gmchen

newtonxihuan posted on 2024-7-18 13:44 Do you want to simulate or actually test? There is no baud meter for actual testing, so you can only use an oscilloscope to test

It is best to have actual test data, not necessarily a Bode plot, but data on a few key points, including the output waveform of the oscilloscope.

newtonxihuan

gmchen posted on 2024-7-19 09:00 It is best to have actual test data, not necessarily a Bode plot, data of a few key points will do, including the output waveform of the oscilloscope

The input signal effective value is 1V, the output signal effective value is 760mv, the frequency is 2khz-300khz, very stable, and no distortion, the input is a sine wave

gmchen

newtonxihuan posted on 2024-7-19 14:12 The input signal effective value is 1V, the output signal effective value is 760mv, the frequency is 2khz-300khz, very stable, and no distortion, the input is a sine wave

Most likely there is something wrong with your circuit (including the manufacturing process) or the measurement process (including the measuring instrument).

I just did an experiment according to this circuit, and the test results are completely normal.

newtonxihuan

gmchen posted on 2024-7-19 16:47 Most likely there is something wrong with your circuit (including the manufacturing process). Or there is something wrong with the measurement process (including the measuring instrument). I just followed this circuit...

The passband gain of the low pass is 1, but that of the high pass is not. I made two of the same PCB board, and both have this problem.

newtonxihuan

newtonxihuan

Teacher, this is my homemade instrument amplifier, but the differential input is the ideal result. I set the amplification factor to 3, but the measured result is not (Vp-Vn)*3. What may be the reason? The 500 resistors may not be precise, the range of 499 and 501. The theoretical value of the amplification factor is 3, but it is actually 3.2-3.8. The smaller the differential signal, the larger the factor. When Vp-Vn=0, the no-load output reaches 20mv, which is not a fixed value.

gmchen

There is a problem: the common-phase input terminals of the two common-phase amplifiers in the previous stage should have a resistor connected to ground. You can refer to the connection method in the data sheet. Without this resistor, the bias current of the op amp has nowhere to be discharged.

In addition, when asking questions, you must describe all the conditions completely. Here are the conditions that need to be clarified: Are the input and output signals AC or DC? What are their amplitudes? If it is AC, what is the frequency? Also, what is the power supply voltage?

newtonxihuan

newtonxihuan

The input is AC, the power supply voltage is positive and negative 5V, the input is 1khz, the effective value is 30mv, the other is 1khz, the effective value is 10mv, according to the theoretical calculation, it is 20*3 (amplification factor) = 60mv, but it is about 80mv, mainly because of the amplitude problem, the waveform is not distorted

gmchen

newtonxihuan posted on 2024-7-28 15:16 The input is AC, the power supply voltage is positive and negative 5V power supply, the input is 1khz, the effective value is 30mv, the other is 1khz, the effective value is 10mv, according to theoretical calculation...

1. Is it a simulation or an actual circuit?

2. What is the resistance of the resistor RG?

newtonxihuan

gmchen posted on 2024-7-28 15:43 1. Is it a simulation or an actual circuit? 2. What is the resistance value of the resistor RG?

Actual circuit, rg adjustable resistor, resistance is 1k

To change the magnification

gmchen

Suspect:

1. Whether the gain is calculated incorrectly.

2. A previous post said that when the input is 0, the output is 20mV. Is this 20mV an AC signal or a DC signal? If it is DC, then it may be an offset output. If it is AC, is the amplifier self-excited?

gmchen

newtonxihuan published on 2024-7-28 15:45 Actual circuit, rg adjustable resistor, resistance is 1k used to change the amplification factor

Since it is a variable resistor, how do you determine the design gain = 3?

newtonxihuan

gmchen posted on 2024-7-28 16:49 Since it is a variable resistor, how do you determine the design gain = 3?

It is set to 3. This is the instrumentation amplifier, which can be set to 3.

newtonxihuan

gmchen posted on 2024-7-28 16:28 Suspicion: 1. Is the gain calculated incorrectly? 2. The previous post said that when the input is 0, the output is 20mV. Is this 20mV an AC signal or a DC signal...

The offset voltage is not that large, and the amplitude of the self-oscillation is not that small.

newtonxihuan

newtonxihuan posted on 2024-7-28 17:20 The offset voltage is not that large, and the amplitude of the self-oscillation is not that small

The gain formula is simple. If the input is slightly larger, the gain is close to 3.

gmchen

newtonxihuan posted on 2024-7-28 17:18 It is set to 3. This is an instrumentation amplifier. It can be set to 3

I don't understand what the setting of 3 means.

No matter what setting you make, there must be a certain resistor value, right? Why don't you just calculate a resistor value based on the gain being 3 and try installing it?

gmchen

newtonxihuan posted on 2024-7-28 17:22 The gain formula is very simple. If the input is slightly larger, the gain is close to 3

This is quite interesting. If the input signal is small, the gain will be increased? I don't quite understand.

Can you send me a waveform?

qhr22

Hello teacher, may I ask, do the resistance and capacitance values and the type of op amp have any effect on the measurement results?

明灯23

I bow down to you. Your explanation is professional and detailed. I have learned a lot. Thanks again for sharing.