The concept of shortness and discontinuity
Due to the large voltage amplification of op amps, the open-loop voltage amplification of general-purpose op amps is generally above 80 dB. The output voltage of the op amp is limited, generally at 10 V to 14 V. As a result, the op amp has a differential mode input voltage of less than 1 mV, and the two inputs are approximately equipotential, which is equivalent to a "short circuit". The greater the open-loop voltage amplification, the closer the potential of the two inputs will be to equality.
"Imaginary short" means that when the analytical op amp is in a linear state, the two inputs can be treated as equipotential, a characteristic called a false short circuit, or virtual short for short. Therefore, it is clear that the two inputs cannot be truly short-circuited.
Due to the large differential mode input resistance of the op amp, the input resistance of general-purpose operational amplifiers is generally more than 1m. As a result, the current flowing into the op amp input is often less than 1UA, which is much smaller than the current in the external circuit of the input. Therefore, the two inputs of the op amp can usually be regarded as an open circuit, and the larger the input resistance, the closer the two inputs are to the open circuit.
"False break" means that when analyzing the working principle of the op amp circuit, please forget about what co-amplification, reverse amplification, what adder, subtractor, and what differential input ...... for a whileIgnore the formulas of those input-output relationships for ...... momentThese concepts will only interfere with you and make people more confused and ignore circuit parameters such as input bias current, common-mode rejection ratio, offset voltage, etc., which is something designers need to consider. What we understand is the ideal amplifier (in most design processes, it is not a problem to analyze the actual amplifier as if it were the ideal amplifier).
Application of imaginary shortness and discontinuity
Next, I will start to use the two "plate axes" - "virtual short" and "virtual break", and I am going to start chattering.
Figure 1 op amp co-directional end grounding = 0V, reverse end and co-direction end are short, so it is also 0V, reverse input terminal input resistance is very high, virtual, 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 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
Solve the junior algebraic equation above.
vout = (-r2/r1)*vi
This is the input-output relation of the legendary reverse amplifier.
In Figure 2, vi and v- are short, then
vi = v- …a
Because of the false break, there is no current input and output at the reverse input, and the current through R1 and R2 is equal, let this current be i, which is obtained by Ohm's law:
i = vout/(r1+r2) …b
vi is equal to the partial pressure on r2, i.e.
vi = i*r2 ……c
Obtained by ABC.
vout=vi*(r1+r2)/r2
This is the formula for the legendary co-directional amplifier.
In Figure 3, it is known from the imaginary shortness:
v- = v+ = 0 ……a
From the imaginary break and Kirchhoff's law, the sum of the currents passing through R2 and R1 is equal to the current passing through R3, therefore.
v1 – v-)/r1 + v2 – v-)/r2 = (vout – v-)/r3 ……b
Substituting the A formula, B becoming.
v1/r1 + v2/r2 = vout/r3
If we take r1=r2=r3, then the above equation becomes.
vout=v1+v2
This is the legendary reverse adder.
If there is no current flowing in the same direction of the op amp because of the false break, the current flowing through R1 and R2 is equal, and 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 imaginary shortness: v+ = v- ....c
If r1=r2 and r3=r4, then it can be deduced from the above equation.
v+ = (v1 + v2)/2 v- = vout/2
So vout = v1 + v2 is also an adder.
Figure 5 shows that the current through R1 is equal to the current through R2, and the current through R4 is equal to the current through R3.
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 imaginary shortness: v+ = v- ....e
So vout=v2-v1 is the legendary subtractor.
In the circuit in Figure 6, the voltage at the reverse input is equal to the co-directional end, and the current through R1 is equal to the current through C1 from the imaginary short. Current through R1.
i=v1/r1
Current through C1.
i=c*duc/dt=-c*dvout/dt
So vout=((-1 (r1*c1)) v1dt
The output voltage is proportional to the integration of the input voltage to time, which is the legendary integration circuit. If v1 is a constant voltage u, then the above equation is transformed as.
vout = -u*t/(r1*c1)
t is the time, then the vout output voltage is a straight line from 0 to the negative supply voltage over time.
In Figure 7, the current through the capacitor C1 and the resistor R2 is equal from the imaginary short, and the voltage at the same and reverse ends of the op amp is equal from the imaginary short. Then:
vout = -i * r2 = -(r2*c1)dv1/dt
This is a differential circuit. If v1 is a DC voltage that is added suddenly, then the output vout corresponds to a pulse in the opposite direction to v1.
