, they will not be able to find the north! I have interviewed at least 100 electronic applicants with a college degree or above, and I can analyze the operational amplifier circuit I have provided a little better than 10 people! Other major graduates can even be imagined.
Today, chip-level maintenance teaches you two invincible tricks. These two tricks are clearly written in the textbooks of all operational amplifier circuits, that is, "virtual short" and "virtual broken ", however, if you want to use it in an efficient way, you must have a deep foundation.
Concepts of virtual short term and virtual break
Generally, the open-loop voltage magnification of general-purpose operational amplifiers is 80 DB or above. However, the output voltage of the op amp is limited. 10 V ~ 14 V. Therefore, the input voltage of the op amp is less than 1 MV, equivalent to equipotential at the two input ends "Short Circuit ". The larger the open-loop voltage magnification, the closer the potential of the two inputs is to be equal.
"Virtual short" refers to the analysis of the Operational Amplifier in a linear state, the two inputs can be considered as an equipotential, this feature is called false short circuit, short virtual short. Obviously, the two inputs cannot be short-circuited.
Due to the large input resistance of the differential mode of the operational amplifier, the input resistance of the general-purpose operational amplifier is above 1 Mbit/s. Therefore, the input current of the operational amplifier is usually less than 1 UA, which is far less than the current of the input circuit. Hence Generally, the two inputs of an operational amplifier are regarded as open circuits. The larger the input resistance is, the closer the two inputs are to open circuits. "Virtual disconnection" means that when the analysis op-amp is in a linear state, the two inputs can be considered as an equivalent open circuit. This feature It is called a false open circuit, or virtual disconnection for short. Obviously, the two inputs cannot be truly broken.
When analyzing the operational principle of the op-amplifier circuit, please forget about what is in the same direction amplification, reverse amplification, what is the addition, subtraction, what is the differential input ...... Forget the formulas for the input/output relationships ...... These things will only interfere with you, making you even more confused. Please ignore the input bias current, common mode rejection ratio, offset voltage, and other circuit parameters for the moment. This is something the designer should consider. What we understand is the ideal amplifier (in fact, the actual amplifier is used as the ideal Amplifier for analysis during maintenance and most design processes ).
Well, let's take two "board axes" ------ "virtual short" and "virtual broken" and start "ding Jie niu.

Figure 1 op amp. Op. grounding at the same end = 0 V. The reverse end and the same end are short, so it is also 0 V. The input resistance at the opposite end is very high, and there is a virtual disconnection, almost no current injection and outflow, then R1 and R2 are equivalent to series. The current flowing through each component in a series circuit is the same, that is, the current flowing through R1 is the same as the current flowing through R2. Current I1 = (Vi-v-)/R1…… flowing through r1 ...... Current of a flowing through R2 I2 = (V-Vout)/R2 ...... B V-= V + = 0 ...... C I1 = I2 ...... D. Solve the above junior high school algebra equation vout = (-R2/R1) * VI. This is the input and output relationship of the legendary reverse amplifier.

Figure 2 VI and V-virtual short, Vi = V -...... A. Due to virtual disconnection, there is no current input and output at the reverse input end. The current is equal through R1 and R2, and this current is set to I, obtained by Ohm's Law: I = Vout/(R1 + R2) ...... B vi is equal to the partial pressure on R2, that is, Vi = I * R2 ...... C uses the abc formula vout = VI * (R1 + R2)/R2. This is the legendary formula for the same-direction amplifier.

Figure 3: virtual short note: V-= V + = 0 ...... A is known from the theory of virtual disconnections and kierhov's law that the sum of current through R2 and R1 is equal to the current through R3, so (V1-v-)/R1 + (v2-v -) /r2 = (Vout-v-)/R3 ...... B is substituted into expression A. Type B is changed to V1/R1 + v2/r2 = Vout/R3. If R1 = R2 = R3 is obtained, the above formula is changed to vout = V1 + V2, this is the legendary reverse calculator.

See Figure 4. Because of the virtual disconnection, if there is no current flowing through the same end of the op amp, the current flowing through R1 and R2 is the same, and the current flowing through R4 and R3 is the same. So (V1-V +)/R1 = (V +-V2)/R2 ...... A (Vout-v-)/R3 = V-/R4 ...... B: V + = V -...... C If R1 = R2, R3 = R4, you can export v ++ = (V1 + V2) from the preceding formula) /2 V-= Vout/2. Therefore, vout = V1 + V2 is also a calculator!

Figure 5 from the virtual disconnection, the current passing through R1 is equal to the current passing through R2, and the current passing through R4 is equal to the current of R3, so there is (v2-V +) /R1 = V +/R2 ...... A (V1-v-)/r4 = (V-Vout)/R3 ...... B. If R1 = R2, V + = v2/2 ...... C If R3 = R4, V-= (Vout + V1)/2 ...... D from virtual short knowledge V + = V -...... E so vout = V2-V1 this is the legendary subtraction.

In figure 6, the voltage at the reverse input end is equal to that at the same direction, from virtual disconnection, and from the R1 current to the current through C1. Current through R1 I = V1/R1 current through C1 I = C * Duc/dt =-C * dvout/dt so vout = (-1/(R1 * C1)) the ∫ v1dt output voltage is proportional to the time point of the input voltage, which is the legendary integral circuit. If V1 is a constant voltage U, the above conversion is vout =-u * t/(R1 * C1) T is time, the Vout output voltage is a straight line from 0 to the negative supply voltage that varies by time.

In Figure 7, the current of the capacitor C1 and resistance R2 is the same from the virtual transient, and the voltage at the same direction is the same as that at the reverse end. Then: vout =-I * r2 =-(R2 * C1) dv1/dt this is a differential circuit. If V1 is a sudden addition of DC voltage, the output Vout corresponds to a pulse in the opposite direction of V1.

