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Operational amplifier application summary 1

2022-07-07 03:57:00 【liht_ one thousand six hundred and thirty-four】

The content includes empty short and empty broken understanding , Reverse phase 、 In phase proportional operational amplifier circuit , Voltage follower , Correlation arithmetic circuit （ Add 、 reduce 、 ride 、 except 、 integral 、 differential 、 Logarithm and exponent ）, Differential amplifier circuit ,I/V、V/I Conversion circuit , Voltage rise circuit ,F/V Conversion circuit , Active power filter , Phase shift circuit, etc . Purple text is a hyperlink , Click to automatically jump to relevant blog posts . Continuous updating , Originality is not easy. ！

Catalog ：

An overview of deficiency and discontinuity

One 、 Inverse proportional operation amplifier circuit

Two 、 In phase proportional operational amplifier circuit

3、 ... and 、 Voltage follower

Four 、 adder

5、 ... and 、 Subtracter

6、 ... and 、 Integral circuit

7、 ... and 、 Differential circuit

8、 ... and 、 Differential amplifier circuit

Nine 、I/V Conversion circuit

Ten 、V/I Conversion circuit

11、 ... and 、 Voltage rise circuit

Twelve 、F/V Conversion circuit

13、 ... and 、 Constant voltage source

fourteen 、 Logarithmic and exponential circuits

1、 Logarithmic circuit

2、 Exponential circuit

15、 ... and 、 Multiplication circuit

sixteen 、 Active power filter

1、 low pass filter

2、 High pass filter

seventeen 、 Monostable circuit

eighteen 、 Op amp as comparator ( Operational amplifier and comparator )

1、 The selection basis of the bias resistance of the comparator

2、 Maximum input voltage of comparator

nineteen 、 Dead zone circuit

twenty 、 Phase shift circuit

1、 Phase advance phase shift circuit

2、 Phase lag phase shift circuit

The 21st 、 Voltage source

1、 Standard voltage source based on comparator

2、 Voltage source based on op amp follower

appendix

1、 Integrated operational amplifier circuit Proteus Simulation

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An overview of deficiency and discontinuity

Because the voltage amplification of the operational amplifier is very large , The open-loop voltage amplification factor of general purpose operational amplifier is 80 dB above . The output voltage of the operational amplifier is limited , Generally in 10 V～14 V. Therefore, the differential mode input voltage of the operational amplifier is insufficient 1 mV, The two inputs are approximately equipotential , amount to “ A short circuit ”. The larger the open-loop voltage magnification , The closer the potentials of the two inputs are equal .“ Deficiency ” It refers to when the analysis operational amplifier is in the linear state , The two inputs can be regarded as equipotential , This characteristic is called false short circuit , Short for short . Obviously, the two inputs cannot be really short circuited .

Because the differential mode input resistance of the operational amplifier is very large , The input resistance of general-purpose operational amplifier is 1MΩ above . Therefore, the current flowing into the input of the operational amplifier is often insufficient 1uA, Far less than the current of the circuit outside the input . so Generally, the two inputs of the operational amplifier can be regarded as open circuit , And the greater the input resistance , The closer the two inputs are to the open circuit .“ Virtual break ” It refers to when analyzing that the op amp is in a linear state , The two inputs can be regarded as equivalent open circuit , This feature It's called a false open circuit , Short for virtual break . Obviously, you can't really open the two inputs .

In the process of analyzing the circuit , Don't worry about other features of op amp for the time being , It is analyzed according to the characteristics of virtual short and virtual break . Of course , If the op amp does not work in the amplification area , It does not meet the conditions of virtual short and virtual break , This method cannot be used to analyze . Such as comparator .

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One 、 Inverse proportional operation amplifier circuit

Voltage parallel negative feedback

shortcoming ： The input resistance is less than R1;

advantage ： Common mode voltage ＝(U+ + U-)/2=0, Low output resistance ;

characteristic ： Falsely .

The signal voltage passes through the resistor R1 Add to the inverting input of the op amp , Output voltage Vo Through the feedback resistor Rf Feed back to the inverting input of the op amp , Constitute a voltage parallel negative feedback amplification circuit .

The in-phase terminal of the operational amplifier is grounded 0V, Inverting end and in-phase end “ Deficiency ”, So too. 0V, The input resistance of the inverting input terminal is very high “ Virtual break ”, There is almost no current injection and outflow , that R1 and Rf Equivalent to a series of , The current flowing through each component in a series circuit is the same , That is, flow through R1 The current and flow through Rf The current is the same .

