Why Does My Op Amp Only Amplify by a Factor of 5 Despite Changing Resistors?

[yungman]

First let me thank all of you, all of you are very kind too me;) and sorry for my bad english;)
as I said before I am busy a little bit and I can't work on my project these pas days, I will continue on Saturday, so please give some time to test your ideas and tell you the results;)
and about mr.yungman's comment, let tell you about the circuit, (I don't want to hide anything, I appreciate your help) the circuit is an ultrasonic receiver, we have another circuit that is transmitter, it works perfectly and it sends ultrasonic waves,it has a transducer too
receiver circuit detect these ultrasonic waves(so its input is only 40 khz not a band) first we amplify input,then with two diodes and one capacitor we convert it DC voltage(I am not sure that convert is right verb for here!) after that we have a comparator, CA3140, it compares pin 2 with 3 and if 3 is bigger than 2 its output(pin6) becomes 6.8V .after that we have 2stages of transistors to amplify current because we want to drive a relay so we need high enough current.
about the transducer, it does not have datasheet, I asked someone and he told me this, so I can't help you with this. is there any question yet?!

If you don't have the data sheet, then we have to reverse engineering a little. Follow my post on #29. I want you to tell me what is R1 and R2 now. Then change R2 to 1/10 the value and tell me the output of the first amp. Then change to 1/20 value and tell me the output. From that, I can tell what is the transducer output.

You need to get a new op-amp. OP37 is a place to start. When you see signal at pin 2 of the op-amp, that is a tell tale sign your op-amp is running out of steam.

Regarding to noise. I want you to verify where the noise come from.
1) Disconnect the transducer, then short the input on the capacitor side to ground to verify whether the noise is still there. If so, the noise is from the electronics. I don't think so from my experience.
2) Re-connect the transducer. This time cover the transducer so it does not detect any signal. Look at the noise. If the noise is still there just as strong, your noise source is the transducer. Again I don't think so.

IF both 1 and 2 are negative, your problem is the environment. The only thing you can do is using band pass filter. I see you are rectifying the signal, so phase is not important. There is a lot you can do to filter. We use parallel LC tank at the front end and tune at 40KHz.

Like Jim Hardy said, why do you limit yourself to 9V? Is it a hand held device?

Just do all these first, get you amp working, characterize the detector. Then we work on the noise. It should be easy.

 

[yungman]

skeptic, I am sorry but I don't understand what you mean by power supply noise?! how can I check this?!

We don't know what your power supply is at this point. If you are using a battery, don't worry about it. But if you are using any other way to power up the circuit, you need to use a scope probe to probe at the 9V point and look at what kind of noise you have.

 

[jim hardy]

"If you are using a battery, don't worry about it."

but tack a capacitor across it just in case. Physically close to the 741's.
A couple microfarads should do. Even just another 0.47 would be okay.

 

[skeptic2]

skeptic, I am sorry but I don't understand what you mean by power supply noise?! how can I check this?!

Because you gave your voltage measurements in volts peak to peak, I assume you were using a scope. Just switch the input to AC and put the probe on the power supply +9V. Turn the voltage knob (cautiously) to a sensitive level (5 mV peak to peak). Does the trace look noisy? If it does, you might add a forward biased diode in series and a large capacitor after it in parallel. Other smaller capacitors can be added right at the op amp supply terminals.

It sounds like an interesting project. I hope you'll let us know when you get it working how far your range is.

 

[nadersb]

gentlemen, some new data: :D
I changed power supply to 20 and here is the results:
op1: pin2:26mV pin6:540mV so gain is:20.76
op2: pin2:500mV pin6:6.1V so gain is:12.2
and with 9V on the power supply we have these voltages:
op1: pin2:24mv pin6:355mV so gain is: 14.79
op2: pin2:330mV pin6:4V so gain is:12.12

about mr.yungman comment I changed the R2 resistor to 1/10 and 1/20 value and I connected transducer instead of signal generator and here is the results:
R2=150k op1:pin2=240mV pin6=3.6V
R2=75k op1: pin2=240mV pin6=2.2V

 

[nadersb]

and about noise:
I checked DC power supply with o'scope and it doesn't look noisy, it works perfectly.(it is not battery it is a digital DC power supply)
as you told me I shorted the input to the ground and there was no noise, after that I connected the transducer but I covered it with my hands, still there was no noise, but without the cover, we had noise! its frequency is about 400Hz
so you tell me that add a parallel LC tank at the end of the pin 6 of the second op amp?!

 

[nadersb]

I don't change the op amp yet, but tomorrow I will. let see what happens if I use op37.
again thank you all for paying attention:)

 

[skeptic2]

so you tell me that add a parallel LC tank at the end of the pin 6 of the second op amp?!

