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Can someone help me with this differential amplifier circuit? The op amp is overheating [Copy link]

 
This post was last edited by Ablikim on 2019-8-20 11:33

R2 is a simulated load

R7, 8, 9, 10 are voltage divider resistors

The voltage across R1 is about 0.038V

After a while, the op amp heated up a lot, the output voltage became unstable, and then there was no voltage output (0V).

Please tell me what is the reason

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The circuit design is risky. You can use an addition circuit. The bias term is divided by 12V. One end of R1 is grounded and the upper end is added.  Details Published on 2019-8-21 23:33

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This circuit is a simulation circuit diagram

What does the actual circuit look like?

The upper left part is 54.5V?

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In the actual circuit, if R8 or R10 is open, R7 or R9 is short, and the values of the four resistors are wrong, the op amp may burn out because the input voltage exceeds the allowable common-mode range.

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maychang published on 2019-8-20 13:32 In the actual circuit, if R8 or R10 is open, R7 or R9 is short, and the values of the four resistors are wrong, the op amp may burn out because the input voltage exceeds...

Is it that serious?

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The input of the LM358 op amp is a PNP transistor. When the power is applied to the circuit, the reverse breakdown of the transistor base-emitter will produce what consequences. Moreover, the transistors in the integrated circuit are isolated from each other by PN junctions. If the voltage on both sides of the PN junction used for isolation becomes positive, the isolation effect  Details Published on 2019-8-20 15:06
 
 
 
 

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qwqwqw2088 posted on 2019-8-20 12:01 This circuit is a simulation circuit diagram. What does the actual circuit look like? The upper left part is 54.5V?

The upper left part is an electric car battery

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What does R2 simulate load mean? What does the whole circuit want to achieve? Can R8 and R10 be connected in this way to simulate the past? How to determine the values of these resistors?  Details Published on 2019-8-20 15:40
 
 
 
 

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Ablikim posted on 2019-8-20 14:19 Is it that serious? Hero

The input end of the LM358 op amp is a PNP transistor. It is easy to imagine what the result will be when the base-emitter reverse breakdown of the transistor is applied with power in the circuit.

What's more, the transistors in the integrated circuit are all isolated from each other by PN junctions. If the voltage on both sides of the PN junction used for isolation becomes positive, the isolation effect will disappear, and it is difficult to predict what the result will be.

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Ablikim posted on 2019-8-20 14:20 The upper left part is an electric car battery

What does R2 simulate load mean? What does the whole circuit want to achieve?

Can the simulation be done by connecting R8 and R10 like this? How to determine the values of these resistors?

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"What is the whole circuit trying to achieve?" I guess the purpose of this circuit is to divide the voltage between R7R8 and R9R10, measure the voltage across R1, and indirectly measure the current in R1.  Details Published on 2019-8-20 16:54
 
 
 
 

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qwqwqw2088 posted on 2019-8-20 15:40 What does R2 simulate load mean? What does the whole circuit want to achieve by connecting R8 and R10 in this way? How to determine the values of these resistors?

“What is the whole circuit trying to achieve?”

I guess the purpose of this circuit is to divide the voltage of R7R8 and R9R10, measure the voltage across R1, and indirectly measure the current in R1.

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maychang posted on 2019-8-20 16:54 "What is the whole circuit trying to achieve?" I guess the purpose of this circuit is to divide the voltage of R7R8 and R9R10, measure the voltage across R1, and indirectly measure...

Yes

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maychang posted on 2019-8-20 16:54 "What is the whole circuit trying to achieve?" I guess the purpose of this circuit is to divide the voltage of R7R8 and R9R10, measure the voltage across R1, and indirectly measure...

This is the current detection part of a charger circuit. My boss asked me to do it, but I didn't get the basics right. I used the differential op amp according to the formula. After the circuit was actually connected, it was normal for the first half an hour. After about half an hour, the value was unstable, and then the op amp output 0V. It was hot when I touched it. I don't know what the reason is? Please enlighten me, hero.

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The input voltage is too high, it would be strange if the op amp doesn't heat up.

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The OP can refer to this circuit, which is a low-end charging and discharging current sampling circuit

image.png (90.52 KB, downloads: 0)

image.png

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微信图片_20190821090621.png
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LuJianchang posted on 2019-8-21 09:08 The original poster can refer to this circuit, which is a low-end charge and discharge current sampling circuit

Got it, I'll take a look, thank you!!!

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Weilin Power Supply published on 2019-8-21 02:05 The input voltage is too high, it would be strange if the op amp does not heat up.

The voltage at both ends has been divided to about 5V

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The circuit design is risky. You can use an addition circuit. The bias term is divided by 12V. One end of R1 is grounded and the upper end is added.
This post is from Analog electronics
 
 
 
 

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