Analog Electronic Circuits Chapter 4 Integrated Operational Amplifier Circuits
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Analog Electronic Circuits Chapter 4 Integrated Operational Amplifier Circuits Section 1 Learning Requirements
Section 2 Constant Current Source in Integrated Operational Amplifier
Section 3 Differential Amplifier Circuit
Section 4 Integrated Circuit Operational Amplifier
Section 5 Main Parameters of Integrated Circuit Operational Amplifier
Section 6 Introduction to Field Effect Transistor
Section 1 Learning Requirements
1. Master the circuit structure and basic characteristics of the basic mirror current source, proportional current source, and micro current source.
2. Master the definition and characteristics of differential mode signals and common mode signals.
3. Master the circuit structure and characteristics of the basic type and constant current source type differential amplifier, be able to skillfully calculate the static operating point of the circuit, and be familiar with the connection methods of the four circuits and the phase relationship between the input and output voltage signals.
4. Skillfully analyze the amplification characteristics of the differential amplifier for differential mode small signal input and the common mode rejection ratio. Be able to calculate A VD , Rid , Ric , Rod , Roc , and K CMR .
5. Be familiar with the main technical indicators of the operational amplifier and the general circuit structure of the integrated operational amplifier circuit.
Learning focus:
Master the analysis methods of basic circuits of integrated operational amplifiers
Learning difficulty:
Analysis of the internal circuit of the integrated operational amplifier
Integrated circuit introduction
: Integrated circuits are made on a small piece of P-type silicon wafer substrate, with multiple transistors (or FETs), resistors, and capacitors, combined into circuits with specific functions.
The structural characteristics of integrated circuits:
1. Direct coupling is adopted.
2. In order to overcome the temperature drift caused by the direct coupling method, temperature compensation is adopted-the input stage is a differential amplifier circuit.
3. BJT or FET is widely used to form a constant current source, replacing the large resistance R, or used to set the static current.
4. The composite tube connection method is used to improve the performance of a single tube.
Integrated circuits are divided into two major parts: digital and analog.
return Section 2 Constant Current Source in Integrated Operational Amplifier
1. Basic mirror current source
The circuit is shown in Figure 6.1. The parameters of T 1 and T 2 are exactly the same, that is, β 1 =β 2 , I CEO1 =I CEO2 . From the circuit, we can see that V BE1 =V BE2 , I E1 =I E2 , I C1 =I C2
 When β>>2, 
I R =I REF is called the reference current. It can be seen from the above formula that when R is determined, I R is determined, and I C2 is also determined accordingly. We regard I C2 as the mirror image of I R , so Figure 6.1 is called a mirror image constant current source .
Improved circuit 1:  Figure 6.2 is a basic mirror current source with a buffer stage. It is an improvement on the shortcomings of the basic mirror current source. The difference between the two is that transistor T3 is added . Its purpose is to reduce the shunting effect of I B on IR of transistors T1 and T2 , improve the mirror accuracy, and reduce the impact of insufficient β value.
 Improved circuit 2:  Figure 6.3 is a mirror current source with an emitter resistor, where Re1 = Re2 . The inputs of the two tubes are still symmetrical, so
 If Re1 of this circuit is not equal to Re2 , then IC2 is proportional to the ratio of (Re1 , Re2 ) , therefore, this current source is also called a proportional current source.
2. Micro current source 
The circuit is shown in Figure 6.4. When IR is constant, I C2 can be determined as: 
It can be seen that I 0 can be controlled by using the base-emitter voltage difference V BE of the two tubes . Since the value of V BE is small, a small working current can be obtained by using a small resistance Re2 - a micro current source.
Example: The circuit is shown in Figure 6.5. 
Given that the parameters of BJT are the same, find the relationship between each current source and the reference current.
 3. Main Application of Current Source - Active Load
As mentioned above, increasing R c can improve the voltage gain of the common-emitter amplifier circuit. However, R c cannot be too large, because the cost of manufacturing large resistors in the integrated process is too high, and, under the condition of constant power supply voltage, the larger R c is, the smaller the output amplitude is. So, can we find a component to replace R c , whose dynamic resistance is large, so that the voltage gain is increased, but the static resistance is small. So as not to reduce the output amplitude? Naturally, we can consider the transistor constant current source. Since the current source has the characteristics of small DC resistance and large AC resistance, it is widely used as a load in analog integrated circuits - active load, as shown in Figure 6.6.  return
Section 3 Differential Amplifier Circuit Basic concepts: 
Figure 6.7 shows a linear amplifier, which has two input terminals, connected to signals vi1 and vi2 respectively ; the signal at the output terminal is v o . In the ideal case where the circuit is completely symmetrical, the output signal voltage can be expressed as 
: Where A VD is the differential mode voltage gain of the differential amplifier. It can be seen that any signal common to the two input terminals of the circuit will not affect the output voltage. But in general, the actual output voltage depends not only on the differential mode signal v id of the two input signals , but also on the common mode signal v ic of the two input signals , which are respectively:
When the two input signals are represented by differential mode signals and common mode signals, there are:
When the differential mode signal and the common mode signal exist at the same time, for the linear amplifier, the superposition principle can be used to calculate the total output voltage, that is, 
Where  is the differential mode voltage gain,  called the common mode voltage gain.