Is= (Vs- V-)/R1

If= (V- - Vo)/Rf

V- = V+ = 0

Is= If

To solve the need to Vo==(-Rf/R1)*Vi

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Two 、 In phase proportional operational amplifier circuit

Voltage parallel negative feedback

shortcoming ： According to the deficiency ,U+＝Vs,U-=Vs, Common mode voltage ＝(Vs+Vs)/2=Vs;

advantage ： Input resistance =Vs/Ii, According to the virtual fault flow Rs Current Ii=0, High input resistance , Low output resistance .

According to the deficiency vN= vP=Vs, The input resistance of the inverting input terminal is very high “ Virtual break ”, There is almost no current injection and outflow , that R1 and Rf Equivalent to a series of , The current flowing through each component in a series circuit is the same , That is, flow through R1 The current and flow through Rf The current is the same .Vs/R1=(Vo-Vs)/Rf, have to vo=(1+Rf/R1)Vs

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3、 ... and 、 Voltage follower

R25 Antireflection , Operational release 5、6 Theoretically, the voltage of the pin is the same , And the input impedance is infinite ！ So the current of the input signal is mainly through R28 The inflow , That is, the voltage at the input point is WK-in Point formation , Theoretically, there will be no current flowing in R25, without R25 Then the signal will 100% Reflect on WK-in On , If the internal resistance of the signal source is very large , That is to say, the carrying capacity is very poor , The reflected signal will be in R28 A strong emission oscillation is formed near the input point, that is “ echo ” Such noise will make the output signal quality very poor after amplification ,R25 and C12 The access can be put in 5pin The reflected signal is effectively absorbed , High frequency reflected signals pass through C12 To the ground （AGND）R25 Block the reflected signal from 5pin The input end of the . that R25 Why 20K Well ？ This could be an empirical value ,R25 It will affect 5pin After all, the op amp is not ideal. It also reflects a lot of signals , Small, like a wire, can't block the reflected signal . You usually get R28 Of 2-3 Like this .R28、R25、R27 The selection of operational impedance depends on the operational impedance of the amplifier .

Voltage parallel negative feedback

advantage ： High input resistance , Low output resistance .

LM358 Connect it into a follower , Conduct Impedance conversion . The input impedance is very large , Minimize the impact on the previous input .

At this time, the amplifier has the function of current amplification , No voltage amplification .V1 voltage ＝V2 voltage .

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Four 、 adder

Due to the virtual short circuit , Net input voltage of op amp vI=0, The inverse end is virtual .

vI=0,vN=0.

Input current at the inverting end iI=0 The concept of , adopt R2 And R1 The sum of the currents is equal to passing through Rf The current of (Vs1 – V-)/R1 + (Vs2 – V-)/R2 = (V- –Vo)/Rf

If you take R1=R2=R3, from a,b The solution of the two formulas is -Vout=Vs1+Vs2

The minus sign in the formula is caused by inverse input , If another stage of inverse circuit is connected , The minus sign can be eliminated .

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5、 ... and 、 Subtracter

Know from the falsehood , adopt R1 The current of is equal to passing through R2 Current , By the same token R4 The current of is equal to R3 Current , There are ：

(V2 – V+)/R1 = V+/R2

(V1 – V-)/R4 = (V- - Vout)/R3

If R1=R2, be V+ = V2/2

If R3=R4, be V- = (Vout + V1)/2

From the empty short knowledge V+ = V-

therefore Vout=V2-V1 This is the subtracter .

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6、 ... and 、 Integral circuit

Use the integral operation circuit to integrate different input signals , You can change the waveform .

For constant direct flow integral operation , It can improve the linearity of output voltage ;

Integrate the square wave , It can output triangular waves , The waveform is transformed ;

Integrate sine quantity , The output frequency can be the same , But amplitude 、 Cosine quantities with different initial phases .

The simulation example can be seen in Baidu online disk “ be based on Multisim10 Simulation of integral operation circuit of ”.

From the empty short knowledge , The voltage at the reverse input is equal to that at the same direction , Know from the falsehood , adopt R1 The current and current passing through C1 The current is equal to .

adopt R1 Current i=V1/R1

adopt C1 Current i=C*dUc/dt=-C*dVout/dt

therefore Vout=((-1/(R1*C1))∫V1dt The output voltage is proportional to the integral of the input voltage over time , This is the integral circuit .

if V1 Is a constant voltage U, Then the above formula is transformed into Vout = -U*t/(R1*C1) t Is time , be Vout The output voltage is a line from 0 To the straight line of negative power supply voltage varying with time .

