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I am making a VU meter, and I need to rectify the audio signal for the IC comparator chip. I found this circuit in the datasheet, but I am unsure of how it works.

enter image description here

I understand that the LF351 op amp is needed to overcome the threshold voltage of the diodes, but what is the purpose of the LM307 op amp? Is it just a buffer? Why not just use another LF351?

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  • \$\begingroup\$ if this is feeding directly into a comparator circuit, have you considered a window comparator circuit instead of precision rectifier -> comparator? depending on your specific needs it might be small, cheaper, more accurate \$\endgroup\$ – JonRB Dec 16 '18 at 23:14
  • \$\begingroup\$ @JonRB Probably using an LM3915 or something like that. \$\endgroup\$ – Spehro Pefhany Dec 16 '18 at 23:32
  • \$\begingroup\$ ah, fair enough. \$\endgroup\$ – JonRB Dec 16 '18 at 23:34
  • \$\begingroup\$ There are better designs that are non-inverting single supply that work down to a few mV analog.com/en/design-center/reference-designs/… \$\endgroup\$ – Sunnyskyguy EE75 Dec 17 '18 at 4:31
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    \$\begingroup\$ @JonRG Yes, this is feeding into an LM3915. \$\endgroup\$ – Sal M Dec 17 '18 at 22:26
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This is a rather tricky schematic; to understand it, you need to consider 2 case whether the input signal (after C1) is positive or negative.

schematic

simulate this circuit – Schematic created using CircuitLab

When it is positive, D2 is blocked and you can remove it and study the simplified schematic. OA1 behaves as an inverting amplifier with gain -1 whose output is \$V_2\$: \$V_2 = - V_{IN}\$. Then OA2 is an inverting summer: \$V_{OUT} = - {R_5 \over R_4} V_{IN} - {R_5 \over R_3} V_2 = - V_{IN} - 2 V_2 = V_{IN}\$.

When it is negative, D1 is blocked and you can remove it and study the simplified schematic. OA1 behaves as a follower with input to ground and output at \$V_1\$: \$V_1 = 0\$. Then OA2 is an inverting summer: \$V_{OUT} = - {R_5 \over R_4} V_{IN} - {R_5 \over R_2 + R_3} V_1 = - V_{IN} - V_1 = -V_{IN}\$.

As for the choice of OA1 and OA2, when $V_{IN}$ crosses zero, the output of OA1 needs to go quickly from -0.6V to 0.6V. You need an op amp with a hich slew rate. There is no such requirement for OA2.

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In answer to your question, the most likely answer is that the LF351 was more expensive so they used a cheaper op-amp where high slew rate was not required. The LM307 is/was a 741 type of op-amp, slow and semi-precision. Slew rate is less than 1/20 that of the LF351.

It's important to have a high slew rate amplifier in the LF351 position because there the output of the input op-amp should move quickly from -0.7V to +0.7V (ideally instantly) since one diode or the other should always be conducting.


To see how it works, consider when the AC-coupled input is > 0, say V1. Current flows through R1 and R2 and D1 so the left side of R3 is at -V1, thus the current flowing into the node at pin 2 of the LM307 is Vin/R4 - Vin/R3 + Vout/R5, so the output voltage (ignoring C2 for the moment) is +Vin

enter image description here


When the AC-coupled input (right side of C1) is < 0, call it -Vin, D2 conducts (keeping the op-amp output from straying too far from ground, reducing the demands on slew rate) and the current flowing into the node at pin 2 of the LM307 is -Vin/R4 + Vout/R5, so Vout = -Vin (again ignoring C2).

enter image description here


C2 acts to average the output voltage with a time constant of C2*R5.

In the second case there is a resistance R2 + R3 between pin 3 of the LM307 and pin 2 of the LF351. Since both nodes are always at virtual ground there is (ideally) no effect from that 200K resistance.

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  • \$\begingroup\$ When the AC-coupled input is positive (V1), you mentioned that the left side of R3 is at -V1. Why is that? Is it because the gain of the 351 is -1, making the voltage in the feedback loop through D1 negative? \$\endgroup\$ – Sal M Dec 17 '18 at 22:43
  • \$\begingroup\$ Yes, exactly. The gain is -1, measured at the right side of R2. The gain from the output of the LF351 will have a slightly different value and a relatively large offset, but we don't care much about that. To drive the inverting input of the LF351 to exactly 0V the voltage at that point must be -V1 (assuming no op-amp bias current and resistors exactly matched). \$\endgroup\$ – Spehro Pefhany Dec 17 '18 at 22:47
  • \$\begingroup\$ Awesome I understand now. I ordered the components, should be a while before they get here though. Thanks for the help! \$\endgroup\$ – Sal M Dec 18 '18 at 0:09
  • \$\begingroup\$ When wiring up the LF351 amplifier, do I need to use a dual supply voltage rail? Or can I just connect the VCC- pin to ground? I assume I can connect it right to ground because no signal is being amplified right? \$\endgroup\$ – Sal M Dec 21 '18 at 20:55
  • \$\begingroup\$ No, it has to swing negative by the maximum positive input voltage, so if the maximum peak input voltage is, say, 7V, you should have a negative rail of about -10V or better (the LF351 needs about 3V more for the output stage). \$\endgroup\$ – Spehro Pefhany Dec 21 '18 at 21:10

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