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Bits
01-11-2012, 01:52 AM
So I think this circuit is considered a "offset voltage adjuster" or "voltage follower" I could be wrong. My problem is I don't know what the equation comes out to be. I was reading some old schematics and this is something that struck my curiosity. Any ideas?

image of op-amp.
88431

0-11V is the range of voltage on pin 3
0-5V is the range of voltage on pin 2 but may exceed that in other configurations, perhaps ranging from 0-11V.

Phil Pilgrim (PhiPi)
01-11-2012, 01:58 AM
You should be able to derive the I/O relationship. Just remember that the output in your circuit will drive the negative input to the point that it equals the positive input.

-Phil

Bits
01-11-2012, 02:08 AM
Phil - please re-word that statement. Only because the output can not make the input change, right?

Just remember that the output in your circuit will drive the negative input to the point that it equals the positive input.

This is miss worded or am I over analyzing this?
Perhaps this circuit will drive the output until the two inputs are equal?

Dr_Acula
01-11-2012, 02:20 AM
Say you put 6V on the + input and 4V on R2. The op amp will adjust its output so that the input to the - pin is equal to the + pin, so both pins will have 6V on them. We don't know what the output is yet, but reading reading from left to right along the resistors, you should have 4V => 15k => 6V => 30k => x volts. My op amp theory is a bit rusty but I'm guessing 10V output?

Bits
01-11-2012, 02:36 AM
Thinking out loud here...

...So the output is used to change the inputs? Perhaps a sort of feedback circuit, if you will, for driving some other device that is tied to the - pin? There is no equation for this circuit because the inputs have to change with respect to the output? So Phil is correct and i just spit on his reply?

frank freedman
01-11-2012, 02:40 AM
So I think this circuit is considered a "offset voltage adjuster" or "voltage follower" I could be wrong. My problem is I don't know what the equation comes out to be. I was reading some old schematics and this is something that struck my curiosity. Any ideas?

image of op-amp.
88431

0-11V is the range of voltage on pin 3
0-5V is the range of voltage on pin 2 but may exceed that in other configurations, perhaps ranging from 0-11V.

Looks like a form inverting summing ckt. . If the + input would be tied to gnd, it would be a simple inverting amplifier with a gain of about 2.

Check in http://www.ti.com/lit/an/slod006b/slod006b.pdf for an excellent op amp guide/tutorial.

FF

frank freedman
01-11-2012, 02:47 AM
Thinking out loud here...

...So the output is used to change the inputs? Perhaps a sort of feedback circuit, if you will, for driving some other device that is tied to the - pin? There is no equation for this circuit because the inputs have to change with respect to the output? So Phil is correct and i just spit on his reply?

In an inverting amp configuration, the current from the output would sum at the - input with the current from the input to achieve a 0V level assuming the + pin is grounded. Called a virtual ground. It will have a gain of 2. with the reference input not at zero, the - pin will not be at 0V, so the offset will not give an inverted output 2X the input value. The offset will shift the output value. Also. if you added more branches into the - input node, you would have a sum of all the voltages into the inputs*2 +/- the input offset in the + pin. Try using TI's Tina TI or Linear Devices spice implementations and play around with the ckt using both DC and AC input wave forms. They both come with lots of active device models to work with. Cheap to since no smoke to let out. (well maybe virtual.....)

FF

Dr_Acula
01-11-2012, 03:16 AM
There is no equation for this circuit because the inputs have to change with respect to the output?

There is an equation - it is output = -R1/R2. It is just a standard inverting amplifier. Normally in an inverting amplifier the + input is tied to ground, or tied to a virtual ground which is usually half the supply voltage.

In the calculation I did in post #4, we can call the reference ground the voltage on the + input which I arbitrarily chose to be 6V.

So our input voltage was 4V, but relative to our 'ground' of 6V, this is -2V. The gain is -30k/15k which is minus 2. So multiply -2x-2 and we get +4. But this is +4 again relative to our 'ground' of 6V so we need to add that to the 6V, which gives an output of 10V.