The best place to put a parallel LC tank is in place of the feedback resistor of op1.

 

[yungman]

gentlemen, some new data: :D
I changed power supply to 20 and here is the results:
op1: pin2:26mV pin6:540mV so gain is:20.76
op2: pin2:500mV pin6:6.1V so gain is:12.2
and with 9V on the power supply we have these voltages:
op1: pin2:24mv pin6:355mV so gain is: 14.79
op2: pin2:330mV pin6:4V so gain is:12.12

about mr.yungman comment I changed the R2 resistor to 1/10 and 1/20 value and I connected transducer instead of signal generator and here is the results:
R2=150k op1:pin2=240mV pin6=3.6V
R2=75k op1: pin2=240mV pin6=2.2V

Are you sure you measure on the first op-amp? Something is really really wrong in the data. Pin 2 is the summing junction. It should read 0V if the amp is running in closed loop under normal condition. As the frequency approach the amp limit, you'll start to see voltage on pin2 because the loop gain is dropping.

It is normal to see in the original measure of 26mV in the original R2 as the amp run out of steam. But when you change R2, the feed back resistor from 1.5M to 150K, you decrease the closed loop gain by 10. Closed loop frequency response should increase by 10 times which from my estimate is 100KHz. You should see less voltage at pin 2. and you get lower voltage at pin 6 of op-amp 1. But your data show opposite. Check your original R2 to verify it is 1.5M. You really need to ohm out your complete circuit to make sure all the value is correct, all the connection is correct.

We cannot work on a circuit that has some major problem. Make sure R2 we are talking about is the resistor connected from pin 2 to pin 6 of the first op-amp. Make sure the original R2 is 1.5M and the new one you replace is 150K and 75K...by using an ohmeter after you lift one end up( no connection). I don't know how you breadboard the circuit, if you turn the IC up side down, you might miscount the direction of the pins. Make sure you double check everything. Far as I concern, we are back to square one again.

 

[nadersb]

no we DON'T! I think you misunderstand sir! when I changed R2 to 150k I also changed AC supply, AS I wrote in my previous comments! with 1.5M we used signal generator with around25 mV but with 150k and 75 k we used transducer! as I said we have another circuit that is transmitter, if you bring these two transducers near to each other you will see strong input signal, so what is the problem?! I don't understnad

 

[nadersb]

I am not as amateur as you think sir, I know which pin is which! and how could be possible that our circuit is wrong and we have wrong connectios but we have good gain?! is it miracle or something?! :D with all due respect I think there is something wrong with your theory sir!

 

[yungman]

I am not as amateur as you think sir, I know which pin is which! and how could be possible that our circuit is wrong and we have wrong connectios but we have good gain?! is it miracle or something?! :D with all due respect I think there is something wrong with your theory sir!

If you are experienced, I would expect you to know to use the same circuit to compare. You gave two result from two totally different condition and expect others to help you. People here are trying VERY HARD to help you and I think your comment is very uncalled for. Three pages into this and you are not going very far.

I know my op-amp theory very well, just trying to help you only. I think someone else is more suitable to help you.

 

[skeptic2]

Before we get into an argument over nothing, could you post your schematic nadersb, including the points where you are making measurements and the values?

 

[jim hardy]

Yungman has a point.

So much signal voltage at pin 2 summing junction says it's not operating as an operational amplifier.
HOWEVER - it is bad practice to measure at the summing junction. Scope lead capacitance delays feedback wrecking your phase margin. It is possible in fact likely that measuring pin 2 upsets the opamp so badly you are getting bogus numbers. As soon as you connect the test prod to pin 2 the circuit changes - test leads have capacitance(especially 1X scope prods) and opamps detest capacitance at their summing junctions.

Perhaps this disagreement is a simple misunderstanding.

When you say "signal at pin 2" are you reading directly ON pin 2,
or are you readig ON LEFT SIDE of the 10K resistor that GOES to pin 2?

There's a BIG difference. It's okay to measure left side of that resistor but not right side.

It is possible that your scope probe, when connected to pin 2, is disrupting the circuit throwing it into oscillation and giving a false reading. Capacitance at summng junction has that effect. When you remove the probe the circuit reverts back to normal.

You might clarify which you're measuring, and if you ARE measuring right on pin 2, watch output voltage at same time (other scope channel or a second voltmeter)and i bet it goes haywire while your scope is connected.
Dont measure pin 2 , measure opposite end of that 10K instead.

