Simple DC power supply design with over-current protection
Currently, a variety of DC power products flooded the market, power technology is relatively mature. However, due to cost considerations, performance requirements for the power supply situation is not very high, over-current protection can be used with integrated voltage regulator circuit, the same can meet the product requirements. Over-current protection circuit for power circuit an indispensable part of the control method according can be divided into shutdown mode and limit, rather than DC power shutdown mode should be adopted.
Over-current protection circuit must first have a current sampling areas, common practice is a small series resistor or Hall element to obtain the current signal. As the relatively large size of the Hall element is very expensive, so consider using a small resistor in series approach.
1 Work
The LM317 voltage regulator with overcurrent protection circuit shown in Figure 1, an integrated regulator circuit is generally divided into five parts, namely the exchange step-down circuit, rectifier circuits, filter circuits, regulator circuits, protection circuits. AC 220V step-down voltage by the power transformer rectified to DC voltage Vin, the voltage through the filter circuit input to the integrated voltage regulator input, output in the integrated voltage regulator available 1.25 ~ 37V DC voltage. Works and the part of the voltage waveform diagram shown in Figure 2.
Simple DC power supply design with over-current protection
The following analysis of the working process of the protection circuit.
1.1 The protection of integrated voltage regulator
In order to obtain a higher output voltage, LM317 regulator regulator side and to the resistance R2 between the value of pressure drop is often greater in R2 than both ends and then a 10μF capacitor C3, to be effective in inhibiting output ripple. When the input or output short circuit, the capacitor C3 of the discharge will have an impact on the R1 voltage, will endanger the regulator reference voltage circuit, so take both ends of the R1 and diode D3 to protect the regulator.
Regulator output capacitor can also work without, as regulator in the 1:1 depth to work under negative feedback, when the output load capacitance of a value, the regulator may appear self-excitation phenomenon . Therefore, the regulator access to 0.1μF input capacitor C1, output access 1000μF electrolytic capacitor C5, the provision of adequate current supply at the same time to prevent possible high-frequency oscillation and reduce noise and improve load transient response. When the input shorted, C5 through the regulator to adjust tube discharge, C5 is bigger, the impact of the discharge current is large, the internal voltage regulator's output through the discharge transistor, the output transistor emitter potential anti- to breakdown. To this end, the regulators at both ends and then diode D2, through the input short circuit C5 D2 discharge, protect the regulator.
1.2 Overcurrent
Over-current protection circuit shown in Figure 3, R5 for the sampling of small resistance. When the power of work, the regulator output positive DC output voltage, the motor started. As the DC motor start transient current iout large (approximately rated current of 8 to 10 times), iout flows through a small resistor R5, and R4 by the charge on C4. By setting the R4, C4's value is greater than the charge time Υ motor start-up time δ, V2 (9013) is off, the motor start to steady state, the current back to the operating current. Once the electrical short circuit or block transfer, the voltage across the capacitor C4 to the conduction voltage V2, the V2 turn, force the regulator output voltage of the reduced reference voltage 1.25V.
Simple DC power supply design with over-current protection
2 Circuit Design
2.1 The choice of an integrated voltage regulator
When choosing an integrated regulator, should take into account performance, use and price aspects. Performance mainly based on the size of the load voltage and current, adjusting the level of rate stability and range of width to work the election. Because of its output voltage LM317 Series adjustable voltage regulator while its higher precision, high ripple rejection ratio and good output voltage temperature characteristics, and has been widely used.
Set the power output of total power P0, load rated voltage U0, the output current rating of I0 = P0/U0, for stable operation of the circuit, also need to consider some design margin (generally taken more than 10%). LM317 series voltage regulator as the main parameters listed in Table 1, according to the calculated current value, select the appropriate regulator.
Simple DC power supply design with over-current protection
2.2 rectifier circuit
Bridge rectifier filter circuit to determine the rectifier diode and filter capacitor value.
2.2.1 Rectifier Diode Selection
Select diode to diode reverse voltage VRM based and forward current IF. As the capacity of the larger filter capacitor, diode conduction angle smaller, pulsed current through the diode the greater the extent, therefore, the amplitude of rectifier current must be considered. The average current flowing through the rectifier ID = Ii / 2, Ii = IR2 + I0, IR2 = IR1 + Iadj ≈ 0.01A (type in Ii for the regulator's input current, IR1, IR2, Iadj were flowing through R1, R2, and adjust the end of the current), then ID = (0.01 + I0) / 2. Taking into account the impact of capacitor charging current, forward current and generally the average current of 2 to 3 times.
Maximum reverse voltage of diode current protection DC power supply with a simple design, the formula for the power transformer secondary voltage U2 RMS, Ui for the rectifier output voltage (input voltage regulator). To ensure stable operation of voltage regulator LM317 input voltage Ui and the output voltage U0 is generally the difference between the range of 5 ~ 15V, take Ui-U0 = 1OV, may Udmax = 1.2Ui = 1.2 (U0 +10) = 12 +1.2 U0. May consider a certain design margin.
