Op amp gain

1. Sep 4, 2011

hello,
I have a strange problem!
I wanted to amplify my signal and i used 2op amps, each of them has a feedback resistor between (-) pin and output( pin 2 and 6 in 741) the ratio of resistors(r2/r1) is 150, but my signal only amplify 5times! I change the R2 resistor but nothing changed!! i put different resistors but all of them had the same results!! I changed my op amp but still the same results! the second op amp worked perfectly but the first stage has this problem!
does anybody have an idea whats wrong with the op amp?! you can see the circuit in the link below:( the value of capacitors are 0.47uf and the first feedback resistor is not 1M it is 1.5M)
http://www.2shared.com/photo/a_JmZITS/Untitled.html

2. Sep 4, 2011

skeptic2

With no signal input, what is the DC voltage at each of the op amp outputs?

3. Sep 4, 2011

after a few second they become zero.

4. Sep 4, 2011

yungman

I don't see anything wrong of the circuit. Basically it is powered by 9V and the +ve input biased at 4.5V midway. All stages are AC coupled so no DC offset problem to talk about.

1) First posibility is: My question is what is you frequency input to see only gain of 5. Remember 740 has gain bandwidth of 1MHz, if you have first stage gain of 150, the -3dB bandwidth is only 1MHz/150 which is less than 10KHz. The circuit talk about Ultra Sound signals, what is the frequency?

The second stage has closed loop gain of 10, the frequency response using a 741 is about 100KHz so it is very different.

Give me the input frequency before we talk more.

2) What is the output impedance of the detector/receiver at the left side of the schematic that drive the first op-amp? It is a current devices like a detector diode or photo detector? If so, your analysis is wrong. That is a current devices and the first stage is actually a TRANSIMPEDANCE amp with the transimpedance gain of 1M ohm or 1uA per volt. There is inconsistency with this theory as you change the R2 and nothing happened. But don't dismiss this just yet because it might still be a combination of bandwidth limited and detector characteristic. Please give me the kind of detector you use to be sure.

Last edited: Sep 4, 2011
5. Sep 4, 2011

skeptic2

If your output becomes zero after a few seconds then perhaps your 4.5 V bias is the problem. Can you check pin 3 on both op amps and see if they have 4.5 V on them? I suspect your circuit isn't wired exactly like the diagram.

Once you are able to get a constant 4.5 V at the output of each op amp, why don't you try a gain of 15 instead of 150 and see what happens?

6. Sep 5, 2011

I use ultrasound transducer, and input signal is 40khz with 30 mv amplitude(pick to pick)
there is a 33ohm resistor parallel with it as you see in the image that i have sent.
what if I change the op amp, and try different model that is compatible with this frequency?! do you have any suggestion which model would be suitable?!

7. Sep 5, 2011

I double checked my connections, and I dont think that (+) pin's bias has any problem, but I will check it again and I will inform you ;)

8. Sep 5, 2011

skeptic2

How did you decide on 33 ohms? What happens to your gain if you replace it with 100k ohms? What values are your capacitors? If your signal is 40 kHz, try using a 270k feedback resistor for each op amp.

9. Sep 5, 2011

because we use signal generator instead of transducer most of the time and its resistor is 50 ohm so we chose a resistor that is near to 50 ohm .
each capacitor is 0.47uF.
I didnt change the value of input resistor, I will change it and inform you what happens;)

10. Sep 5, 2011

skeptic2

I understand now about the 33 ohm resistor. Still I would prefer a value of 100k or higher at least when using a transducer. The lower the value, the more it will lower the Q and sensitivity of the transducer. A gain higher than you need can cause the op amp to oscillate if the right phase shift occurs. A good rule of thumb is that 60 dB of gain is the most you can achieve on the same board at the same frequency. It is possible to get positive feedback through the power supply, ground or other connections with gains that high.

It still is important that you have 4.5 V at the output of each op amp with no signal in.

Last edited: Sep 5, 2011
11. Sep 5, 2011

yungman

I don't see 33ohm, can you double check, I cannot see the entire picture of the signal detector.

