Active Low Pass Filter Conundrum

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aaronbeans

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Hi Guys,

This is my first post so please bear with me. I have come across a circuit which i believe is a low pass filter using an op amp, however it is slightly irregular in the way the feedback is connected and I dont understand why. Referring to the circuit schematic, i dont fully understand what the purpose of R1 is and why the feedback capacitor (C1) and resistor (R2) are connected at different sides of R1.

My initial thoughts were that R1 was for bias current correction but I would of assumed that the feedback components would still of been connected together to one side of R1. Any help on explaining the purpose of R1 and why the feedback is connected as it is would be much appreciated!



Thanks
Aaron
 

Incomplete schematic, what's the driving source? If it's a voltage source, R2 would be useless and the circuit acts as an integrator rather than a low-pass.
 
FvM said:
Incomplete schematic, what's the driving source? If it's a voltage source, R2 would be useless and the circuit acts as an integrator rather than a low-pass.
Click to expand...

Hi,

Sorry the driving source is the output from an analogue gyro - the gyro provides a voltage output of approx 20uV/deg/Second. Hope this helps.

Thanks
Aaron

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Hi,

The input is the to the circuit is an analogue gyro, which provides 20mV/deg/Second.

Hope this helps.

Thanks
 

I have added a couple of extra parts on the schematic (sorry i didnt think they were anything to with the circuit). In terms of gyro characteristic what do you need to know. I have added the bias level and sensitivity on the schematic.



Makes sense now you say it is an integrator because it is connected to the output of the gyro! Still not sure on R2 though.

As the output of the gyro is biased to 2.5V when there is no rotation, surely that would cause the integrator to saturate if it was just left for any period of time?

Thanks

Thanks
 

I am trying to find out some more information about the gyro as its quite old and finding the documentation is proving tricky. I will post when I find it.

On a side note - right at the beginning you mentioned that R2 is useless if it is driven by a voltage source - why is that and if it was a current signal would R2 become useful? i.e. if the circuit was connected to a sensor with a current output rather than voltage?

Thanks
 

Ok thanks, so with a current input does the circuit still operate as an integrator? Please correct me if I am wrong but i thought op-amps worked on voltage inputs and not current, therefore shouldnt there be a conversion to voltage via a resistor before the op amp circuit?

Sorry about my lack of knowledge on this circuit am i finding it hard to analyse it at the moment!
 

The purpose of R2 is unclear without detailed informations about the gyro characteristic. Or it's just an erroneous or wrongly copied schematic.

I would expect a reset means for the integrator.

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Still wondering about the gyro output impedance. Without this information, the discussion is meaningless I fear.

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In the first case, R2 would be connected between tow low impedance and haven't any effect on circuit behaviour, in the second case, R2 would provide feedback.

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No, in case of a current source the circuit effectively acts as I/V converter.

In both cases, the design is questionable in my view. But It's generally problematic if you neither knew the intended purpose nor the circuit parameters.
 

I thought i knew enough about the circuit to analyse it, but from talking to you i realise i need to find out more! I will look into the surrounding circuits more carefully and gain a better understanding of those then look into this one again.

Thanks for your help, it was very useful!

Aaron
 

I think its a current to voltage converter with a built in LPF. The reason is that looking at the input the current that gets sent in comes from - Vin, so its input impedance will be 500 ohms. The voltage source sits behind 68 MF caps, with a Dv/Dt of 20mV/sec, means that the caps transfer very little current. i.e. they are a constant current generator. Inside the loop, the difference between the Iin-Iout = I 6.2M (actual input current), but this is working with the .1 MF cap to perform a LPF. T =.6S which is consistent with the Dv/Dt of the input, i.e. it performms a little rool off at the working frequency.
Frank
 

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