Useful Tips | Solve operational amplifier problems using virtual short and virtual open

Latest update time：2021-09-02 18:32

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运算放大器组成的电路五花八门，令人眼花瞭乱，是模拟电路中学习的重点。在分析它的工作原理时倘没有抓住核心，往往令人头大。为此本人特搜罗天下运放电路之应用，来个“庖丁解牛”，希望各位从事电路板维修的同行，看完后有所斩获。

遍观所有模拟电子技朮的书籍和课程，在介绍运算放大器电路的时候，无非是先给电路来个定性，比如这是一个同向放大器，然后去推导它的输出与输入的关系，然后得出Vo=(1+Rf)Vi，那是一个反向放大器，然后得出Vo=-Rf*Vi……最后学生往往得出这样一个印象：记住公式就可以了！如果我们将电路稍稍变换一下，他们就找不着北了！偶曾经面试过至少100个以上的大专以上学历的电子专业应聘者，结果能将我给出的运算放大器电路分析得一点不错的没有超过10个人！其它专业毕业的更是可想而知了。

今天，芯片级维修教各位战无不胜的两招，这两招在所有运放电路的教材里都写得明白，就是“虚短”和“虚断”，不过要把它运用得出神入化，就要有较深厚的功底了。

The concept of short and long

Since the voltage amplification factor of the op amp is very large, the open-loop voltage amplification factor of general-purpose op amps is generally above 80 dB. However, the output voltage of the op amp is limited, generally between 10 V and 14 V. Therefore, the differential input voltage of the op amp is less than 1 mV, and the two input terminals are approximately equal in potential, which is equivalent to a "short circuit". The larger the open-loop voltage amplification factor, the closer the potential of the two input terminals is to being equal.

"Virtual short" means that when analyzing the linear state of the op amp, the two input terminals can be regarded as equal in potential. This characteristic is called a false short circuit, or virtual short for short. Obviously, the two input terminals cannot be truly short-circuited.

Since the differential input resistance of the op amp is very large, the input resistance of general-purpose op amps is generally above 1MΩ. Therefore, the current flowing into the input terminal of the op amp is often less than 1uA, which is much smaller than the current of the circuit outside the input terminal. Therefore, the two input terminals of the op amp can usually be regarded as open circuits, and the larger the input resistance, the closer the two input terminals are to being open circuits. "Virtual disconnection" means that when analyzing the linear state of the op amp, the two input terminals can be regarded as equivalent to open circuits. This characteristic is called a false open circuit, or virtual disconnection for short. Obviously, the two input terminals cannot be truly disconnected.

When analyzing the working principle of the operational amplifier circuit, first of all, please temporarily forget about the same-direction amplification, reverse amplification, adder, subtractor, differential input... Forget about the formulas of the input-output relationship... These things will only disturb you and make you more confused; please also temporarily ignore the circuit parameters such as input bias current, common-mode rejection ratio, offset voltage, etc., which are things that designers need to consider. What we understand is the ideal amplifier (in fact, in maintenance and most design processes, there is no problem in analyzing the actual amplifier as an ideal amplifier).

Okay, let's grab the two "axes" - "empty shortness" and "empty break", and start "butchering the cow".

In Figure 1, the same-direction end of the op amp is grounded = 0V, the reverse end and the same-direction end are virtually shorted, so it is also 0V, the input resistance of the reverse input end is very high, virtually disconnected, and there is almost no current injection and outflow, then R1 and R2 are equivalent to being connected in series, and the current flowing through each component in a series circuit is the same, that is, the current flowing through R1 and the current flowing through R2 are the same. The current flowing through R1 I1 = (Vi - V-)/R1 ……a The current flowing through R2 I2 = (V- - Vout)/R2 ……b V- = V+ = 0 ……c I1 = I2 ……d Solving the above junior high school algebraic equation, we get Vout = (-R2/R1)*Vi This is the input-output relationship of the legendary reverse amplifier.

In Figure 2, Vi and V- are virtually shorted, then Vi = V- ……a Because of the virtual disconnection, there is no current input or output at the reverse input end, and the currents passing through R1 and R2 are equal. Suppose this current is I, and from Ohm's law, we have: I = Vout/(R1+R2) ……b Vi is equal to the divided voltage on R2, that is: Vi = I*R2 ……c From formula abc, we get Vout=Vi*(R1+R2)/R2. This is the formula for the legendary co-directional amplifier.

