12V to 3V circuit, 12V to 5V is the easiest way

LEDs are flow control devices, so it is ideal to calculate their rated current.

A resistor can be directly connected in series, the battery is 12V, and the LED is 3V, so the voltage of the resistance is 12-3=9V, and the LED working current: by P=UI, that is: I=P U=1 3, therefore, the current of the resistance is also 1 3A, then the resistance value is:

r=u i=9 (1 3)=27 ohms, and the resistor power is p=ui=9*(1 3)=3w.

Therefore, use a resistance of 27 ohms, the power of 3W resistor, if you can't find it, you can choose multiple series connections, but it should be noted that the resistance value can only be greater than 27, the power must be greater than 3W, otherwise, it will lead to disaster ......

1: Use a three-terminal buck IC, or use an adjustable voltage buck IC, MC34063.

2. Circuit: A conductive circuit composed of metal wires and electrical and electronic components, called a circuit.

Add a power supply to the input of the circuit to generate a potential difference at the input end, and the circuit can work. Some phenomena can be seen intuitively, such as voltmeter or ammeter deflection, light bulb light, etc.; Some may require a measuring instrument to know if it is working properly. According to the nature of the current flowing, there are generally two types.

The circuit through which direct current passes is called a "DC circuit", and the circuit through which alternating current passes is called an "alternating current circuit".

With a voltage converter chip.

LM7108 is recommended

It can support step-down within 36V, fluctuation of 5%, drive capacity is not clear, or it may be 5A) The recommended circuit is in the chip manual.

Personally, I am not optimistic about this method of taking out the voltage with a resistor divider If this 3V is used for power supply, if there is a circuit behind it, it is easy to cause impedance mismatch problems, and it is a good choice to get a voltage conversion chip.

Make four diodes, in series.

Or find a 27 ohm resistor and put it in series.

It can be controlled with resistors and polar capacitors.

1. There are two ways to reduce 12V to 5V:

1. The voltage division method of the voltage generator is too costly and the space utilization rate is poor.

2. Resistance series voltage divider method.

2. Solution analysis.

1. Steps: First calculate or measure the current of the 5V electrical appliance, and then calculate the size of the voltage divider resistor. Assuming that the current of a 5V applier is 12-5=7V, the voltage on the divider resistor is 12-5=7V, and the resistance of the divider resistor is r=7 ohms).

2. The circuit diagram is as follows:

3. Use the integrated voltage regulator tube IC step-down method.

1. Commonly used integrated voltage regulator IC components, such as voltage regulator tube LM7805, three-terminal voltage regulator tube TLM317, etc.

2. Formula: vo=(.

3. The circuit diagram is as follows:

4. Summary: In the case of considering the cost unchanged, the first method is more suitable.

A 12V voltage requires a current of 3W and a resistance of 48 ohms. Namely.

i=p u=3 12= r=u i=12 ohms) If the voltage becomes 3V, a series resistance of 3 ohms is required, ie.

i=p u=3 3=1(a) r=u i=3 1=3 (ohms) Of course, the power of the resistor should also reach 3W and above, and the resistor will be burned out if it is not allowed to do so.

1. The simplest is a power resistor, a 5V regulator diode, the load is connected in parallel on the 5V regulator, which belongs to the parallel regulator, the efficiency is low, the resistance and the regulator tube consume a lot, the output power is very related to the power current of the regulator tube, and it is usually used in tens of mA small current places.

2. Series resistance and 10-12 silicon diodes (the diode can be replaced by a 7V power regulator tube, and the current output is not as large as the diode), which is a simple series voltage regulator circuit, and the advantage is that if the load current is large.

The current value depends on the forward conduction current of the diode, the current can reach the amperage level (the power supply needs to output 12V sufficient current), if it is a fixed current, you can choose the resistance value of the power resistor, reducing the number of diodes in series.

V three-terminal regulator, can be said to be a very simple circuit, the regulator model is different, the output current value amperes, three pins, in, ground, out, respectively connected to the power line. The disadvantage is that the efficiency is low, the output current is large, and the power consumption on the regulator block = (12-5V x The current is large, and a heat sink needs to be installed.

Brother, add a 7805, you don't have to calculate.

The parameters of its transformer are as follows:1. According to the information on the official website of Suning Tesco, the input voltage: AC100-240V50 60Hz, and the output voltage: DC12V.

