Offset amplifier's input to prevent opamp clipping

Question

We can do this:

With AC coupling and biasing op-amp input.
Summing using buffered resistor
Using sum and difference inherited in OpAmp and sum signal with desirably created virtual ground

1 is good but I want DC coupled circuit. 2 is good but I my circuit have variable resistance and low summing resistor attenuate signal and use power. High resistor in the other hand create noise. 3 seems to be good but I don't know how to apply it in the circuit below.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

I've heard unity non inverting input is reference, I don't know why but if this is then input is shorted to reference and the only variable is reference then there is no way to use virtual ref on it, Then 3 is not implementable without summing. But the opamp must be able to sum two signal not the summing mention in 2. using buffered resistor. OpAmp designed to summ and difference but why I can not do that?

What output do you want for, say 0V input? What output do you want for 3V input? — Math Keeps Me Busy, Jun 11 '22 at 12:10
I ask because it is not clear what you are trying to achieve. Not knowing what you are trying to achieve means one does not know what would be or wouldn't be a solution. — Math Keeps Me Busy, Jun 11 '22 at 12:29
Passive components by definition cannot buffer. Only active components like op-amps can so if you want to buffer then you need an active component that can properly accept the input signal. Therefore, you cannot you use passive components to modify a signal so that it can be accepted by a buffer while simultaneously preserving the buffering effect of the circuit. It's illogical and doesn't make sense. — DKNguyen, Jun 11 '22 at 15:43
@DKNguyen you mean this is an OpAmp in buffer arrangement and you can not expect it more than buffering, interresting and thanks, it's first step of answer but also do you know why? And what is the right way? — mohammadsdtmnd, Jun 11 '22 at 15:50
If you want buffering then the right (only) way is to use a buffer opamp that can accept the signal then do whatever you want with it afterwards. If you want buffering then the signal needs to go straight into the buffer with no other paths. If you stick passive components between signal source and buffer that produce alternate paths away from the buffer input then you lose buffering. — DKNguyen, Jun 11 '22 at 15:51
@DKNguyen I can not directly buffer since it's outrange. I hope to level shift it by inherent OpAmp sum ability, but seems sum is called non-inverting amlification and difference is called inverting amplification, doesn't it? And the solution is the first picture here: https://electronics.stackexchange.com/a/37096/152533 .If so then I've got answered my question. Though I've like to dive more on buffering, why it's reference is not viewable, what the hell has happened? (this is the last part of question :) ) — mohammadsdtmnd, Jun 11 '22 at 16:00
Then don't use something out of range. You're not listening: buffering means feeding directly to an active device which means that device cannot be out of range. You either live with the divider impedance or buffer with an active component that is in range. — DKNguyen, Jun 11 '22 at 18:41

score 3 · Answer 1 · answered Jun 11 '22 at 16:38

Amplifying a signal around Ground with a single supply is a nice application for a JFET. As they have a negative turn on voltage, you can easily make an op-amp with a very large input offset voltage. The offset voltage will be roughly the pinch-off voltage of the JFET + 0.7 V.

When you wire it up as a voltage follower, it will add its offset to the input:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

score 1 · Answer 2 · answered Jun 11 '22 at 20:50

You want the 1.5V input mapped to 6V output, with gain 1.

First, we can mirror the input around the average of these two voltages, performing the mapping but with gain -1. The "mirror" - or simply the reference voltage is:

$$V_{ref1} = {1.5+6 \over 2}{\,\rm V} = 3.75{\,\rm V}.$$

The first stage is an inverting amplifier with gain -1 and 3.75V reference voltage.

The second stage should restore the overall gain to +1 by inverting the voltage again, this time around the midpoint

$$V_{ref2} = 6{\,\rm V}.$$

schematic