Operational amplifier circuit analysis of virtual open and virtual short

The concepts of virtual shortness and virtual break

Since the voltage amplification factor of the operational amplifier is very large, the open-loop voltage amplification factor of general general-purpose operational amplifiers is above 80 dB. The output voltage of the op amp is limited, generally between 10 V and 14 V. Therefore, the differential mode input voltage of the op amp is less than 1 mV, and the two input terminals are approximately at the same potential, which is equivalent to a "short circuit". The greater the open-loop voltage amplification, the closer the potentials at the two input terminals are to the same value.

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

Since the differential mode input resistance of the operational amplifier is very large, the input resistance of general general-purpose operational amplifiers is 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 external circuit at the input terminal. Therefore, the two input terminals of the op amp can usually be regarded as open circuit, and the greater the input resistance, the closer the two input terminals are to open circuit. "Virtual break" means that when the operational amplifier is in a linear state, the two input terminals can be regarded as equivalent open circuits. This feature is called a false open circuit, or virtual break for short. Obviously the two input terminals cannot be truly disconnected.

When analyzing the working principle of the op amp circuit, first of all, please temporarily forget about co-directional amplification and reverse amplification, what adders and subtractors, what differential input... Forget about the formulas of the input-output relationship for the time being... these things are just It will disturb you and make you more confused; please also ignore circuit parameters such as input bias current, common mode rejection ratio, offset voltage, etc. for the time being. These are things that designers should consider. What we understand is an ideal amplifier (in fact, during maintenance and most design processes, there is no problem in analyzing the actual amplifier as an ideal amplifier).

Okay, let's grab two "axes"--"Xu Duan" and "Xu Duan" and start "Pao Ding Jie Niu".

1) Reverse amplifier:

figure 1

Figure 1: The non-inverting end of the op amp is grounded = 0V, the reverse end and the non-inverting end are virtual short, so they are also 0V. The input resistance of the reverse input end is very high, virtual open, and there is almost no current injection and outflow, then R1 and R2 are equivalent Therefore, in series, 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.

Current flowing through R1: I1 = (Vi - V-)/R1…………a

Current flowing through R2: I2 = (V- - Vout)/R2…b

V- = V+ = 0………………c

I1 = I2……………………d

Solve the above junior high school algebra equation to get Vout = (-R2/R1)*Vi

This is the input-output relationship of the legendary inverting amplifier.

2) Co-directional amplifier:

figure 2

In Figure 2, Vi and V- are virtual short, then Vi = V-...a

Because of the virtual break, there is no current input or output at the reverse input terminal. The currents passing through R1 and R2 are equal. Let this current be I. According to Ohm's law: I = Vout/(R1+R2) ……b

Vi is equal to the partial pressure on R2, that is: Vi = I*R2...c

From the abc formula, we get Vout=Vi*(R1+R2)/R2. This is the formula of the legendary co-directional amplifier.

3) Adder 1:

image 3

In Figure 3, we know from the virtual short: V- = V+ = 0…a

According to virtual break and Kirchhoff's law, the sum of the currents passing through R2 and R1 is equal to the current passing through R3, so (V1 – V-)/R1 + (V2 – V-)/R2 = (V-–Vout) /R3……b

Substituting into equation a, equation b becomes V1/R1+ V2/R2 = Vout/R3. If R1=R2=R3, then the above equation becomes -Vout=V1+V2. This is the legendary adder.

4) Adder 2:

Figure 4

Please see Figure 4. Because of the virtual break, no current flows through the same end of the op amp, so the currents flowing through R1 and R2 are equal, and similarly, the currents flowing through R4 and R3 are also equal.

Therefore (V1 – V+)/R1 = (V+ – V2)/R2…a

(Vout – V-)/R3 =V-/R4…b

From the virtual short, we know: V+ = V- ……c If R1=R2, R3=R4, then from the above formula it can be deduced that V+ = (V1 + V2)/2 V- = Vout/2, so Vout = V1 +V2 It’s also an adding machine, haha!

5) Subtractor

Figure 5

Figure 5 shows from the virtual disconnection 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 virtual short, we know that V+ = V-…e

So Vout=V2-V1 This is the legendary subtractor.

6) Integral circuit:

Figure 6

In the circuit in Figure 6, we know from the virtual short that the voltage at the reverse input terminal is equal to the non-directional terminal.

From the virtual break, the current through R1 is equal to the current through C1.

Current i=V1/R1 through R1

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

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

If V1 is a constant voltage U, then 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 with time from 0 to the negative power supply voltage.

7) Differential circuit:

Figure 7

In Figure 7, we know from the virtual disconnection that the currents through capacitor C1 and resistor R2 are equal.

It is known from the virtual short that the voltages at the non-inverting terminal and the reverse terminal of the op amp are equal.

Then: Vout = -i * R2 = -(R2*C1)dV1/dt

This is a differential circuit.

