Op Amp Pair - Add Matched Non-Inverting

** Is that so ?

Looks like a mess that cannot work to me

Have you actually built it ?

.... Phil

Reply to
Phil Allison
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** It is not up to anyone's standards.

You are trying to bias BOTH inputs ( one fixed and one adjustable ) and you have two feedback resistors in parallel. WTF ??

** Have you looked at the app notes for the LM358 ???

Or any simple op-amp schems on the web ?

... Phil

Reply to
Phil Allison

I'm assuming you have these connections: wire pin4 to 0V wire pin8 to +12V

what you have is a virtual ground at pin2: the voltage on that pin will be maintained equal to the voltage on pin3 by the action of the op-amp so with 27K above and 15K below and 1 2 12V supply it will be about 4V

That means the input looks like a 15K resisor into a 4v supply.

easuest way to get a non-inverted output with the same sort of input is a unity gain inverter driven off the output of the first op amp

10k pin2 to pin6 10k pin6 to pin7 wire pin5 to pin3.
umop apisdn
Reply to
Jasen Betts

for a non inverting output, connect output to - input and then drive the + input with our variable voltage. The output will represent the voltage at the + input..

This type of config is used as a buffer..

In your inverting unit, it looks like your output gain is less than

1? You can achieve that with putting a R diver on the + input so that it shows less than what you have on the input. If you want gain with non inverting operation then you need to have a feed back R from the output to the - input and another R from the - input to common. This voltage divider will give you gain. If for example you put two of the same values on the - input, one from the output and one going to common, it'll give you a gain of 2.


Reply to
Maynard A. Philbrook Jr.

R1 R2 | ___ ___ ===-|___|---------+|___|------+ GND | | | | | | | | | |\| | +-+|-\ | | >+-------------+ Output Input --------------+|+/ + |/|

R1 and R2 sets the diver ratio which gives you gain.

gain = (R1+R2)/R1

That is a non inverting gain stage or what ever you want to do with it.

If you want to adjust gain for less than 1 then you can have R1 connect to the positive rail. In your prior example, I suppose you could connect a pot to R1 instead of the ground so that the wings of the pot will connect to ground and the positive rail, this will allow you to ajust your gain either way.

The trick is to first calculate your Max forward gain when the pot is all the way to COMMON/Grd, then when you ajust to the position side this will shift the gain in the backwards direction.

Of course you can also set up a single OP-amp to do differential input, meaning you can input to both the - and + from out side sources via R dividers..

R1 R2 - ___ ___ Input --------|___|---------+|___|------+ | | | | | | | | | |\| | R3 +-+|-\ | ___ | >+-------------+ Output Input ----|___|--+--+|+/ + + |/| | .-. | | | | R4 '-' | === GND

That one will allow you to use both inputs to differentiate between the two inputs.


Reply to
Maynard A. Philbrook Jr.

You mean it has built-in redundancy?

I will change the unnecessary components and learn what it does.


Kevin Foster

Reply to
Kevin Foster

Yes. The design may not be up to your standards, but it does work.

I am looking for something similar with a non-inverted output. Or something better, in which case I can improve the existing design.

Kevin Foster

Reply to
Kevin Foster

The circuit diagram linked below is an inverting op amp.

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I would like a second one that provides non-inverted output.

I have the general idea, but would like them to match in configuration and performance as closely as possible.

Can someone please advise the best way to achieve this?

As you can tell from my drawing, I am still learning.

Thank you.

Kevin Foster

Reply to
Kevin Foster


If this is homework, getting the answer here will do you no good. Sure you might get an A on that assignment but when you have to put your big boys draws on and solve a circuit problem yourself with no help, you will not have developed that level of understanding.

EE'ing is not about solving individual text book problems. Its about solving a continuum of problems using the skills obtained from school and head lumps from figuring things out the hard way.

A few questions about your application

1) What is the end application for your circuit? 2) Why do the two need to be matched so well? 3) What is the specification for matching? 1% .1% 4) Over what frequency range must they be matched? DC? MHz? GHz? 5) Is phase important or amplitude only?
Reply to
Mook Johnson

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