Positive feedback problem of Lm339 differential comparator

shaorc

Positive feedback problem of Lm339 differential comparator [Copy link]

 

[If you don't understand, ask] As the title says, in order to prevent jitter interference at the signal input near the threshold voltage, a positive feedback is added to form a hysteresis comparator to suppress oscillation. This positive feedback is R22 in the figure. [1] Why does this eliminate jitter interference at the input near the threshold voltage? [2] How is the value of this true feedback resistor calculated?

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For a single-limit comparator with positive feedback of the hysteresis comparator, two threshold voltages, i.e. the threshold width of the hysteresis comparator, must be determined. The calculation of the resistor is generally explained in the Schmitt trigger circuit.

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There is a discussion thread on this topic in the forum https://en.eeworld.com/bbs/forum ... 2%C6%F7%B5%E7%C2%B7 In addition, search for "hysteresis comparator" on Baidu Encyclopedia or find a textbook on op amps to find content on this topic

maychang

【1】Why does this eliminate the jitter interference of the input near the threshold voltage? After adding positive feedback, the comparator cannot flip at the original flip level (threshold voltage) after flipping, but flips at another level (threshold voltage). When the interference signal amplitude is less than the difference between the two flip voltages, the comparator cannot flip. This eliminates the interference.

maychang

【2】How do you calculate the true feedback resistor? Calculate based on the two threshold voltages you need, R68, and the comparator output amplitude.

shaorc

maychang posted on 2018-10-9 17:32 [2] How do you calculate the true feedback resistor? Calculate based on the two threshold voltages you need, R68, and the comparator output amplitude.

In the figure below, I want to calculate the output voltage This is the calculation process. It turns out that Vout is more than 1000 volts. Did I calculate it wrong? Or is it the output high level at this time, that is, the power supply voltage of 16V? Because the positive feedback gain is very large

shaorc

maychang posted on 2018-10-9 17:31 [1] Why does this eliminate the jitter interference of the input near the threshold voltage? After adding positive feedback, the comparator cannot flip back to the original flip voltage after flipping...

This is a document I found. I understand the concept and the hysteresis principle, but it does not mention the threshold voltage or the threshold voltage algorithm. I want to know the threshold algorithm. I remember that when I looked at the hysteresis comparator, the threshold voltage was obtained by output voltage division. I don’t know how it is calculated here

maychang

shaorc posted on 2018-10-9 20:33 In the figure below, I want to calculate the output voltage. This is the calculation process. It turns out that Vout is more than 1000 volts. Did I calculate it wrong? Also...

The comparator does not work in a linear state, and there is no virtual short. The high level of this circuit output is roughly 1.1V, which is the forward voltage drop of the optocoupler light emitting diode.

shaorc

maychang published on 2018-10-9 21:09 The comparator does not work in a linear state, and there is no virtual short or virtual break. The high level output of this circuit is roughly 1.1V, that is, the forward direction of the optocoupler light emitting diode...

So? How does the comparator determine whether it is working in a linear state? In addition, the output in the figure has a pull-up resistor R69 to 16v. In this case, the output high level is also 1.1V? Without virtual short or virtual break, how to calculate the value of the positive feedback resistor R22?

maychang

shaorc posted on 2018-10-10 07:51 So? How does the comparator determine whether it is working in a linear state? Also, the output in the figure has a pull-up resistor R69 to 16v. In this case, the output...

"How does the comparator determine whether it is working in a linear state?" The comparator never works in a linear state. If it works in a linear state, it is not a comparator, but an amplifier.

maychang

shaorc posted on 2018-10-10 07:51 So? How does the comparator determine whether it is working in a linear state? In addition, the output in the figure has a pull-up resistor R69 to 16v. In this case, it also outputs a high level of 1.1V?

"The output in the figure has a pull-up resistor R69 to 16v. In this case, it also outputs a high level of 1.1V?" Yes. LM339 is an open collector output, that is, pin 13 is the collector of an internal transistor, and the collector of the transistor is not connected to any circuit in the chip except pin 13. 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.

maychang

shaorc posted on 2018-10-10 07:51 So? How does the comparator determine whether it is working in a linear state? In addition, the output in the figure has a pull-up resistor R69 to 16V. In this case, the output...

"How to calculate the value of the positive feedback resistor R22 without virtual short and virtual open?" Calculate from the output voltage and the threshold voltage you require and R68.

shaorc

maychang published on 2018-10-10 09:28 "How can we calculate the value of the positive feedback resistor R22 without using virtual short and virtual open?" Calculate from the output voltage and the threshold voltage you require and R68.