2. Output current: 3A, output power: 36W.

3. Shell protection level: 44, dimensions: 47474mm, plug sensitivity and specifications: .

This specification of power transformer is suitable for the need for 12V voltage and 3A current bridge equipment, such as routers, cameras, audio, etc., its main application is to provide stable DC current for the equipment, to ensure the normal operation of the equipment. For equipment that requires the use of power transformers, choose the corresponding specifications of power transformers according to the needs of the equipment at the time of purchase.

This is a buck circuit, which is a buck circuit, and it turns out to be like this:

First of all, the 6 pins produce a rectangular wave with an amplitude of 12V and an adjustable duty cycle, which is filtered by the inductor L2 and the capacitor C15 to form a + voltage, and R19 is the feedback resistor, to detect whether the voltage on the C15 is, if it is higher, the chip reduces the duty cycle to make the voltage drop, and in the same way, if it is lower, the duty cycle is increased to increase the voltage.

The latter is the power supply filtered by the inductor and capacitor, and the voltage is more stable.

The core of the buck circuit is based on the fact that narrow pulses of equal impulse but different shapes produce the same effect on the inertial link. Similar to the momentum theorem.

This is a synchronous DC-DC core. L6 and C33 are for filtering and have nothing to do with the 12-turn principle.

In the first stage, the upper MOS tube inside the chip is turned on, the voltage at the end of the chip is 12V, and the C15 is charged through the inductor L2;

In the second stage, the lower MOS tube inside the chip is turned on, the voltage at the PAHSE terminal of the chip is 0V, and the energy in L2 is charged to the capacitor C15 through C15 and the lower MOS tube inside the chip.

The feedback terminal FB and the feedback resistors R19 and R18 of the chip control the on-time of the upper and lower MOS tubes inside the chip, so that the output voltage is equal to.

When the en's of the chip is high, the chip starts to work, and the low level does not work;

The capacitor C28 on the boot side is a bootstrap boost, which provides a driving voltage to the upper MOS transistor inside the chip.

The circuit diagram of the 12V switch regulated power supply is as follows, and the outline package and pins are also marked above.

Among them, the voltage regulator uses LM2576ADJ, which is a very commonly used switching regulator with adjustable output voltage, which is easy to buy, and its output voltage is mainly determined by the ratio of R1 and R2, and the calculation formula is also written in the figure, where VOUT is the output voltage, VREF is the internal reference voltage of the voltage regulator (equal, substitute VOUT= into the formula, get R2=, in order to ensure the accuracy of the output voltage, it is recommended that R2 choose an adjustable resistor, leaving appropriate room for adjustment to the circuit. The capacitance of the electrolytic capacitor has been marked in the figure, and the withstand voltage value of the electrolytic capacitor can be selected as 25V when the input voltage is 12V.

Finally, it is better to drive the LED with a constant current circuit than with a constant voltage circuit, but you don't give the rated current of the diode used here, so you have to provide you with a constant voltage circuit.

Hope it helps.

You can go and buy a DC-DC converter that converts 12V DC to 3V DC. You can also use an LM317 three-terminal regulator to make a simple circuit and adjust the potentiometer to make the output voltage of LM317 3V, so that the 12V voltage can be stably reduced to 3V voltage.

The two diodes in the diagram are designed to protect the three-terminal regulator and can also be left unused.

It's very simple. The AC 12V output of the 12V AC transformer needs to be rectified before it can be changed to DC to charge the battery. Between the rectifier output and the "ground", it is enough to connect a hundreds or even thousands of micromethods electrolysis.

1) When the left output is high, the right side gets the voltage of about 4V, and when the left output is low, the right side gets the left and right voltage, so the logic level should be able to meet the input requirements of the logic circuit of the right 5V power supply, so as to realize the level conversion;

2) For the chip output of the left circuit, if you can, set it to the open-drain output, then you can remove the diode, and only need a pull-up resistor to get the output (input) voltage with a high level of 5V;

3) In fact, there are many single-chip microcomputer chips that can be set to their output as push-pull output mode, open-drain output mode, etc., to see if the chip you use has this function, so that the circuit connection, the realization of level conversion is much simpler;

Irrationality. When the GPIO outputs, the 1N4148 will turn on, so the output cannot reach 5V.

There is no load capacity, the IO is high, the output is to pull up 5V, the IO is low, and the diode is turned on to pull the 5V level back to about the left.