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### Topic: Confused by how opamp is supposed to work (Read 3239 times)previous topic - next topic

#### giants_fan3

##### May 04, 2011, 01:31 pmLast Edit: May 04, 2011, 01:39 pm by giantsfan3 Reason: 1
I am having some trouble understanding and getting this opamp to work. My first opamp... Maybe I'm missing a fundamental principle of opamps...?

Situation
Very simple: I have a small signal (mV or maybe tens of mV). I would like to amplify it so that it's larger and so I can read it through an ADC later.

Method
I'm using an Analog Devices OP177 opamp.

Datasheet-recommended application circuit for
"Differential amplifier":

It's very simple according to the data sheet: (http://www.analog.com/static/imported-files/data_sheets/OP177.pdf)
[Application circuit above is from datasheet]

My circuit
For testing, I'm simulating a small input signal by using a potentiometer connected to a DC supply of +10 V. The pot is set to output a voltage of about +10mV (with respect to GND)...

So +10 mV goes to IN+ pin of opamp, and GND goes to IN- pin of opamp.

I supply the opamp with 10V for supply V+ pin, and GND for supply V- pin, even though the app circuit recommends -10 V for supply V-...
Could that have been an issue?

Thus, my final circuit:

Problem
I expect a gain of 100 (b/c I used 1 MOhm and 10KOhm resistors).
So V_output should be 10 mV * 100 = 1 V.
But instead at the output, I'm getting 9.5 Volts (!?!)
i.e., nearly the supply voltage with which I'm powering the opamp.

Note: When I changed the supply voltage, same problem; e.g. if supply voltage is 15 V, I'm getting opamp output of 14.5 V.

Why is the opamp not giving me the expected correct output, even though there's nothing complicated here?

#### bubulindo

#1
##### May 04, 2011, 02:14 pm
Don't quote me on this... but if I remember correctly the Op Amp cannot put out the same voltage it is supplied with. The device has losses and some voltage drops inside too... so, having that difference towards the maximum is, on my perspective, nothing to be worried about.

what I don't quite get is the voltage divider on the input of the Op Amp... compensating the +1 on the gain of the Op Amp?
This... is a hobby.

#### giants_fan3

#2
##### May 04, 2011, 03:16 pm
@bubulindo:
but even if that were true (that the opamp won't output at levels close to the voltage driving it), the output I'm getting is FAR different from what I should be getting. A gain of 100 would imply an output of around 1 V; I'm getting near 9.5 V, and even more surprisingly, the output's not changing when I adjust the potentiometer...

and the voltage divider on the input is just the potentiometer I'm using to simulate an input signal (of around 10 mV)... perhaps I didn't draw correctly.

#### bubulindo

#3
##### May 04, 2011, 03:39 pmLast Edit: May 04, 2011, 03:51 pm by bubulindo Reason: 1
I mentioned the output from the potmeter because that would be an obvious cause to your problem. Say you had 1 V in your output, and the Op Amp would behave the way it is now. I have no idea about your knowledge of electronics, therefor I hinted at that possibility.

Can you make another drawing? The way I see it, you are applying 10mV to a voltage divider formed by R3 and R4 that drops the voltage by 0,01 (1-0,99) to the output of the op amp. This way it's ok.
However, if you are applying the 10V straight to R3, what you are seeing is as it should be. By my calculations (and I'm not the expert) if you put 10V on R3, you'll get 10*0,01 V on the op amp input. And that will saturate the op amp.

So, like I said, get rid of R3 and R4, make sure the potmeter outputs 10mV, put the potmeter straight in to the + pin on the op-amp and give it a try like that.
Where did you assemble this? If it's on a breadboard, double check the connections, or assemble it again in another place on the same (or another) breadboard. Maybe some connection isn't quite complete on the feedback loop and you get this.

Then the obvious tips for you to sort this out are: checking the same schematic with a different OP177 or simplify the schematic and see if the op amp is actually working. Or see if you didn't wire stuff the wrong way around.

I'm still amazed you can get 10mV out of the potmeter fed by 10V. :S
This... is a hobby.

#### Jack Christensen

#4
##### May 04, 2011, 03:51 pm
Something must not be right, wiring or something.  Assume you have verified the input is indeed 10mV.  I'm not familiar with the particular op-amp you are using, but this is pretty basic, any garden-variety op-amp ought to work just fine.  And it should be independent of supply voltage, as long as the voltage is somewhat greater than the expected output voltage; in this case there should be more than enough headroom.

