maychang

shaorc posted on 2018-10-10 14:09 Wow, I understand a lot of things at once. I have a few more questions about your reply. [1] The following figure is the figure in the first post. According to the method you mentioned, the calculation...

[4] A question from this morning. When working in the nonlinear region, it is a comparator, and in the linear region, it is an amplifier. The appearance of the op amp in the figure is the same. How to determine the working state? That depends on the relationship between the output and the input. The amplifier may also work in a nonlinear state, which is usually called saturation.

shaorc

maychang posted on 2018-10-10 14:22 【4】A question this morning, when working in the nonlinear region it is a comparator, and in the linear region it is an amplifier. The operational amplifiers in the diagram have the same appearance, how can we determine...

This is the only question left. Can we look at the difference this way? If both ends have variable input values, such as differential, it is an amplifier; if one end is a fixed value and the other end is a variable input, it is a comparator. And you said to look at the relationship between input and output, which I don't know.

maychang

shaorc posted on 2018-10-10 20:33 Only one question is left. Can we look at the difference this way? If both ends have variable input values, such as differential, it is an amplifier; and if one end is a fixed value, the other...

You can't look at it this way. Amplifiers often have two input terminals, one input terminal is fixed, and the other input terminal is variable. Such amplifiers are often called single-ended amplifiers. Both non-inverting amplifiers and inverting amplifiers belong to this category.

maychang

shaorc posted on 2018-10-10 20:33 Only one question is left. Can we look at the difference this way? If both ends have changing input values, such as differential, it is an amplifier; and if one end is a fixed value, the other...

The so-called "looking at the input-output relationship" is to see whether the output changes with the input. When the input changes gradually, the output also changes gradually, which is an amplifier. When the input changes gradually but the output does not change much (the output has only two specific values), it is a comparator.

shaorc

maychang posted on 2018-10-10 20:49 The so-called "looking at the input-output relationship" is to see whether the output changes with the input. When the input changes gradually, the output also changes gradually, which is an amplifier. The input gradually changes...

Thank you for your answer all day, I have written it down and understood it

shaorc

maychang published on 2018-10-10 20:49 The so-called "looking at the input-output relationship" is to see whether the output changes with the input. When the input changes gradually, the output also changes gradually, which is an amplifier. The input gradually...

In this figure, does the pull-up resistor R69 provide a high level function? The following is what you said before that the LM339 is an open collector output, and the 13th pin is the collector of an internal transistor. The collector of the transistor is not connected to any circuit in the chip except the 13th pin. When the internal transistor of this chip is turned off, the optocoupler relies on R69 to supply current, and the current size is about (16V-1.1V)/2000Ω. The voltage drop of the light-emitting diode in the optocoupler at this current is about 1.1V (infrared light-emitting tube). When the internal transistor is turned on, the voltage drop of the transistor is much less than 1.1V, so there is almost no current in the optocoupler. If the internal transistor is turned off, it is calculated according to your statement (16V-1.1V)/2000Ω. At this time, the photosensitive diode is turned on, and the comparator outputs a high level. Therefore, an external pull-up resistor is necessary and is an important guarantee for providing a high level? Secondly, the power supply connected to the pull-up resistor is selected as 16V here, in order to be exactly equal to the power supply of the op amp, for the purpose of convenience , in fact, other values can also be selected, such as 18V, 20V, etc. So, the value of the pull-up resistor is to ensure that the current calculated by the above formula is less than the rated current of the photosensitive diode?

maychang

shaorc posted on 2018-10-11 10:38 In this figure, does the pull-up resistor R69 provide a high level? The following is what you said before: LM339 is an open collector output, and pin 13 is an internal...

The pull-up resistor R69 is required, otherwise the comparator cannot output a high level. The pull-up resistor can be connected to other power supplies, such as 18V, 20V, etc., but it cannot exceed the maximum voltage specified in the LM339 manual (the manual specifies a maximum of 36V). The same as the chip power supply voltage is for convenience.

maychang

shaorc posted on 2018-10-11 10:38 In this figure, does the pull-up resistor R69 provide a high level function? The following is what you said before. The LM339 has an open collector output, and pin 13 is an internal...

"So, the value of the pull-up resistor is to ensure that the current calculated by the above formula is less than the rated current of the photodiode?" The pull-up resistor is combined with the power supply voltage. Of course, it is necessary to ensure that the current in the optocoupler is less than the maximum allowable value, but at the same time, it is also necessary to ensure that the current in the optocoupler is greater than the minimum value required for conduction. If R69 is selected as 1 megohm, the current in the optocoupler is too small and the optocoupler cannot work properly.

shaorc

maychang posted on 2018-10-11 11:03 "So, the value of the pull-up resistor is to ensure that the current calculated by the above formula is less than the rated current of the photodiode?" Pull-up resistor combined with power supply...

