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Resistor network fork DC level shifter

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Junus2012

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Hello,

The circuit below are used to scale and level shift the bipolar AC signal of +-10 V to be +-2.5 V. Clearly seen from the voltage divider between R1 and R3 the scale is simply achieved as

10 kOhm/(10 k hm+30 kOhm) X 10 = 2.5.

For the level shifting, a reference voltage of 2.5 is then added with the resistor R2.

My question is about the value of the resistor R2, how to choose its value?, its looks for me it is the parallel result of R1&R3 but why?


pasivelevelshifter.JPG


Thank you
 

A comment on your sentence re resistor divider:

The input signal amplitude is not just divided between R1 & R3 as in your calculations. It is instead divides btwn R1 & R2//R3 (R2 in parallel w/ R3). This is because both GND and the 2.5V DC reference from opamp are not moving, so from the point of view of small signal calculations they are both AC grounds (or AC 0V).

=> the signal is attenuated ~1/8th:
+-10V --> +-1.25V (around a 1.25V bias point)
 
Dear friends, I have a question.

1. If I want to the shifting without scaling, it means in the circuit from post #1 I have to remove R1 & R3, what will be the consideration for the value of R2? I think I should choose it high so it will neglect the effect of the source impedance.

2. What is the consideration for the opamp used in the circuit with regard to dynamic performance? should it be carefully considered with regard to the signal dynamic properties or only DC performance is enough

Thank you
 

1. If I want to the shifting without scaling,
If you want no scaling, they you also have no shifting with the shown circuit.
You need to add an external op amp if you want shifting without scaling.
2. What is the consideration for the opamp used in the circuit with regard to dynamic performance? should it be carefully considered with regard to the signal dynamic properties or only DC performance is enough
Entirely depends upon the ADC sampling rate and the signal characteristics.
How fast is the signal level changing, and how soon do you need to know what the change is?
 
Hi

for an active "amplification" plus "DC shift" circuit it´s easy to use the standard "differential amplifier ciruit".

1667347545534.png


Use 2 x 2 resistor values:
R1 = R2
R3 = R4

Where Gain = R3/R1
V1 = inverting source
V2 = noninverting source
VRef = output reference

-----
Some examples:
you want to amplify a -0.5V ..+0.5V signal to a 0... +3.0V signal (ADC_VRef = 3V). Non inverting.
-->
* so the total input span is 1V
* the total output span is 3V
--> you need again of 3
* input center is 0V
* output center is 1.5V (DClevel shift)
-->
Let R1 = R2 be 10k
then because of gain R3 = R4 = 3 x 10k = 30k
for the level shift connect Vref to 1.5V

but maybe you don´t have 1.5V by hand, but the 3V ADC_VRef.
So you may replace "R3= 30k connected to 1.5V" by an equivalent circuit:
To get 1.5V from 3.0V you may use a resistive voltage divider: 1:1:
So R3a = R3b, R3a connected to 3.0V, R3b conected to GND.
Now to get the equivalent to 30k: R3a = R3b = 60k (because R3a and R3b act as if they where connected in parallel)

Klaus
 
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