How does PWM boost output, about the question of PWM based boost circuit

No matter how modulated, simply outputting the PWM pulses through a continuous passive network cannot be boosted.

PWM is just a square wave pulse with adjustable width, which can be boosted or bucked, depending on how you design the circuit.

The basic principle is: the voltage at both ends of the inductor is equal to the inductance multiplied by the current change rate, when the current change rate is very large, for example, a certain current value drops to zero instantly, and the circuit is disconnected by the switch that turns on the inductor to achieve it, and the induced potential generated at both ends of the inductor is quite high. If this induced potential is released to the load, it is clear that the load will be subjected to a much higher voltage than the original power supply.

From this, we can also know that as long as it is a square wave pulse, it can boost the voltage, which is not the same concept as PWM.

When the inductor is energized, the current cannot be abruptly changed, and the on-time of the PWM pulse determines the height of the current in the inductor, the higher the height of the inductor, the higher the current change rate and the greater the induced potential of the same change time. This is why PWM pulses regulate and stabilize voltage.

From the above, it can be seen that there are 5 basic components in the circuit that uses PWM pulse boost: 1. Low-voltage DC power supply. 2. Inductance.

3. Controlled electronic switch (triode of certain power, base input PWM control pulse). 4. One-way backstop element (rectifier diode). 5. Filter element (electrolytic capacitor).

The switching power supply of many electrical appliances also uses a switching transformer in order to generate multiple sets of DC voltages, and most of them also have an isolation function.

By loading the output PWM into the base of the triode, the power supply current is connected by the inductor and forming a loop by the triode, and a higher voltage can be obtained.

The higher the frequency, the smaller the inductance can be, but the switching frequency increases, and the switching loss is also large, so a compromise is required.

This is a similar circuit, the IGBT for boosting is better, the PWM waveform of the single-chip microcomputer generally cannot directly drive the MOS tube, and the driving current is insufficient, and an amplification circuit and isolation circuit are generally added.

"Can the boost drive circuit be replaced by a photocoupler? ”

Do you want to say, "Is it possible to couple the drive signal with an optocoupler"? PC817 used for general switching power supply may not be suitable, and the response speed is not enough.

PWM frequency, 5kHz - 50kHz can be, inductance parameters, 50 uh - 300uh, high frequency The inductance is smaller, the frequency is low, the inductance is larger, the greater the power The enameled wire of the inductor coil The wire diameter should be thicker, and the capacitance, about 1000 uf, the higher the frequency can be smaller.

You can refer to the circuit of the boost circuit chip 34063.

Find a dedicated IC....More convenient... Here's what we do. Hehe.

I don't understand it, the role of the microcontroller in this circuit is mainly to provide PWM. In the line of the boost circuit chip 34063, a single-chip microcomputer is rarely used to provide PWM, and the external capacitor can run.

Use an OC gate output driver chip and connect a pull-up resistor to the power supply you want at the output terminal.

There are no more than two methods, closed-loop control: sampling output voltage one" is fed back to the single-chip microcomputer.

AD 1 adjusts the PWM output lock voltage. Open-loop control: disassemble and stabilize the pressure before load, such as adding a voltage stabilizing tube.

Then adjust the PWM output of the Luming finger to slightly higher than the start control of the stabilizer tube.

It should be that the general optocoupler drive current is very small 50mA maxIf your driving frequency is very high, because the S-D electrode capacitance is relatively large and the current is small, it will not be able to reach the driving voltage if it is not fully charged at once. Therefore, it is generally necessary to use a drive source with a relatively large current, such as a transistor and MOS drive device.

You don't need to push the 540 directly with the op amp single-chip microcomputer, and compare it after adding AD sampling, if the voltage is large, reduce the temporary-to-empty ratio.

Go directly to the third-order and fourth-order RC filter circuit, and it seems that you already have M10 software. If you want to output a fast change in voltage, then increase the PWM frequency and reduce the time constant T of RC, but the speed and stability are mutually restrictive, so if there is a high requirement for the change speed, it is not recommended to use the RC filter circuit.

The resistance is 10k, the capacitor is 104, if you calculate it, calculate the specific size of the resistor capacitor, if you don't know how to calculate, use m10 to try it yourself.

Also, remember to add a transmitter at the end, this signal is too weak.

The output v1 is followed by an op amp, and then a resistor is connected to the positive side of the capacitor. The negative pole of the capacitor is to the ground. The positive electrode of the capacitor is a stable DC voltage v.

Connect V to pin 3 of U1 again.

Without a single-chip microcomputer with a 34063 chip, it can be made into a step-up circuit, or a step-down circuit.