Input impedance inverting amplifier topology questions - Page 1
Forum section: electronics / beginners

author: trevortjes
author_rank: Contributor
timestamp: February 16, 2020, 04:22:13 pm
content: Hey I stumbled upon this problem while trying to teach about operational amplifiers and more specifically the input impedance/resistance of an inverting opamp. The textbooks will all say that the input impedance is equal to the input resistor. The problem I am having with this is that a virtual ground does not have the properties of a real ground and since the input of the inverting input is very high, all current must flow through both resistors and then into the opamp. Which makes the input resistance R1+R2.Now because I wanted some proof of my theory I tried simulating it. And I got 1 result that confirms my theory, and another that debunks it. The first result is attached as input impedance 1. The spice simulation tells us that indeed the input resistance is R1+R2 like I thought it was. Then in the second attachment I replaced the ground of the opamp with a supply of -10V. Now the textbook explanation applies, input resistance = input resistor.I'm starting to understand that the virtual ground kind of forces a current to flow from the input voltage source, through the input resistor to the virtual ground, but because the current isnt allowed to travel into the input, it's forced to go through the input resistor, kinda like a current source. But that still leaves me questioning the results of the spice simulations.

author: StillTrying
author_rank: Super Contributor
timestamp: February 16, 2020, 05:15:29 pm
content: In the first one the op amp output would need to go to -1V to keep the virtual ground at 0V, it can't without a -Ve supply.

author: mikerj
author_rank: Super Contributor
timestamp: February 16, 2020, 05:19:27 pm
content: With the first example the non-inverting input and the -ve rail are tied to at 0v. This means a positive voltage on the inverting input will try to drive the output below 0v, but it cannot i.e. you have saturated the output of the amplifier. In this condition your input impedance is R1+R2, but you don't have an amplifier any more.In the second case you have given the amplifier a negative rail so the output can swing below 0v. This means the inverting and non-inverting input remains at the same voltage and your input impedance is R2.

author: trevortjes
author_rank: Contributor
timestamp: February 16, 2020, 06:55:34 pm
content: StillTrying and mikerj, that's indeed something I was thinking about. Thanks for bringing light to the situation!

author: David Hess
author_rank: Super Contributor
timestamp: February 16, 2020, 07:07:41 pm
content: The input impedance is only controlled within the bandwidth of the operational amplifier which can be shown because at high frequencies where the open loop gain has fallen, the difference between the inverting and non-inverting input must increase to produce the same output voltage.This does *not* apply to current feedback operational amplifiers. For them, the inverting input is a buffered output from the non-inverting input and the differential voltage remains small even with no feedback at all.

author: gbaddeley
author_rank: Regular Contributor
timestamp: February 18, 2020, 09:28:31 am
content: The - input is practically at ground because the + input is tied to ground, and the feedback loop ensures that at DC and LF conditions the - & + inputs are at the same potential. If you vary Rf you will observe it has no effect on input resistance, it stays fixed at Ri. Sorry, the text books are correct.