2.2.2 Design of filter capacitor
Electrolytic capacitor C1 filter selection principle is: whichever is greater than the discharge time constant RLC1 charge cycles 3 to 5 times, the pressure must be greater than the value of Uc pulse peak voltage. The bridge rectifier circuit, the ripple voltage peak 2U2, C1's charge cycle AC power cycle T is equal to half of that simple design with over-current protection DC power supply rectified the formula for the equivalent RL load resistance, and RL = Ui / Ii = (10 + U0) / (0.0l + I0), substituted C1 value can be determined by the formula
2.3 Power transformer design
In the series regulator circuit to determine the transformer secondary voltage is very important. If there is surplus and the second to do a higher voltage will increase the adjustment of the loss of control, so that the radiator was increased accordingly. Therefore, to design excellent power, transformer parameters are to be adjusted several times. Offer a comprehensive reference to explain the history of the power transformer design of the various points din, this will not go into. Approximate calculation method used here to determine the U2 and I2.
U2 = Ui/1.2 = 0.83 (10 + Uo)
I2 = (1.5 ~ 2) Ii = (1.5 ~ 2) × (0.01 + Io)
2.4 The integrated voltage regulator circuit design
To ensure the regulator can in no-load work. The current flowing through resistor R1 can not be too small. Generally take IR1 = 5 ~ 10mA, so R1 = VREF/IR1 = 1.25 / (5 ~ 10) × 10-3 ≈ 120 ~ 240Ω, the formula for the regulator reference voltage VREF. The output voltage U0 and VREF, R1, R2 have the following relationship:
Un = VREF + (IR1 + Iadj) R2 = (1 + R2/R1) VREF + IadjR2 (1)
Adjust the resistance R2, you can change the size of the output voltage. As Iadj small (only 50μA), so equation (1) can be written as Uo = (1 + R2/R1) VREF = 1.25 (1 + R2/R1) (2)
From (2) obtain R2 = (0.8U0-1) R1.
2.5 Protection Circuit Design
The choice of circuit protection diode is relatively simple, as long as the guarantee to meet the breakdown voltage and impulse current two requirements on it. The main function of R3 is to limit the transistor's base-level current, generally take 1 ~ 2kΩ. Here to talk about the over-current protection circuit.
2.5.1 Start State
Motor starts charging time Υ must meet start time is greater than δ, V2 does not turn the motor to start properly. As the starting current large, generally rated current of 4 to 7 times, can be regarded as constant, set I = 5I0. According to Figure 4 known.
Simple DC power supply design with over-current protection
Simple DC power supply design with over-current protection
2.5.2 Protection Status
Motor load set to run at the rated condition, motor current I0 to have stabilized. Electrical short circuit or block transfer, the current suddenly increases to short-circuit current Is, capacitor C4 starts charging. Consider some design margin, to take the protection current limit IG <IS, equation (3) the same set up as a class-wide response to this equation, initial condition uc4 (0 +) = I0R5, mandatory component uc4 (∞) = IGR5 , solution was
Simple DC power supply design with over-current protection
2.6 Thermal Design
Regulator the maximum allowable power dissipation depends on the chip, the maximum junction temperature TjM, when T <TjM to work properly when the voltage regulator. Therefore, the regulator of the heat capacity of the stronger, the lower the junction temperature, it can withstand the larger power. Regulators cooling capacity depends on its resistance to the semiconductor device can be reduced after the heat sink increases the total thermal resistance. If so Rθ1 said the device from the junction to case thermal resistance, Rθ2 that case to the heat sink from the surface of the device thermal resistance, RθA that surface from the junction to heatsink thermal resistance, then RθA = Rθ1 + Rθ2. If so Rθd said heat sink to ambient thermal resistance, Rθ 'that after the heat sink increases the total thermal resistance junction to the air, then Rθ' = RθA + Rθd. Integrated voltage regulator set the maximum allowable junction temperature for the TjM, the maximum ambient temperature of TAM, increase power consumption of the device after the radiator PD, have the relationship (9).
Simple DC power supply design with over-current protection
Rθd obtained after the equivalent thermal resistance of heat sink through the investigation and the relationship between material thickness and surface area of the manual. The scope of available surface area. Table 2 lists several common package thermal resistance.
3 results
AC supply voltage 220 (1 ± 1O%) V, Output rated voltage U0 = 24V, rated power P0 = 15W, rated current I0 = 0.625A, motor start-up time δ = 50ms, to allow short time ts = 500ms, to protect current setting value Ic = 2A, the maximum TAM = +45 ℃ ambient temperature conditions. To design the circuit parameters in Table 3.
Simple DC power supply design with over-current protection
Selected by the design parameters of Table 3 after assembly, was determined for the following technical specifications to achieve:
Output characteristics of Un = 24V, In = 0 ~ 1A:
Voltage stability Sv ≤ 5 × 10-6;
Load S1 ≤ 5 × 1O-5;
Temperature coefficient α ≤ 1 × 10-5 / ℃.
Motor can start properly, when the motor locked rotor, the overcurrent protection function properly.
4 Conclusion
In summary, the use of LM317 integrated voltage regulator designed for low-power DC power supply, the circuit structure is simple, high efficiency, low cost, good performance of the output voltage has a good prospect.