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So I stumbled across a rather strange ideal op-amp circuit that I've never encountered before. I've done enough problems regarding inverting amplifiers, non-inverting, summing, difference, and such, but I haven't done one that involves two amplifiers together. The problem asked to find both Vo and V4, so I figured that perhaps doing one amplifier at a time would help. enter image description here Assuming that the voltage difference between the inputs for both inverting and non-inverting terminals was negligible, I framed the left inverting amplifier to include V1, R1, R2, and R3. I attempted at finding the output voltage v_o1, at the end of the left amplifier, through the formula *v_o1 = -(R2/R1)V1, which ended up as -15 V.

However, when attempting to find Vo, I encountered the first roadblock: I know that since the current going into the inverting terminal is zero, i1 = i2 + i3. By using the node-voltage method with this, I identified i2 = (0 - v_o1)/R2 and i3 = (0 -Vo)/R3. This is where I'm unsure if this is all correct. I presumed that the i2 and i3 branches that stemmed off from the node is equal to i1, but could I be wrong about this? I ended up with Vo = 50 V, but I don't fully know whether this is right, or if I'm missing some passive sign somewhere.

As for (b) on finding V4 across the resistor, I think I may had solved this incorrectly. Switching to the second amplifier at the right I assumed that, because we're dealing with ideal amplifiers, the currents from the inverting/non-inverting terminals are zero, such that i4 + i5 = 0. However, considering that Vo might not be correct, I went with my result and ended up with V4 = 90 V. I'm also worried about this because instead of including the whole two amplifiers together, I'm only focusing on one at a time for each part, and that I might very well be missing key variables to include, such as v_o1. Am I wrong in saying this? It would be much appreciated if anyone can provide some insight or advice on this tricky two-stage amplifier circuit. King Regards

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – clabacchio Dec 4 '18 at 9:42
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schematic

simulate this circuit – Schematic created using CircuitLab

The schematic shows how the whole circuit should/could be interpreted. Of course, I have assumed that the second opamp is used as a non-inv. amplifier (gain A). Only in this case, the circuit can be used as a linear amplifier.

Regarding the output Vout1 it can be treated as a classical inverting configuration. For gain calculation you can set Vn to zero (Vn is the voltage at the inv. terminal). Because this voltage is produced from two voltage sources (V1 and Vout1) you have to use superposition rules Vn=Vn1+Vn2 (Vn1 caused by V1 with Vout1=0 and Vn2 caused by Vout1 with V1=0)

If the gain Vout1/V1 is found, it will not be a problem to find the wanted ratio Vout2/Vin because of Vout2=Vout1*A with A=(1+R5/R4).

EDIT/Comment:

Realizing that R3 enters the feedback path with a value reduced by the factor A (because the current through R3 is A times larger if compared with R2), you immediately can write down the result (making use of the well-known inverter formula):

G1=Vout1/V1=-(R2||R3eff)/R1 with R3eff=R3/A

G2=Vout2/V1=A*G1

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  • \$\begingroup\$ Oh okay, I can see now what you were referring to earlier about the second amp acting as a non-inverting amp in that configuration. But with that schematic, one of the things I'm having a hard time following is the ain calculation set up. If Vn is the voltage at the inverting input terminal, then with Vn1 = -Vn2, are you saying that Vn2 does not account for Vout2? And based on the configuration, we already know what Vout1 and Vout2 (or Vo) are too right? What exactly does that gain (Vout1 / V1) represent though? Apologies for the many questions I've asked, as this concept is all new to me. \$\endgroup\$ – Heavywake Dec 4 '18 at 9:57
  • \$\begingroup\$ In my terminology, we have Vn2=f(Vout1) for V1=0, OK? Hence, we have a simple voltage divider that consists of R1 and the feedback block, which in our case contains two parallel elements (R2||Reff). What is Reff? This is R3 in series with a constant gain A. Hence, the total driving voltage for R3 is Vout1*A. That`s all you have to know. Finally, when you set Vn1=-Vn2, both quantities (V1 and Vout1) appear in one equation so that you are able to find the ratio Vout1/V1. As you can see, the whole Task consists of two voltage divisions only. \$\endgroup\$ – LvW Dec 4 '18 at 11:35
  • \$\begingroup\$ Okay, so Vn2, being a function of Vout1, for V1, includes the feedback block on which R2||Reff. And with the driving voltage of R3 being the product of gain A and Vout1, that is what leads to Vn2. Being a function of it. That I can understand. And so in the case for V1 =0, what we're left with is Vout1 being the primary variable that drives the gain. In that sense, is Vout the key variable we need to solve for then? \$\endgroup\$ – Heavywake Dec 4 '18 at 12:06
  • \$\begingroup\$ Vout1 is the primary variable (driving voltage) during computation of Vn2 - that´s correct. However, during computing of Vn1, the input signal V1 is the primary variable. And with Vn=Vn1+Vn2=0 both variables appear inthe equation. These are basics! It is a simple application of the superpositon principle. One additional hint: Looking at the R3 branch (and making up your mind) it is easy to see that R3 enters the parallel combination of both feedback resistors with the value R3/A. Knowing this, you immediately can write down the equation Vn1=-Vn2. \$\endgroup\$ – LvW Dec 4 '18 at 13:27

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