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alternating analog signal

hallo, 
i would be happy to your help in a little issue.
i plainning a card which use tiva MCU.
the MCU has analog input that measures voltage between 0-3V, the voltage range i want to measure is between -10v to +10v. the highest frequency is about DC-100KHz.
may someone know's about component that can alter the analog signal linearly to the range of 0to 3v?
i know i can do it with amp-op and resistors, but i want to make it with Dedicated component.
thank's alot!

  • Depending on impedances and the availability of a stable 6V reference;

  • You'll still need an opamp as a voltage follower on the output(although, then you might as well use the opamp instead of a voltage divider with resistors, to level shift the signal), as the impedance of the adc input is in the range of 10K-30K ohms(if my memory is correct), and the above is more in the range of 660K ohm.

    .

    Also, the OP is asking for a 'dedicated component', I would say use Google, bit I doubt you'll find a dedicated component just for doing that...

  • Applaud both responders - I too line up w/classic op-amp due to it's flexibility & great availability.

    Believe that the ADC input is less than 1K ohm - thus op-amp should prove suitable.

    Poster's "usually" are inexpert - thus their goals/desires may not be well thought/considered.  I'd ask - "Who would design/build/market such a tightly restricted, "dedicated component?"  Especially when low-cost, uber-fliexible - readily available op-amps - in every conceivable grade - haunt disty shelves...

    Bending to the "will" of posters does not always make sense - especially when "NO" justification for requirement is provided. 

  • thank you very much! great answer! 

    i calculate the Vadc with superposition and what i found is:

    Vadc=1.5+Vin*0.3

    did my calculate correct?

    if it is, it seems to me that the upper resistance suppose to be of 17k, did i wrong?

  • Maybe this explains the working a little bit better. Don’t ask me how to calculate but the 7k is correct.

    First you need to attenuate the input voltage from +/-10V to +/-3V (or 20V to 6V). Then add linear to the attenuated input voltage (+3V - -3V) 0-3V.

    If using a single rail unity gain amplifier, the output of the Op-Amp can’t go to zero volt, then attenuate the voltage a little bit more. 

  • shrag boker said:
    if it is, it seems to me that the upper resistance suppose to be of 17k, did i wrong?

     I don't know from where it come that circuit but it is ok, voltage divider reduce 10V to 3V and -10 to -3V from partition rule:

     Vout= Vin*Rout/(Rout+Rin) = 10*3.07/(3.07+7) = 3.07V

     The second part I don't know why use two 2MOhm Resistor and 6V reference, why not use a 3V reference and 1MOhm resistor?

     Again from Thevenin Law the two resistor equals 1MOhm and equivalent generator equal 3V, more close to TIVA power too.

     The two resistor (Generator equivalent and from voltage divider) are a voltage divider too, so imposed 1.5V from generator and +-1.5V from input analog.

     All is OK than impedance seen from part connected to ADC input, here impedance equal the two 1MOhm resistor so it equals 500KOhm, too high to drive an ADC input!

     So an operational fit better this application, feed high impedance divider to operational non inverting input with positive offset too then set gain on inverting input.

  • Leo Bosch said:

    If using a single rail unity gain amplifier, the output of the Op-Amp can’t go to zero volt, then attenuate the voltage a little bit more. 

     To obtain a working circuits a rail to rail input output OPamp is forever needed.

     Output can go to zero voltage and input work when voltage is in power supply rails range. You can do a different simple circuit using an inverting circuits feeding signal and power rail voltage too in summing mode.

     

                                           |\

                       GND--------| +\ 

                                           |     >---------+------- Out

    AC---R1R1R1----+----|  -/                 |

    Ref --R2R2R2----+     |/                    |

                                      \--RFRFRF---- /

    Vout = RF * [(AC/R1)+(Ref/R2)]

     If example RF= 1MOhm, Vref= 3V and AC +-10V then R2 can be 2MOhm and R1 (10/1.5)MOhm  s-> 6.666MOhm