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Why does the op-amp output voltage saturate?
Vout = Vin (1 + R2/R1) According to this positive feedback topology – made an error here, it’s Vout = Vin (-R2/R1) (edit) Now I read in a tutorial – http://www.electronics-tutorials.ws/systems/feedback-systems.html , that If the input voltage Vin is positive, the op-amp amplifies this positive signal and the output becomes more positive.
What should the gain be of an op-amp?
A real op-amp can’t output at a voltage outside the range of it’s supply voltage, so it’s being driven to the rails hard. What you can do (and I’ll add some schematics later) is add negative feedback to the circuit, which decreases the gain to a more manageable level. For a straight difference, you want the gain to be 1.
Why does op amp amplifier’s voltage difference zero?
Therefore the conditions that allows us to consider that v+=v- is negative feedback combined with a Vout below VCC and Vout over VEE (negative power supply). This is a bit simplistic explanation of course, but it allows you to analyze circuits with ideal OPAMP.
Why does opamp 741 have an in built offset voltage?
In short this is due to imperfections in the opamp that are naturally present. Opamp 741 has an in built offset voltage, that is present between the inverting and non inverting inputs. This small offset voltage when amplified by the open circuit gain of the opamp saturates the output voltage.
What does saturation mean in linear amplification mode?
Saturation is another word for clipping where the amplitude of the output is no longer able to work in the linear amplification mode and an increase in input no longer increases the output. If you plot it on the oscilliscope it looks like someone took a pair of scissors to the top of the waveforms, hence the name “clipping”.
What happens when the input voltage Vin is positive?
If the input voltage Vin is positive, the op-amp amplifies this positive signal and the output becomes more positive. Some of this output voltage is returned back to the input by the feedback network. Thus the input voltage becomes more positive, causing an even larger output voltage and so on.