OK, thank you for your supplement. I calculated the high and low threshold voltages of the comparator in the first post. As shown in the figure, the threshold voltage difference is too small. The input fluctuation is difficult to be less than the threshold voltage difference. How can it play the role of hysteresis comparator? Did I calculate it wrong?

maychang

shaorc posted on 2018-10-11 16:04 OK, thank you for your supplement. I calculated the high and low threshold voltages of the comparator in the first post. As shown in the figure, it seems that the threshold voltage difference is too small. The input...

I did not calculate it wrong, it is indeed small. The reason is (1) The output end changes little, the maximum is 1.2V, the minimum is 0.1V. (2) Compared with R22 and R68, R22 is too large and R68 is too small. If the circuit does not change, you can consider reducing R22.

shaorc

maychang posted on 2018-10-11 18:03 I did not make a mistake in my calculation. It is indeed quite small. The reason is that (1) the output terminal has a small change range, with a maximum of 1.2V and a minimum of 0.1V. (2) R22 is too large compared to R68.

Thank you

shaorc

maychang published on 2018-10-10 09:26 "In the figure, the output has a pull-up resistor R69 to 16V. In this case, the output is also a high level of 1.1V?" Yes. LM339 is an open collector output, ...

Here I have to dig deeper into the internal 13-pin transistor of the LM339 that you mentioned. As shown in the figure, the one connected to Vo on the right is what you said. If the transistor is turned off, the output is high level, that is, the optical coupler behind it will be turned on. But in the figure in my first post, the same direction end is the reference value 5V, and the reverse end is the signal input end. Is it true that the input of the reverse signal end is less than 5V at this time, which will cause the transistor to turn off and output a high level?

maychang

shaorc posted on 2018-10-12 08:51 Here I have to dig a little deeper. I have seen the internal 13-pin transistor of the LM339 you mentioned. As shown in the figure, the one connected to Vo on the right is what you said if...

"In the first post, the same-phase end is the reference value of 5V, and the reverse end is the signal input end. Is it true that the input of the reverse signal end is less than 5V at this time, which will turn off the transistor and output a high level?" Yes. It can be seen from the markings of the same-phase input end and the reverse-phase input end, and it can also be deduced step by step from the structure diagram.

shaorc

maychang published on 2018-10-12 10:19 “In the first picture, the same direction end is the reference value of 5V, and the reverse end is the signal input end. Is it true that the input of the reverse signal end is less than 5V at this time, which will make the transistor...

Oh, it means that when the reverse end is less than the same direction end, after the internal circuit operation, the gate signal given to the transistor is a low level, but the transistor is not connected at this time, so that the output vo is a high level?

maychang

shaorc posted on 2018-10-12 10:52 Oh, it means that when the reverse end is less than the same direction end, after the internal circuit calculation, the gate signal given to the transistor is a low level, but the transistor is not connected at this time...

"The gate signal given to the transistor is a low level" The LM393 output tube is a bipolar tube. The bipolar tube is called the base, not the gate.

maychang

shaorc posted on 2018-10-12 10:52 Oh, it means that when the reverse end is less than the same end, after the internal circuit operation, the gate signal to the transistor is low, but the transistor is not connected at this time...

"But the transistor is not connected at this time, so that the output vo is high?" Yes. The collector is open-circuited, the output tube is turned off, and the pull-up resistor causes a high level.

shaorc

maychang published on 2018-10-12 11:09 "But at this time the transistor is not connected, so that the output vo is a high level?" Yes. The collector is open-circuited and the output tube is turned off, and the pull-up resistor causes a high level...

Good morning, I have a question to verify. According to the voltage of the star connection and the triangle connection, the motor parameters in this figure only have a phase voltage of 160v, so can it not be used for 380VAC mains?

maychang

shaorc posted on 2018-10-22 08:57 Good morning, I have a question to verify the motor parameters in this figure. According to the voltage of star connection and triangle connection, only 160v phase voltage...

The motor is marked with a star connection rated voltage of 280V and a delta connection rated voltage of 160V. Obviously, it cannot directly use 380V AC power. There are Fn 103Hz and Fmax 603Hz on the nameplate, but I don’t know what it means.

maychang

Shaorc posted on 2018-10-22 08:57 Good morning, I have a question to verify. According to the voltage of the star connection and the triangle connection, the motor parameters in this figure only have 160v phase voltage...

The motor is also marked "Asynchronous servo motor", which is an asynchronous servo motor. Is it a variable frequency motor?

shaorc

maychang posted on 2018-10-22 09:57 The motor is also marked "Asynchronous servo motor", which means it is an asynchronous servo motor. Is it a variable frequency motor?

It is like this, the 380VAC three-phase power input is rectified into 540V voltage, and the 540V is supplied to the driver. The driver then controls and inverts the output UVW three-phase voltage to the motor in the picture. Since it is 380vac input, it should also be 380vac output, so the motor in the picture cannot be used for 380v AC power supply. Can I understand it this way?