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Hi All. I just started playing around with Circuit Lab and put together an inverting OpAmp circuit. The simulation shows it working properly. It uses an LM741 OpAmp with matched resistors. With a 5V- input to the inverting OpAmp input, I should be seeing 5V+ on the output. I'm seeing that in simulation, but on the breadboard, I'm seeing exactly 28V+. The link for the project is below. I'd be eteranlly grateful if someone with more knowledge of analog circuits could shed some light.

Thanks in advance...

Thanks very much for the great info. Every circuit example I came across showed a bipolar power supply to the 741, but when I saw it working in simulation I thought maybe I got lucky. Not having a bipolar power supply on hand, I tried a positive PS. I had a feeling the problem was related to that.

I'll also look into the other software you suggested. More accurate modeling would be nice.

Thanks again....

Just out of curiosity...How would you go about creating a +-15V supply from a 18VDC? I was looking at something like this....

http://www.mouser.com/Search/ProductDetail.aspx?R=RB-1515Dvirtualkey56830000virtualkey919-RB-1515D

or this...But this one is expensive....

http://www.mouser.com/Search/ProductDetail.aspx?R=ASA00CC18-Lvirtualkey54040000virtualkey826-ASA00CC18-L

@rrc1962,

Rather than go to the trouble of buying or building a split supply, you can run any opamp off a single supply but you have to create a local bias supply that, to the opamp, looks like a split supply.

Here's one way to do it:

Wouldn't that limit the output range to half of Vcc? What I have is a positive ground device that needs to interface, via an analog input, to a negative ground system. I'm using the opamp to sort of rectify the ground potential of the analog signal. Basically, I have an 18VDC supply and up to 10V of analog signal. I need to input -10V to the inverting opamp and get +10V out relative to the ground of the receiving system.

If I split the 18V supply and create a local bias, wouldn't that mean I'll only see 9V on the output?

I'm actually leaning towards using board mounted split supply with AC input. AC power will always be available in this particular application. The cost is higher, but it's easier.

"Wouldn't that limit the output range to half of Vcc?"

Absolutely.

Not just to

Vcc/2

but to

Vcc/2 - Vx

where

Vx

is how close your chosen opamp can swing to the rail for a given load current.

My example would work for you if you could set the supply to 30V and connect your input circuit ground to the midrail net. That would give you a maximum output of about +12V with respect to midrail.

Basically you use a floating 30V supply and connect the midrail net to your system ground. To see how this works: move the ground symbol from the negative terminal of the supply (the opamp negative rails) and on to the midrail net and re-run the sim.

In effect that gives the opamps +/-15V supplies.

It all sounds a bit baffling without drawing the whole circuit to show how the supplies work but if you have two opamps then it's cheaper than buying two or a dual supply PSU.

An alternative is to buy a rail to rail input opamp or an opamp that has a common mode input range that includes ground (i.e. the opamp's negative rail). It doesn't have to have a rail to rail output.

If you try the link below it should show you a selection of devices offere by Farnell (Newark in the US):

http://uk.farnell.com/jsp/search/browse.jsp?N=2031+203333+110106794&No=25&Ntk=gensearch&Ntt=opamp&Ntx=mode+matchallpartial&locale=en_UK&appliedparametrics=true&getResults=true&originalQueryURL=/jsp/search/browse.jsp%3FN%3D2031%2B203333%26Ntk%3Dgensearch%26Ntt%3Dopamp%26Ntx%3Dmode%2Bmatchallpartial%26No%3D0%26getResults%3Dtrue%26appliedparametrics%3Dtrue%26locale%3Den_UK%26divisionLocale%3Den_UK%26catalogId%3D%26skipManufacturer%3Dfalse%26skipParametricAttributeId%3D%26prevNValues%3D2031%2B203333