Can this be done at the input end of the optocoupler?

sky999

Can this be done at the input end of the optocoupler? [Copy link]

 

Figure 1 is the circuit in use, but because the external input terminal is in high impedance state when there is no input, it is sometimes unstable. I want to connect a 50K resistor directly between pins 1 and 2 to make it like Figure 2 to increase stability. I wonder if this is possible. Thank you.

qwqwqw2088

What is the external input here?

It is the IO port driver of the MCU

Adding a resistor here doesn't make much sense, but it can still help

You still need to check why it is unstable, control program, IO port settings, etc.

Or add a freewheeling diode at the input end.

dcexpert

There is no problem with the high impedance state itself, but see if there are other factors affecting the input end.

daemondong

The most correct use of optocouplers is isolation. Logically, the 24V power supply on the board should not be used, and the positive and negative terminals should be input externally at the same time.

sky999

qwqwqw2088 posted on 2021-9-27 07:49 What is the external input here? It is the IO port driver of the MCU. Adding a resistor here is not very meaningful, but it works a little bit. You still need to check it...

The external input is usually an N-channel MOS tube. Would adding a pull-up resistor help stabilize it? The reason for instability may be that the external line is too long. Should a freewheeling diode be added here? ? I don't quite understand. What role does it play?

sky999

dcexpert posted on 2021-9-27 08:42 There is no problem with the high impedance state itself. Check whether there are other factors affecting the input end.

The line is too long, so a resistor is added to stabilize it. I am curious about how to calculate the power consumption of the 6.8K resistor above when it is turned on?

damiaa

If the IO port needs a pull-up resistor, the pull-up resistor directly pulls the IO port to the power supply.

qwqwqw2088

sky999 posted on 2021-9-27 09:10 External input is generally an N-channel MOS tube. Adding a pull-up resistor will have a stabilizing effect, right? The reason for instability may be that the external line is too long. Add a freewheeling diode here...

The screenshot on the first floor is small, and the optocoupler is cut halfway. I thought the external connection was controlled by MCU. I guessed wrong, no diode is needed.

Got it, the external input is an N-channel MOS switch

sky999

damiaa posted on 2021-9-27 09:13 If the IO port needs to be pulled up by a pull-up resistor, the pull-up resistor is directly pulled from the IO port to the power supply.

This is a finished circuit board. I just drew a simple schematic diagram. I can't pull the power supply. Because the distance between pin 1 and pin 2 is just right for a resistor, I plan to connect a resistor directly between pin 1 and pin 2.

sky999

qwqwqw2088 posted on 2021-9-27 09:55 The screenshot on the first floor is small, and the screenshot of an optocoupler is cut in half. I thought the external access was controlled by the MCU. I guessed wrong. I don’t need a diode. The external input...

This is a finished circuit board. I just simply drew the principle.

lcofjp

You can try this, no problem.

dcexpert

sky999 posted on 2021-9-27 09:11 The line is too long, add a resistor to stabilize it. I am curious about how to calculate the power consumption of the 6.8K resistor above when it is turned on?

Adding a resistor will not help much and will reduce the power of the LED. The power consumption can be calculated by the voltage drop across the LED and the power of R2.

sky999

dcexpert posted on 2021-9-27 11:17 Adding a resistor will not help much and will also reduce the power of the LED. The power consumption can be calculated by the voltage drop across the LED to calculate the power of R2.

What I mean is, when they are on, R2 and R3 are in series, part of the power of R2 itself is given to the LED, and the other part is consumed by R3. How do we calculate the total power consumption of R2?

baopodao

R3 is used to shunt the current of the interference signal, which will have a certain effect when the interference is small, so it will not affect normal use. If the input signal response speed is not required, capacitors can also be considered for filtering.

damiaa

sky999 posted on 2021-9-27 10:23 This is a formed circuit board. I just drew the schematic diagram. I can't pull the power supply because the distance between pin 1 and pin 2 is just enough to put a resistor...

This also has a pull-up effect. You can try it.

chunyang

The method of parallel resistors is meaningless. Diodes and LEDs are current-type devices. As long as the front stage is guaranteed to be in a high-impedance state in the non-driven state, the LED will never malfunction. The malfunction must be a design problem of the front stage driver. The original poster should provide the specific circuit of the front stage.

dcexpert

sky999 Published on 2021-9-27 11:42 What I mean is that conduction means that R2 and R3 are connected in series. Part of the power of R2 is given to the LED, and the other part is consumed by R3. Then R2...

Within a certain range, the voltage drop of the LED is basically constant, so the voltage across R2 can be calculated, and thus the power on R2 can be calculated.

sky999

baopodao posted on 2021-9-27 13:57 By using R3 to shunt the current of the interference signal, it will have a certain effect when the interference is small, which will not affect normal use. If the input signal...

OK, thanks

sky999

damiaa posted on 2021-9-27 15:13 This also has a pull-up effect. You can try it.

OK, thanks!

sky999

chunyang posted on 2021-9-27 16:20 The method of parallel resistors is meaningless. Diodes and LEDs are current-type devices. As long as the front stage is guaranteed to be in a high-impedance state in the non-driven state, the LED will never malfunction...

Thank you, the pre-stage input is not fixed, there may be many different circuits