In Figure 3, from the virtual short circuit we know that: V- = V+ = 0 …a From the virtual short circuit and Kirchhoff’s law, we know that the sum of the currents through R2 and R1 is equal to the current through R3, so (V1 – V-)/R1 + (V2 – V-)/R2 = (Vout – V-)/R3 …b Substituting into formula a, formula b becomes V1/R1 + V2/R2 = Vout/R3 If R1=R2=R3, the above formula becomes Vout=V1+V2, which is the legendary adder.

Please see Figure 4. Because of the virtual short circuit, no current flows through the op amp's same-direction terminal, so the current flowing through R1 and R2 is equal, and similarly the current flowing through R4 and R3 is also equal. Therefore (V1 – V+)/R1 = (V+ - V2)/R2 ……a (Vout – V-)/R3 = V-/R4 ……b From the virtual short circuit, we know that: V+ = V- ……c If R1=R2, R3=R4, then from the above formula we can deduce that V+ = (V1 + V2)/2 V- = Vout/2 Therefore Vout = V1 + V2 It is also an adder.

From the virtual short circuit in Figure 5, we know that the current through R1 is equal to the current through R2. Similarly, the current through R4 is equal to the current through R3, so (V2 – V+)/R1 = V+/R2 …a (V1 – V-)/R4 = (V- - Vout)/R3 …b If R1=R2, then V+ = V2/2 …c If R3=R4, then V- = (Vout + V1)/2 …d From the virtual short circuit, we know that V+ = V- …e So Vout=V2-V1 This is the legendary subtractor.

In the circuit of Figure 6, we know from the virtual short circuit that the voltage at the reverse input terminal is equal to the voltage at the same direction terminal, and from the virtual short circuit, we know that the current through R1 is equal to the current through C1. The current through R1 i=V1/R1 The current through C1 i=C*dUc/dt=-C*dVout/dt So Vout=((-1/(R1*C1))∫V1dt The output voltage is proportional to the integral of the input voltage over time, which is the legendary integration circuit. If V1 is a constant voltage U, the above formula is transformed into Vout = -U*t/(R1*C1) t is time, then the Vout output voltage is a straight line that changes from 0 to the negative power supply voltage according to time.

In Figure 7, we know from the virtual short circuit that the currents through capacitor C1 and resistor R2 are equal, and from the virtual short circuit, we know that the voltages at the same end and the reverse end of the op amp are equal. Then: Vout = -i * R2 = -(R2*C1)dV1/dt This is a differential circuit. If V1 is a DC voltage suddenly added, the output Vout corresponds to a pulse in the opposite direction to V1.

Figure 8. From the virtual short circuit, we know that Vx = V1 …a Vy = V2 …b From the virtual short circuit, we know that there is no current flowing through the input of the op amp, so R1, R2, and R3 can be regarded as connected in series, and the current passing through each resistor is the same, current I=(Vx-Vy)/R2 …c Then: Vo1-Vo2=I*(R1+R2+R3) = (Vx-Vy)(R1+R2+R3)/R2 …d From the virtual short circuit, we know that the current flowing through R6 is equal to the current flowing through R7. If R6=R7, then Vw = Vo2/2 …e Similarly, if R4=R5, then Vout – Vu = Vu – Vo1, so Vu = (Vout+Vo1)/2 …f From the virtual short circuit, we know that Vu = Vw …g From efg, we get Vout = Vo2 – Vo1 …h From dh, we get Vout = (Vy –Vx)(R1+R2+R3)/R2 In the above formula, (R1+R2+R3)/R2 is a constant value, which determines the amplification factor of the difference (Vy –Vx). This circuit is the legendary differential amplifier circuit.