A new solution to the principle of integral circuit -- Amplifier and capacitor “ Three changes ”.

Compare the same phase composed of resistance and arithmetic circuit 、 Inverse operational amplifier circuit , For the integral amplifier composed of capacitance and operational amplifier , How to understand and master in principle , Ordinary people often find it more difficult . The feedback resistor in the inverting amplifier , Replace with capacitors , It becomes an integral amplifier circuit as shown in Figure 1 . For resistance , It seems to be something more practical , The output state of the circuit can be seen at a glance , Replace with capacitors , Due to charging 、 Uncertainty of discharge , Capacitance is also a relatively “ virtual ” Object of , Its circuit output state , It's a little hard to figure out .

Figure 1

Composition of integrating circuit and signal waveform diagram Want to understand its output state , We must first understand the temperament of capacitance . The basic function of a capacitor is to charge 、 discharge , It's an energy storage element . Sensitive to changing voltage （ A strong reaction ）, Insensitive to DC （ Even indifferent ）, It has the characteristics of AC and DC isolation . For people who see that everything in the world is resistive , Capacitance can also be regarded as a variable resistance , Thus, the mystery of the output of the integrating circuit can be solved . According to the law of conservation of energy , Energy cannot be generated for no reason , Nor can it disappear for no reason , The theorem that the voltage at both ends of the capacitor cannot change suddenly is derived from it .

1） Charging moment , Charge has not yet accumulated between the two plates of the capacitor , Therefore, it can maintain the original state of zero voltage at both ends , But the instantaneous charging current is the maximum , It can be equivalent to a very small resistance or even a wire , If the capacitor is short circuited at the moment of charging , Not too bad , For example, in the main circuit of frequency converter , Limited current charging measures shall be taken for loop capacitance , It's just that ;

2） During capacitor charging , Over time , The charging voltage increases gradually , And the charging current gradually decreases , It can also be considered that the equivalent resistance of the capacitor changes from the minimum to the maximum ;

3） After the capacitor is fully charged , The voltage at both ends is the highest , But the charging current is basically zero , At this time, the capacitance is equivalent to the maximum resistance , For DC , It can even be equivalent to an open circuit , Infinite resistance .

4） Summarize the above , During capacitor charging , There is equivalent to the minimum resistance or conductor 、 Equivalent to the resistance from small to large 、 Equivalent to three states such as maximum resistance or open circuit . It is this changing characteristic of capacitance , The integral amplifier circuit can be transformed into three identities as shown in Figure 2 .

Figure 2

In the working process of integrating circuit “ Three changes ” See Figure 2 .

1） Voltage follower . In the input signal t0（ Forward jump ） moment , The capacitor charging current is maximum , Minimum equivalent resistance （ Or as a conductor ）, The circuit immediately turns into a voltage follower circuit , From the virtual ground characteristics of the circuit , The output is still 0V.

2） Inverting amplifier . In the input signal t0 After the moment, during the flat top , The capacitor is in a relatively gentle charging process , Its equivalent RP Experience less than R、 be equal to R And greater than R Three stages of , So in the amplification process , Under the action of amplification characteristics , In fact, it has experienced inverse attenuation 、 Reverse phase 、 Three small processes such as inverse amplification . Whether it's attenuation 、 Inverse or inverse amplification , At this stage , The integrating circuit actually acts as a linear amplifier .

3） In the second half of the input signal level term , The charging process of the capacitor is over , The charging current is zero , The capacitance is equivalent to an open circuit , The integral amplifier is amplified from closed-loop to open-loop comparison , The circuit then becomes a voltage comparator . At this time, the output value is the negative power supply value . It is said that people will change their faces , In fact, the circuit can also be transformed . Under the control of capacitance , The amplifier instantly changes three identities . Can see through these three identities of the integral amplifier , Integral amplifier “ Real body ” There is no hiding . amplifier , In fact “ Zoom in and compare , It's easier to zoom in ” Dancing and playing in the circle .

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7、 ... and 、 Differential circuit

Know from the falsehood , Through capacitance C1 And resistance R2 The current is equal , From the empty short knowledge , The voltage at the same end and the opposite end of the operational amplifier is equal .

be ： Vout = -i * R2 = -(R2*C1)dV1/dt, This is a differential circuit .

