# Network sensor and ADC design

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#### kentman234

##### Newbie level 5
Hello,

can you please help me with the network design here? I need some hints to get started.

Thanks a lot.

#### barry

I would first write the node equations. You know the following:

1) For an input of -5V, the voltage at the ADC input should be 0V
2) For an input of +3.3V the ADC input should be 3.3V
3) the network provides an attenuation of 3.3/8.3
4) the network provides an offset of +5V

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Points: 2

#### kentman234

##### Newbie level 5
I would first write the node equations. You know the following:

1) For an input of -5V, the voltage at the ADC input should be 0V
2) For an input of +3.3V the ADC input should be 3.3V
3) the network provides an attenuation of 3.3/8.3
4) the network provides an offset of +5V

I didn't really get the third point. Can you explain more about it?

I think the offset of +5V adds to +3.3V but I don't know where the voltages are located. I am confused about this and I also don't know what the next step would be, I mean the general procedure of approaching the solution.

#### barry

The input range is 8.3 volts (-5 to +3.3); the output range is 3.3 (0 to 3.3V). Thus, the gain (attenuation) is 3.3/8.3.

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So here's a hint: Write your node equations and solve for Vadc (input to the ADC). Then solve the equation for Vin=-5 and Vin=+3.3. That should give you the values of the two unknown resistors.

#### FredCailloux

##### Junior Member level 1
Here is another hint. When your sensor output -5V your ADC should read Zero. Since you have R2(pin1) at +5Vdc and and R2(pin2) at Zero when the ADC read Zero, this imply that you have the same voltage drop between R2 and (R1+470), so, this tells you right away that R1 = R2-470. Now you can find another equation that will bring a gain has described by the attenuation (3.3/8.3) and you have the two necessary equations to calculate your two unknowned R1 and R2.

kentman234

### kentman234

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Points: 2

#### KlausST

##### Super Moderator
Staff member
Hi,

Input voltage range is -5.0V to 3.0V = 8.0V in total. Mind the "0".

--> Thus the gain should be : 3.3/8.0

Klaus

kentman234

### kentman234

Points: 2
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Points: 2

#### barry

Hi,

Input voltage range is -5.0V to 3.0V = 8.0V in total. Mind the "0".

--> Thus the gain should be : 3.3/8.0

Klaus

Right. I'm just so used to seeing 3.3V everywhere...

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Points: 2

#### kentman234

##### Newbie level 5
Here is another hint. When your sensor output -5V your ADC should read Zero. Since you have R2(pin1) at +5Vdc and and R2(pin2) at Zero when the ADC read Zero, this imply that you have the same voltage drop between R2 and (R1+470), so, this tells you right away that R1 = R2-470. Now you can find another equation that will bring a gain has described by the attenuation (3.3/8.3) and you have the two necessary equations to calculate your two unknowned R1 and R2.

And what about the second equation? Because I find it strange to use Thevenin's theorem here. I had a very similar example on the book with input impedance known but with amplifier included instead of ADC. Also AC voltage was considered to be the signal of interest so DC voltage to be as AC ground. In that example Thevenin's theorem was used but again with input impedance known. Here I don't know which direction to follow.

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I have finally wrote the node equations by setting vin to be 3V and thus getting 3.3V at the ADC input. After solving a number of equations I got the current values then the resistor values, (R2=1400 R1=930) which are identical to the values stated in the solutions section. Thank you all very much for guiding me through this.

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