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AD8429 noise problem

Question asked by ghooooosttt on Aug 22, 2014
Latest reply on Aug 26, 2014 by ghooooosttt

Hello,

I want look at noise of amplifier AD8429. I made a schematic (see Fig 1) and I did some experiments.

schematic.png

Fig 1. Schematic

 

EXPERIMENT 1

 

Device is power off.

Output noise is about 518uV (see waveform on Fig 2).

I think, this is own noise of measurement channel, that is induced on the coaxial cable and PCB.

tek00015.png

Fig 2. Own noise of measurement channel

 

EXPERIMENT 2

 

Device is power on.

Rg resistor is not placed. Gain is 1.

Output noise about 4.62mV (see waveform on Fig 3). 

I don't understand why did I get too large output noise?

In theory, I think:

1. No Source Resistance Noise, because the inputs connected to ground.

2. No Current Noise, because the inputs connected to ground.

3. Voltage Noise is sqrt((Output noise/Gain)^2+(Input noise)^2+(Noise of Rg resistor)^2) = sqrt((45/1)^2+1^2) = 45 nV/sqrtHz

4. Total Noise Density is 45 nV/sqrtHz.

5. Frequency bandwidth is 15 MHz.

6. Tatal output voltage noise is sqrt(15 MHz)*45 nv/sqrtHz = 174 uV

 

tek00014.png

Fig 3. Total output noise (gain is 1)

 

EXPERIMENT 3

 

Device is power on.

Rg resistor is 120R. Gain is 50.

Output noise about 4.42mV (see waveform on Fig 4). 

Why does the noise not change, when I change the gain?

In theory, I think:

1. No Source Resistance Noise, because the inputs connected to ground.

2. No Current Noise, because the inputs connected to ground.

3. Voltage Noise is sqrt((Output noise/Gain)^2+(Input noise)^2+(Noise of Rg resistor)^2) = sqrt((45/50)^2+1^2+(4*sqrt(0.12))^2) = 1.93 nV/sqrtHz

4. Total Noise Density is 1.93 nV/sqrtHz.

5. Frequency bandwidth is about 2 MHz.

6. Tatal output voltage noise is sqrt(2 MHz)*1.93 nv/sqrtHz = 2.73 uV

 

tek00016.png

Fig 4. Total output noise (gain is 50)

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