Gains should be reported as (voltage on pin 6)/(voltage on LEFT side of resistor connected to pin 2) . Not (pin 6) / (pin 2) because pin 2 is the op-amp's most ticklish spot.

A sanity check on pin 3 would be a good idea just to make sure the capacitor is not allowing any AC there. Measring pin 3 might upset the opamp a little but the cap should hold it reasonably still. If you see more than a couple millivolts, might be a board layout problem. Arrange your signal common so that the cap on pin 3 is commoned real close to transducer and far from power supply negative lead..

I hope you guys find this is indeed a mis-communication because i am interested in learning from all of you.

That's another advantage of building something instead of simulating iit, you develop real world skill with real world test equipment.

hope this helps.

I found it interesting that the gain increased when you raised the voltage.
But i think we have two or three things going on in your circuit.
How good we'll all feel when we've got them straightened out!
As one of my old country Math professors used to say - "once you get the problems nailed down you can usually stomp them to death"

Keep at it, for Mother Nature loves to make us work hard for the lessons she teaches us .

old jim

 

[minijh]

Hi every one! New player is in! :D I'm Nader's teammate on this little project. He is on a trip so I'm taking his turn.
First I wanted to thank everyone including "yungman", "skeptic2" & Jim. For sophomore students of EE -who have been busy analyzing complex circuits on paper or via simulation- getting involved with circuits in the real world (!) seems a little different; Actually more than a little! ;) So thanks a lot for helping us learning & having fun in combination. I guess I better apologize for any misunderstanding unwittingly occurred. As Sceptic wisely mentioned it was over nothing! :)

 

[minijh]

@Jim:
Nice points you mentioned abt pin 2 of opamps indeeeeeeed! But I have this question: when you say "signal on pin2 means our opamp isn't in its operational state", How strong a signal should be to say so? I mean is a signal with p-p of 10 mvs considered as "no signal"? :D
About what you asked about our measurements, as you asked us to measure pin 2 we did so,directly on pin2! :D Reading before the resistor is just equivalent to reading pin6 of the previous opamp in ac analysis ?! Correct me if I’m wrong :)
I can’t say anything but I just loved what you said & I hope the misunderstanding is over:
“I hope you guys find this is indeed a mis-communication because i am interested in learning from all of you.
That's another advantage of building something instead of simulating iit, you develop real world skill with real world test equipment.”

 

[jim hardy]

" Reading before the resistor is just equivalent to reading pin6 of the previous opamp in ac analysis ?! Correct me if I’m wrong :)""
"

welll for all practical purposes, Yes.
i see a capacitor in between pin 6 and 10K resistor so the DC value could be different on pin 6 side of that cap, but in your circuit only by a few millivolts. EDIT: And you did say ac analysis... my bad.

""About what you asked about our measurements, as you asked us to measure pin 2 we did so,directly on pin2!"
oops we should have said "signal to pin 2 measured at junction of 10K and cap"



""But I have this question: when you say "signal on pin2 means our opamp isn't in its operational state", How strong a signal should be to say so? I mean is a signal with p-p of 10 mvs considered as "no signal"? ""

let's think for a moment.
Thinking is exact and measurement is approximate.
The basic principle of the opamp is that it will hold the two inputs equal.
That is true for both inverting (your arrangement) and non-inverting (where input signal goes to pin 3 not 2).

In an exact world, your circuit would hold pin 3 at exactly 4.5 volts, with no variations because of the RC filter 10K and the capacitor.
The opamp would attempt to hold pin 2 at exactly the same voltage.
In a real world there is likely some teeny bit of signal frequency showing up at pin 3 from parasitic capacitances or ground loops or something. The opamp will obediently try to reproduce that at pin 2.
So look at pin 3, it's okay to put the scope there because of that big cap already on pin 3. If you see 10 mv there the opamp is entitled to put 10 mv on pin 2 as well. But you can't look at pin 2 for reasons already mentioned.
I will be surprised if there's as much as a millivolt at pin 3, let alone ten millivolts.

Now more directly to your question:
return to thinking mode:
What is open loop gain of opamp?
something like 10^6 at low frequency.
So in an inverting configuration, the signal at pin 2 should be [output signal] / 10^6
which is going to be in microvolts.
Remember you can only get 4v of signal in your circuit, and to produce that takes only 4/10^6 which is 4 microvolts at pin 2.
If you have more, the circuit is failing to be an ideal inverting amplifier.

As you raise frequency the opamp's gain falls off so the voltage at pin 2 required to produce same output swing becomes larger. Read that appnote on slew rate limiting.

i too hope the misunderstanding is over.

Let us know how it works measuring on upstream side of 10K at pin 2.
And whether gain is affected by supply voltage.



