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OpAmp Circuit assistance - :(

Discussion in 'General Electronics Discussion' started by Disabled Engineer, Feb 18, 2016.

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  1. Disabled Engineer

    Disabled Engineer

    Feb 18, 2016
    Once apon a time, my brain worked...

    What I am trying to do in a nutshell is this;

    I have an HBLED driver, claiming it runs on 0-10VDC with a "Linear" increase of output power for the LED(s)

    **WHAT is actually the inputs are**

    1.75VDC to 9VDC

    I know im missing this simple circuit in my head, but when working with OpAMps, I used to having 0V as my low value and 5 or 10V max. In this circuit for it to be linear, I need to have 1.75V = Zero and 9V =10V

    I would copy and paste the circuit I drew, but it is not legible..,

    Ideas? Dual OpAmps for input
    comparitor for switching resistor values???
  2. AnalogKid


    Jun 10, 2015
    What is the opamp part number?
    What is the power source for the circuit?
    What is the input to the circuit?

    "I would copy and paste the circuit I drew, but it is not legible.."

    You're an engineer, it says so right in your name. Make it legible and post it.

  3. Disabled Engineer

    Disabled Engineer

    Feb 18, 2016
    Im using a simple dual Linear R-R Opamp..

    The schematic is unreadable. Basically Im trying to adjust the upper & Lower range of the input to the amp.

    I was simply using a non-inverted 2X gain circuit taking 0-5V up to 0-10. The actual LED driver is not 0-10V linear, its upper limit is 9V and lower is 1.75V.

    Im not sure if my brain is allowing me to be clear enough (sorry)

    Sorry, the power source is a 0-5V linear voltage, opamp single supply

    Here's what I was using before realizing the LED driver didnt use the entire 0-10V swing.. I was trying to do this WITHOUT pushing the inputs/Outputs into a DAC & Micro - which would be easier, but hopefully unneeded

    Attached Files:

    Last edited: Feb 18, 2016
  4. Disabled Engineer

    Disabled Engineer

    Feb 18, 2016

    give me a few to redraw this mess

    I edited the prior post with scan
  5. AnalogKid


    Jun 10, 2015
    A couple of circuit approaches, depending on things only you know. If the power source to the opamp is greater than 10 V AND stable, then it can be divided down to form the DC offset needed to shift the output 1.75 V.

    Where does the 0-5V control signal come from, and what is its output impedance.

    9 - 1.75 = 7.25. 7.25 / 5 = 1.45. We will round this up to 1.5 to make things easier.

    To have a non-inverting forward gain of 1.5, an opamp needs two resistors in the feedback loop: 10K series leg from the output to the inverting input, and 20K shunt leg from the inverting input to GND.

    To have an output that is offset 1.75 V above GND, the input needs to be offset by 1.75 / 1.5 = 1.167 V. Since the opamp's negative power rail is GND, the input offset must be done at the non-inverting (+) input. This is inconvenient and messy because that also is the input for the control signal. This is why the output impedance of the control signal source is important - it is part of the resistor calculations for the offset voltage.

  6. Disabled Engineer

    Disabled Engineer

    Feb 18, 2016
    Input impedance is customer based system, not more than 10K Ohm, My testing impedance is via a 12bit DAC @ 1 Ohm

    I know it seems messy in my head.

    THANKS very much for you continued help.

    It will be easier (for me possibly) to just use (switch) customer equipment to a Micro & 5V DAC and just MAP the output as needed..
  7. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    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.

    If you are getting on in years, and your brain isn't firing on all twelve cylinders like it did during youth, you may have a dietary deficiency. Avoid whole grains and carbohydrates and make sure you have enough vitamins and minerals in your daily diet. If possible, get in some light exercise every day. Walking for a half hour or so every day is recommended.
  8. Disabled Engineer

    Disabled Engineer

    Feb 18, 2016

    You a man among men!

    Thanks. Ill just make up the circuit and use the programmable power supply I have to switch between 0-5V, perfect.

    Regards to you final paragraph, I wish the brain was the only issue. I usually do not want to go into detail regarding my issues, but its not age so much as BiPolar Depression, PTSD, SMall Personality disorder. This coupled with my 9 right knee surgeries and 8 shoulder surgeries, heart surgeries, Back issues, & More - Its a never ending cycle. I have days where I cannot get out of bed for hours because of pain, pain causes increase in depression, depression increases pain.....Etc.

    I used to be VERY active, worked as a Paramedic after the Army, then engineering, then I started to literally fall apart.

    Sorry for all of that, I do appreciate the sentiments.
    I'm excited to test the circuit theory.
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