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When we have an OpAmp as the one below: enter image description here

This was extracted from an exercise in which V2 was an AC power source. However, I have tried to imagine what would happen if it was DC. It is obviously a non-inverting OpAmp, but according to my book, this is a Buffer Gain = 1 system because C1 blocks any current from flowing through R1. Could someone explain this? I am confused since I do not see how C1 can affect the node at which we apply Kirchhoff's Current Law.

If this is the case, where we have a Buffer, then I know VA = V2. But since we have C2, right after this capacitor, would we have VB = 0 for the same reason?

I am very confused and I could use some help with both questions (in bold above). Thank you!

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  • \$\begingroup\$ C1 blocks any current from flowing through R1 More precise would be: "C1 blocks all DC current ...". So for DC C1 is an open, R1 in series with an open: you can leave out R1. That leaves only R2 and source V2 connected to the opamp. \$\endgroup\$
    – Bimpelrekkie
    Mar 14, 2018 at 13:41
  • \$\begingroup\$ With very large open-loop voltage gain, the opamp will make its output pin's average Vout be the same as V2. \$\endgroup\$
    – analogsystemsrf
    Mar 14, 2018 at 16:20

2 Answers 2

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If V2 is DC, we do not have any AC source on the circuit. Neglecting the transients, when you only have DC sources the capacitors act as open circuits, while the inductors are closed circuits.

Think it like that: you connect your DC sources, turn them on, caps charge up, current ramps up in the coils, and at a certain point steady state is reached: dV/dt = 0 means i = 0 in caps, and conversely V = 0 in inductors -> open circuits, closed circuits.

Now, if you remove C1 and C2 from your schematic, you can remove also R1 because one end is not connected. No current can flow in R2 because no current can flow in the inverting input of the amplifier. No current in a resistor means that the voltage across it is 0, so you can see it as a short circuit -> you have your buffer, and as you know VA = V2.

Apply the same reasoning of R2 to RL and you also get VB = 0.

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  • \$\begingroup\$ Then, whenever I have DC sources, will all capacitors act as open circuits, no matter where they are placed in the circuit? \$\endgroup\$
    – Bee
    Mar 14, 2018 at 19:29
  • \$\begingroup\$ Also, if V2 was an AC source, would the voltage VA = VB? Because there is no voltage drop across C2 right? \$\endgroup\$
    – Bee
    Mar 14, 2018 at 19:58
  • \$\begingroup\$ VA = VB only for frequencies above the pole introduced by C2, that should be at around 1 / (2 pi C2 RL) Hz \$\endgroup\$
    – Vladimir Cravero
    Mar 14, 2018 at 23:59
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The DC gain is 1 (to the output of the op-amp) is 1. That output voltage (equal to the DC value of V2) is blocked by C2 so the voltage at RL is zero, but it's important because the op-amp output can only swing within the power rails (and often not nearly to the rails). To see this, ignore C1 (and therefore R1) and you just have a unity gain buffer.

The AC gain (assuming high enough frequency that the capacitors act as shorts) is +(1 + R2/R1). For this to be true |Xc1| << R1 and |Xc2| << RL.

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