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This is not an assignment I'm prcaticing for my mids Can someone solve this problem for me. Will the node betweeb R4 and R3 be equal to Vi? If so what case will there be a virtual ground and what case will a node be in some reference voltage like Vi.

How I went about solving it i. Used nodal equation in node between R3 and R4 , R3 and R2 and R2 and R1 and then substituted value of Vi until there were only two V terms one Vi and one Vo to get the expression of vo and Vi. However my expression was too long and now I doubt my method.


In addition to Andy`s answer (which I completely agree upon) I like to point to the fact that both opamps are to be considered as IDEAL (watch the infinity symbol). Hence, you can rely on the fact that the difference between both opamp input terminals is zero (Vin+=Vin-). Together with known formulas for a simple voltage divider and the ideal inverting opamp circuit this should give you enough mathematical relations to solve the problem.

Comment regarding "virtual ground": This principle applies to inverting circuits only (pos. input node grounded as for A2). In the case under discussion, this principle merges into the assumption Vin+=Vin- for the amplifier unit A1.


To answer your first question about the node between R4 and R3 you have to decide if the feedback is positive or negative. I'll leave that up to you but if you conclude that the feedback is positive then the answer is a certain "no" and both op-amp outputs will be pushed against the power rails in saturation.

If you decide the answer is "yes" then, as a first approximation, the voltage at the node of R4 and R3 will be equal to Vin. If you took a more detailed look into op-amp behaviour the answer would almost certainly be "not quite" or "no". This is because there are offset voltages to consider and leakage currents from the inputs to consider - these make the perfect op-amp "imperfect" and you'll find that there will be a small error voltage that can rage from a microvolt to several millivolts. Non-infinite open loop gain also erodes the chances of both inputs being equal.

If you then looked deeper into the workings of the op-amp, you should conclude (op-amp dependant) that at relatively high frequencies, the feedback becomes positive and in fact the configuration you have will likely "sing" at some obscure frequency in the high tens of kHz to several MHz.

But, assuming perfect op-amps, you assume the two input voltages are equal and work backwards from that presumption....

The output from A2 will be

\$V_{IN}\times \dfrac{R_3+R_4}{R_4}\$

And, the output from A1 will be smaller than the output of A2 by the factor R2/R1

From this you should be able to solve your question.

  • \$\begingroup\$ I honestly didn't understand ur explaination. I'm very new to this subject can you please explain these things in simpler terms? \$\endgroup\$ – Shivji Sep 6 '14 at 8:59
  • \$\begingroup\$ Do you mean if there is positive feedback then there will be unstabiity in the gain and eventually the output will shift to saturation according to A/(1+AB) ? And if it is negative feedback then the node will remain in Vi/ \$\endgroup\$ – Shivji Sep 6 '14 at 9:03
  • \$\begingroup\$ Also In reference to positive feedback which amplifier are you talking about ? the first one or the second one \$\endgroup\$ – Shivji Sep 6 '14 at 9:03
  • \$\begingroup\$ @Shivji Positive feedback causes instability and in the simple case of DC will cause both op-amps to end-stop against the power rails - both contribute to positive feedback and both will likely end-stop - it's the overall configuration of the circuit (and not the op-amps by themselves) that creates a positive feedback scenario. \$\endgroup\$ – Andy aka Sep 6 '14 at 9:28

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