Current source circuit

　　In the manufacturing process of integrated circuits, it is much easier to make various types of transistors on silicon wafers than to make resistors, and the area of ​​silicon wafers occupied is also much smaller. Therefore, in addition to being amplifiers, triodes in integrated circuits are used as constant current sources or active loads in large quantities to provide a suitable static operating point for the amplifier and increase the amplifier's gain. The following first introduces the constant current source and active load circuits in integrated circuits. Basic current source circuit 1. Mirror current source circuit The circuit shown in Figure 7-2-1 is a typical mirror current source circuit. 　　The working principle of this circuit is: when the circuit is completely symmetrical, the current IR on the resistor R _can be used as the reference current of the circuit. According to the node potential method, the expression of this current can be obtained as:    　　   Under the condition of β»2, we can get     (7-2-1) by shifting the terms 　　. It can be seen from the above formula that when the values ​​of Vcc and R are determined, the collector current of transistor T0 _has a certain value IR _. Due to the symmetry of the circuit, the collector current of transistor T1 _is in a mirror relationship with the collector current of transistor T0 _, and also has a certain value IC _. The 　　mirror current source circuit has a simple structure and a wide range of applications, but it has the disadvantage that when _IC is large, the power consumption on resistor R is also large. The improved method is to add resistor _Re to the emitter of the two transistors , so that the relationship of the mirror current becomes a proportional relationship, forming a proportional current source circuit. 2. Proportional current source circuit The proportional current source circuit is shown in Figure 7-2-2.       The working principle of this circuit is: From the structure of the circuit, we can know     (7-2-2) According to the current equation of the transistor, we can get    According to the symmetry of T ₀ and T _1, we can get  Substituting the above formula into formula 7-2-2, we can get:  When β»2, we have I _C0 ≈I _E0 ≈I _R , I _C1 ≈I _E1 , and substituting these relations into the above formula we get In a certain range, IR _≈IC1 , _the logarithmic term in the above formula can be ignored, then      (7-2-3) Compared with formula 7-2-1, it can be obtained that, under the same IC1 _, a larger R can be used to reduce the value of IR _and reduce the power consumption of R. At the same time, Re0 _and Re1 _are the emitter resistances of the two transistors, and the introduction of current negative feedback makes the output current of the two transistors more stable. 3. Multi-channel current source circuit The　　integrated operational amplifier is a multi-stage amplifier circuit. Because the static operating points of each stage of the amplifier are different, multiple current source circuits are required. In the manufacturing process of the integrated circuit, multiple proportional current source circuits can be combined together to form a multi-channel current source circuit, as shown in Figure 7-2-3.     The working principle of this circuit is: when the reference current IR _is determined, according to formula 7-2-3, different bias currents IC1, IC2 and IC3 can be obtained _by selecting _different Re1 _{, Re2} and _Re3 . _The multi -channel current source circuit can also be composed of MOS tubes. The multi-channel current source circuit composed of MOS tubes is shown in Figure 7-2-4.　  　 4. Amplifier with current source as active load　　From the previous knowledge, it is known that the open circuit voltage gain of the common emitter or common source amplifier circuit is or. From the expression of voltage gain, it can be seen that the voltage gain of the amplifier is proportional to R _C or R _d . To increase the voltage gain of the amplifier, the resistance value of Rc or R _{d must be increased while β and g m} remain unchanged . 　　Rc or R _d becomes larger. To keep the static operating point of the transistor unchanged, the DC supply voltage of the circuit must also be increased, which will cause the power consumption of the integrated circuit to increase. In order to solve this problem, in the integrated circuit, a current source is used as an active load to replace Rc or R _d . 　　Using a current source as an active load can achieve the amplifier to obtain a suitable static operating point current while the power supply voltage remains unchanged. For AC signals, a large equivalent resistance r _ce or r _ds can be obtained to replace Rc or R _d . The differential amplifier circuit using a current source as an active load is shown in Figure 7-2-5.   

　　In the figure, transistors T1 _and T2 _form a differential amplifier; transistors T3 _, T4 _and T5 _form a current source circuit to provide the differential amplifier with a suitable static operating point current; T6 _and T7 _form the active load of the differential amplifier. This circuit can not only make the differential amplifier have a suitable static operating point current, but also have a large equivalent resistance rce for AC signals _{to replace R C} in the original circuit and obtain a large voltage amplification factor.