Have set up a voltage divider inputting 1 volt into a LM741 op amp to act as a voltage follower. For some reason I get 2 volts at the output. Have checked everything I can think of and can’t figure out why.
\$\begingroup\$
\$\endgroup\$
3
-
7\$\begingroup\$ WHY are people still using the LM741? It was old and out-specced by 10-cent op-amps in 1990! \$\endgroup\$– Connor WolfCommented Dec 23, 2013 at 2:56
-
\$\begingroup\$ What's the uA741's appeal? \$\endgroup\$– RedGrittyBrickCommented Dec 23, 2013 at 13:24
-
\$\begingroup\$ Because as a beginner I saw it in just about every book and beginner type article so that's what I used. Now I know its not the correct type for me I have moved on \$\endgroup\$– bigshopCommented Feb 24, 2014 at 18:52
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
8
This is an easy problem to understand if you look at the datasheet.
The LM741 needs headroom at each rail to operate. What this is saying is for +/- 15V supplies, the output could swing as little as +/- 12V. If you're running it single supply like that, I guarantee it's slamming the negative rail. Upgrade to a modern op amp. For this application, I would look for something CMOS with a rail to rail output. A couple examples: TLC2272, LMC6462.
answered Dec 23, 2013 at 2:09
-
\$\begingroup\$ Thanks for your reply. Im very much a beginner so could you expand on your answer some more. Can you suggest what better type of op-amp I might look at?. Im using this to reduce the voltages going into my daq card. \$\endgroup\$– bigshopCommented Dec 23, 2013 at 2:16
-
\$\begingroup\$ Then take the op amp out entirely. Your DAQ card most likely has a high impedance input. \$\endgroup\$ Commented Dec 23, 2013 at 2:24
-
\$\begingroup\$ My daq card is only 144k input impedance and it loads the signal too much that's why I need to add the follower. \$\endgroup\$– bigshopCommented Dec 23, 2013 at 2:30
-
2\$\begingroup\$ You should really edit this question to reflect what you're actually trying to do. There is probably a better way to do this than attenuating a signal by a factor of 100. \$\endgroup\$ Commented Dec 23, 2013 at 2:39
-
1\$\begingroup\$ Im measuring DC signals. Have changed my resistors now to attenuate 10:1. After reading Matt Young answers I now see why my circuit would not work. I now know I have the wrong type of op-amp for my application. By changing the power supply to the current op-amp to +10v and -10v I can now get the expected result. \$\endgroup\$– bigshopCommented Dec 23, 2013 at 5:29