Question about obtaining the input network transfer function of an inverting OPA - Page 1
Forum section: electronics / projects

author: opa627bm
author_rank: Regular Contributor
timestamp: March 24, 2021, 04:02:08 am
content: Hi all,I am trying to review some of the fundamentals that has been long forgotten since school.by reading thishttps://peer.asee.org/a-method-for-obtaining-the-transfer-function-of-inverting-and-non-inverting-op-amp-circuits-based-on-classical-feedback-theory.pdfI can follow the non-inverting section pretty well, but when comes to the inverting section:page 8,The input network transfer function is defined as - RF/(RF+RA), and feedback transfer function still RA/(RF+RA)Since now Vin is a term in the equation,Vi-Vout/(RA+RF) = Vi-V_ /RA = V_ - Vout)/RFI know the gain is simply -RF/RA , but i am trying to make sense of it.any input will be appreciated, thank you !My question is now did the author come out with - RF/(RF+RA) as for the input network transfer function..thank you !Regards,Li

author: ledtester
author_rank: Super Contributor
timestamp: March 24, 2021, 07:40:18 am
content: If we naively translate the inverting op-amp circuit to a transfer network we might begin with something like:The key to figuring out what goes in the ? ? ? area is to realize that the formula for the inverting input is$$\text{Inverting input} = \frac{v_i R_f + v_o R_a}{R_a + R_f}$$and so we can fill in the ? ? ? as follows:Note how the minus sign in the \$v_i\$ transfer function is canceled by sending it the minus input of the new summing node.Now we can short circuit the top summing node because we are just adding 0 -- i.e. the output of the lower summing node can go directly to \$G(s)\$ -- and that leads to the diagram in the paper.Correction:We can short circuit the lower summing node to \$G(s)\$ with a multiplier of -1 since the lower summing node goes the negative input of the upper summing node.

author: opa627bm
author_rank: Regular Contributor
timestamp: March 24, 2021, 09:03:11 pm
content: Hi Ledtester,thank you so much for the reply.I can get the V_ transfer function using superposition.but how to add the correct sign (input network (some people call it alpha ) is negative) The feedback network (some people call it beta ) ,how to separate the combined term correctly into alpha and beta and with correct sign ?Thank you for the help!Regards,LiPS: sorry the pic I attached is rotated for some reason, from my viewer it is in the correct orientation.

author: ledtester
author_rank: Super Contributor
timestamp: March 25, 2021, 12:14:23 am
content: I'm not sure exactly what you're asking about, but perhaps this will help...The equation \$V_A = V_B B + V_C C\$ is the same as this network:Of course, you can decide to put the minus sign on C rather than B if you also switch the inputs to the summing node.You have:$$\begin{align}V &= \frac{ V_{\text{IN}} R_F + V_{\text{OUT}} R_A } { R_F + R_A } \\&= V_{\text{IN}} \frac{R_F}{R_F+R_A} + V_{\text{OUT}} \frac{R_A}{R_F+R_A} \\\end{align}$$So set:$$\begin{alignat*}{2}V_B &= V_{\text{IN}}, &B = \frac{R_F}{R_F+R_A} \\V_C &= V_{\text{OUT}}, &C = \frac{R_A}{R_F+R_A} \\\end{alignat*}$$and apply the diagram.

author: opa627bm
author_rank: Regular Contributor
timestamp: March 25, 2021, 04:45:33 am
content: Hi Ledtester,Sorry for not making my question clear.I have used PDF this time, that should me the orientation correct.My question is, how to map the terms in the V_ transfer function with correct sign to the input network transfer function and feedback network transfer function. and how to determine the correct sign on the summing block (+） and (-)Regards,Li

author: ledtester
author_rank: Super Contributor
timestamp: March 25, 2021, 09:21:22 am
content: Both of these are equivalent - you can decide to use either one. Note carefully where the + and - signs are on both the transfer functions and the summing node inputs.