Please help solve ramp for op amps!

[ilbs]

I think it has more to do with:

vout=V2(gainNon-inv) + V1(gainInv)

I will also look more .... but tomorrow 

[c4757p]

gain for non inverting is:

(r2/(r1+r2))(1+rf/ra)

...the hell'd you get that from? Fix that equation and your problem is solved...

[ilbs]

don't know how... that's why I'm here....

[c4757p]

Just Google "noninverting amplifier"! Then click through to the almighty Wikipedia, where you see this:




Try your math again with that. Your R1 and R2 have nothing to do with Gain2, the amplifier gain - they only affect Gain1, the attenuator gain.

[ilbs]

try my math again with what???

you said my gains are right!
and you said rf is badly calculated!

so if it has nothing to do with gain 2, so then
it has to do with gain1 which is
0.4722222 so is it:

V2(1+0.472222)
9(1.47222)
13.24

so what does 13.24 have to do with this....
look.... I'm am obviously trying here...
I am not being lazy!!! I just don't get
this stuff the way you are trying to
explain this.

Does anyone have any detailed
help on this please....

thank you

[c4757p]

Gain2 = 1 + RF/Ra

Your R1/R2 are correct. Though you might want to try picking values closer to standard ones.

[ilbs]

if gain2 = 2

and if you are saying that gain2 is:

(1+rf/ra)

then
ra should = 10k

and

rf should also = 10K

therefore:

if r2 = 4722, then va should = 4.24999998
and therefore vb should also = 4.24999998
which works out to:

Vra and vrf = 0.75000V and therefore:
5-0.75000-0.75000 = 3.5vdc

Is this correct.... I am not at the lab
so I can only test this tomorrow

[c4757p]

If I remember correctly, yes. I didn't save my data, but it looks good.

[ilbs]

One last question,

concerning:

You already know Vgnd (it's 5V), so we can work backwards:

(Vin Gain1 - Vgnd) Gain2 = (Vout - Vgnd)
(9 Gain1 - 5) Gain2 = (3.5 - 5)
9 Gain1 Gain2 - 5 Gain2 = -1.5

Why are we presuming that Vgnd is 5Vdc, is it because we know the inverting input is 5V??

Because I tried setting Vgnd to 6 Vdc and that didn't work anymore   

[c4757p]

The inverting input is the virtual ground. Note that in the usual noninverting arrangement, that point would go to real ground, not a voltage input.

[ilbs]

c4757p,

While we are in a lengthy thread, I may as well ask. I have no trouble with algebra, I have done up to college math including calculus. But my usual problem
is defining the equations. Like many (I'm sure) coming up with the equations can be real tough.

How do you guys come up with these equations.... I mean there must be some sort of a rationalization for one to come up with:

(Vin x G1 - Vgnd)(G1) = Vo-Vgnd

In the grand scheme of things (and correct me if I am wrong) the above formula is an intention to make a point in the circuit equal to another point in
the circuit while knowing that these two points MUST be equal,   right?

In other words, I am sure there are other solutions ... but I don't see any other... I think this one would be wrong:

(Vin x G1 - Vgnd)(G1) = V1-Vgnd

are there any other equations we could of written for this?

[c4757p]

I have no trouble with algebra, I have done up to college math including calculus. But my usual problem
is defining the equations.

That's a common position to end up in. It often takes some experience and intuition to know.

How do you guys come up with these equations.... I mean there must be some sort of a rationalization for one to come up with:

(Vin x G1 - Vgnd)(G1) = Vo-Vgnd

In the grand scheme of things (and correct me if I am wrong) the above formula is an intention to make a point in the circuit equal to another point in
the circuit while knowing that these two points MUST be equal,   right?

Yep, you got it! It's just a description, in "algebra language", of what I expect to see at a bunch of points in the circuit. The piece-by-piece explanation that I gave a couple posts ago was pretty much exactly my thought process when I wrote it.

are there any other equations we could of written for this?

Plenty! You could have gone the pure Ohm's Law route which I originally (foolishly) suggested - I forgot, with that many resistors it often ends up a bit complicated. You'd end up with a system of multiple equations, which I'd probably just toss into MATLAB for a solution

[ilbs]

c4757p,

I learned a lot in this thread 


Thanks for your help