Amplification 5.5V-8.5V to 0-3.3V Range

I have a input voltage of say... 5.5V to 8.5V range. I need the final output at 0V to 3.3V range so that I can connect this to a ADC.

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I am unable to figure out how this can be achieved.

You need an gain and offset control using a positive supply. very accurate called Vref.

If you want to scale i out range from 3 to to 3.3 from some odd reason, then the non-inverting gain is 3.3/3 instead of 1.

But Non-inverting gain is always 1+ |Av-| and inverting can it be 0. So you must attenuate the input and use a small inverting gain like 0.5 or 1 then attenuate by the resulting negative gain from Vref= +X

So if + pré-attenuation is 0.5 then +gain of 2 is ok and you use a Vref of 2.5 for Vin-. and an inverting gain of -1 so overall Av+ =1/2 x 2= 1 For 0-3 out assuming inputs work up to 8.5/2 with a 5V supply.

So with only 3.3V now you see you need + input attenuators of 1/3 with rail to rail input and output Op Amp and a Different inverting gain >2 or .... you do the math and be careful.

You can drop voltage through a series zener diode. To get 5V drop, string a combination of diodes/ leds/ zeners. Or whatever adds up to 5V.

Many options depending on the source attributes. If
it has a high transition density and reasonable 1/0
balance then capacitively coupling it to a resistor-feedback
inverter (or Schmitt, better yet) input could be easy
sleazy. If it has high source drive then perhaps a 5.2V
zener against a resistor load to Gnd, same Schmitt input.
For slow signals an instrumentation amplifier would do
what you want (A=1.1, offset input ~7V). Application
defines approach.

I donot want a high and a low.

I want analog signals. that is... as input voltage varies from 0.9 to 1.5V, the output of the circuit should vary from 0.9V to 3.3V or 0v to 3.3V whichever is possible

Hi,

for good performance I recommend a "difference amplifier circuit".

one opamp, two equaly resistors Ri (at input side), two equal resistors Ro (at output side). one 5.5V reference signal.

your input voltage range is: 8.5V - 5.5V = 3V
your output voltage range is: 0V to 3.3V = 3.3V

--> you need a gain of V_OR / V_IR = 3.3V / 3.0V = 1.1

* And you need to connect the input reference to 5.5V
* and you need to connect the output refrence to GND

***
5.5V reference:
It depends on your circuit how/where to get it.

Klaus

tiwari.sachin said:

I donot want a high and a low.

I want analog signals. that is... as input voltage varies from 0.9 to 1.5V, the output of the circuit should vary from 0.9V to 3.3V or 0v to 3.3V whichever is possible

Click to expand...

Did you not understand my solution?

SunnySkyguy said:

Did you not understand my solution?

Click to expand...

Maybe a circuit with pots, so he can dial in a offset and a gain.
:lol:

RobertFalbo said:

Maybe a circuit with pots, so he can dial in a offset and a gain.
:lol:

Click to expand...

Maybe or calculate the exact gain and offset as I did and do it right 1st time with 0.1% R values or as needed.

It's worth noting that for an analog application, a
zener drop will introduce mucho noise.

So far I don't see any real specification for accuracy
and with supplies typically 5% recommended, maybe
1-2% delivered on a decent DC-DC, there may be no
point to putting a fine point on it.

SunnySkyguy said:

Maybe or calculate the exact gain and offset as I did and do it right 1st time with 0.1% R values or as needed.

Click to expand...

Yes that would be easy, if he had a fixed values for input, 1st he wanted 5.5 to 8.5, then just before my last post he changed it to 0.9 to 1.5.

That's why I suggested pots.