If V1 It's a sudden DC voltage , The output Vout Correspond to one direction and V1 Opposite pulse .

The OP AMP in the upper circuit can amplify the AC signal , At the same time, the network is a high pass filter , The phase lag of the signal 90°, The system may be unstable , So as to enter the situation of self-excited oscillation . The common differential circuit will be this type ：

In the improved differential circuit , Input resistance and feedback capacitance are added , It is believed that this kind of circuit will often be seen in actual analog signal processing , It is precisely because of the introduction of these two elements , Make the signal produce 90° Phase shift of , such , It can keep the system stable .

But the circuit is not perfect , It is affected by the frequency of the input signal , When the frequency is too high , It will become an integral circuit .

An obvious feature of op amp is that it is easily affected by bias current , In order to minimize the influence of differential circuit , Usually we will add a resistor at the positive and negative input terminals , Limit the bias current , Typical circuits are as follows ：

Sometimes a bias resistor is added to the forward input , The size is equal to the size of the feedback resistance .

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8、 ... and 、 Differential amplifier circuit

All the amplification circuits mentioned above have an obvious feature , That is, they just amplify a certain potential point , The other potential point is grounded by default . And sometimes we There is no grounding at both ends of the potential that needs to amplify the voltage , Suitable for differential amplifier circuit .

chart 1 Basic circuit form (R1=R3,R2=R4)

Differential amplifier , According to the input 、 Different output methods , It can be divided into two terminal inputs 、 Double ended output ; Double ended input 、 Single ended output ; Single ended input 、 Double ended output , Single ended input 、 Single ended output and other circuit forms , Among them, for the differential amplifier composed of operational amplifier device circuit , Double ended input 、 The circuit form of single ended output is widely used .

The circuit advantages of differential amplifier ： Amplify differential mode signal and suppress common mode signal , In terms of anti-interference performance “ Something extraordinary ”, This is inseparable from its circuit structure . You can use two triode circuits to build one, as shown in the figure 1 Medium a circuit , Explain the circuit characteristics of the differential amplifier .

（1） Amplifier circuit for single power supply , Its output （ namely Q1\Q2 Of C extremely ） Static working point is 1/2Vcc Most suitable , It can guarantee its maximum dynamic output range . as long as RC1、RB1 The value of equal offset element is appropriate , Can make UC1、UC2 The static voltage of is 2.5V, That is, static differential output voltage 2.5V－2.5V＝0V;

（2） The circuit design should make Q1、Q2 The static working parameters of are consistent , The two constitute “ Mirror image ” circuit ,RE Is the current negative feedback resistance , Its DC resistance is small , The dynamic resistance is extremely （ The current flowing is almost constant ）, To improve the differential performance of the circuit .

（3） When IN+=IN- when , Or when the signal voltage of both rises and falls synchronously ,OUT+、OUT- The terminal voltage is also rising and falling synchronously , And rise 、 The decrease is equal , Its input differential output value will still be 0V. For example, two input signals are generated on a static basis Q1、Q2 The same increment of base current , Then the collector voltage will drop , Such as by 2.5V Reduced to 1.5V when , be UC1-UC2=1.5V-1.5V=0V, This shows that the circuit ignores the common mode input signal , It has excellent anti-interference performance .

as everyone knows ,RS485 Communication circuit , Is to use differential bus transmission , It has a strong anti-interference effect .

（4） When IN+、IN- The input signal has relative changes on a static basis , namely IN+-IN-≠0 when , Such as IN+ When the input voltage changes in the positive direction ,OUT- Will change in the negative direction （ meanwhile OUT+ Will change in a positive direction ）, Make the two outputs deviate in reverse 2.5V Produces a signal output . When OUT- by 1.5V,OUT+ by 3.5V when , At this time, it produces 2V Signal voltage output .

It shows that the circuit effectively amplifies the differential mode signal . Differential amplifier is a selective amplifier , Ignore common mode interference , Amplify the useful differential mode signal .

chart 1 Medium b circuit , It is a differential amplifier composed of operational amplifier devices . The picture clearly shows , Whether the input signal is 2.5V or 5V, as long as IN1=IN2,OUT The end is 0V. From this point of view and significance , When the bias element of the differential amplifier R1=R3,R2=R4 when , also IN1=IN2 when , Its output “ Falsely ”.

