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For amplifying the output of a vibration sensor.

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Newbie level 6
Feb 13, 2013
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I am posting the circuit diagrams of both the amplifiers:-

1.circuit of AD 620 instrumentation amplifier,designed for getting a gain of about 106.

2.circuit of non inverting amplifier using op amp ic 741,designed for getting a gain of about 101.

The gain equations of both the amplifiers are written inside a box near both the circuit diagrams iam attaching with this post.

I tried both the above amplifiers for amplifying the output of a vibration sensor or analyzer.

But when i gave the input from the function generator both the circuits amplified well.

For the non inverting amplifier using 741 ic,when i gave the input from the function generator at first i was getting a gain of about 50 and 68 when the input signal frequency was about 9.7 KHz.When i changed the input signal frequency to about 1 KHz i got the gain of about 100.But the original gain designed for this amplifier circuit was 101.When i gave the sensor output signal to the input of the non inverting amplifier it gave a gain of about 450.

When i used AD620 amplifier circuit,I get a gain value of about 100 when i was giving input signal from the function generator.When i gave the output signal from the vibration sensor as the input to the AD620 amplifier it also gave about a gain of about 450.

Why is the gain becoming very high when i give the sensor output as the input to the amplifier?

Which amplifier should i use for my need(for amplifying the sensor output)?

Should i make any modifications to my circuit.If then please suggest those modifications clearly.

Somebody please help me and reply fast.


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the LM741 IC has a gain bandwith product of 1,000,000. Therefore at a gain of 100, the bandwidth would be at most 10k. This would mean an actual gain of 3dB less is 1/2 times less which is why you get a gain of 50 or so.

What is the frequency of output of the sensor? What is its amplitude? Does your sensor have a high output impedance?

In your circuit for the AD620, why do have an RC filter at the inputs? The cut off filter for it is 1.59Hz.

The frequency of the output of the vibration sensor changes according to the variations in the vibrations in the surroundings of the sensor.Its amplitude is in the range of 10 or 20 mvs to 1 or 2 volt mostly.How to find whether the sensor have a high output impedance?

RC filter at the inputs is provide a dc return path for the input bias currents.I get the correct gain when i gave the input from the function generator only when connected that rc filter at the input.Until that was not given i was not getting correct gain and also i was not getting AC output .I was getting DC output those times with a very high gain of about 450 then.

If the amplitude of the sensor output is 1V, then a gain of 100 would make the op-amp output saturate.

Try an input RC filter with the R = 100k as given in the data sheet.

I think it’s a little thing called Gain Bandwidth Product:–bandwidth_product

BW = A*GBW where BW is the bandwidth, A is the gain, and GBW is the Gain Bandwidth Product.

In my example, when the input signal freq was 9.7KHz i got a gain of ~50 for a GBW of 9.7K * 50 = 485KHz, which confirms with the low end of the spec for a 741. So, if the GBW of my 741 is 485KHz then at 1KHz, we can expect a maximum gain of 485KHz/1KHz = 485.

If i insert a low pass RC filter in between the vibration sensor output and the 741 op amp input will the gain become alright?I think some times the vibration sensor output will be at very high frequencies.So a low pass RC filter may remove these high frequency components.

In which circuit should i use the RC filter with R=100K at the input of the op amp.In 741 circuit or AD620 circuit?

I have mentioned that it is the GBW that is causing the low gain at high frequency in my first reply.

It does not work the way you have mentioned. You are using frequency to specify the gain while it is the other way round.
Closed loop Gain Acl is fixed by you by setting the resistors. The gain bandwidth product is the open loop gain multiplied with the open loop bandwidth. ie GBW = Aol * ω3dBol

When you fix the closed loop gain, you reduce the over all gain to get an increase in bandwidth.
ω3dBcl = Aol * ω3dBol / Acl

Take a look at the first figure in the link you put about GBW.

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Any high frequency components from the sensor will have less gain due to the GBW concept mentioned before. You do not need a separate LPF.
You already have a high pass filter between the sensor and the AD620 amp to provide the bias currents. Can the sensor provide the bias currents? If so, then you can ditch this filter.

Do you have a link for the data-sheet of the sensor?

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