It is like this. First, shouldn't the threshold voltage be set as shown in the figure below, by connecting a bidirectional voltage regulator to the output, and the voltage regulation value is given to the input end through voltage division as the threshold voltage, but the output in the figure in my first post does not have this voltage regulator. In this case, how to determine the threshold value? Second, according to the figure in my first post, the output voltage of my comparator is either 0 or 1.2v? Third, I saw you said it twice, and you have to cooperate with R68, but you can't use virtual short and virtual open. I am really confused and can't calculate it.

maychang

shaorc posted on 2018-10-10 11:51 That's right, first, shouldn't the threshold voltage be set like the following figure, by connecting a bidirectional voltage regulator to the output, and the voltage regulation value is given to the input end through voltage division...

"First, shouldn't the threshold voltage be set like the following figure, by connecting a bidirectional voltage regulator to the output, and the voltage regulation value is given to the input end through voltage division as the threshold voltage, but the output in the figure of my first post does not have this voltage regulator, so how to determine the threshold value?" Such a voltage regulator is used in the circuit, of course, it is calculated according to the voltage regulation value of the voltage regulator. Now there is an optocoupler in the circuit, of course, it is calculated according to the forward voltage drop of the optocoupler light-emitting tube. As for the bidirectional use of the Zener tube in the figure on the 13th floor, that is because the figure on the 13th floor uses dual power supplies, and the comparator is a push-pull (totem pole) output, which can output in both directions, while the LM339 in the first post uses a single power supply and has an open collector output.

maychang

shaorc posted on 2018-10-10 11:51 That's right, first, shouldn't the threshold voltage be set like the figure below, by connecting a bidirectional voltage regulator to the output, and the voltage regulation value is given to the input end through voltage division...

"Second, according to the figure in my first post, the output voltage of my comparator is either 0 or 1.2v?" Almost. The output cannot be zero, but the LM339 can be as low as 0.1V when the output is low.

maychang

shaorc posted on 2018-10-10 11:51 That's right, first, shouldn't the threshold voltage be set as shown in the figure below, by connecting a bidirectional voltage regulator to the output, and the voltage regulation value is given to the input end through voltage division...

"Third, I saw you said it twice, and you have to cooperate with R68, but you can't use virtual short and virtual break, I'm really confused, I can't calculate it" To calculate the threshold voltage, you need to calculate the input voltage at the time of flipping. Virtual break and virtual short are established at the moment of flipping, because the comparator is in a linear working state during the flipping process. Calculate the lower threshold voltage (the lower of the two threshold voltages) in Figure 6, starting from the output high level. First determine the output voltage 1.2V, assuming the input voltage is V1, then ref is 5V, 1.2V and 5V are divided by R68 and R22, and the comparator's in-phase input terminal is 4.990V. When the comparator's inverting input terminal voltage rises to 4.990V, the comparator flips and the output decreases. When calculating the upper threshold voltage, just set the output low level to 0.1V, and the rest of the calculation is the same as above.

shaorc

maychang posted on 2018-10-10 12:24 "Third, I saw you mentioned it twice, and you have to cooperate with R68, but you can't use virtual short and virtual break. I'm really confused and I don't know how to calculate it." Calculation...

Wow, I understand a lot of things at once. Regarding your reply, I have a few more questions [1] The figure below is the figure in the first post. According to the method you said, calculate the lower threshold voltage value. Assume that the input voltage is V1, then ref is 5V, 1.2V and 5V are divided by R68 and R22, and the comparator non-inverting input terminal is 4.990V Here Vout is 1.2V, so the calculation process is (5-1.2)*R22/(R22+R68), which is not 4.990V. [2] It is obvious that LM339 is also powered by a single power supply, but how can we tell that the op amp on the 13th floor is powered by a dual power supply? Is it because it uses a bidirectional voltage regulator? [3] LM339 has an open collector output, and the output on the 13th floor is a push-pull output. Is this determined by the output peripheral circuit? [4] A question this morning, it is a comparator in the nonlinear region and an amplifier in the linear region. The op amps have the same appearance in the figure. How to determine the working state?

maychang

shaorc posted on 2018-10-10 14:09 Wow, I understand a lot of things at once. Regarding your reply, I still have a few questions [1] The figure below is the figure in the first post. According to the method you mentioned, the lower threshold voltage value is calculated...

[1] The figure below is the figure in the first post. According to the method you mentioned, the lower threshold voltage value is calculated... This formula calculates the voltage across R22, plus 1.2V to get the voltage between the common-mode input terminal and ground.

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...

[2] It is obvious that the LM339 here is also powered by a single power supply, but how can we tell that the op amp on the 13th floor is powered by a dual power supply? Is it because it uses a bidirectional voltage regulator that we can infer it? The text on the 13th floor shows that the output has a negative value of -Uz, so it is judged that it uses a dual power supply.

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 picture below is the picture in the first post. According to the method you mentioned...

[3] LM339 has an open collector output, and the output of the 13th floor is a push-pull output. Is this determined by the output peripheral circuit? For LM339, look at the manual, and for the 13th floor picture, look at the output variation range (the output has negative values).