How are the two filtering formulas on the datasheets of AD620 and AD623 derived?

hf12345

How are the two filtering formulas on the datasheets of AD620 and AD623 derived? [Copy link]

 

Question 1: How was the first formula on the AD620 graph derived? Question 2: How was the formula in the circle on the AD623 graph derived?

gmchen

Let's first talk about how the formula in the first figure is derived: Since the two input terminals of the differential amplifier circuit are completely symmetrical, the half-circuit method can be used to solve the differential signal parameters. Split CD into two capacitors in series - the midpoints of the two capacitors are grounded, and the capacity of the two capacitors should be 2CD. Then this 2CD is connected in parallel with CC to form a first-order low-pass network with R, and its -3dB frequency is that formula. Let's talk about the formula in the second figure. In fact, as long as we consider that R1=R2 and C3 is much larger than C1 and C2, the second formula is the first formula.

hf12345

gmchen posted on 2018-8-29 08:21 Let me first explain how the formula in the first figure is derived: Since the two input terminals of the differential amplifier circuit are completely symmetrical, the half-circuit method can be used to solve the differential mode signal...

According to your idea, I roughly drew the circuit diagram. I wonder if it is correct? If so, why is the capacitance doubled?

maychang

hf12345 posted on 2018-8-29 10:55 According to your idea, I roughly drew the circuit diagram. I wonder if it is correct? If so, why is the capacitance doubled?

No. In your diagram, 2CD and CC are connected in parallel on the left, and 2CD and CC are connected in series on the right. So the => sign in the diagram does not hold.

hf12345

maychang posted on 2018-8-29 11:00 That's wrong. In your diagram, 2CD and CC are connected in parallel on the left, and 2CD and CC are connected in series on the right. So the => sign in the diagram is not true.

Oh, it is indeed wrong. Why does the CD capacitance double? How to explain it?

maychang

hf12345 posted on 2018-8-29 11:07 Oh, it is indeed wrong. Why does the CD capacitance double? How to explain it?

Two 2CD connected in series, its capacity is exactly CD.

hf12345

maychang posted on 2018-8-29 11:52 Two 2CDs connected in series have a capacity of CD.

I understand! I have another question. Why do we need to divide CD into two capacitors with midpoints when analyzing? What is the basis?

maychang

hf12345 posted on 2018-8-29 12:34 I understand! I have another question. Why do we need to divide CD into two capacitors and connect them at the midpoint during analysis? What is the basis?

"I have another question. Why do we need to divide CD into two capacitors and connect them at the midpoint during analysis? What is the basis?" That is because of what gmchen said on the second floor: "Since the two input terminals of the differential amplifier circuit are completely symmetrical, the half-circuit method can be used to solve the differential signal parameters." The formula for calculating the differential signal is in the oval box in the first post. For the differential signal, it is symmetrical to the ground. So if we divide CD into two, each of which is 2CD, the connection point between the two capacitors is the AC ground potential, so this point is still equivalent to the original circuit after being grounded. Then calculate according to the calculation method of the single-ended circuit.

maychang

hf12345 posted on 2018-8-29 12:34 I understand! I have another question. Why do we need to divide CD into two capacitors and connect them at the midpoint during analysis? What is the basis?

If you don't want to divide it into two halves, you can still calculate it according to the single-ended circuit. The first post AD620 circuit (marked Figure 42) has two input terminals on the left that are input signals. The calculation of FilterFreq is to calculate at which frequency the voltage amplitude between the two input terminals of the instrumentation amplifier drops to 0.707 times the voltage amplitude between the two input terminals. Then from the perspective of the two input terminals, there are two resistors R and capacitor CD connected in series between the two input terminals, and two capacitors CC are connected in parallel at both ends of capacitor CD, and the relationship between the two CC is in series (the two CC are connected together through the "ground"). Obviously, the capacitance of capacitor CD after connecting two CC in parallel is (CD+1/2CC), and the total resistance is 2R. The result is the same as the formula in the ellipse, except that the constant factor 2 is moved outside the denominator brackets, becoming 2R(CD+1/2CC).

hf12345

maychang posted on 2018-8-29 13:23 If you don't want to divide it into two halves, you can calculate it according to the single-ended circuit. First post AD620 circuit (marked Figure 42) The two inputs on the left...

