Op Amps: Difference between the + and - inputs. - Page 1
Forum section: electronics / beginners

author: ruku
author_rank: Contributor
timestamp: July 05, 2011, 06:22:31 pm
content: So I'm reading through TI's excellent "Op Amps for Everyone" in attempts to better my skills with Op Amps. I do have one question though. I entirely don't understand the difference between the + and - input terminals.So I can work that out no problem, but I don't understand why the - input has to be on top. Shouldn't you get the same result with the + on top?

author: Unknown
author_rank: Unknown
timestamp: July 05, 2011, 06:26:45 pm
content: Read into the term 'positive feedback'.

author: johnmx
author_rank: Frequent Contributor
timestamp: July 05, 2011, 06:59:21 pm
content: It’s very simple. An opamp will change its output until both inputs are at the same potential, i.e. the potential differential between + and – terminals is zero. And the output follows the following rules:-if + terminal is more positive than the – terminal, the output will raise;-if - terminal is more positive than the + terminal, the output will fall;

author: Zero999
author_rank: Super Contributor
timestamp: July 05, 2011, 07:00:59 pm
content: Look up Schmitt trigger.

author: ruku
author_rank: Contributor
timestamp: July 05, 2011, 07:03:42 pm
content: Alex and Hero999, I did a quick read. Thanks for pointing me in the right direction. johnmx, I'm still chewing on that one...So when the feedback goes to the positive terminal, it adds to the response to get more. When the feedback goes to the negative terminal, it takes away from the response to get less.I just looked over all the applications of Op Amps and realized that the ones pertaining to analog signals used the - terminal. That's so the op amp's don't ring or become unstable when handling signals?Also, the Schmitt triggers and oscillators use the positive terminal for feedback. I think I get it now, but the "ideal op amp rules" are messing with my head. Theoretically, shouldn't the circuit in my first post behave the exact same way with the terminals switched?

author: jahonen
author_rank: Super Contributor
timestamp: July 05, 2011, 07:27:49 pm
content: The very idea behind an opamp is such that we have fairly crude amplifier (with relatively high distortion and high output impedance) with very much higher gain than we finally need. Negative feedback is then applied to reduce the gain to required level. At same time the gain drops, distortion, output impedance and other non-linearities reduce.You can't just switch + and - inputs, you'll need negative feedback to get linear output. For math inclined people, this is quite obvious when thinking about basic formula of the opamp, V_out = G_raw*(V_i+ - V_i-). If you apply some circuit theory (calculate V_out when R_g and R_f are present) and take a limit of output voltage when G_raw approaches infinity (an useful exercise), you'll get a familiar formula for an inverting amplifier, V_out = -U_in* R_f/R_g. If you make it other way around, the result will be infinite instead.Regards,Janne

author: ejeffrey
author_rank: Super Contributor
timestamp: July 05, 2011, 08:56:20 pm
content: The idea of an op-amp is that you treat it in the limit of infinite gain. If the output is finite (i.e., in range), the input difference must be zero. This part works whether you have the + / - terminals the right way around or not. Both versions are in 'equilibrium' -- the input to the amplifier is zero, so it doesn't change the output.However, if the input terminals are reversed, you have an unstable equilibrium. Like a pencil standing on its point, any tiny fluctuation causes the error to continuously increase, and the device will 'hit the rail' (or oscillate). In your example circuit, if you flipped the + and - inputs, a slight increase to the signal input would increase the voltage on the positive input terminal. That would cause the output voltage to rise. This causes the amplifiers + terminal voltage to rise further, leading to ... and then the output is stuck outputing the maximum voltage possible. Likewise, if the fluctuation is negative, the output will end up at the most negative voltage possible.So, regardless of whether your op-amp circuit is inverting or non-inverting with respect to the input signal, the feedback signal always has to go to the negative terminal.Positive feedback is usually undesirable. At DC, it causes amplifiers to immediately slam into one rail or the other. If the feedback only happens at high frequency, perhaps due to unwanted capacitive coupling, it causes the amplifier to oscillate. So positive feedback is used when you want something to switch very fast and cleanly (the schmitt trigger) or when you want an oscillator -- the wien bridge oscillator uses a combination of positive and negative feedback to make an oscillator.

author: Mechatrommer
author_rank: Super Contributor
timestamp: July 05, 2011, 11:29:21 pm
content: Quote from: ruku on July 05, 2011, 06:22:31 pmdo you mean the drawing?