Good luck with your project.
Yungman and i both think your 741 is not quite up to the task of high gain at 40khz.
Here's a faster opamp, two to a package and widely available from hobbyist sites:
http://www.national.com/profile/snip.cgi/openDS=TL082
be aware you may need to raise power supply to 12 volts.
should give you better gain. There's way faster ones yet out there.

We learn by doing. Hats off to you guys.

And thanks for the kind words.

 

[minijh]

1) Thanks a lot for clarification Jim, I'll test the circuit just the way you said & I'll inform you.

2) "Now more directly to your question:
return to thinking mode:
What is open loop gain of opamp?
something like 10^6 at low frequency.
So in an inverting configuration, the signal at pin 2 should be [output signal] / 10^6
which is going to be in microvolts.
Remember you can only get 4v of signal in your circuit, and to produce that takes only 4/10^6 which is 4 microvolts at pin 2.
If you have more, the circuit is failing to be an ideal inverting amplifier."

What you just said shot the 741 in head! :D we thought it would be better to get rid of him as soon as possible! Poor 741! so we went for some other ops including op37 as yungman advised. Do we have to change anything? I mean R1 or R2 or..? because when we substitute 741 with the new fellow, the circuit wrecked badly!

3) By the way, after your advice we tuned the dc supply up to 20 volts, with 30mvs of signal generator as the input, we got near 6 volts out!
4) Nice technics Skeptic & Yungman said abt searching for the source of noise! we checked them all, as Yungman wisely predicted it was the environment. We also tried to detect its frequency with oscope, it was some 200Hz. What do you suggest for out LC Tank filter? For maximum efficiency of the circuit-maximum board-, what would be the best frequency to put the bandwidth around? To Strongly pass 40ks & kill 200s?

5) Hats off to you Jim! You've been extraordinary helpful! :)

P.S: @Skeptic: Sry for my delay, Now that everything got crystal clear I'll check for all the pins of the circuit & get you know of it. Our laboratory was closed these days,I have to wait till tomorrow. :)

 

[skeptic2]

What you just said shot the 741 in head! :D we thought it would be better to get rid of him as soon as possible! Poor 741! so we went for some other ops including op37 as yungman advised. Do we have to change anything? I mean R1 or R2 or..? because when we substitute 741 with the new fellow, the circuit wrecked badly!

3) By the way, after your advice we tuned the dc supply up to 20 volts, with 30mvs of signal generator as the input, we got near 6 volts out!

4) What do you suggest for out LC Tank filter? For maximum efficiency of the circuit-maximum board-, what would be the best frequency to put the bandwidth around? To Strongly pass 40ks & kill 200s?

What do you mean, "...the circuit wrecked badly"? What did the circuit do?

Is the 6 volts out at the output of the first op amp or the second? If the second, what is the output from the first op amp? Are the outputs clean sinewaves or are they distorted or noisy?

You might start with a tank circuit with 100uH and 160nF. My preference is to strongly pass 40 kHz, that way you eliminate everything you don't want.

 

[minijh]

@Skeptic2:
I meant there was no amplification, like when you use ops out of order or something like that. :-?? We checked for some 2 more op37 but still same result. So the problem couldn't be the op-amps themselves.

6volt out the circuit, output of the second one. I'll check tomorrow for sure, I mean in less than 12 hours! :D

Thanks for your comment on filtering.

Tomorrow!The big day!
:)

 

[skeptic2]

The only problem with using a tank circuit is its low impedance. For instance with the values I gave you the Q of the inductor will be under 30. Since the reactance of the inductor is 25 ohms, the maximum impedance of the tank circuit would be 750 ohms. That would mean to get a gain of 25 you'd need to replace the 10 K resistor at the input with a 30 ohm resistor.

 

[minijh]

Finally, I got into laboratory & took this data out the circuit before they threw me out of there! :D

@Jim:I checked for pin 3 in ac status, there was some like 0.1 mv p-p! as you had told me so, earlier :)

op741/ DCSup=20v/ Vin(p-p)=30mv/ R(feedback)op1=150k

op1:
pin2(B4 Resistor):30 mv (As expected! :D)
pin3:10 v
pin6:320 mv (ac)
=> gain=10.6

op2:
pin2(B4 Resistor):320 mv (as expected)
pin3:10 v (not directly measured)
pin6:5 v (ac)
=> gain=15.62

 

[minijh]

What takes attention is the difference in gain of each stage. It seems stage 1's got bugs down there! :D

 

[jim hardy]

To see your progress is very heartening. Thanks!


first stage gain is ten , lower than desired;; and
second is fifteen, higher than desired ?

okay first things first
First stage
i'd wager first stage is slew rate limited.
Will your signal generator allow you to drive with a square wave instead of sine?
Try that and look at output. The more it resembles a sawtooth od triangle wave, the more you are seeing the opamp limited by how fast it is capable of moving its output.
Repeat but with a 400 hz squarewave not 40 khz, output should be a more faithful reproduction.

