Re: inverting configuration of op-amp, output wave forms
Hi,
I personally don´t like the phrase "phase shift by 180 degree" in this case, because phase shift (in my eyes) has something to do with time delay.
With an inverting OPAMP circuit we don´t have considerable time delay.
The signal voltage just has opposite sign.
+ at input becomes - at output
- at input becomes + at output
Therefore you may see it statically.
Let´s imagine just two resistors in a string. Let´s name them Ri and Rf. (Don´t think of an OPAMP)
Let´s assume both have same value. Let´s say 1kOhms
Now connect the free end of Ri to +1V.
Now assume you have a variable voltage supply connected to the free and of Rf.
What voltage is necessary, that the center of both resistors become 0V?
You can use this method, because the OPAMP input is high impedance, it doesn´t (considerably) influence the voltage at the center node. The OPAMP just "senses" the input voltage and adjusts/regulates the output voltage.
Klaus
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Added:
I just recognized you wrote "non inverting". The above explanation still was for inverting.
**
Now for non-inverting:
From Post#2:
* the OPAMP tries to regulate the output in a way that the same voltage (GND, 0V) is at the inverting input, too.
--> it regulates the output voltage in a way, that the inverting input gets the same voltage than the non-inverting input.
That´s the key.
Lets assume there are the two resistors Rg and Rf. Rg is connected to GND, RF is connected to the output. The center node is connected to the inverting input.
Now V_in is at the noninverting input.
The OPAMP regulates the inverting input to the same voltage. Means the voltage across Rg is the same as V_in.
I_Rg = V_in / Rg.
the same current is through Rf. The voltage across Rf is: V_Rf = I_Rg x Rf.
To calculate the output voltage: you need to add the the voltage across Rf to the voltage at the inverting input. V_out = V_in + V_in * Rf / Rg
= V_in * ( 1+ Rf / Rg)
A = 1 + Rf / Rg
Klaus