Schematic diagram - How does this circuit use two op amps to achieve a constant current output of 1mA?

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Schematic diagram - How does this circuit use two op amps to achieve a constant current output of 1mA? [Copy link]

 

As shown in the figure above, the red box shows the combination of two op amps outputting a 1mA constant current to the RTD (PT100) below.

My questions:

I don't quite understand the principle of how this two op amp combination achieves constant current.

My analysis:

I tried to analyze it myself, but I felt that I didn't get to the point.

I can only roughly explain why it is 1mA.

But it still doesn't explain why the constant current?

The following figure shows my own analysis process:

So I would like to ask about the principle of constant current in this circuit. Please give me some advice, thank you!

maychang

If 2.5kΩ and 100Ω are connected in series, the current is close to 2.5kΩ, so the current is 1mA. Then why do we need two op amps? Wouldn't it be easier to connect the upper end of R RREF directly to 2.5V V REF ?

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maychang posted on 2022-9-26 20:09 If 2.5 kilo ohms and 100 ohms are connected in series, it is close to 2.5 kilo ohms, so the current is 1mA, then why do we need two op amps? Connect the upper end of RRREF directly to the 2.5V V ...

Yes, teacher, I think so too, so this analysis is not valid. We can only roughly calculate that the current is 1mA. Follower A2 should play a positive feedback role. When the resistance of PT100 increases, the voltage fed back to the positive phase end of A1 increases, making the output voltage also increase. Since I=U/R, the voltage and resistance increase, so the current is constant. But we can only make a qualitative analysis. The specific design process is not very clear.

maychang

Plakatu published on 2022-9-26 20:44 Yes, teacher, I think so too, so this analysis is not valid. We can only roughly calculate that the current is 1mA. Follower A2 should be the starter...

See the picture below.

For the convenience of description, the non-inverting input terminal of op amp A2 is marked as A, and the output terminal of op amp A1 is marked as B. Because op amp A1 is a follower, its output terminal potential is the same as the non-inverting input terminal, also marked as A. The voltage of point A to ground is VA , and the voltage of point B to ground is VB .

maychang

Placatto published on 2022-9-26 20:44 Yes, teacher, I think so too, so this analysis is not valid. We can only roughly calculate that the current is 1mA. Follower A2 should be...

The two resistors R1 and R2 at the inverting input of op amp A1 are both 25 kilo-ohms, and the two resistors R3 and R4 at the non-inverting input are also 25 kilo-ohms (not necessarily 25 kilo-ohms, but the four resistors must be equal). This forms a bridge circuit, with R1 to R4 being the four arms of the bridge. For such an amplifier, the potentials of the non-inverting input and the inverting input must be equal (the two inputs are "virtually shorted" when the op amp works linearly). Of the four resistors, the right end of R1 is grounded and the voltage is zero, the left end of R2 is VB , the right end of R4 is connected to a 2.5V reference voltage, and the left end of R3 is VA . Based on the above conditions, the equation of the bridge circuit is listed, and it can be seen that VB must be 2.5V higher than VA . RRREF is 2.5 kilo-ohms, so the current flowing through RRREF is 1mA.

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maychang posted on 2022-9-26 21:05 The two resistors R1 and R2 at the inverting input of op amp A1 are both 25 kilo-ohms, and the two resistors R3 and R4 at the non-inverting input are also 25 kilo-ohms (not necessarily 25 kilo-ohms, but required...

Does the bridge circuit refer to the differential amplifier circuit? The information I found has a circuit that is similar to the circuit in the question, but the connection method of the resistor at the common-phase end is slightly different.

xutong

3-wire rtd is a bit complicated

54chenjq

This is the recommended solution for high-precision RTD measurement from Microchip, the voltage of VRref is 2.5V, the resistance is 2.5K, and the current is 1mA. The accuracy of the entire constant current source is greatly affected by the accuracy of the resistor.

