EE Question - $50 Bounty! - Op Amp Design

[MoltenVoltage]

Hi all you electrical engineers.

I have a problem and am willing to pay $50 for the solution.  Payment will be made by PayPal, so you will need to be able to receive money that way.

I need to take a 1.1 to 2.4 volt DC input and, using one half of a rail to rail dual op amp (8-pin), make that into a clean 0 to 9 volt DC output.  This design also needs to have a gain control which will allow the upper limit of the DC output (at 2.4 volts in) to be set at any value between 0.5 through 9 volts DC. To clarify, the gain control will adjust the output range from anywhere between 0 through 0.5 volts, on up to 0 through 9 volts.

Finally I need the exact same thing except with the output inverted in a unique way.  2.4 volts in will equal 0 volts output and 1.1 volts in will equal 9 volts output at the max setting, BUT the gain control needs to adjust the LOWER LIMIT of the output such that the 2.4 volt input can be set to correspond to an output level between 0 and 8.5 volts.

Hopefully this is clear. To rephrase, I need the gain control for the inverted design to allow me to control the output range.  The inverted design output will always go between 9 volts and some lower limit (between 0 and 8.5 volts).  9 volts will be output when the input is 1.1 volts, and the lower limit will be output when the input is 2.4 volts.  The lower limit will be adjusted by the gain control.

Here is a TI app note that describes all the equations to find the component and Vref values, but it is way over my head:
http://focus.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=slyt189&fileType=pdf

The first person to post working schematics for both designs will receive payment.  Partial or imperfect solutions will not receive any payment.

Once I have confirmed that the solution works (which will be within a day), I will promptly send $50 U.S. to the PayPal account you identify.

Thank you, and please let me know if you have any questions by asking here on the forum!   

[iaresee]

This better not be your homework.

That PDF tells you how to solve for the resistor values.

You say 1.1 V - 2.4 V input to produce 0 V - 9 V output. That's two data points:

Vin	Vout
1.1 V	0 V
2.4 V	9 V

The data pair says:

0 = m(1.1) + b

The second data pair says:

9 = m(2.4) + b
m = (9 - b)/2.4

Two variables. Two equations. We have all we need. Substitute and solve:

0 = ((9-b)/2.4) 1.1 + b
-b = (9.9 - 1.1b)/2.4
-2.4b = 9.9 - 1.1b
(-2.4 + 1.1)b = 9.9
b = 9.9/-1.3
b = -7.6153846153846153846153846153846

And now m:

m = (9 - (-7.6153846153846153846153846153846))/2.4
m = 6.9230769230769230769230769230767

So slope (m) is positive but it doesn't cross at (0,0) (b <  0) so you're going to use Case 2 in that PDF to figure out what Rf, Rg, R1 and R2 should be. You know:

m = 6.9230769230769230769230769230767
|b| = 7.6153846153846153846153846153846
Vref = 9V

I honestly don't know how you'd do an inverted signal using that same setup. I don't think it's possible. But I'm probably wrong.

[MoltenVoltage]

No, its a very real-world problem.

Using a halfway point instead of zero (I need a linear output) I came up with these equations:

4.5 = m(1.75) + b
9 = m (2.4) + b

therefore m = 6.9231
and b = -7.6154

same as you.

That is as far as I could get.

I have NO idea how to calculate the resistor and Vref values because the equations and other EE references are entirely unfamiliar.

How do you know that Vref = 9v?  Wouldn't the Case 2 schematic have simply said +V instead of Vref if it were to be the max voltage?

Thank you for the feedback, Ian!

[R.G.]

With what precision?

[MoltenVoltage]

With what precision?

Not sure what aspect you are asking about.

The output needs to be proportional to the input

Obviously there will be a bit of variance based on resistor tolerances, etc.

This is for a voltage-controlled guitar effect and I need it as linear as reasonably possible.

[Eb7+9]

This is for a voltage-controlled guitar effect and I need it as linear as reasonably possible.

by the way you pose the problem I think this is what you're looking for :



pure op-amp algebra so very linear ...

issues like temperature drift and offsets can be improved - solution assumes ideal op-amps
input source on second circuit should be buffered, first one has hiZ input so no need for buffering there

solving for Vref in second circuit involves two linear equations in two variables ...

resistor and Vref values are rounded off to produce <5mV error max provided 9v supply is well regulated
btw, what you call gain I'm calling "scale"'

[MoltenVoltage]

Thank you for the prompt reply!

