6
\$\begingroup\$

I am trying to build a basic 2 input summing amplifier using radio shacks 741 op amp. I have a feeling I damaged my IC chip through my wacky experimentation's. But lets run through the math just in case I am missing something. After all, I am new to this!

If I power the op amp with a voltage supply of +-5V and I set one input to 1.28V and the other input to 0 the voltage output should be -1.28V. Simple node voltage calculations and a PSpice simulation confirm this prediction. For some reason I am getting an output of 1.92 V without attaching anything to my inverted and non inverted inputs. So the 1.92 is coming directly from the supply. I was under the impression that the supply did not effect the sum. Where am I going wrong? Or is the chip busted, playing tricks and causing me severe electonxiety.

Any advice? Thanks in advance.

Turns out I can not add images, I am two points shy! Here is the link, sorry for the inconvenience.

http://www.itssimplydesign.com/summing_op_amp.jpg

\$\endgroup\$
  • 2
    \$\begingroup\$ Show your schematic, please. \$\endgroup\$ – Brian Carlton Oct 21 '11 at 23:23
  • \$\begingroup\$ The link to the schematic is up. \$\endgroup\$ – atomSmasher Oct 21 '11 at 23:39
  • 7
    \$\begingroup\$ basic 741 opamp answer: there's always a better op-amp to use \$\endgroup\$ – Jason S Oct 22 '11 at 2:23
  • \$\begingroup\$ Information only: An LM324 quad or LM358 dual opamp is about as good as a 741 and "more or less" operates from a single supply. Usually a good alternative in new designs. \$\endgroup\$ – Russell McMahon Oct 22 '11 at 2:40
6
\$\begingroup\$

I agree with Oli:

If you have another chip, try that to make sure this one is not blown.

If they both do the same thing, then this smells like a case of phase-reversal. Op-amps have a valid common-mode input range. In the case of an LM741 (datasheet from www.national.com), the +/- 15V supply range yields a guaranteed +/- 12V common-mode input range. Outside that range, it used to be common for the output gain to sometimes reverse sign, causing positive feedback rather than negative feedback and the output to saturate at the rail until it came back into its valid input range.

Modern op-amps are carefully designed to not have this problem, because it's confusing and nasty to debug. Which is one of several reasons why my tongue-in-cheek comment ("basic 741 opamp answer: there's always a better op-amp to use") is good advice. There are plenty of cheap and good op-amps that have much better performance specs than the 741.

This webpage seems to indicate the 741 doesn't suffer from phase-reversal -- the ones that do are JFETS -- so maybe it is just your chip that's busted -- but I'd use the LM358 instead.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your help! I am going to get a new chip tomorrow. I will be sure to check back in with an update. Thank you for the links! I am not sure about phase reversal, it seems like it could be the problem if I had more voltage going in. The max voltage I tried to send through the +- pins was 3.3V. I'm not sure how that relates to the bounds of the cheapo radio shack components! \$\endgroup\$ – atomSmasher Oct 22 '11 at 4:17
  • 1
    \$\begingroup\$ @atomSmasher - I strongly recommend, if you are buying things for hobbyist usage, you never buy just one of anything. Spare ICs are dirt cheap. \$\endgroup\$ – Connor Wolf Oct 22 '11 at 11:39
  • 1
    \$\begingroup\$ @Fake Name - I am starting to realize this! I am a new EE / CPE major who likes to try things out before they are covered. Sometimes things get messy. Believe me, from now on I will take your advice. \$\endgroup\$ – atomSmasher Oct 22 '11 at 14:41
  • 1
    \$\begingroup\$ "The max voltage I tried to send through the +- pins was 3.3V" -- DING DING DING -- we have a winner. Change it to +/- 9V from a pair of batteries and you should probably see a correctly working circuit. \$\endgroup\$ – Jason S Oct 22 '11 at 16:39
4
\$\begingroup\$

As Olin mentions, you should get -1.28V on the output with the circuit shown. With nothing attached to the two summing inputs, it should be 0V.

Just to be sure - when you say nothing attached to the inverting and non-inverting inputs, don't you mean the two summing inputs? (i.e. where V1 and V2 enter. The non-inverting input is the + pin with a 1k resistor to ground, if you don't have this attached then it is likely to do strange things)

As well as the small cap across the feedback resistor, adding a decoupling cap on both power pins (say 100nF from pin to ground) would be a good idea.

Check resistor values in case you have inadvertently added some gain by using e.g a 15k feedback resistor accidentally. Make sure you have the inverting and non-inverting inputs the right way round. Check for continuity and shorts and make sure the supply voltages are correct and stable.

If you have another chip, try that to make sure this one is not blown.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your help! I am going to get a new chip tomorrow. I will be sure to check back in with an update. When I mentioned nothing was attached I was referring to the summing inputs, which are currently running into my +- pins. The thought was if nothing was running into the +- pins there should be no output. But my voltmeter is reading well over a volt. \$\endgroup\$ – atomSmasher Oct 22 '11 at 4:14
2
\$\begingroup\$

If everything is set up right, you should get -1.28V on the output from the circuit as shown. However, there are differences between real opamps and theoretical ones.

First, make sure this opamp can run properly from 10V of supply. I think it can, but you need to check. It will also need some headroom, but you are operating with plenty on either side so that shouldn't be it.

Are you sure the opamp is stable? Add a small capacitor, like a few 10s of pF between the output and the inverting input as close to the chip as possible. Does that change the output voltage? If so, the output was oscillating before.

You should also have a bypass cap accross the power leads close to the chip.

Your schematic doesn't show pin numbers, so that's a easy thing to mess up when wiring. Add the pin numbers, then double check your wiring. Probe the pins right at the package to make sure they are the right voltage. The positive input should be 0V, and the negative input very close to it.

If all else fails, try another chip. Maybe this one got damaged.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the excellent suggestions. I hooked the network up to a 9V battery (the closest I can get to a 10V supply) and it did the same thing. However, I probed the chip at the package and it is reading 1.92 at the input. (with 5v and 9v) But I don't have a source hooked up to the output! The smallest capacitor I have is 1000The 5v is the only source hooked up. 3.3 microF. Ill try and pick up another chip tomorrow and test it out. I will let you know what happens. Thanks again. \$\endgroup\$ – atomSmasher Oct 22 '11 at 0:09
  • \$\begingroup\$ @atomSmasher - I'm a bit confused with your exact setup now - how exactly did you hook the 9V battery up? Are you using 9V for one rail and 5V for the other? What do you mean by "source hooked up to the output"? and "5V is the only source hooked up"? What pin to? It would help to understand if you adjust the schematic to your exact circuit setup with pin numbers and any relevant voltage readings/observations noted. \$\endgroup\$ – Oli Glaser Oct 22 '11 at 3:16
  • \$\begingroup\$ I have tried several configurations in an attempt to get a different output reading. The main issue arose when I tried to amplify a 3.3V (invert and noninvert pins) dc signal using a +-5V voltage supply(Vin and Vout). When I measured the voltage on the output it was reading 1.92V. So I disconnected the 3.3V signal and 10k resistors and measured the voltage of the output again, it still read 1.92V. I thought that maybe my supply was too small, so I tried hooking up a nine volt battery in place of the 5V supply, still had the same results. \$\endgroup\$ – atomSmasher Oct 22 '11 at 4:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.