Let's analyze a circuit that everyone is familiar with. Many controllers receive 0~20mA or 4~20mA current from various detection instruments. The circuit converts this current into voltage and then sends it to ADC to convert it into a digital signal. Figure 9 is such a typical circuit. As shown in the figure, when the 4~20mA current flows through the sampling 100Ω resistor R1, a voltage difference of 0.4~2V will be generated on R1. From the virtual cutoff, we know that if there is no current flowing through the input end of the op amp, the current flowing through R3 and R5 is equal, and the current flowing through R2 and R4 is equal. Therefore: (V2-Vy)/R3 = Vy/R5 ………a (V1-Vx)/R2 = (Vx-Vout)/R4 ………b From the imaginary short circuit, we know that: Vx = Vy ………c When the current changes from 0~20mA, then V1 = V2 + (0.4~2) ………d Substituting equations cd into equation b, we get (V2 + (0.4~2)-Vy)/R2 = (Vy-Vout)/R4 ………e If R3=R2, R4=R5, then from ea we get Vout = -(0.4~2)R4/R2 ………f In Figure 9, R4/R2=22k/10k=2.2, then from equation f Vout = -(0.88~4.4)V, that is to say, the 4~20mA current is converted into a voltage of -0.88 ~ -4.4V, which can be sent to the ADC for processing.

Current can be converted into voltage, and voltage can also be converted into current. Figure 10 is such a circuit. The negative feedback in the figure above is not directly fed back through a resistor, but is connected in series with the emitter junction of the transistor Q1. Don't think it is a comparator. As long as it is an amplifier circuit, the rules of virtual short and virtual open are still in line!

From the virtual open, we know that there is no current flowing through the input of the op amp,

then (Vi – V1)/R2 = (V1 – V4)/R6 ……a
Similarly, (V3 – V2)/R5 = V2/R4 ……b
From the virtual short, we know that V1 = V2 ……c
If R2=R6, R4=R5, then from the abc formula, V3-V4=Vi

The above formula shows that the voltage across R7 is equal to the input voltage Vi, then the current through R7 I=Vi/R7, if the load RL<<100KΩ, then the current through Rl and through R7 are basically the same.

Here is a complicated one, haha! Figure 11 is a three-wire PT100 preamplifier circuit. The PT100 sensor leads to three wires of the same material, wire diameter and length, and the connection method is shown in the figure. A 2V voltage is applied to the bridge circuit composed of R14, R20, R15, Z1, PT100 and its wire resistance.

Z1, Z2, Z3, D11, D12, D83 and capacitors play a filtering and protection role in the circuit. They can be ignored in static analysis. Z1, Z2, Z3 can be regarded as short circuits, and D11, D12, D83 and capacitors can be regarded as open circuits. From the resistor voltage divider, we know that V3=2*R20/(R14+20)=200/1100=2/11 ……a From the virtual short circuit, we know that the voltage of the 6th and 7th pins of U8B is equal to the voltage of the 5th pin V4=V3 ……b From the virtual short circuit, we know that there is no current flowing through the 2nd pin of U8A, so the current flowing through R18 and R19 is equal. (V2-V4)/R19=(V5-V2)/R18 ……c From the virtual short circuit, we know that no current flows through the third pin of U8A, V1=V7 ……d In the bridge circuit, R15 is connected in series with Z1, PT100 and the line resistance. The voltage divided by PT100 and the line resistance in series is added to the third pin of U8A through resistor R17, V7=2*(Rx+2R0)/(R15+Rx+2R0) …..e From the virtual short circuit, we know that the voltages of the third pin and the second pin of U8A are equal, V1=V2 ……f From abcdef, we know that (V5-V7)/100=(V7-V3)/2.2 Simplified, we get V5=(102.2*V7-100V3)/2.2, that is, V5=204.4(Rx+2R0)/(1000+Rx+2R0) – 200/11 ……g The output voltage V5 in the above formula is a function of Rx. Let's look at the influence of line resistance. The voltage drop generated on the lowest line resistance of Pt100 passes through the middle line resistance, Z2, R22, and is added to the 10th pin of U8C. From the virtual break, we know that V5=V8=V9=2*R0/(R15+Rx+2R0) ……a (V6-V10)/R25=V10/R26 ……b From the virtual short, we know that V10=V5 ……c From formula abc, we get V6=(102.2/2.2)V5=204.4R0/[2.2(1000+Rx+2R0)] ……h From the equation group composed of formula gh, we know that if the values of V5 and V6 are measured, Rx and R0 can be calculated. Knowing Rx, we can know the temperature by looking up the pt100 graduation table.

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