Simple summing amplifier

Discussion in 'Homework Help' started by hang-on, Dec 21, 2014.

1. hang-on Thread Starter New Member

Oct 18, 2014
25
1
Hi,

I'm looking for help on constructing a very simple summing amplifier using an op amp. I have built a small circuit (please see attached schematic), but it is not behaving like I thought it would.

When I put my voltmeter to the two lines I need to add together, it reads 0,72 V (black probe to ground, red probe to left leg of the 1K resistor, marked with a red pen). I would expect the summing amplifier to perform 0,72V + 0,72V = 1,44V, right? When I measure the output from the op amp (black probe to ground, red probe to LED pin), I get 0,6V?

Have a nice weekend!

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2. bertus Administrator

Apr 5, 2008
15,806
2,389
Hello,

The voltage you are measuring is dependend on the input resistance of your meter.
There is no path to ground at the input.

The circuit would give an output of -(V1 + V2) .

Have a look at the following page of the eBook:
http://www.allaboutcircuits.com/vol_3/chpt_8/8.html

Bertus

3. hang-on Thread Starter New Member

Oct 18, 2014
25
1
Thank you! It is a good page in the ebook - I will build the passive averager and proceed up to the shown op amp summer circuit.

4. hang-on Thread Starter New Member

Oct 18, 2014
25
1
Oh, this day is not going so well with regards to electronics. I'm trying to build the passive averager circuit mentioned on the page in the eBook, but I'm not sure I'm during it right (see attached schematic). Or maybe I'm doing my measurements wrong.

When I measure voltage, I touch the negative pin on the LED, directly connected to ground, with the black probe. I touch the legs of the various resistors with the red probe.

In my understanding, I have three input signals: 1,92V + 1,92V + 3,22V. The sum, divided by three, is 2,35V. I thought that amount of voltage would go into the LED, but at the positive pin I only got 1,92V.

I'm doing something really basic very wrong, but I don't know what?!

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5. Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Your problem is that you do not understand how LED work. Diode is a nonlinear device and its forward voltage drop varies only a little with the current.
Forward voltage drop for a diode is almost constant vs current see the example

6. hang-on Thread Starter New Member

Oct 18, 2014
25
1
You are right. I am in the dark here What should I do with my simple circuit in order to measure the correct averaged output? Should I replace the LED with a resistor? How should I correctly measure the output from the averager?

7. Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
The correct circuit should look like this

And Vout ≈ (V1 + V2 + V3)/3 but this circuit will work only if R2 and R4 and R6 are much larger than voltage divider resistors.

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8. hang-on Thread Starter New Member

Oct 18, 2014
25
1
Thanks a ton. I'll build this soon!

9. WBahn Moderator

Mar 31, 2012
18,088
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First, please try not to post 1.8MB image attachments when something about one percent of that size will do very nicely. The following is only 24KB.

Your schematic shows an LED connected to the output but the voltage you are measuring, 0.6V, is more consistent with it being a silicon diode.

What is your opamp being powered by (in other words, what are your supply rails)? If this circuit were operating in the linear region, then the output of the opamp would be negative and the diode would be reverse biased. But that's only possible if it is powered by a negative rail. If it is being operated from a single supply, then you might be seeing 0.6V as a result of that being the lowest output voltage that the opamp can deliver. What opamp are you using?

With 0.72V at the two points you show, the current in each of the input legs would be (5V-0.72V)/100kΩ=42.8μA. That would put the voltage at the inverting input of the opamp at about 677mV. At that point the two currents merge and you would have about 85.6μA going through the feedback resistor which would bring the output down to

Vout = 677mV - 85.6μA·1kΩ = 592mV, which agrees closely with your measured output.

So, my suspicion is that you are powering an opamp that is not rail-to-rail from a single-ended supply.

10. hang-on Thread Starter New Member

Oct 18, 2014
25
1
Ok. I'm sorry for posting the unoptimized images. And yes - you are right. I'm working with a quad op amp on a single supply (LM-124N series I believe). good call! Thanks for the calculation and explanation. I have something to work with now!

11. WBahn Moderator

Mar 31, 2012
18,088
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As a quick fix, power your opamp with +/- 9V supplies (from two 9V batteries, for instance) if you don't have a suitable bench supply. There are also ways to create a split supply from a single supply, but it's nice to keep things simple at first.

Also, either get rid of the LED or put a suitable current limiting resistor in series with it. If your supplies are 9V, then to limit the max current to no more than about 15mA you could use a 330Ω or 470Ω resistor. If you put two LEDs in parallel but of opposite polarity, then you can have a quick visual indication of whether the output is positive or negative and you only need one current limiting resistor in series with the combination.

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12. hang-on Thread Starter New Member

Oct 18, 2014
25
1
Thanks for the additional info, WBahn! Much appreciated. I did not realize the difference between powering the op amp with a single supply vs. dual +/- supply. I have actually bought some ICL7660 voltage converters (along with the op amps) to give me some negative voltage, but then I realized that the op amp could take a single supply, and then I just went with that. Now I'll bring back the negative voltage.

I'm in way to deep on this one, but I'm motivated to built myself at least something like a test circuit/demoboard for op amp experimentation. When I level-up, so to speak, I dream about using op amps to convert the RGB signal from the video chip in a SEGA Master System to YPbPr output. But that is another story

13. WBahn Moderator

Mar 31, 2012
18,088
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Any opamp can be powered from a single supply since all the opamp sees is a voltage at the positive power pin that is so many volts higher than the voltage at the negative power pin. What matters is where the voltages at the other pins are relative to these two voltages. Traditional opamps are designed to be used in situations where the input and output signals are roughly centered between the supply rails and the easiest way to accomplish this is with two supplies using the common point between them as the signal reference. Opamps that are designed to be used with signals that are very near the negative supply rail are what generally get referred to as single-supply opamps.