Second stage has more input swing so can drive output at a faster slew rate.
But why so much gain? are you still using 100K / 1K there?
any green stripes on those second stage resistors?
brown black orange = 10K
brown black yellow = 100K
brown green yellow = 150K
brown black green = 1meg
brown green green = 1.5 meg

and if you are using 1% resistors there's a third stripe for third digit before the multiplier band...
------------

did you say the circuit 'wrecked' with op37?
that looks like a wonderful opamp.
http://www.analog.com/en/all-operat...amplifiers-op-amps/op37/products/product.html
Fast opamps are sensitive to capacitance at their inverting input and board layout(long leads) can get you.

See if a few picofarad capacitor in parallel with your 1 meg feedback resistor settles it down.
Tack one across the .47 that's on pin 3 also - capacitors that are made from a rolled up foil have some parasitic inductance so need a high frequency bypass. I don't know what style is your 0.47uf .

old jim

 

[skeptic2]

I suspect that if you are using a 741 for the 2nd op amp and are getting 5V pk-pk out that you are seeing a triangle wave instead of a sine wave. That is a sign you are slew rate limited. That shouldn't be true of op amp 1 though. You must have some other problem with op amp 1.

 

[minijh]

Good news every one! Using your advice, we checked OP-37 for our circuit- but we had to change resistors- and we got 9 volt out!cheers! :)
(what we did was playing with resistors, triggering the signal and here are the results:)

DC Sup=12/OP-37/Vin(p-p)=25 mv

OP1
R1=10k/R2=180k
pin2=25 mv
pin3=6v
pin6=0.5 v

OP2
R1=100k/R2=1.8M
pin2=0.5v
pin3=6v
pin6=9v

As we have a relay to be turned on, we took 9v dc out with two diodes and a capacitor. Regarding this, Is it important to check for slew rate or something?
*by the way as Jim said, I checked the signals for they were sines, just as signal generator produced.

What we have achieved in our project up to now:
1. Producing 40KHz pulses with p-p 0f 5v and send via a transducer(transmitter),
2. Receive the transmitted signal within a second transducer (receiver),
3. Amplifying with gain of some360,
4. Rectifying the signal,
5. Detecting the signal using a comparator3140,
6. Amplifying current the relay needs,
7. Tik Toks on relay! :)
8. Do whatever you want we relay's output...

There is only one thing left for better efficiency, and that is filtering. As you have suggested before it is better to do so before amplification; using a band-pass filter. We need low image impedance there -Correct me if I’m wrong- We checked the Rin of our circuit with an ohmmeter turning off the supplies & we got some 22Ks, Isn’t it a little big for our purpose?
Thanks every one
:)

 

[skeptic2]

1. You could make 1 op amp a low pass active filter with gain and the other one a high pass active filter with gain.

2. You could use a Twin T notch filter in the feedback loop of the 1st op amp which would give you a bandpass characteristic.

3 There are probably other active bandpass filter designs on the internet.

 

[jim hardy]

congratulations on your success !

Hmmmm - Active filters are fascinating because they are such a direct application of math.
They are a whole field unto themselves.
My experience is limited, Yungman is FAR more conversant than i.

Once upon a time i needed sharp filters with gain
and stumbled across this IC
and built my filters (22 and 27 khz) , measured Q almost 100 had to tweak down to ~50 to get bandwidth to cover the FM signals i was decoding. (Telephone touch tones FM'd onto ultrasonic carriers. )

In other words:
"Here's a great filter that can be built by an amateur"
http://www.national.com/ds/LM/LM359.pdf
the Biquad on page 22/23
and someplace National has an appnote on active filters with a discussion of this one.
AN72 mentions it but i think there's a better one somewhere in their library.
http://www.national.com/an/AN/AN-72.pdf

it's called 'biquad' because both numerator and denomoinator of transfer function have a quadratic in them
so there's poles and zeroes galore to tinker with.
The formulas in the datasheet work.
I hand picked my resistors and capacitors to achieve the values called for by formulas.
worked quite well.

Just a thought...

old jim

ps keep your eyes peeled for those old 1970's catalogs - paperbacks several inches thick.
National, Signetics, RCA, TI et al
Packed with information, they are. You find them in junkshops and Ebay.