Op amps can do lots of things. They can invert a signal. They can amplify a signal. They can amplify and invert at the same time. They can add two signals together. They can give you the difference between two signals. The can change the impedance of a signal with a simple voltage follower.

Grab some op amps and resistors and a breadboard and try out a few circuits. Op amps are great fun!

Bits
01-11-2012, 04:17 AM
Thanks guys and know I have op-amps in front of me now. Thanks and sorry Phil for disagreeing with you.

Phil Pilgrim (PhiPi)
01-11-2012, 04:32 AM
'Nothing wrong with a healthy dose of skepticism! :)

-Phil

jdolecki
01-11-2012, 09:40 AM
Texas Instruments has a great design book called "Op Amps for everyone"

http://focus.ti.com/docs/analog/catalog/resource/appnoteabstract.jhtml?familyId=78&abstractName=slod006b

sylvie369
01-11-2012, 11:57 AM
Chapter 9 of the Propscope text also has some really good information on op amps. It's available here:

Lawson
01-11-2012, 03:10 PM
Only because the output can not make the input change, right?

This statement would only be true if R1 was left out of the schematic. Without R1 the op-amp circuit becomes a comparator circuit. (aka an open loop Op-amp with a high output V/uS limit) With R1 there is strong negative feedback in the circuit, and this negative feedback sets the properties of the circuit. Every op-amp circuit you see will have a component in the same position as R1, understanding what that component does makes it easier to understand the rest of the circuit. (usually a resistor, capacitor, or some combination of the two)

Lawson

Loopy Byteloose
01-11-2012, 05:28 PM
You might want to download a copy of "Op Amps for Everyone" by Mancini. The 3rd edition is 2009 and you might have to pay for it, but earlier editions are easily gotten on line for free.

If that tome doesn't suit you. I learned a great deal via audio preamp and audio mixer DIY projects at Elliot Sound Products. This is all created by Rod Elliot in Australia and he apparently is an electronic educator that runs a side business in creating audiophile-quality solid-state DIY projects. Everything he sells can fit into an envelop for easy shipping world-wide and he has some generalized op-amp boards for people that want to explore their use. But the real goodie is his website that is extreme well-written and informative. When you buy a product, you get all of it on a CD and access to information not provided without password on the regular site.

Try here ==> http://sound.westhost.com/

The main thing about op amps is that they are usually used with very low level signals, so having a good board rather than building on a breadboard eliminates a lot of hazards of construction. He also tends to use dual op-amps for stereo sound as the dual op-amp inherently matches the left and the right channel by having been manufactured on the same silicon at the same time.

Simply put, I love visiting his site just to read. And having built a few of his projects successfully makes me want to read more.

Having said all that, I think you will find most of his applications are for a dual voltage power supply of +15 and -15. I see you want to use a single power supply and many people here try to use a single power supply with only +5 volts. Originally, I did so too; but when you get deep into the specifications of op-amps, dual power supplies and high voltages make the results easier to attain. In your case, you can easily get the 0-11 volts - but it is not going to be much in Watts. That needs an additional power stage on op amps output, which can be a power transistor or MOSfet. (And for going from digital to a higher digital voltage; it may be rather silly to bother with an op-amp at all.)

Phil Pilgrim (PhiPi)
01-12-2012, 04:14 AM
Here's how I had hoped things would progress from my initial reply to Bits' inquiry. Start with the annotations in this schematic and the assertion that the opamp's output will keep the voltage on the two input pins equal:

http://forums.parallax.com/attachment.php?attachmentid=88507&d=1326345295

So we know from the initilal premise that:

V+ = V- = Vi

Also, the current into the V- node has to equal the current out of the V- node. So:

(Vo - Vi) / R2 = (Vi - Vb) / R1

From here it's just algebra. Solving for Vo,

Vo / R2 - Vi / R2 = (Vi - Vb) / R1
Vo / R2 = (Vi - Vb) / R1 + Vi / R2
Vo = (Vi - Vb) R2 / R1 + Vi

-Phil