Double ended input 、 Why does the output end of the single ended output differential amplifier appear “ Falsely ” Characteristics ？

chart 2 Working state diagram of differential amplifier

1） Upper figure a circuit ,IN1 be equal to IN2 The state of .

（1） Because of the... At the input “ Virtual break ” characteristic , The in-phase input is in high resistance state , Its input voltage value only depends on R1、R2 Partial pressure value , by 2V. In phase input 2V The voltage can be regarded as the input comparison reference voltage ;

（2） Because of the difference between the two inputs “ Deficiency ” characteristic , It can be further deduced that its inverting input , namely R3、R4 Series voltage dividing circuit , Its b spot =a spot =2V. This is the feedback voltage . The control purpose of the amplifier is to make the feedback voltage equal to the reference voltage ;

（3） from R1=R3,R2=R4 The condition is known , The amplifier output is only in “ Falsely ” state , That is, the output is 0V, To satisfy b spot =a spot =2V, This can lead to a working characteristic of the differential amplifier .

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2） Upper figure b Medium （1） circuit ,IN1 Greater than IN2 The state of .

（1） At this time, the voltage at the in-phase input is higher than that at the inverting input , The output voltage changes in a positive direction , Its R3、R4 The current direction in the bias circuit is shown in the figure ;

（2） from R3、R4 The resistance ratio is known ,R3 electric current ＝(2.8V-1.5V)/10k,R4 Voltage drop at both ends 1.3V/10K*40K=5.2V, The output voltage is 2.8V+5.2V＝8V.

（4） At this time, the input voltage difference is IN1-IN2=2V, The output voltage is 8V. obviously , The differential voltage amplification factor of the differential amplifier =R4/R3 yes 4 Double voltage differential amplifier . From this, we can deduce the differential input magnification of the differential amplifier (IN1-IN2)*R4/R3＝-OUT.

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3） Upper figure b Medium （2） circuit ,IN1 Less than IN2 The state of .

At this time, the voltage at the input terminal of the reverse in-phase is higher than that at the input terminal of the same phase , The output voltage changes in the negative direction , Its R3、R4 The current direction in the bias circuit is shown in the figure . Again , In accordance with the R3、R4 The ratio of resistance value can be inferred , Under this input condition , The output voltage is -8V, The circuit still amplifies the input differential signal 4 times .

From the work of the circuit （ fault ） For state judgment , Direct measurement R3、R4 Partial voltage state of series circuit , as long as R3、R4 Series partial pressure is established , Then the circuit is basically （ At least op amp chips ） It's good ; The voltage amplification factor of the circuit is also obtained ; Just measure the input voltage difference （R1、R3 Left terminal voltage difference ）, Then measure the output voltage for comparison , Then the quality of the peripheral bias circuit , Will also come to a clear conclusion .

From the empty short knowledge Vx = V1 ---->a

Vy = V2 ---->b

Know from the falsehood , No current flows through the input of the operational amplifier , be R1、R2、R3 It can be regarded as series connection , The current through each resistor is the same , electric current I=(Vx-Vy)/R2 ---->c

be ：Vo1-Vo2=I*(R1+R2+R3) = (Vx-Vy)(R1+R2+R3)/R2 ---->d

Know from the falsehood , Flow through R6 And flow through R7 The current is equal to , if R6=R7, be Vw = Vo2/2 ---->e

In the same way R4=R5, be Vout – Vu = Vu – Vo1, so Vu = (Vout+Vo1)/2 ---->f

From the empty short knowledge ,Vu = Vw ---->g

from efg have to Vout = Vo2 – Vo1 ---->h

from dh have to Vout = (Vy –Vx)(R1+R2+R3)/R2 In the above formula (R1+R2+R3)/R2 Is the fixed value , This value determines the difference (Vy –Vx) The magnification of . This circuit is the differential amplification circuit .

When using differential amplification circuit , One thing needs special attention , Not only |k*（U1-U2）|<15（ Preferably less than 13V about , Get better results ）, and Un And Up It should also be less than 15v, Otherwise, the amplification will not work in the linear region , Cause the circuit to work abnormally .

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Nine 、I/V Conversion circuit

It is also a current amplifier .

Many controllers accept signals from various detection instruments 0~20mA or 4~20mA electric current , The circuit converts the current into voltage and then sends it to ADC Convert to a digital signal . Pictured 4~20mA Current flows through the sample 100Ω resistance R1, stay R1 There will be 0.4~2V Voltage difference of .

