I am all for using micros and DACs, but I think the

@AnalogKid has the least troublesome solution. Basically you use your original circuit but place a resistor between the 0 to 5V source and the non-inverting input to the op-amp. Call this resistor Rin. Then you connect another, much larger resistor, also to the non-inverting input. Call this resistor Ros. It will be used to apply an offset voltage to bring the output up to 1.75 V when the input is 0 V. The other end of this offset injecting resistor is connected across a potentiometer and the potentiometer is driven between ground and a positive voltage from the op-amp V+ power rail. Now all you have to do is select resistor values!

This circuit is far from optimum, but it works "gud enuf" to control your LED driver. As

@AnalogKid said, first thing is to calculate the gain of the non-inverting op-amp configuration. You want the output to vary from 1.75 to 9.0 V, a change of 7.25 V, when the input varies from 0 to 5 V, a change of 5.0 V. The gain required to do this will depend on how much the 0 to 5 V input is attenuated by Rin, Ros, and Rpot at the non-inverting input of the op-amp. Plug in some values to see how this works out: If Rin = 10 kΩ, Ros = 100 kΩ, and Rpot = 10 kΩ (5 kΩ if pot is centered) then we have a voltage divider that attenuates the input by 105/115 = 0.913 (pot centered) or 100/110 = 0.909 (pot wiper at ground position). This assumes the other end of the pot is connected to V+ but in a practical circuit there would be a resistor between the pot and V+ to limit the adjustment range to less than one volt. This resistor will raise the effective resistance of the pot as the pot is adjusted toward that voltage, so the attenuation of the input signal will be somewhat less. For practical purposes, we can "guesstimate" the input signal will be attenuated by 0.9 no matter what position the offset potentiometer is adjusted to. So, instead of the non-inverting input of the op-amp varying from 0 to 5.0 V, it now varies from 0 to 4.5 V. That means the non-inverting circuit gain needs to be 7.25 / 4.5 = 1.61

The only thing remaining is to figure out how much offset voltage is required to make the output 1.75 V when the input is 0 V. Clearly, with a gain of 1.61, this will require the non-inverting input to be offset by 1.75 V / 1.61 = 1.06 V. The voltage divider that was in effect for the input signal is now working in reverse for the offset voltage produced by the pot. We have Ros = 100 kΩ in series with Rin = 10 kΩ and (presumably) a low source-resistance for the 0 to 5 V input signal. So, we get about 10 / 110 = 0.09 of the pot wiper voltage appearing at the non-inverting op-amp input. To get that 1.06 V offset at the non-inverting op-amp input will require 1.06 V / 0.09 = 11.66 V.

Hmmm. That is pretty darn close to an assumed 12 V for V+, so maybe Ros = 100 kΩ is a little bit large. Or maybe Rin = 10 kΩ is a little bit small. You could increase the value of Rin and decrease the value of Ros to bring the pot voltage down to about half the V+ voltage while still providing about one volt offset at the non-inverting op-amp input. This will change the attenuation of the input signal (making the attenuation larger) so the non-inverting op-amp gain will have to increase, which means the offset voltage will need to decrease. These "what if" decisions are easy to model with a circuit simulator. Or set up a spread-sheet to automagically calculate the values. Unfortunately, I don't trust spread-sheet software, even if I write the cell formulas myself. It is too easy to get plausible but wrong answers because the underlying math is hidden in the cells and may be just wrong.

Or leave the resistor values "as is" and use, say, a 1 kΩ pot connected between +V and a 10 kΩ resistor to ground. That will limit the adjustment range at the wiper to about 11 to 12 V, which should make it fairly easy to set the output to 1.75 V. The smaller value (1 kΩ versus 10 kΩ) will have a minimum effect on the attenuation of the input signal so you can keep the same gain = 1.61. If the op-amp feedback resistor is Rf and the resistor from the inverting input to ground is Rx, then a value of Rf = 10 kΩ and Rx = 16 kΩ will yield a nominal gain of (1 + Rf/Rx) = 1.625 (instead of 1.61). You could make one of these resistors variable to trim the gain to produce 9 V output with 5 V input. Note however that the gain adjustment interacts with the offset adjustment: you will have to alternately apply 0 V and 5 V while adjusting gain and offset for 1.75 and 9 V output.

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