1. Basic differential amplifier circuit
1. Basic Circuit 
The basic differential amplifier is shown in Figure 6.8. In the figure, T1 and T2 are transistors with the same characteristics, the circuit is symmetrical, and the parameters are also symmetrical. For example: V BE1 = V BE2 , R c1 = R c2 = R c , R b1 = R b2 = R b , β 1 = β 2 = β. The circuit has two input terminals and two output terminals.
2. Working principle
(1) When vi1 = vi2 = 0, that is, in static state, since the circuit is completely symmetrical: I c1 = I c2 = I 0 /2, R c1 I c1 = R c2 I c2 , V o = V c1 -V c2 = 0, that is, when the input is 0, the output is also 0 .
(2) When adding a differential signal, that is, vs1 = -vs2 = vsd / 2, from the circuit point of view, the increase of vB1 increases iB1 , which increases iC1 , which decreases vC1. The decrease of vB2 decreases iB2 , which decreases iC2 , which increases vC2 . From this, we can deduce that: vO = vC1 - vC2 = 2vC1 . Each change v is not equal to 0, so there is a signal output. If a common-mode signal is added to the input end, that is, vs1 = vS2 , due to the symmetry of the circuit and the constant current source bias, ideally vO = 0, no output. This is what is called "differential"; that is, there is a difference between the two input ends, and the output end changes.
3. Principle of suppressing zero drift
In a differential circuit, whether it is a change in temperature or a fluctuation in the current source, it will cause changes in the IC and VC of the two transistors . This effect is equivalent to adding a common-mode signal to the two input terminals. Ideally, v o remains unchanged, thereby suppressing zero drift. Any signal that has the same effect on the base of the two tubes of the differential amplifier is a common-mode signal . Common examples are:
(1) vi1 is not equal to -vi2 , and the signal contains a common-mode signal;
(2) interference signal (usually acting on the input terminals at the same time);
(3) zero drift.
In actual situations, it is difficult to achieve complete symmetry and an ideal constant current source for the two tubes, but the output drift voltage will also be greatly reduced. In summary, amplifying differential-mode signals and suppressing common-mode signals are the basic characteristics of differential amplifiers. Usually, we are interested in differential-mode input signals. For this part of the useful signal, we hope to obtain the largest possible amplification factor; while the common-mode input signal may reflect the drift signal caused by temperature changes or some interference signal that enters the amplifier circuit along with the input signal. For such common-mode input signals, we hope to suppress them as much as possible and not amplify and transmit them.
4. Calculation of main technical indicators 
(1) Estimation of static operating point
I C1 =I C2 =I c =I O /2
V C1 =V C2 =V cc -I c R c
I B1 =I B2 =I c /β=I B =I/2β (2) Differential mode voltage gain and input and output resistance
The differential amplifier circuit has two input terminals and two output terminals. Similarly, the output is also divided into double-ended output and single-ended output. In combination, there are four connection methods: double-ended input and double-ended output, double-ended input and single-ended output, single-ended input and double-ended output, and single-ended input and single-ended output. 
(a) Dual-input dual-output circuit
Differential input: v i1 =-v i2 =v id /2, then when i C1 rises, i C2 falls.
If the circuit is completely symmetrical, then △i C1 =△i C2 , because I O does not change, so ve =0, the circuit can be represented by Figure 6.9.
From the above calculation, it can be seen that when the load is completely symmetrical in the circuit, in the case of dual input and dual output, A VD = A V1 , it can be seen that the circuit uses double the components in exchange for the ability to suppress zero drift.
Differential input resistance R i : Equivalent resistance seen from the two input terminals R i =2r be
Differential output resistance R 0 : Equivalent resistance seen from the two output terminals R 0 =2R C
R 0 , R i is twice that of a single tube.
(b) Dual-input single-output circuit
For differential signals: Since the C pole of the other transistor is not used, V 0 is only half of the dual output.
