TL431
1 Introduction Power Supply Widely used in the circuit module, such as the popular LM78, ASM117 series, the majority of only three pins, simple structure, easy to use. Usually the pressure difference between input and output is greater than 2V to work properly, but the low voltage Battery Powered occasions. General increase in the power supply 2 V Power supply Need to increase by 20% to 40% (corresponding to the output 3 V or 5 V) the cost and size. Use Switching Power Supply Principles of production of low-voltage differential module (such as the MAX605), the input voltage and output voltage is almost equal to the output stability under voltage, but the Force needs such as circuit Power supply Ripple is relatively small. Tackle this problem. Discrete devices used to design a low dropout voltage regulator power supply circuits. Circuit devices use conventional devices and low cost. Simple. The experiment tested the actual circuit, has good load characteristics and voltage stability.
2 circuit works Figure 1 shows the low level DC power supply circuit diagram. The circuit is a voltage reference, voltage amplification and current amplification and other three sessions were. Among them, the reference voltage generated by the TL431, the circuit connection according to Figure 1, when the current through the R0 in the 0.5 ~ 10mA obtained with a stable 2.5 V reference output.
Specific output voltage values determined by the operational amplifier UA, the advantages of using the same phase amplifier input impedance at its maximum, may well be TL431 2.5V output voltage and circuit isolation after the class, to not load change; operational amplifier and resistors R3 and R2 form part of scaling can be carried out as required on the reference voltage output scaling, but can not exceed the maximum output voltage of the op amp's supply voltage.
Current amplification using two Transistor , UA through the driver control to adjust pipe to adjust pipe VQ2 VQ1, form part of feedback for current amplification, the output voltage regulation, in order to achieve regulated output. UA diode voltage VD in the op-amp output to adjust the tube VQ2 base emitter voltage of a negative, so VQ2 immediate access to cut-off state, current Ic2 quickly lower, VQ2 the VCE result VQ1 increase the base voltage increases, so VQ1 The base current IB reduced, thus reducing the output current IC Q1 ( IB); and vice versa. RL is the output load, C0 and C1 is a filter Capacitance .
Three main parameters of the circuit design
Version 3.1 design the control
Control link circuit equivalent diagram shown in Figure 2 and Figure 3, in which Figure 2 shows the voltage gain ratio diagram, Figure 3 for the current schematic to enlarge.
According to Figures 2 and 3, control link circuit equation can be drawn:
The formula, Irg UA for the output of op amp 1 output control current.
From (2) known, Irg VQ2 Rights by controlling the control current IC2 VQ1 the base current, IB1, R8 control adjustment tube VQ2, then control VQ1 output current IC1, VQ2 form with VQ1 series negative feedback without further amplification VQ1 output current IC1, IC1 with the R8 on diversion. Circuit output voltage Vcc is 5V, driving rated load is 350 , 7V power supply is the standard output of the battery. Op amp selection LM358, taking R1, R2 to 10k , TL431 current range is 100 ~ 150 mA, use R1 = 3k , to meet the requirements. VCC = (1 + R2/R1) x2.5 = 5V. R8 and R9 properly selecting the resistance value, so VQ1 and VQ2 are working in the linear region.
3.2 select and adjust the tube operating point of the set of static
Adjustment tube VQ1 and VQ2 main parameters: (1) reverse voltage VCEO not less than 2 times Vin margin; (2) the maximum allowable current ICM, not less than 2 times the output current I0 of the margin; (3) Power dissipation PCM should the power loss of the safe area; usually for safe and reliable parameters to be selected according to the actual value of the times. Set the appropriate static operating point (that is, to determine the base static current Ih, emitter current Ie, the collector-emitter static voltage of a Uce), can ensure the accuracy of the output steady while making the smallest adjustment of the loss of control. The right to adjust the static working point of the first requirements management work in the enlarged state, and secondly to meet the power and load fluctuations cases, Ib, Ie, Ucc as small as possible to reduce the loss. Set the static work point to select the appropriate drive control VQ1 and bias resistors R8, R9. VQ1 static working point:
The formula, Irg control for the op amp output signal, Vin as the supply voltage, Vcc to the output voltage of 5V, RL for the rated load 200 , VD is the diode turn-on voltage of 0.7V.
From (3) and (4) determine VQ2 parameters, and then calculate the resistor R9:
Use magnification 1, 2 in the 30-80 adjustment between the pipes, the larger adjustment of magnification control in smaller power consumption, but the stability of the lower, where is 50 selected to design the power supply in 5.2 ~ 9V between the fluctuations in order to prevent the burning of high voltage adjustment control VQ2, plus about 1k resistor R8 to limit protection.
3.3 The over-current protection circuit
Figure 3, the resistor Ri and the composition of over-current protection transistor VQ3 link. Output current is too large, the voltage on the sampling resistor Ri is greater than 0.7V, VQ3 turn, force the tube to adjust the base voltage Vbe lower power output until the close. R4 = 0.7/kIC. Where, LC is the output current, K is the maximum over-current coefficient value is usually about 1.5. R7 = (Vcc-Uce3) / Ie3 Vrg/Ic3, restrictions Ic3 not be too big to avoid over-current damage VQ3.
4 test Figure 4 for the design of a DC power supply module, input power to DC 5 ~ 9V battery group; were characteristic of the circuit design for power supply and load characteristics test, which tests the load characteristics of the battery to 6.5V analog input actual working environment. Figure 5 records the results of its tests. Output ripple test data, when the power supply input voltage 5-11V, output ripple is 5 ~ 8 mV.
5 Conclusion
Can be seen through analysis and testing, the DC power regulator module with high precision regulator, the load characteristics of good features, and simple circuit, additional interfaces can be used to monitor the actual power supply P0, the circuit has been put into production, through the practice Inspection of the circuit design and reliable performance, low power consumption, single power supply can satisfy the demand of the application.
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Very good article on the use of the TL431. I have used this device in the past and it works great.
Thank You,
GuruSantiago
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