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The role of capacitors in op amp circuits [Copy link]

 

I would like to ask you guys, what is the function of capacitor C1 (39pF) in the figure?

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Put it at B, not at A  Details Published on 2021-3-20 15:41

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Such a small capacitor is probably to ensure the stability of the op amp.

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Thanks, senior. Is there any particularity in the value of R2 (330 ohm in the figure)? It and C1 together form a compensation circuit? I searched for a long time in the spec but couldn't find relevant information. Thanks!  Details Published on 2021-3-18 18:47
Thanks, senior. Is there any particularity in the value of R2 (330 ohm in the figure)? It and C1 together form a compensation circuit? I searched for a long time in the spec but couldn't find relevant information. Thanks!  Details Published on 2021-3-18 15:24
 
 

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maychang posted on 2021-3-18 14:49 Such a small capacitor is probably to ensure the stability of the op amp.

Thanks, senior. Is there any particularity in the value of R2 (330 ohm in the figure)? It and C1 together form a compensation circuit? I searched for a long time in the spec but couldn't find relevant information. Thanks!

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Sorry! I don’t know that either.  Details Published on 2021-3-18 15:38
 
 
 
 

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xiaxingxing posted on 2021-3-18 15:24 Thank you, is there any particularity in the value of R2 (330 ohm in the figure)? It and C1 together form a compensation circuit? I have been looking for it in the spec for a long time...

Sorry! I don’t know that either.

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Thank you, senior. I have another question. As shown in the figure below, what are the advantages and disadvantages of placing the laser at A and B? The similar constant current sources I made before all placed the laser at B. Thank you!  Details Published on 2021-3-18 16:38
 
 
 
 

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maychang posted on 2021-3-18 15:38 Sorry! I don’t know this either.

Thank you, senior. I have another question. As shown in the figure below, what are the advantages and disadvantages of placing the laser at A and B? The similar constant current sources I made before all placed the laser at B. Thank you!

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To put it at A, it depends on the output capability of your op amp. Unless the op amp output is rail-to-rail, it is usually less than the power supply voltage. If you connect it at A, the voltage driving the triode is equal to the sum of the sampling resistor, the load voltage and the voltage driving the triode. If you put it at B, the dynamic range will be larger.  Details Published on 2021-3-18 18:06
To put it at A, it depends on the output capability of your op amp. Unless the op amp output is rail-to-rail, it is usually less than the power supply voltage. If you connect it at A, the voltage driving the triode is equal to the sum of the sampling resistor, the load voltage and the voltage driving the triode. If you put it at B, the dynamic range will be larger.  Details Published on 2021-3-18 17:20
 
 
 
 

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xiaxingxing posted on 2021-3-18 16:38 Thank you, senior. I have another question. As shown in the figure below, what are the advantages and disadvantages of placing the laser at A and B? The constant current sources I made before all put...

To put it at A, it depends on the output capability of your op amp. Unless the op amp output is rail-to-rail, it is usually less than the power supply voltage. If you connect it at A, the voltage driving the triode is equal to the sum of the sampling resistor, the load voltage and the voltage driving the triode. If you put it at B, the dynamic range will be larger.

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OK, got it, thank you sir!  Details Published on 2021-3-22 15:15
 
 
 
 

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xiaxingxing posted on 2021-3-18 16:38 Thank you, senior. I have another question. As shown in the figure below, what are the advantages and disadvantages of placing the laser at A and B? The constant current sources I made before all put...

The currents at point A and point B are very similar (the difference between the collector current and emitter current of the transistor is only the base current), and the constant current performance is also very similar. However, when it is placed at point B, as mentioned on the sixth floor, the requirements for the output swing amplitude of the op amp are less stringent.

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maychang posted on 2021-3-18 14:49 Such a small capacitor is probably to ensure the stability of the op amp.

The capacitor is connected in parallel to the inverting input and output terminals. Is the Miller effect used to compensate the phase?

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It is used to stabilize the op amp and reduce noise.

https://wenku.baidu.com/view/0d75ed9ca3c7aa00b52acfc789eb172ded63990f.html

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OK, thanks sir. I'll go and study.  Details Published on 2021-3-22 15:14
 
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R2 should be the feedback current, right? It should be for this feedback current

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玩板看这里:

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EEWorld测评频道众多好板等你来玩,还可以来频道许愿树许愿说说你想要玩的板子,我们都在努力为大家实现!

 
 
 

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Put it at B, not at A
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okhxyyo published on 2021-3-19 10:15 It is used to stabilize the op amp and reduce noise. You can read this article https://wenku.baidu.com/view/0d75ed9ca3c7aa00b52acfc789eb1 ...

OK, thanks sir. I'll go and study.

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Alas, published on 2021-3-18 17:20 To put it at A, it depends on the output capability of your op amp. Unless the op amp output is rail-to-rail, it is usually less than the power supply voltage. You connect it at A and drive the three-stage...

OK, got it, thank you sir!

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