A1- = A1+ ; (0)

A1O = 2A1- ; (1)

A2 - = A2+ = A2O = 2A1+ - 2.5v (2)

VRref = A1O - A2+ = (1) - (2) = 2.5V

In addition: Please note that the parameters of the MCP609 op amp, IB, VOS and slew rate are

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54chenjq Published on 2022-9-27 00:05 This is the recommended solution for high-precision RTD measurement of microchip. The voltage of VRref is 2.5V, the resistance is 2.5K, and the current is 1mA. The accuracy of the entire constant current source is affected by the resistance...

I understand! Thank you!

普拉卡图

Thanks to the big guys on the 5th and 8th floors for their ideas

Drawing on the ideas from the 5th and 8th floors, and my own analysis above,

I can also solve this problem using a linear equation.

In fact, you don't have to set the following point to 0V, just set it to X and eliminate it.

It can be analyzed like this:

maychang

Placatto posted on 2022-9-27 08:56 Thanks to the big guys on the 5th and 8th floors for their ideas. Drawing on the ideas on the 5th and 8th floors and my own analysis above, I can also solve the linear equation with one variable...

"In fact, you don't have to set the point you follow to 0V, just set it to X and eliminate it at the end."

It is not zero, but floating (relative to the ground).

The purpose of this circuit is to float this point, but always maintain 2.5V across R RREF .

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maychang posted on 2022-9-27 09:43 "In fact, you don't have to set the following point to 0V, just set it to X and eliminate it at the end." It is not zero, but floating (relative to the ground). ...

Yes, let’s disassemble it again. Is the design in the picture below the bridge circuit you mentioned ?

If the connection is like this, the voltage difference between AB will be equal to Vin .

That is the core essence of constructing this constant current source.

You can directly refer to and apply it in your future designs, right?

普拉卡图

Placatto posted on 2022-9-27 12:51 Yes, let's disassemble it again. Is the design in the figure below the bridge circuit you mentioned? If it is connected like this, the voltage difference between AB will be equal to...

It feels quite appropriate to understand this circuit as amplifying in the same phase ratio, hahaha!

maychang

Placatto posted on 2022-9-27 12:51 Yes, let's disassemble it again. Is the design in the figure below the bridge circuit you mentioned? If it is connected like this, the voltage difference between AB will be equal to...

"Let me disassemble it again. Is the design in the picture below the bridge circuit you mentioned ?"

Yes. The bridge circuit refers to these four resistors and the op amp.

maychang

Placatto published on 2022-9-27 13:01 It feels quite appropriate to understand this circuit as amplified in the same phase ratio, hahaha!

"It seems appropriate to understand this circuit as amplifying in phase ratio, hahaha!"

Of course you can understand it that way.

jojoee

I have been talking for a long time but haven't gotten to the point. This is a well-known Howland circuit. The cleverness of the Howland architecture lies in the fact that it not only has negative feedback, but also uses positive feedback to generate a high output impedance to approximate the internal resistance of an ideal current source.

maychang

jojoee posted on 2022-9-28 10:25 I haven't gotten to the point after talking for a long time. This is a well-known howland circuit. The cleverness of the HOWLAND architecture lies in that it not only has negative feedback, but also uses positive feedback to generate a...

The circuit in the first post is not a Howland circuit.

The picture below is the Howland circuit.

Look carefully, are the two circuits the same?

maychang

Figure 17 is reproduced from Sergio Franco's "Circuit Design Based on Operational Amplifiers and Analog Integrated Circuits" page 56.

jojoee

maychang published on 2022-9-28 13:39 The 17th floor diagram is copied from Sergio Franco's "Circuit Design Based on Operational Amplifiers and Analog Integrated Circuits" page 56.

It means that you have little experience and have not learned the essence. The core of this Howland architecture is both positive and negative feedback, using positive feedback to generate high output impedance.

jojoee

snoa474a(1).pdf (175.05 KB, downloads: 9)

2022-9-28 14:12 上传

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This is TI's AN-1515 A Comprehensive Study of the Howland Current Pump, which provides some practical examples of howland