If V1 is a suddenly added DC voltage, the output Vout corresponds to a pulse in the opposite direction to V1.

8) Differential amplifier circuit

Figure 8

From virtual short, we know that Vx = V1……a

Vy = V2……b

It can be known from the virtual break that there is no current flowing through the input terminal of the op amp, then R1, R2, and R3 can be regarded as series connection, 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

It can be known from the virtual circuit that the current flowing through R6 is equal to the current flowing through R7. If R6=R7, then Vw = Vo2/2...e

In the same way, if R4=R5, then Vout – Vu = Vu – Vo1, so Vu = (Vout+Vo1)/2…f

From the virtual short, 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 fixed value. This value determines the amplification factor of the difference (Vy–Vx).

This circuit is the legendary differential amplifier circuit.

9) Current detection:

Figure 9

Analyze a circuit that everyone touches more. Many controllers accept 0~20mA or 4~20mA current from various detection instruments. The circuit converts this current into voltage and then sends it to the ADC to convert it into a digital signal. Figure 9 is such a typical circuit. As shown in Figure 4, the current of 20mA flows through the sampling 100Ω resistor R1, and a voltage difference of 0.4~2V will be generated on R1. From the virtual break, we know that if no current flows through the input terminal of the op amp, the currents flowing through R3 and R5 are equal, and the currents flowing through R2 and R4 are equal. Therefore:

(V2-Vy)/R3 = Vy/R5…a

(V1-Vx)/R2 = (Vx-Vout)/R4…b

We know from virtual short: Vx = Vy……c

The current changes from 0~20mA, then V1 = V2 + (0.4~2)…d

Substituting equation cd into equation b, we get (V2 + (0.4~2)-Vy)/R2 = (Vy-Vout)/R4...e

If R3=R2, R4=R5, then Vout = -(0.4~2)R4/R2...f from ea

In Figure 9, R4/R2=22k/10k=2.2, then f-type Vout = -(0.88~4.4)V,

That is to say, the 4~20mA current is converted into a -0.88~-4.4V voltage, which can be sent to the ADC for processing.

Note: If the current in Figure 9 is reversely connected, Vout = +(0.88~4.4)V,

10) Voltage and current conversion detection:

Figure 10

Current can be converted into voltage, and voltage can be converted into current. Figure 10 is such a circuit. The negative feedback in the above picture is not directly fed back through the resistor, but is connected in series with the emitter junction of transistor Q1. You should not think that it is just a comparator. As long as it is an amplifier circuit, the law of virtual short and virtual break is still consistent!

From the virtual break, we know that no current flows through the input terminal of the op amp,

Then (Vi – V1)/R2 = (V1 – V4)/R6…a

In the same way (V3 – V2)/R5 = V2/R4…b

From virtual short, we know V1 =V2...c

If R2=R6, R4=R5, then V3-V4=Vi is obtained from the abc formula

The above formula shows that the voltage at both ends of R7 is equal to the input voltage Vi, then the current through R7 I=Vi/R7, if the load RL

11) Sensor detection:

Figure 11

Here comes 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 as shown in the figure. A voltage of 2V is applied to the bridge circuit composed of R14, R20, R15, Z1, PT100 and their line resistance. Z1, Z2, Z3, D11, D12, D83 and each capacitor play a filtering and protection role in the circuit. They can be ignored during static analysis. Z1, Z2, Z3 can be regarded as short circuit, and D11, D12, D83 and each capacitor can be seen. To open the way. From the resistor voltage division, V3=2*R20/(R14+20)=200/1100=2/11…a

From the virtual short, we know that the voltage of pins 6 and 7 of U8B is equal to the voltage of pin 5, V4=V3...b

It can be seen from the virtual disconnection that no current flows through the second pin of U8A, so the current flowing through R18 and R19 is equal. (V2-V4)/R19=(V5-V2)/R18 ……c

It can be seen from the virtual disconnection that no current flows through the 3rd pin of U8A, V1=V7...d In the bridge circuit, R15 is connected in series with Z1, PT100 and the line resistor. The voltage obtained by the series connection of PT100 and the line resistor is added to U8A through the resistor R17. Pin 3, V7=2*(Rx+2R0)/(R15+Rx+2R0) …..e

From the virtual short, we know that the voltage of pin 3 and pin 2 of U8A is equal, V1=V2...f

From abcdef, (V5-V7)/100=(V7-V3)/2.2 Simplify to 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 of the above formula is a function of Rx. Let’s look at the influence of line resistance. The voltage drop generated on the bottom line resistor of Pt100 passes through the middle line resistor, Z2, and R22, and is added to the 10th pin of U8C.

From the virtual conclusion, 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 the formula abc, we get V6=(102.2/2.2)V5=204.4R0/[2.2(1000+Rx+2R0)] ……h

From the equation set composed of the formula gh, if the values ​​of V5 and V6 are measured, Rx and R0 can be calculated. Knowing Rx, the temperature can be known by looking up the pt100 scale.