Your answer is awesome!

hf12345

gmchen posted on 2018-8-29 08:21 Let’s first talk about how the formula in the first figure is derived: Since the two input terminals of the differential amplifier circuit are completely symmetrical, the half-circuit method can be used to solve the differential mode signal...

Let’s talk about the formula in the second figure. In fact, as long as we consider that R1=R2 and C3 is much larger than C1 and C2, the second formula is the first formula. ```` Master, I finally understand it, thank you very much! ! !

hf12345

maychang posted on 2018-8-29 13:23 If you don't want to divide it into two halves, you can calculate it according to the single-ended circuit, which is also feasible. First post AD620 circuit (marked Figure 42) The two inputs on the left...

I have another question for you. Under what circumstances do you need to add the above filtering circuit to reduce radio frequency interference? I want to make a pressure monitoring system now, and I am going to refer to the 5V single-power supply pressure monitoring circuit recommended in the official datasheet of AD620, as shown in the figure above. However, from the circuit point of view, the output of the sensitive bridge is directly connected to the two differential inputs of the op amp, and the front end does not use the above RF interference suppression circuit to filter first. So in general, under what circumstances is this RF interference circuit needed? Can you give an example? Thank you

hf12345

maychang posted on 2018-8-29 13:23 If you don't want to divide it into two halves, you can calculate it according to the single-ended circuit, which is also feasible. The first post AD620 circuit (marked Figure 42) The two inputs on the left...

My personal understanding is that the radio frequency interference will be more serious in places with many wireless devices, so when designing a pressure monitoring system, you need to consider adding a differential low-pass filter to the front end to suppress the interference of radio frequency signals. However, if there is no more electromagnetic interference in the use environment, then when designing, you can use the circuit diagram recommended by the official manual of ad620, and there is no need to add a differential low-pass filter to the front end. I wonder if my understanding is correct?

gmchen

hf12345 posted on 2018-8-29 22:43 I would like to ask you another question. Under what circumstances do you need to add the above filtering circuit to reduce radio frequency interference? I want to make a pressure monitoring system now...

Most pressure detection systems are used in industrial sites. According to my experience, the interference in these places mainly comes from the power grid, especially the start-up of the motor, equipment with thyristors, and a large number of electronic ballasts and other switching loads, which will form serious interference signals in the power grid. Therefore, this kind of equipment must be protected in two aspects: The first is to exclude interference from the equipment, that is, there must be a more complete power filter at the place where the power is introduced. I worked on several factory technical improvement projects several years ago. At that time, most of the power supply filters on the market had poor anti-interference capabilities. It took a long time to find a relatively ideal power supply filter. The second is to connect the filter circuit to the circuit. Since most of the pressure sensor signals change slowly, the cutoff frequency of the filter can be made relatively low (below 50Hz), which can eliminate most of the interference.

maychang

hf12345 posted on 2018-8-29 22:48 My personal understanding is that the RF interference will be more serious in places with many wireless devices, so when designing a pressure monitoring system, you need to consider adding...

Interference is everywhere, but it is strong in some places and weak in some places. It is not necessarily the case that the RF interference is strong in places with many wireless devices. In the substation of the power industry, there are few wireless devices, but the RF interference is quite strong. This is because the voltage of various devices in the substation is very high, and the closing and opening operations are frequent, which generates strong RF interference. The interference in the electric furnace steelmaking plant is also very strong. Whether your pressure monitoring system needs to add differential low-pass filtering at the front end depends not only on the electromagnetic environment, but also on the distance from the sensor (strain bridge) to the instrument amplifier. If the sensor and amplifier are on the same circuit board, it may not be necessary to add a low-pass filter between the sensor and the amplifier. If the distance is a little far, especially if the sensor and amplifier are placed in different housings and the signal is transmitted by twisted pair, low-pass filtering is necessary.

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