Know from the falsehood , No current flows through the input of the operational amplifier , Then flow R3 and R5 The current is equal to , Flow through R2 and R4 The current is equal to .

so ：(V2-Vy)/R3 = Vy/R5 ----a

(V1-Vx)/R2 = (Vx-Vout)/R4 ----b

From the empty short knowledge ： Vx = Vy ----c

The current flows from 0~20mA change , be V1 = V2 + (0.4~2) ----d

from cd Type in b Formula (V2 + (0.4~2)-Vy)/R2 = (Vy-Vout)/R4 ----e

If R3=R2,R4=R5, By e-a have to Vout = -(0.4~2)R4/R2 ----f

In the figure R4/R2=22k/10k=2.2, be f type Vout = -(0.88~4.4)V, That is , take 4~20mA The current is converted into -0.88 ~ -4.4V voltage , This voltage can send ADC To deal with . notes ： If the current in the figure is reversed Vout = +(0.88~4.4)V.

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Ten 、V/I Conversion circuit

Current can be converted into voltage , Voltage can also be converted into current , The anti-interference ability of current transmission is better . The above figure shows such a circuit . The negative feedback of this figure is not directly fed back through resistance , Instead, a triode is connected in series Q1 Emission junction , Don't think it's just a comparator . As long as it's an amplifier circuit , The law of virtual short and virtual break still conforms to .

Know from the falsehood , No current flows through the input of the operational amplifier , be (Vi – V1)/R2 = (V1 – V4)/R6 ----a

Empathy (V3 – V2)/R5 = V2/R4 ----b

From the empty short knowledge V1 = V2 ----c

If R2=R6,R4=R5, By abc Formula V3-V4=Vi

The above formula explains R7 Voltage at both ends and input voltage Vi equal , Through R7 Current I=Vi/R7, If the load RL<<100KΩ, Through Rl And through the R7 The current is basically the same .

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11、 ... and 、 Voltage rise circuit

Just be an ordinary follower , Add the signal to Vcc/2 It's OK on the Internet OK, Consider expansion AD The accuracy can be slightly enlarged .

V+=2.5V,LM358P The operational amplifier is short ,V-=2.5V,Uo-2.5=2.5-Ui,Uo=5-Ui.

If the signal frequency is low, you need to increase the capacitance , The capacitance value is initially calculated by the turning frequency 1/(2*3.14*R*C).

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Twelve 、F/V Conversion circuit

The measured frequency signal is transported into A, NAND gate B、C、D Zoom in , Clipping , Shaping into a regular waveform with the same frequency as the measured signal , As the input signal of monostable trigger circuit .

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13、 ... and 、 Constant voltage source

VD1 The pressure drop of the pipe after conduction Uz Basically unchanged , Input like this A1 The voltage at the inverting input is Uz, This is a stable DC voltage . According to the closed-loop gain formula of integrated operational amplifier （ According to the previous empty short empty break ） The output voltage can be calculated Uo：Uo＝R3/R2*Uz.

because Uz Stable , resistance R2 and R3 Stable , So output Uo Stable , explain A1 With constant voltage output characteristics .

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fourteen 、 Logarithmic and exponential circuits

1、 Logarithmic circuit

Logarithmic operation Ui Not less than 0, Otherwise, the triode BE Knot reverse bias cut-off , There is no feedback loop .

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2、 Exponential circuit

Exponential operation Ui Not less than 0, Otherwise, the triode BE Knot reverse bias cut-off , There is no feedback loop .

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15、 ... and 、 Multiplication circuit

Division composed of analog multiplier 、 Calculation of square root and RMS detection circuit ：

1） Division operations

Here Ui2 Must be greater than 0

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2） Square root operation

Here Ui1 Must be less than 0

Here Ui1 No matter positive or negative

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3） RMS detection circuit

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sixteen 、 Active power filter

1、 low pass filter

1） First order low pass filter

Of the last formula (1+Rf/R1) It can magnify , Strong load carrying capacity .
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2） Second order low pass filter

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2、 High pass filter

The position of the resistance and capacitance in the above figure are reversed .

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seventeen 、 Monostable circuit

Static , capacitance C1 After charging , Operational release A1 Forward terminal voltage U2=V+,A1 Output high level . When the input voltage Ui Turn to low power , diode D1 Conduction , capacitance C1 adopt D1 Rapid discharge , send U2 Suddenly fell to 0V, Cause at this time U1>U2, So op amp A1 Output low level . When the input voltage becomes high , diode D1 end ,V+ the R3 Give the capacitor C1 Charge , When C1 The voltage on the is greater than U1 when ,A1 The output goes high again , Thus ending a monostable trigger .

