1. Non-inverting amplifier circuit

Imaginary short: The ideal magnification of the op amp is 100,000 times. Generally, the output is more than 3V or 5V. The op amp amplifies the voltage difference of the input signal. If it is amplified by 100,000 times, it can only mean that the voltage difference of the input signal is very close. , approximately equal, we call it virtual short (with the relationship of the magnification of 100,000 times, only U+ is approximately equal to U-, which is called virtual short)

Virtual break: Because the input internal resistance of the op amp is infinite, the current on the external resistor is very small and approximately 0, so the input end of the op amp can be regarded as disconnected, which is called virtual break

Imaginary short: U1=U2,

Virtual break: (U2-0)/R2=(Uout-U2)/Rf

U2*Rf=(Uout-U2)*R2

U2*Rf=Uout*R2-U2*R2

Uout=(U2*Rf+U2*R2)/Rf

=U2*(Rf+R2)/Rf

=U2*(1+R2/RF)

=U1*(1+R2/RF)

Summary: The resistance value of the feedback resistor Rf should not be greater than 300K, otherwise the current on the feedback branch will be small and easily interfered. The resistance value of the op amp circuit requires 1% accuracy

Summarize:

1. The same-direction amplifying current also uses the virtual short and virtual breaking characteristics of the op amp, Ohm’s law and the characteristics of the current on the same branch to be equal everywhere to list the formula for conversion and finally draw the conclusion.

2. Advantages: The requirements for the input signal source are extremely low, it can be a large internal resistance signal source, and the signal input loop current is small

3. Disadvantages: The output current capability of the signal source is poor, the anti-interference ability is poor, and it is easy to be interfered

4. In order to pursue the magnification, many people increase Rf and reduce Ra at the same time. When reducing Ra, is it necessary for the Ui to have a stronger load capacity, that is, the current of the input signal should be larger, and at the same time, it also has certain effects on the input signal. requirements, not all signals are suitable, so the output waveform will eventually be distorted

Second, the reverse amplifier circuit

1. Inverting amplifier circuit: the input is positive, the output is negative, the input is negative, and the output is positive

2. The above figure discusses the situation of the positive and negative power supply pairs, otherwise the output of the negative power supply will be cut off.

3. According to the false conclusion, U2=0, and from the false short: U1=U2=0

Practical application of the reverse amplifier circuit: (raise the threshold circuit)

Calculation method:

Summarize:

1. From the results, a 2.5V DC voltage is added to the voltage difference, raising 2.5V, and the remaining magnification can be obtained by resistor matching.

2. Advantages: The signal source has strong output current capability and strong anti-interference ability.

3. Disadvantages: high requirements for the input signal source, requiring the signal source to provide a large current

3. Op amp follower circuit

analysis Summary:

1. Op-amp follower circuit: Assuming that the output is 0 at the beginning, the difference between U2 and Ui is adjusted through the negative feedback line, and then the output increases to reduce the difference between the input terminals. Finally, after several adjustments, the output is approximately Uout=Ui, which tends to in stability

2. Because the ideal magnification of the op amp is hundreds of thousands of times, plus negative feedback, if you want to stabilize, the input terminals need to be approximately equal

3. Op-amp follower circuit: After several cycles, it reaches a balanced state. Rf resistance: generally take 100R to destroy the oscillation, and 10K is also possible.

