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I have the following circuit, with an electret microphone and a TL082 op-amp:

opamp circuit

I was asked to calculate the load effect of the amplifier. However, I don't understand: I thought that the whole point of op-amps was that they had a very low output impedance, so that the voltage Vo2 was maintained constant by the current supply from the op-amp no matter the load. In that case, the load regulation would be zero. Is this correct?

ocrdu
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Clerni
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    Isn't the question about the loading effect of the INPUT impedance of the op-amp circuit? – John D Apr 10 '23 at 18:57
  • Hint: the positive input is grounded, so the negative input will want to be a "virtual ground" as well. This means it will always be very near 0 VDC, likely with micro-volts of signal. When incoming positive happens, virtual ground is biased ever-so-slightly positive, so the output must swing negative (a lot, due to high value of R2) to bring the virtual ground back to 0 V. Inverse for the negative input. C2 is there to "slow down" how fast the output can change - this essentially makes the feedback path a low-pass filter. C2 may also help stabilize the op-amp and prevent oscillation. – rdtsc Apr 10 '23 at 20:14
  • @JohnD, I kind of see where you are going. Normally the load effect is defined as \$s_i = \frac{\Delta V_{out}}{\Delta i_{out}}\$ (at least, this is the definition I was given in class). That is why I thought that it was talking about the output. How would you define it for the input? – Clerni Apr 11 '23 at 05:48
  • @rdtsc, I have computed the gain and the amplifier is a band pass filter with pass band between \$18 Hz\$ and \$22.7kHz\$. How can this help me to compute the load effect? – Clerni Apr 11 '23 at 05:49
  • Your definition is for load regulation, which is a specific loading effect. In this case @rdtsc gives you the hints you need. How does adding the op-amp circuit affect the voltage at Vo1? I.e. what would be the voltage (vs frequency) at Vo1 if the op-amp weren't present, and what will be the voltage after adding it? – John D Apr 11 '23 at 15:36
  • Please clarify with the instructor whether they are interested in opamp output impedance, or input impedance. You are right, the opamp should have a fairly low output impedance, but this is not zero; the TL082 datasheet should give a value for output impedance (expressed in Ohms.) – rdtsc Apr 11 '23 at 17:09
  • Thank you @JohnD, for the hints. Please, check my answer. – Clerni Apr 12 '23 at 06:58
  • Thank you @rdtsc, for the hints. Please, check my answer. – Clerni Apr 12 '23 at 06:59

3 Answers3

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It's true that an op-amp's inputs have such a high resistance that generally (not always) we can ingore any current they sink or source. That is, input current may be considered to be zero.

However, the two common op-amp amplifier configurations, non-inverting and inverting "look" very different from the perspective of the input source. Look at the red arrows below, indicating current flow from/to the output of the op-amp:

schematic

simulate this circuit – Schematic created using CircuitLab

The non-inverting setup, on the left, never draws any current from the input source V1. That's because it has no involvement in feedback from the op-amp's output.

By contrast, on the right we have the inverting configuration, and as you can see the input source V2 is now responsible for sinking/sourcing any and all feedback current! That's why the inverting configuration "loads" its own input.

This loading is also trivial to calculate, since we know one thing about op-amps with negative feedback; the ouptut always adjusts itself in such a way as to equalise the potentials at its two inputs. That means, in both configurations, \$V_Q=V_P\$. For the inverting arrangement \$V_P=0V\$, so \$V_Q=V_P=0V\$, giving Q the name "virtual earth".

With Q known to be at zero potential, the input source V2 can consider R1b to be a resistor straight to ground, and so R1b is the effective "input resistance" of this amplifier circuit. Source V1 in the non-inverting circuit (left) does not "see" any such load, since it's not being asked to source or sink current from/to the op-amp's output.

This has significant implications for sources V1 and V2. In the non-inverting configuration, source resistance of V1 has no effect (so long as it's far less than the input resistance of the op-amp), since no current is flowing through that source resistance to influence what potential actually arrives at the op-amp's input. In the following (left) circuit I show the source's resistance as RSa, and the potentials at its left and right ends must be equal, with no current through it:

schematic

simulate this circuit

On the right, though, for the inverting setup, it's clear that any source resistance RSb is in addition to R1b. The usual gain expression \$-\frac{R2b}{R1b}\$ is no longer correct. It becomes \$-\frac{R2b}{R1b+RSb}\$, a consequence of all feedback current flowing via both RSb and R1b.

Simon Fitch
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1

Your preamp gain is only 4.5 times but should be 50 to 200 times.
Texas Instruments recommends using a transimpedance amplifier like this: https://www.ti.com/lit/ug/tidu765/tidu765.pdf (Link added Apr14, 2023) electret mic preamp

Audioguru
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  • Audioguru - Hi, (a) Please [edit] your post to provide the [required reference link](/help/referencing) for the source of the image you adapted, to avoid this being deleted for that reason. (b) Please also add some explanatory text explaining how this post answers the original question. Multiple site members flagged this as "not an answer" as there is no explanation. While the explanation might be obvious to *you*, the fact that multiple site members flagged it as NAA means it is not obvious *enough* how this is an answer to the question. Thanks. – SamGibson Apr 13 '23 at 21:56
  • My mention of "transimpedance amplifier" says how the circuit works. – Audioguru Apr 14 '23 at 20:02
1

Thank you, @JohnD and @rdtsc, for the comments and hints:

In order to find the load effect of the input impedance of the OpAmp circuit, we need to know the input impedance if the OpAmp: $$Z_i = \frac{\Delta V_{o1}}{\Delta i_{in}}$$

As \$i_{in} = \frac{V_{o1}}{\frac{1}{jwC_1} + R_1} = \frac{jwC_1V_{o1}}{1+jwC_1R_1}\$, we have:

$$Z_i = \frac{\Delta V_{o1}}{ \frac{jwC_1\Delta V_{o1}}{1+jwC_1R_1}} = \frac{1+jwC_1R_1}{ jwC_1} = R_1 + \frac{1}{jwC_1}$$

For a large enough \$w\$, this will simply become:

$$Z_i \approx R_1 = 22\ k\Omega$$

We also need to find the output impedance of the sensor together with the conditioning circuit. Using Thévenin, we fin that the impedance seen from the \$V_{o1}\$ looking away from the OpAmp (the sensor is now an open circuit) is \$R_{th} = R_3\$. We take the Thévenin voltage as \$V_{th}\$. We can now replace the source and conditioning circuit by a voltage source \$V_{th}\$ and a resistor \$R_{th} = R_3\$ in series.

  • Without load, we will have \$V_{o1} = V_{th}\$
  • With a load equal to the input impedance of the OpAmp (\$R_1\$), we will have \$V_{o1}\$ equal to the voltage divider of \$V_{th}\$: $$V_{o1} = V_{th}\ \frac{R_1}{R_1+R_3}$$

Therefore, the load effect is \$\frac{R_1}{R_1+R_3} = 0.91\$, as this is the factor by which the voltage \$V_{o1}\$ is reduced on connecting the OpAmp to the circuit.

In order to remove the load effect, we can add a buffer at the input of the amplifier, just before the capacitor \$C_1\$.

Clerni
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