What is the problem of selecting the current limiting resistor between the microcontroller and the o

Generally, the current flowing through the optocoupler light-emitting tube can be calculated by selecting the resistance of 3-10mA. The second problem is that if the number of optocouplers is small (no more than 5 and the parameters must be the same), it can work, but it is not recommended to connect this way, because it may be problematic to open at the same time.

There is a ratio between the input current and the output current. Felt by optocouplers.

Demand comes first. See how much current the load requires. Convert to the elementary level. You're going to have to provide at least that much current.

Insufficient drive capacity? The day after tomorrow can make up for it. For example, add a transistor and resistor.

The output voltage of the microcontroller is the voltage slew rate of the IO port. Basically, it is either up or down in the power supply voltage range. Specially designed IO port. For example, there is no high-level output capability for open drain and open set. If you want to use it, you need to cooperate with another circuit.

The input current of the optocoupler is very small Generally, a few mA can work The single-chip microcomputer can be driven directly, and the resistance is determined according to the voltage of the IO port and the current of the optocoupler.

In general, it depends on the load capacity of the single-chip microcomputer and the current bearing capacity of the optoelectronic device, the ordinary 51 series single-chip microcomputer can provide about 20mA current sink, and the optoelectronic device such as PC817, the voltage drop under normal operation is about, the working current of 20mA is a bit large, you know, the internal emission of the optoelectronic device is a light-emitting diode, with the characteristics of a diode, so when calculating, we generally calculate according to the current of 10mA, that is, 5V minus the optocoupler voltage drop and then divide by the current value, The resistance is 380ohm, and 510 is generally taken, but it generally depends on the actual use of the occasion, such as the speed of the controlled device, and the voltage level.

The correct thing to do is to look at the current transfer ratio of the optocoupler...

There is a ratio between the input current and the output current. Felt by optocouplers.

Demand comes first. See how much current the load requires. Convert to the elementary level. You're going to have to provide at least that much current.

Insufficient drive capacity? The day after tomorrow can make up for it. For example, add a transistor and resistor.

The output voltage of the microcontroller is the voltage slew rate of the IO port. Basically, it is either up or down in the power supply voltage range. Specially designed IO port. For example, there is no high-level output capability for open drain and open set. If you want to use it, you need to cooperate with another circuit.

When only the current-limiting resistor is used, the current-limiting resistor:

r=(；When current limiting + shunt, current limiting resistor:

r=(, shunt resistance:

r=。where: r - current limiting resistor;

VC - the auxiliary power supply voltage of the switching power supply, which is obtained by the rectification filter of the auxiliary winding of the switching transformer;

optocoupler diode forward voltage drop;

if——Forward current of optocoupler diode, check optocoupler data sheet;

IR - the current through the shunt resistor, the designer decides how much current to be distributed;

r=(, shunt resistance:

Because the performance parameters of optocouplers of different brands and models are different, even the optocouplers of the same brand and model are discrete, so the calculated resistance value should be appropriately debugged according to the current transmission ratio and other parameters of the optocoupler in the actual circuit.

If you can't just use a shunt resistor, if you only use a shunt resistor without a limiting resistor, the optocoupler diode will burn.