Differential input resistance: Since the input circuit does not change, Ri = 2rbe , differential
output resistance: R0 = Rc1 . (c) Single-ended input circuit For single-ended input, it is equivalent to grounding b2 in Figure 6.10. When vi > 0 , ic1 increases , which makes ie1 increase and ve increase . Since the b pole of T2 is grounded, vBE2 = 0-ve = -ve , so vBE2 decreases and ic2 also decreases. In the whole process, under the action of single-ended input vi , the current of the two BJTs increases ic1 and decreases ic2 . Therefore, when single-ended input is used, the differential amplifier T1 and T2 still work in the differential state. Single-ended input is consistent with dual-ended input. Summary: ① As long as it is a dual-ended output, whether it is single-input or dual-input, its A VD , Ri , and Ro are the same.
 ② As long as it is a single-ended output, regardless of whether it is a single input or a dual input, its A VD , R i , and R o are the same.
(3) Common-mode voltage gain 
① Double-ended output AVC .
Because v i1 = v i2 , the changes are equal, that is, v C1 = v C2 , so 
In fact, it is not easy to make the circuit completely symmetrical, but even so, the AVC is very small and the amplifier circuit's ability to suppress common mode is still very strong. ② For the common-mode signal of the single-ended output AVC , because the currents on
both sides increase or decrease at the same time, the value obtained at the e-pole is twice i e . ve = 2i e Re , which is equivalent to connecting 2Re resistors to the emitter of each BJT , as shown in Figure 6.11. (Here Re is the AC equivalent resistance of the constant current source) Therefore,  (4) Common mode rejection ratio K CMR
K CMR is a technical indicator to measure the ability of differential amplifier to suppress common mode signals. It is defined as:  The larger the A VD is, the smaller the A VC is. The stronger the common-mode rejection capability is, and the better the performance of the amplifier is. Therefore, the larger the K CMR is, the better. Ideally, the K CMR of the dual-ended output is =∞
. The common-mode rejection ratio of the single-ended output is:  The total output voltage of the double-ended output circuit is: 
The total output voltage of the single-ended output circuit is: 
Example 1: The input stage of the integrated operational amplifier BG305 is shown in Figure 6.12. β 1 = β 2 = 30, β 3 = β 4 = β 5 = β 6 = 50 of each BJT, V BE = 0.7V of each BJT, R b = 100K, R c = 50K, R w = 10K (the sliding end is adjusted to the midpoint), Re = 1K, and RL, that is, R i of the second stage, is 23.2K. Find: (1) The static operating point of the amplifier stage; (2) The differential gain A VD ;
(3) The differential input resistance R i , the differential output resistance R o .
Solution (1) Find the static operating point of the amplifier stage 
Solution (2) Differential amplifier A VD
 Solution (3) Differential input and output resistances R i , R o 
return Section 4 Integrated Circuit Operational Amplifier
Classification of analog integrated circuits
Analog integrated circuits can be roughly divided into:
linear amplifiers, power amplifiers, comparators, multipliers, voltage regulators, (D/A, A/D) converters, phase-locked loop devices, etc.
, linear amplifiers can be divided intogeneral-purposeandspecial-purpose types. Among linear amplifiers, the earliest developed and most widely used is the integrated operational amplifier circuit. Figure 6.13 shows the actual pictures of some operational amplifiers. 
1. Simple integrated circuit operational amplifier
1. Basic block diagram and symbol of integrated operational amplifier (as shown in Figure 6.14) 
2. A simple integrated operational amplifier (as shown in Figure 6.15)
(1) DC analysis:
(2) Calculation of the total gain of the amplifier circuit
return Section 5 Main Parameters of Integrated Circuit Operational Amplifiers
1. Input offset voltage V IO
2. Input bias current I IB 
3. Input offset current I IO  
4. Temperature drift 
5. Maximum differential input voltage V idmax
6. Maximum common mode input voltage V iCmax
7. Maximum output current I Omax
8. Open-loop voltage gain A VO
is the voltage gain of the differential signal without external feedback. 
9. Open-loop bandwidth BW (f H )
10. Unity gain bandwidth BW (f T )
11. Conversion rate SR 
return Section 6 Introduction to Field Effect Transistor (FET)
1. Junction Field Effect Transistor
1. Structure and characteristics of JFET
1. Structure  
The structure of the field effect transistor is shown in Figure 6.18. It uses impurities to diffuse high-concentration P-type regions on both sides of an N-type semiconductor, represented by P + , to form two P + N junctions. Two electrodes are drawn out from the two ends of the N-type semiconductor, called the drain D and the source S respectively . The electrodes of the P regions on both sides are connected together and called the gate G. If a positive voltage is applied between the drain and the source, the majority carriers (that is, electrons) in the N region can conduct electricity. They start from the source S and flow to the drain D. The direction of the current is from D to S, which is called the drain current ID . Since the conductive channel is N-type, it is called an N-channel junction field effect transistor.
