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Currently working on an oscilloscope design that would require each channel to be able to take in between 30VPP, -30V to +30V, which will go to an ADC which has a maximum input of 3.3V.

What types of designs or specific circuits could I implement to offset the voltage where 0 is -30V and 3.3V is +30V ?

ALSO - Yes, I will also be implementing a voltage divider, but the ADC cannot take in negative voltage.

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Inverting converter, -30 -> +30 provide 3.3V -> 0V

enter image description here

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schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Could you please explain the circuit or how you got those resistor values? \$\endgroup\$
    – Beshoy Hanna
    Commented Jan 28, 2020 at 8:37
  • \$\begingroup\$ I don't get quite those values. \$\endgroup\$
    – jonk
    Commented Jan 28, 2020 at 8:42
  • \$\begingroup\$ I did the voltage divider formula twice in my head. @jonk, do you get different results? \$\endgroup\$
    – Jasen Слава Україні
    Commented Jan 28, 2020 at 8:45
  • \$\begingroup\$ Yeah. The center node voltage is \$\frac{V_\text{IN}\cdot R_2\cdot R_3+3.3\:\text{V}\cdot R_1\cdot R_3+0.0\:\text{V}\cdot R_1\cdot R_2}{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}\$. You should be able to compute the two values where \$V_\text{IN}=\pm30\:\text{V}\$. They won't be what you wrote. \$\endgroup\$
    – jonk
    Commented Jan 28, 2020 at 8:50
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    \$\begingroup\$ @Jasen See this. With \$R_3=10\:\text{k}\Omega\$ I get \$R_2=8.9\:\text{k}\Omega\$ and \$R_1=80.9091\:\text{k}\Omega\$. Something like that, anyway. \$\endgroup\$
    – jonk
    Commented Jan 28, 2020 at 9:12
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Texas Instruments has published pdf: "Op Amps For Everyone":

https://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf

I used this pdf when I needed to move the -4 V - +8.8 V signal to 0 V - 2 V. On page 60 is chapter 4.3, which describes step by step how to choose components for the circuit.

enter image description here

The advantage of this circuit is that the function is linear.

The downside is that you must have the voltage to power the amplifier.

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