I am anxious to try out your solution.

I have a few questions about your schematic:

DESIGN 1:
What is the purpose of the 10K trimpot?  To tweak the reference voltage?

Will any NPN transistor do?

I assume the trimpot is needed to account for differences among transistors.  Please confirm.

What is the reference to (325x) above the schematic?

Why is there an arrow through the 500K resistor above the op amp?  Is that a rheostat?
If so, please explain its function in the context of this circuit and how to adjust it.

Where to connect the open circle at the output of the first op amp (the one that says 0v to 9v above it, and comes after the 10K scale potentiometer).  It appears to be unconnected.


DESIGN 2:

How would I buffer the input?


Finally, is the email you provided je@lynx.net or jc@lynx.net?  It is a little hard to read.
Is that also your PayPal account?

I will test the design this evening and hopefully it works!

Thanks again.

- Karl

[Eb7+9]

I made a mistake ...

Vref in the first circuit should be 1.286v and Rin=47k, Rf=278k, and not 325k
total error with roundoffs is more like 10mv ... sorry

I drew in a trimmer from which you could adjust the desired Rf value in the first circuit
(it's not there now ...)

the output on the first op-amp is there if you need/want it - wasn't too clear from the text
that node can be left unconnected as is ... same thing with the second circuit

the 10k trimmer is there to set the reference voltage Vref at the emitter of the NPN (follower) voltage source
the AC Zout of the DC voltage source will be around 30 ohms, relative to the high enough "47k" input resistor will provide extremely good linearity over the intended range ...

by buffered output I mean like the output of an op-amp circuit from the output of the circuit feeding it ...
if in doubt stick another op-amp in non-inv unity gain as buffer (ie., neg-in tied to output same as the output buffers above) between Vin source and input of circuit

if you do end up having input current offset problems you can always insert a resistor in the non-inv leg of the first op-amp stage with value same as Rf (278k in first case, 325k in second) to help cancel it - and that's w/o extra buffering or anything ... depends on how accurate you want this thing to be ...

that's about it

[ACS]

OK, I'll stick my neck out here...

Karl, I need to be blunt on this - are you using this forum for R&D?  I notice that you seem to operate a commercial pedal business at http://moltenvoltage.com/  Personally, I haven't got a clue on how to do what you're asking about, but if you do intend to use the output of this thread in a commercial product, then at best you've mislead the people here by not stating this clearly in your OP.

If this is for some kind of private use, then I unreservedly apologise in advance.

Regards
Aidan

[Taylor]

Ooh! It just got interesting.

Scandal aside, this is an interesting concept: if we learn something from this forum, under what circumstances is it okay to use it later in a commercial design? What level of specificity constitutes commercial use? I pretty much assumed he was planning to use it in a commercial design, or I reckon there wouldn't be $50 attached.

[MoltenVoltage]

OK, I'll stick my neck out here...

Karl, I need to be blunt on this - are you using this forum for R&D?  I notice that you seem to operate a commercial pedal business at http://moltenvoltage.com/  Personally, I haven't got a clue on how to do what you're asking about, but if you do intend to use the output of this thread in a commercial product, then at best you've mislead the people here by not stating this clearly in your OP.

If this is for some kind of private use, then I unreservedly apologise in advance.

Regards
Aidan

Hi Aidan,

I'm not trying to mislead anyone.  I assumed it would be obvious that I want to use this information for a pedal design and I thought it was clear that it would be for a commercial purpose based on the $50 fee for the work, however I agree that my original post could have been more clear. 

I've been posting here for a few months now and have yet to see anyone offer cash for information.  If you feel that is a misuse of this forum, then I am sorry, but the way I look at it, I offer incentive to get information, but then that same information, that I alone pay for, is part of the tremendous database that this forum represents and is freely available to everyone else.

It seems like a win-win to me, but if Aron feels otherwise, I assure you it will never happen again.