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Ask an Expert Forum How it this analog front end circuit working?
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How it this analog front end circuit working?

Sudeep AJ
Sudeep AJ over 2 years ago

image

This circuit has been taken from a "DIY Oscilloscope" project. This circuit is changing the input wave swing to 0V to 3.3V. The circled voltage source is Vsys/2, which is the voltage at which the zero level of the input wave will be shifted to. Vsys is 3.3V here. 

The output is as follows:

image

I wanted to know how is this working and can I use it for my project?

The original circuit with Raspberry PI PICO is as follows.

image

Thank you.

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  • wolfgangfriedrich
    wolfgangfriedrich over 2 years ago +5
    U1 is set up as inverting opamp with gain of -R2/R1 = -0.1 and biased to the voltage on the positiv input 1.65 V (= 3.3V/2) because of the AC coupling cap C1. So the input sin wave with 14.1 V peak to…
  • scottiebabe
    scottiebabe over 2 years ago +4
    Its not the best circuit ever but its definitely worth putting on a breadboard and experimenting with it. With these magnitude of resistors in the feedback path you may find you need a tiny compensation…
  • phoenixcomm
    phoenixcomm over 2 years ago +3
    Sudeep AJ Ok, here we go. I hate to do this but, did you figure out what bandwidth you need? I noticed that you limit your voltage to 3.3v which by the way is a FLOATER in TTL. I personally think you are…
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  • dougw
    dougw over 2 years ago

     As wolfgangfriedrich indicates, this circuit attenuates the input by a factor of 10, but it would work over a wider range of input voltages if it used rail-to-rail output opamps.

    U1-1 might be problematic as a clamping circuit with this particular op-amp, but it can't exceed the rail unless VSYS is a high voltage, so it shouldn't even be needed in this circuit.

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  • Sudeep AJ
    Sudeep AJ over 2 years ago in reply to dougw

    wolfgangfriedrich  dougw . Thank you for your valuable inputs. If I really need rail-to-rail op-amp and input range is only 1.8V. How in simulation it is working then? In simulation it seems like it is serving the purpose, basically taking the input sine of large swing to 0V - 3.3V swing centered around 1.65V so that my Microcontroller can take it properly. What I also observed in the simulation is, if I change the Vsys to 5V(if I want to connect it to Arduino UNO) and make the reference voltage as 2.5V, it serves the purpose too. I am also not understanding the purpose the U2 op-amp and 3 Diodes.

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  • ggabe
    ggabe over 2 years ago in reply to Sudeep AJ

    All 3 diodes are clamping voltages to protect the opamp input from over voltage as well as the MCU. The value of D3 and its opamp buffer, U1:1 is questionable, if the opamp is on a 3.3V rail and the MCU input can go up to 3.3V.

    I agree the LM358 is a bad choice, primarily b/c the low bandwidth and the non linearity it has in the crossover region. The input voltage range is not limited by the output swing, and the inverting input always maintains 1/2Vcc. It would be nice to use the full range up to full Vcc, however this benefit comes if the MCU’s ADC input tolerates up to full Vcc. Not all does. 

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  • Sudeep AJ
    Sudeep AJ over 2 years ago in reply to ggabe

    thank you for your inputs! If LM358 is a bad choice which one do you prefer?  And I am not understanding the notation U1:1 / U1-1, What does that mean??

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  • Sudeep AJ
    Sudeep AJ over 2 years ago in reply to ggabe

    thank you for your inputs! If LM358 is a bad choice which one do you prefer?  And I am not understanding the notation U1:1 / U1-1, What does that mean??

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  • ggabe
    ggabe over 2 years ago in reply to Sudeep AJ

    I misread the schematics - should have said U1.1. The LM358 is a very old part, a better choice would be if you pick something from the 10MHz GBP range, like MCP6291/2/4 (Microchip ships free samples). Works well rail to rail, up to 7V, available in through hole version.

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