2. Characteristics
① vGS <0, Ri is very high;
② Voltage-controlled device;
③ Unipolar device;
3. Output characteristics of N-channel JFET 
When v GS = 0, the channel resistance is minimum and ID reaches maximum.
When v GS < 0, the depletion layer becomes larger, the channel resistance becomes larger, and the corresponding ID decreases . Therefore, the characteristic curve shown in Figure 6.19 is formed. 4. Comparison of the characteristics of JFET and BJT
(1) Field effect tube is a voltage control device: i D is controlled
by V GS . From the output characteristics, the parameter of each different output characteristic curve is V GS . In the constant current region, i D is basically independent of V DS . And the amplification effect of the field effect tube is described by transconductance g m = △i D / △V GS | V DS . The transistor controls i C through i B. The parameter is i B. In the amplification region, i C is basically independent of V CE . The amplification effect is described by the current amplification factor β = △i C / △i B. (2) i G = 0. Therefore, both DC and AC R i are very high. The transistor b and e poles are in a forward biased state, and R i between b and e is relatively small: several thousand ohms. (3) The field effect tube uses a polar multi-pole conduction (unipolar device), which has the characteristics of low noise and little influence by external T and radiation (good temperature stability). (4) Due to the symmetry of the field effect tube, sometimes D--S poles can be used interchangeably. All performance is basically unaffected. It is more convenient and flexible in application. However, if S and substrate are connected together during manufacturing, D--S cannot be interchanged. (5) The manufacturing process of field effect tube is simple, which is conducive to large-scale integration, and it occupies a small area and has a high degree of integration. (6) The MOS field effect tube has a high R i , and the induced charge of the G pole is not easy to discharge. The SiO2 layer is thin, and the equivalent capacitance between the G pole and the substrate is very small. A small amount of induced charge can form a high voltage, which will break down the SiO2 and damage the tube. When storing the tube, G--S should be short-circuited to avoid the G pole being suspended. When welding, the soldering iron shell should be well grounded to prevent the tube from being broken down due to leakage of the soldering iron. (7) Application of field effect tube When the field effect tube works in the constant current region, when the GS voltage changes △V GS , the D pole current changes accordingly △i D. If △i D passes through a larger R L , the △V 0 =△i D R L taken out from R L may be many times larger than △V GS , that is, △V GS is amplified. Therefore, field effect tubes can play an amplifying role in circuits just like transistors.
| | | | Voltage controlled device g m = △i D /△V GS | V SD | Current control device β = △i c /△ i B |v CE | | The input PN junction is reverse biased, i G = 0, and R i is very large | The PN junction at the input end is forward biased, i b ≠ 0, and r be is small | | Unipolar device, one carrier: low noise, good temperature stability | Bipolar devices, two types of carriers: higher noise, poor temperature stability | | Symmetrical structure, DS poles are interchangeable | The structure is asymmetric, and the CE poles cannot be interchanged. | | Can be used as a voltage-controlled resistor with better performance | Poor linear effect | | The manufacturing process of field effect tube is simple and symmetrical, which is conducive to large-scale integration, small area and high integration. | The structure of the transistor is asymmetric, the manufacturing process is more complicated than that of the field effect transistor, and the integration is not high enough. | | 2. MOS Field Effect Transistor 
The input resistance of JFET can reach 10 7 ohms, but in essence, this is the reverse resistance of the PN junction, and there is always a reverse current when reverse biased, which limits the higher requirements for resistance under certain working conditions. At the same time, from the manufacturing process point of view, it is still relatively complicated to highly integrate it.
The insulated gate field effect transistor is made of metal-oxide-semiconductor field effect transistor, called Metal-Oxide-Semiconductor , referred to as MOSFET. The gate of this field effect transistor is isolated by an insulating layer (such as SiO2 ) , so R i is higher and can reach more than 10 9 ohms.
The difference between MOS tubes and JFETs lies in their different conductive mechanisms and current control principles. JFET uses the width of the depletion layer to change the width of the conductive channel to control ID , while OSFET uses the electric field effect on the surface of the semiconductor to change the conductive channel by the amount of induced charge to control the current.
MOS tubes are divided into two categories: N-channel and P-channel. Each category is further divided into enhancement type and depletion type:
Enhancement type: When V GS = 0, there is no conductive channel, I D = 0
Depletion type: When V GS = 0, there is a conductive channel, I D ≠ 0
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