Figure 8. zomkochi vx = V1 ...... A Vy = v2 ...... B is determined by virtual disconnection. If there is no current flowing through the input end of the Op Amplifier, R1, R2, and R3 can be considered as series connections. The current through each resistor is the same, and the current I = (VX-Vy) /R2 ...... C: Vo1-Vo2 = I * (R1 + r2 + R3) = (VX-Vy) (R1 + r2 + R3)/R2 ...... D. It is determined by virtual disconnection that the current flowing through R6 is equal to the current flowing through R7. If R6 = R7, vw = Vo/2 ...... E Similarly, if r4 = R5, Vout-vu = vu-vo1, so vu = (Vout + vo1)/2 ...... F from virtual short knowledge, VU = VW ...... G from EFG vout = Vo-vo1 ...... H is determined by DH vout = (Vy-VX) (R1 + r2 + R3)/R2 (R1 + r2 + R3)/R2, this value determines the magnification of the difference value (Vy-VX. This circuit is the legendary differential amplification circuit.

Analysis of a circuit that people are exposed. Many controllers accept 0 ~ 20mA or 4 ~ 20mA current. The circuit converts the current to a voltage and then sends the ADC to a digital signal. Figure 9 shows a typical circuit. 4 ~ 20mA current flow through the sampling 100 Ω resistance R1, 0.4 ~ will be generated on r1 ~ 2 V voltage difference. From the virtual disconnection, if there is no current flowing through the input end of the Op Amplifier, the current flowing through R3 and R5 is equal, and the current flowing through R2 and R4 is equal. So: (V2-Vy)/R3 = Vy/R5 ...... A (V1-Vx)/r2 = (VX-Vout)/R4 ...... B By virtual short knowledge: vx = Vy ...... C Current from 0 ~ 20mA change, then V1 = V2 + (0.4 ~ 2 )...... D is derived from the CD type into the B type (V2 + (0.4 ~ 2)-Vy)/r2 = (Vy-Vout)/R4 ...... E if R3 = R2, r4 = R5, then vout =-(0.4 ~ 2) R4/R2 ...... In Figure F, R4/r2 = 22 K/10 k = 2.2, then the F-Type vout =-(0.88 ~ 4.4) V, that is, 4 ~ 20mA current is converted to-0.88 ~ -4 V voltage, which can be sent to the ADC for processing.

Current can be converted to voltage, and voltage can also be converted to current. Figure 10 is such a circuit. The negative feedback is not directly fed back by resistance, but is connected to the launch knot of the transistor Q1. Do not think it is a comparator. As long as it is a amplifying circuit, the law of virtual transient disconnection is still consistent!
It is determined by virtual disconnection that there is no current flowing at the input end of the Op Amplifier,
Then (Vi - V1)/R2 = (V1 - V4)/R6 ...... A
Likewise (V3 - V2)/R5 = V2/r4 ...... B
Zhiming V1 = V2 ...... C
If r2 = R6, r4 = R5, the ABC-type V3-V4 = vi
The above formula indicates that the voltage at both ends of R7 is equal to the input voltage VI, the current I = VI/R7 through R7, if the load RL <100kb Ω, the current through RL and through R7 is basically the same.

it's a complicated one! Figure 11 is a three-line pt100 pre-amplification circuit. The pt100 sensor extracts three wires with the same material, diameter, and length. A 2 V voltage is added to a bridge circuit consisting of R14, R20, R15, Z1, pt100, and wire resistance. Z1, Z2, Z3, D11, D12, d83, and various capacitors play a filtering and protection role in the circuit. When static analysis is performed, it can be ignored. Z1, Z2, and Z3 can be considered as short circuits, d11, D12, d83, and each capacitor can be viewed as an open path. It is determined by the resistor partial pressure. V3 = 2 * R20/(R14 + 20) = 200/1100 = 2/11 ...... A is composed of virtual transient Zhi. The u8b 6th, 7-pin voltage, and 5th-pin voltage are equal to V4 = V3 ...... B is determined by the virtual disconnection. If there is no current flowing through the u8a 2nd foot, the current flowing through the R18 and r19 is equal. (V2-V4)/r19 = (V5-V2)/R18 ...... C is disconnected by virtual machines. There is no current flowing through the u8a 3rd Foot, V1 = V7 ...... D In the bridge circuit, R15 and Z1, pt100 and wire resistance are connected in series. The voltage obtained by the pt100 and wire resistance in series is added to the 3rd feet of u8a through the resistance R17, v7 = 2 * (RX + 2r0)/(R15 + RX + 2r0 )..... E. the voltage of the u8a 3rd Foot is equal to that of the 2nd foot, V1 = v2 ...... F by abcdef, (V5-V7)/100 = (V7-V3)/2.2 simplified V5 = (102.2 * V7-100V3)/2.2 I .e. V5 = 204.4 (RX + 2r0) /(1000 + RX + 2r0)-200/11 ...... G output voltage v5 is the RX function. Let's look at the effect of the line resistance. The voltage drop generated on the lowest-end line resistance of pt100 goes through the intermediate line resistance, Z2, R22, and adds it to the 10th foot of u8c, which is known by virtual disconnection, v5 = V8 = V9 = 2 * R0/(R15 + RX + 2r0 )...... A (V6-V10)/r25 = V10/R26 ...... B From virtual short knowledge, V10 = V5 ...... C By expression ABC V6 = (102.2/2.2) V5 = 204.4r0/[2.2 (1000 + RX + 2r0)]… H is composed of the equations of GH. If the V5 and V6 values are determined, the RX and R0 values can be calculated, and the RX values can be obtained. Check the pt100 score table to know the temperature.