One suggestion, the circuit as shown is a differential amplifier.  You really only need a non-inverting amp, and by grounding the inverting input, the diff amp is essentially reduced to a non-inverting amp.  As you can see, the typical non-inverting amp will save a couple resistors.

So one thing I might try is to rewire it as a non-inverting amp.  Another thing, depending on the potentiometer, it might be very hard to get 10mV out when the pot is across 10V.  That's only 0.1%.  Just from a mechanical standpoint, things can get jumpy operating a pot very near one end.  So I might add a fixed resistor in series with the pot.  Put a 1M resistor in series with the 10K pot, and it'll give roughly 0-100mV output, and can be operated away from the "ends".

#### bubulindo

#5
##### May 04, 2011, 04:02 pm

@bubulindo:
but even if that were true (that the opamp won't output at levels close to the voltage driving it)

If you have a look at the manual you supplied, you'll see an interesting parameter called Output Voltage Swing... and when powered from +/-15V you'll get different outputs depending on the load... none of them reach 15V either. So I may have been right.
This... is a hobby.

#### floresta

#6
##### May 04, 2011, 04:44 pmLast Edit: May 04, 2011, 05:13 pm by floresta Reason: 1
It's been a while since I did anything with op-amps but I remember a few things.  Here are some comments and recommendations (not guaranteed):

1- You should not start out with a high gain design, it's too easy to drive it into saturation.
2- You should not try to use a differential amplifier with one input grounded, I believe that you basically wind up with a comparator.
3- You should not try to run a split-supply design with a single supply.
4- You need a special op-amp (rail-to-rail ??) to get output voltages close to the power supply voltage.

I suggest you start out with (search for information about) a single-supply, inverting op-amp circuit.  Here's one possibility: (http://www.ti.com/litv/pdf/sloa058)

Don

 Here are some more references:

http://www.linear.com/pdf/an11.pdf
http://pdfserv.maxim-ic.com/en/an/AN656.pdf
http://www.analog.com/library/analogDialogue/archives/29-3/consider.html

#### westfw

#7
##### May 04, 2011, 05:40 pm
An op-amp is also rated for "rail to rail" capability on its inputs, and it takes special care for an op-amp to be able to accurately amplify signals that are near the voltage extremes of the power supply.

Quote
So +10 mV goes to IN+ pin of opamp, and GND goes to IN- pin of opamp.

I supply the opamp with 10V for supply V+ pin, and GND for supply V- pin, even though the app circuit recommends -10 V for supply V-...
Could that have been an issue?

Oh yes!  The OP177 is designed for a bipolar supply, and is NOT rail-to-rail, and you're operating it with both inputs too close to the "negative" power rail (-V and -V + 10mV, essentially.)  You might get better results with a "logical ground" near 5V (5V into one input, 5V+10mV into the other.)

#### MarkT

#8
##### May 04, 2011, 07:35 pm
The OP177 has a high spec - 25uV offset is about as good as it gets.  It is optimized for precision and designed to be used with dual rail supplies of +/-15V.    If you want this level of precision you may need split supplies - if you don't need this amount of precision a lower voltage single-rail opamp is what you need - there are plenty these days than can operate down to the negative supply on the input and output, but they are all compromises in design (to drive the output to a supply rail you will lose most of your bandwidth for instance).  Depends on your requirements.

The usual strategy is to generate an analog ground at half the supply to reference signals against.  If you want more inspiration read the op-amp sections in "The Art of Electronics" by Horowitz and Hill - very readable, very practical coverage of all these issues.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### giants_fan3

#9
##### May 10, 2011, 05:28 am
Thank you everyone; learned a lot from the various comments.

The solution for the OP177 opamp did turn out to be using dual-rail supply, +10V and -10V.
Got expected amplification; tested out with a known input from a potentiometer.

By the way, in order to convert, e.g., 5V Arduino provision to +10V and -10V for the opamp's dual-rail...
Here is the chip I used that does this job very well: MAX680
MAX680 is a chip that takes 5V and outputs +10 and -10V.

I've been studying this amp and other opamp theory in more detail, so do let me know if anyone uses this amp and has any problems!

#### cowasaki

#10
##### May 13, 2011, 08:45 am
Is there an alternative that doesn't require the split voltages?  The MAX680 isn't a cheap chip especially if you don't need to be particularly clean with the output eg detecting a noise using a mic etc eg using a couple of transistors to build a simple amp.

#### westfw

#11
##### May 13, 2011, 09:22 am
Quote
Is there an alternative that doesn't require the split voltages?

Probably.  It depends on the exact problem; in some cases you may get by with an op-amp that has "rail to rail inputs", and in other cases perhaps a "single supply instrumentation amplifier."

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