If the diode D1 Get rid of , This circuit has power on delay function .

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eighteen 、 Op amp as comparator ( Operational amplifier and comparator )

1、 The selection basis of the bias resistance of the comparator

25 Micro safety is generally no problem , According to the parameter of maximum bias current , When the comparator reverses, the bias current will have 0.25 Changes within microamps , The greater the partial voltage bias current is, the less susceptible it is to interference , Of a comparator “ precision ” The higher , The specific bias current depends on the accuracy requirements .

For example, you need a comparator in 10V reverse , use 2.5 The comparator may be in if the current is biased by microampere 9~11v reverse , use 25 Wei'an will be 9.9~10.1v Time reversal , use 0.25 Milliampere will be 9.99~10.01v Time reversal , And so on , The above is just a rough estimate for illustrative purposes , The specific size must be determined by oneself .

(V+) - (V-) >= X, X When is the output high level ;(V+) - (V-) <= Y, Y When is the output low level .

The offset voltage is 10mV, Then the reference voltage 2.5V, The comparator will be in 2.49V~2.51V Response between ;

The offset voltage is 1mV, Then the reference voltage 2.5V, The comparator will be in 2.499V~2.501V Response between .

Examples of specific components ：

1）LM324 Input offset voltage maximum 5mV

2）OP07 With very low input offset voltage （ about OP07A The maximum is 25μV）, At very low input signal input forward end or reverse end （ such as 20mV） Try to make a comparison .

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2、 Maximum input voltage of comparator

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nineteen 、 Dead zone circuit

When the input signal Vin Enter a certain range （ dead zone ） when , The output voltage is 0; When out of this range , The output voltage of the circuit changes with the input signal .

The figure shows the diode bridge dead band circuit .

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twenty 、 Phase shift circuit

1、 Phase advance phase shift circuit

In electronic circuits, it is often necessary to change the phase of sine wave signals , Such as phase shifting . Using OP AMP and RC The network can build a phase-shifting range 0～180 Degree phase shift circuit . Phase lead ( The output is ahead of the input ) The phase shift circuit of is shown in the figure ：

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2、 Phase lag phase shift circuit

Through adjustment R3, You can change the size of the phase shift , And basically will not affect the amplitude of the output voltage . And phase lag ( Output lags input ) The phase shift circuit of is ：

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The 21st 、 Voltage source

1、 Standard voltage source based on comparator

Q19 It is the reference voltage source APL431, Used to generate 2.5V Reference voltage of .

Generated by the reference voltage source +2.5V The reference voltage passes R61、R50 partial pressure , stay R50 The pressure drop on is equal to 1.05V, So the voltage comparator U2B Positive phase input of 5 The foot voltage is constant 1.05V.

At power on , Voltage comparator U2B Inverting input of 6 Feet due to R51 At low level , here 5 The foot potential is higher than the inverting input 6 Foot potential , Voltage comparator U2B The output of the 7 Pin output high level , Drive FET Q34 Conduction .Q34 After conduction ,+3.3V The voltage passes through Q34 Of D、S extremely 、 resistance R52、R51 Divide the voltage and add it to the voltage comparator U2B Inverting input of 6 foot , Due to the voltage comparator U2B Positive phase input of 5 The foot voltage is constant 1.05V, therefore , When the voltage comparator U2B Inverting input of 6 The pin voltage is higher than 1.05V when , Output terminal 7 The pin will output low level , Make conductive Q34 end .

stay Q34 At the deadline ,U2B Of 6 The pin voltage will drop below 1.05V, At this time, the output terminal will output high level again ,Q34 Conduction . So again and again , You can make the voltage output terminal Q34 Of S The pole voltage is stable at the set value . Output voltage Vout=1.05x(1+R52/R51)≈2.6V, by DDR Memory powered .

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2、 Voltage source based on op amp follower

Make Rf=0,R1=∞, The in-phase proportional op amp above becomes a voltage follower , It has extremely high input impedance , Extremely low output impedance .

For example, the current connection 1 A load , But its load current changes greatly , Up to 200mA. Here's the picture Vout Power the load ,51R Prevent the triode at the moment of power on CE Spikes caused by capacitance .