Figure 8. It is known from the short-sightedness
vx = v1 ……a
vy = v2 ……b
If there is no current flowing through the op amp input, then R1, R2, and R3 can be regarded as series connection, and the current through each resistor is the same.
i=(vx-vy)/r2 ……c
Then: vo1-vo2=i*(r1+r2+r3) = (vx-vy)(r1+r2+r3) r2 ......d
From the imaginary assertion, the current flowing through R6 is equal to the current flowing through R7, and 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 imaginary shortness, vu = vw ......g
Obtained by EFG
vout = vo2 – vo1 ……h
From dh vout = (vy vx)(r1+r2+r3) r2
In the above equation, (r1+r2+r3) r2 is the fixed value, which determines the magnification of the difference (vy vx). This circuit is the legendary differential amplification circuit.
Next, analyze a circuit that has a lot of contact. Many controllers accept 0 20 mA or 4 20 mA currents from various detection instruments, and the circuit converts this current into voltage and then sends it to the ADC to convert it into a digital signal, Figure 9 is a typical circuit. As shown in Figure 4, a 20mA current flows through the sampling 100 resistor R1, and 0 will be generated on R14 2V voltage difference. If there is no current flowing through the op amp input, 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 short term: vx = vy ......c
The current varies from 0 to 20 mA.
v1 = v2 + 0.4~2) …d
It is obtained by substituting the CD formula into the B type.
v2 + 0.4~2)-vy)/r2 = (vy-vout)/r4 ……e
If r3 = r2 and r4 = r5, then it is obtained by e-a.
vout = -(0.4~2)r4/r2 ……f
Figure 9. r4/r2=22k/10k=2.2, then f-formula vout = -(0..)88~4.4) v, that is, converts 4 20mA into -088 ~ 4.4V voltage, this voltage can be sent to the ADC to be processed. Figure 9 is used in scientific instruments and temperature transmitters, and other microcurrent to voltage signals are more common.
The current can be converted into voltage, and the voltage can also be converted into current. Figure 10 is one such circuit. The negative feedback in the above figure is not directly fed back through the resistor, but is connected in series with the transmitting junction of the transistor Q1, so don't think that it is a comparator. As long as it is an amplification circuit, the law of short and short disconnection is still in line!
From the imaginary judgment, no current flows through the input of the op amp, then.
vi – v1)/r2 = (v1 – v4)/r6 ……a
Same for (v3 v2) r5 = v2 r4 ......b
It is known from the short-sightednessv1 = v2 ……c
If r2=r6 and r4=r5, then v3-v4=vi is obtained by abc
The above equation shows that the voltage at both ends of R7 and the input voltage VI are equal, then the current through R7 i=VI R7, if the load Rl <<100K, then the current through Rl and through R7 is basically the same.
Come to a complex.
Figure 11 shows a three-wire PT100 pre-amplification circuit. The PT100 sensor leads to three wires of the same material, wire diameter and length, as shown in the figure. A voltage of 2V is applied to a bridge circuit consisting of R14, R20, R15, Z1, PT100 and their line resistors. Z1, Z2, Z3, D11, D12, D83 and each capacitor play a filtering and protection role in the circuit, and can be ignored during static analysis, Z1, Z2, Z3 can be regarded as a short circuit, and D11, D12, D83 and each capacitor can be regarded as an open circuit. From the voltage division of the resistor, v3 = 2 * r20 (r14 + 20) = 200 1100 = 2 11 ......a
From the imaginary short, the voltage of the u8b pin and the voltage of pin 5 are equal.
v4=v3 ……b
If there is no current flowing through the second pin of U8A, the current flowing through R18 and R19 is equal.
v2-v4)/r19=(v5-v2)/r18 ……c
From the imaginary assertion, there is no current flowing through the third pin of U8A, and v1=v7 ......d
In the bridge circuit, R15 and Z1, PT100 and the line resistor are connected in series, and the voltage obtained by the connection between PT100 and the line resistor is added to the third pin of U8A through the resistor R17
v7=2*(rx+2r0)/(r15+rx+2r0) …e
From the imaginary short, the voltage of the 3rd and 2nd pins of U8A is equal, and v1=v2 ......f
obtained by abcdef, (v5-v7) 100=(v7-v3) 22 Simplification yields v5=(102.2*v7-100v3)/2.2
i.e. v5=2044(rx+2r0)/(1000+rx+2r0) –200/11 ……g
The output voltage v5 is a function of rx, and we look at the effect of the line resistance.
The voltage drop generated on the bottom line resistor of PT100 passes through the middle line resistor, Z2, R22, and is added to the 10th pin of U8C, which is known by the virtual judgment
v5=v8=v9=2*r0/(r15+rx+2r0) …a
v6-v10)/r25=v10/r26 ……b
It is known from the short-sightedness
v10=v5 ……c
Derived from the formula ABC
v6=(102.2/2.2)v5=204.4r0/[2.2(1000+rx+2r0)] h
The equation composed of the equation GH knows that if the values of V5 and V6 are measured, Rx and R0 can be calculated, and Rx is known, and the magnitude of the temperature is known by checking the PT100 indexing table.