085,1086 can be selected, aic1084 is very good; LDO, the low dropout regulator, is a low-voltage differential linear regulatorFor a traditional linear voltage regulator.
What you said is not very detailed. You already have a 12V power port on the board, and now you want a 0V voltage port. If so, it is simple, depending on the power size of your 5V load, choose the appropriate LDO chip or fixed three-terminal voltage regulator can be achieved.
If you want to get a stable 5V, you can use a three-terminal regulator to buck, also known as LDO, there are many models to choose from under the Internet search, commonly used LM7805, AMS117-0, etc., this kind of buck program can be obtainedTo a very stable 5V output. Pay attention to the thermal dissipation power of the IC. This type of circuit has the advantages of stable output and low noise.
DC? If you consider efficiency, you can use DCDC, generally 2359, or other, there are a lot of Baidu search, but it is recommended that you convert twice, otherwise the fever is serious.
A more convenient method is to use the three-terminal voltage regulator integrated circuit 7805 to form a buck voltage regulator circuit to obtain 5V DC output. The stabilized power supply composed of three-end regulated IC7805 has the functions of overvoltage protection, overcurrent protection and overheat protection. The performance is very stable and the output current is very stable≥1A. Only a very small number of peripheral components, the circuit is simple and easy to make.
Because the input and output are DC, there is no need to use capacitive filtering, only one 7812 can be used. The output current of the LM7812 (package pictured above) is 1A, if you need a larger current, you can search for a current expansion circuit online. The LM7812 is protected against overload and overheating. If the LM7812 heats up seriously, install a heat sink.
The simplest is a power resistor, a 5V voltage regulator diode, the load is in parallel on the 5V regulator tube, which is a parallel regulator, low efficiency, large power consumption of the resistance and regulator tube, the output power and the power current of the regulator tube are very related, usually used in tens of MA small current places.
Steps: First calculate or measure the current of 5V electrical appliances, and then calculate the partial voltage resistance. Assume that the current of the 5V appliance is 0.2A, the voltage on the divider is 12-5=7V, and the resistance R of the divider is 7÷0.2=35 (ohms).
The output current is less than 1A, 7805 Install radiator, left 2200UF and 0.1UF, right 1000UF and 0.1UF.
To change the 12V voltage to 5V, in addition to using a transformer, there is a series voltage division resistor method. First calculate or measure the current of 5V electrical appliances, and then calculate the partial voltage resistance.
For the DC voltage of 12V, two diodes can be connected in series at the positive output end (the current should be adapted to the needs), so that the voltage is reduced to 10V, followed by two equivalent resistors in series, and the output is positive or negative 5V.
1, you can use the method of voltage division: with a winding resistor connected to the 12V voltage, in the resistance from the neutral line 47% of the position tapped to lead out a line, this line and the resistance of the neutral line between roughly 5V. However, note that the current flowing through the resistance should be much greater than the current flowing through the 5V tap, so that the 5V tap voltage is relatively stable.
2, v with a resistor partial voltage can be, and then connected to vcc can be. The power supply does not add capacitance, and the reset circuit should add capacitance. Fallen leaves chasing the autumn wind does not move | Posted on 2012-04-02 Report | Comments 1 0 12V is a battery? If it's a battery don't use any filter capacitor.
3, the diode is not partial voltage, but step-down, that is, in the 12V power output series of multiple diodes, can indeed step-down, the disadvantage is that the output voltage stability is not good, and the loss is similar to the linear regulator.
4, the V power supply has passed 7805Converted to 5V. They're just on the same ground. 12V is equivalent to the input voltage, 5V is its output voltage. Therefore, after the MCU, of course, it is 5V to supply power to the MCU.
5, a little basic can try. Generally switching power supplies have a feedback TL431 or voltage regulator shown in the figure there are other ways, but the regulator and 431 are the most commonly used. Replace the regulator or replace the 431 sampling resistance with two identical ones.
6, with switching power supply, linear regulator 7805 used here is too low efficiency, itself to consume 35W power consumption (pressThe LM2596 regulator chip has a variety of output voltage options, including 3V, 5V, 12V, and adjustable (ADJ), with a maximum output voltage of 37V. The chip supports two operating modes: low power mode and normal mode, and can be switched by external control signals.
3, LM2596 switching voltage regulator is a step-down power management monolithic integrated circuit, capable of output 3A drive current, while having good linearity and load regulation characteristics. The output voltage will not be greater than the input voltage. If the output voltage is greater than the input voltage, a booster power supply chip is used. As XL600VT1050.
4. Yes. Using the current international advanced high-frequency modulation technology, its working principle is to extend the voltage and current of the switching power supply, to achieve a wide range of voltage and current regulation.
5, LM2596-5V, LM2596-3 constant pressure output model. To obtain 8V output voltage, please use figure A, to obtain 3V, 5V output voltage, please use figure B.
6, the principle diagram of the adjustable power module of LM2596 is as follows: Input filter circuit: CLCC3 composed of double π filter network is mainlyThe electromagnetic noise and clutter signal of the input power supply are suppressed to prevent interference with the power supply, and also to prevent the high-frequency clutter generated by the power supply itself from interfering with the power grid.