40KHz is too high for 741 with gain of 100. Maybe try something like OP-37 or others that has higher gain bandwidth product. But I think you have bigger problem than that.

Did you say you read 0V at the output of the both op-amps(pin 6)? If so, that is wrong, you should get 4.5V on both. I think Skeptic2 is doing a good job helping you, I don't think you need two people telling you what to do. Good luck.

12. Sep 5, 2011

I will check pin 3, but what if pin 3 is 4.5?! what else could be the problem for zero output?!
and last thing sir, what do you mean by "bigger problem"?! what is it?!
thank you

13. Sep 5, 2011

jim hardy

i know this sounds dumb but ---

National Semconductor's engineers used to field this problem so often they put it into an application note.........

"how to" schematics usually dont show power connections to the IC
and often newbies dont connect power because it's not shown

and i notice that circuit linked doesn't show power to pins 4&7
and it's easy to mis-wire a round can opamp when you're looking at it from the bottom and using a top view wiring diagram
http://en.wikipedia.org/wiki/File:LM741_Pinout_Round.svg

dont be offended by this dumb question its just i have no idea as to your level of expertise.
i'm not offended if you ignore this post. You can even resent me it's okay.

lastly what is your source for 9V?
If it's a small battery place a few microfarads across it - reason is opamps demand a low-impedance power supply which a BL006P battery is not.

as yungman observed skeptic is doing a good job with technical advice.

but I dont mind asking the real dumb questions.
hope it doesn't offend you guys
and i wont meddle further

old jim

14. Sep 5, 2011

:D
dont worry I connected 7 and 4 to Vcc and ground;) I am working with a Dc voltage source.
I will be thankful for your ideas ;)

15. Sep 5, 2011

yungman

What I meant is you have some basic wiring problem. You should get 4.5V at the output on both of the op-amp. If it float to 0V, something is very wrong. I suggest you check your wiring and check for cold solder joint. Get the 4.5V on both output before you worry about the gain. Measure pin 3 is 4.5V, pin 2 should follow pin 6 because output is connect to pin 2 through R2.

Measure and verify pin 7 is 9V and pin 4 is 0V. After you get the correct DC reading, if Skeptic is not around, I'll keep helping you.

16. Sep 6, 2011

I checked it today and pin 6 is around 4.5 in both op amps;) so this problem is solved;) I also changed the input resistor( 33ohm) and it worked, I mean when I used 33ohm ac voltage of pin 2 was about 10 mV pick to pick but now we have much more better signal and its about 25pick to pick, so thank you for your suggestion, but I didnt change the op amps yet, when I do it I will inform you;)
thank you

17. Sep 6, 2011

skeptic2

With the new values how much signal are you getting out of the op amps?

18. Sep 6, 2011

about 4 V pick to pick, but the problem is we amplify noise too, and i think we should add a filter to our circuit!

19. Sep 6, 2011

yungman

I really don't understand what is pick to pick!!! When the op-amp is working, you really should not see any signal on pin 2. It is a summing junction. What I want you to tell me is what is the input to R1, what is the output at pin 6 of the first op-amp.

If your input frequency is 40KHz, if you are using 741 of gain bandwidth product of 1MHz, the -3dB point is about 10KHz. Above 10KHz, the gain will roll off at 6dB per octave. 40KHz is two octave above 10KHz, so your gain will go down 12dB which result of a gain of 25 instead of 100. This mean if you have 1mV at the input of R1, you will have 25mV at pin 6 of the first op-amp.

I do not see the 33ohm in your diagram, so I don't know that part. One thing is the way your circuit is, you are assuming that the output impedance of the detector is very low impedance. If the output impedance is high, your gain is

$$A_{cl}=\frac{R_2}{R_1+R_{det}} \;\hbox { instead of }\; \frac {R_2}{R_1}$$

So the whole thing is not as simple as you think. First you have to show me what kind of detector you use as I asked on my first post. And show me the complete circuit where is the 33ohm.

20. Sep 6, 2011

Averagesupernova

What I fail to understand is why we haven't switched to a better op-amp than the 741. Get a good op-amp and then switch back and forth. It will be a good lesson on frequency response.