4. When designing and circuit, it is necessary to avoid direct connection of pins, which is prone to oscillation

5. Operational amplifier follower circuit characteristics: output Uo=Ui, with enhanced load capacity

4. T-type amplifier circuit

Assuming that the op amp is powered by positive and negative power supplies, as shown in the conclusion of the reverse amplifier circuit in the above figure, if you want to increase the magnification of the op amp, you should increase the Rf resistance and reduce the resistance of the Ra resistance, but if you take Ra to the limit If the value is too small, the power consumption is large, the anti-interference ability is too large, and the Rf value is also taken to the extreme value of 300K. If the magnification is still unable to meet the requirements at this time, and do not want to continue to consider how to continue to amplify on the basis of the first-level amplification, it is necessary to To meet the anti-interference ability of the circuit and increase the magnification, only by increasing the resistance of the Rf resistor and increasing the current of the Rf loop at the same time can it be satisfied at the same time, then divide the RF into 2 and become RF1 and The voltage divider between RF2, RF1 and RF2 is still Uo, and it is ensured that the current on RF1 is equal to the current of the original reverse amplifier circuit, Uo is equal to the original Uout, assuming that Ui is positive and the output Uo2 is negative, then the voltage value of Uo greater than Uo2, the current on the entire feedback loop is generally increased I4=I2+I3, I2 and the original reverse amplification, the feedback current of the circuit remains the same, the anti-interference ability is the same, Ra and Rf1 are also the same as the original The circuit remains the same, and Uo is also the same as the original circuit at this time, the increased Rf2 is the increased impedance in the feedback loop, and the loop current is increased to maintain the anti-interference ability, so the T-type circuit came out as the times require, and the calculation and derivation are more complicated. I will send it as a screenshot, and I will continue to update the actual operation of the op amp circuit on this basis. If there are errors in the text, please point out a lot of common progress.

Summarize:

1. The magnification is ultimately related to the two products selected in the box. The maximum value of RF1 is 300K, and the minimum value of Ra is 10K. If the resistance value is too low, it will clamp the internal resistance of the signal source, or the internal resistance of the signal source will affect the signal quality. The load carrying capacity of the source is very poor. If the value of Ra is too small, it will increase the load carrying capacity of the signal source and affect the signal quality.

Five, op amp addition circuit

1. Calculation idea of ​​op amp adding circuit: Assume Vin1 is calculated, the other two paths assume that the input signal is 0, and the default is grounded

2. Calculation idea of ​​op amp adding circuit: Assume that Vin2 is calculated, the other two paths assume that the input signal is 0, and the default is grounded

3. Calculation idea of ​​op amp adding circuit: Suppose Vin3 is calculated, and the other two paths assume that the input signal is 0, and the default is grounded

Operational amplifier adding circuit formula derivation (no impedance matching)

Six, op amp current detection circuit

analyze:

In the above picture, U1=U2=0V, if Ui is positive and Uo is negative, there are two loops in the above picture, the red loop UI is positive, Uo is negative, in the black loop, the left side of D2 is negative, and the right side passes through RD and RA are pulled up to Ui, they are positive, so D2 is blocked, so D2 can be removed, so the black branch is blocked, and DR and RA/RB are not on the same order of magnitude, internal resistance method analysis, small internal resistance clamps large internal resistance resistance, so RD does not affect UI2, so UI2=1/2Ui, we regard Uo2 as the final output.

If DR fails, both ends are 1/2Ui. According to internal resistance method analysis, failure is equivalent to infinite resistance. 100K appears very small in front of infinity, so Uo2=Ui2=1/2Ui

It is known that when U1=U2=0V and Ui is negative, Uo is positive and Uo2 is also positive, there will be two loops, the black loop exists, and the red loop also exists, because U1=0V, U0 is positive, so D1 is off, and Because the internal resistance to ground of the op amp is infinite, the current on the U1 branch is approximately 0, which can be regarded as 0, then the Rb branch and D1 can be removed, because the positive and negative input terminals of the op amp need input signals, so Rb is reserved, but There is no current in this branch

At this time, there is only one black loop in the loop, because U1=0V, and the RB branch current is 0, so the voltages at both ends of Rb are equal, that is, Ui2=0V, at this time, the currents flowing through Rd and Ra are equal, and the equation

Circuit use: use the current-sensing resistor to detect the motor current sampling, over-current circuit, and inversion can also detect over-current, such as: it will be used when doing PFC (power factor correction), in some circuits, we need to negative treat it as positive, is it easy to deal with this reversal?

Summarize:

1. The capacitor connected in parallel on the negative feedback resistor is used for phase compensation. It is generally pF level, at most a few nF, and a few tens of nF is large.

Reviewing Editor: Tang Zihong

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