In a schematic I've been trying to understand I came across this sub-circuit: inverter-followed-by-buffer

It's an op-amp inverter directly followed by a buffer. VIN comes from a DAC in a microcontroller and this circuit produces a VOUT which is negative VIN. The op-amp is supplied by positive and negative rails (not shown here). So far so good.

But I don't fully see the rationale of using OA2 in this circuit. The only reason I can see is this: Without the buffer (OA2) a sudden load at VOUT would draw a current from VIN until the op-amp OA1 feedback adjusts (about 1µs). With the buffer (OA2) this is not the case anymore. Am I getting this right? Or am I missing something?

  • \$\begingroup\$ Are both resistors definitely 10 kohm? \$\endgroup\$ – Andy aka Mar 10 '16 at 13:16
  • \$\begingroup\$ R1=R2 can be chosen appropriately \$\endgroup\$ – mcmayer Mar 10 '16 at 15:51
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    \$\begingroup\$ Was that what you saw in the schematic. I do ask for a reason. \$\endgroup\$ – Andy aka Mar 10 '16 at 17:14
  • \$\begingroup\$ Yes, it was 10k. \$\endgroup\$ – mcmayer Mar 10 '16 at 18:59
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    \$\begingroup\$ For an inverter with high gain the output impedance of the op-amp can become quite high as frequency rises (due to limitations on op-amp gain-bandwidth-product) and, if a unity gain stage is added, you get a much tighter controlled low output impedance at high frequencies. \$\endgroup\$ – Andy aka Mar 11 '16 at 9:35

You are right. In most cases this is silly, adds offset voltage, and uses another part. Most likely this is just someone's knee jerk reaction, or blindly following a rule of "always buffer the signal" without thinking about it too hard. Not all schematics out there are the result of good design.

There are some subtle advantages to the second buffer-only opamp:

  1. The feedback current thru R2 eats into the total output current capability of OA1. OA2 has all of its current capability available to drive the output.

    In this case with R2 being 10 kΩ, this is a weak argument since the feedback current is so small relative to the capability of most opamps. Sometimes a circuit like this happens because R2 was much lower before, and the second opamp wasn't removed after a design change that raised R2.

  2. OA2 protects the input signal from abuse of the output signal. Vin sees the fixed impedance of R1 only as long as OA1 is acting in closed loop operation. If something loads Vout so that OA1 can't drive it to the desired voltage, then the negative input of OA1 is no longer at 0 V, and the Thevenin equivalent that Vin is driving changes.

    In this circuit, the output of OA2 can be abused without affecting the output of OA1, which in turn won't affect Vin, maybe. The reason I say "maybe" is that some opamps have back to back diodes between their inputs. I didn't look up your opamp, so I don't know whether that is the case here. If so, then abuse of Vout will get back to the positive input of OA2, which will get back to Vin.

    This is again a weak argument since loading a opamp output to the point where it can't drive to the desired voltage is generally running the opamp out of spec.


It doesn't have much effect on the performance, except to make it somewhat slower because there are two poles in the transfer function.

Chances are the designer only needed the one op-amp in the dual and chose to do something benign with the remaining amplifier to keep it out of trouble. This is a common situation with LM324 quad and LM358 dual amplifiers.

There is no common inexpensive equivalent of the LM358 that has a single amplifier- any other parts tend to be more expensive and/or may be limited in some way (such as having lower maximum supply voltage) so if an LM358 is good enough then you may as well use it and waste the 2nd amplifier.


The "buffer" is just there to, as the name implies, "buffer" the output.

Since OA1 is part of a feedback network, some of it's output is used already (lost through R2 and R1.) Which means OA1 has less drive capability. So if you were to connect OA1 to some other part of a circuit, unintended things could happen. OA2 simply "follows" or "buffers" the output of OA1, and it has zero output loading, so has full drive capability. This "buffering" is commonly seen and used, and makes the operation of the circuit more robust and reliable.

Also, buffers matter in terms of delay. In both digital and analog circuit design, high-speed signals can be significantly delayed by circuit elements. Sometimes, multiple buffers are used - seemingly with no purpose - except to introduce a delay. This is usually done so that two signals "meet up again" in the time domain.

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    \$\begingroup\$ Ok I see. But if I a assume a VIN=10V the feedback loop of OA1 requires 1mA. Then the OA2 buffer looks like overkill to me. \$\endgroup\$ – mcmayer Mar 10 '16 at 12:56
  • \$\begingroup\$ For this circuit, it likely is. But this also depends on the op-amp used; if the op-amp could only drive 5mA lets say, then the feedback resistor is already consuming 20% of it's output capability. Further loading may cause it to skew the signal; as the output cannot drive properly, the feedback input will contain this error. With a buffer added, there is both more output drive available, and loading that output does not affect the operation of OA1. Win-win. :) \$\endgroup\$ – rdtsc Mar 10 '16 at 15:01

When the power is on, there is supposed to be little difference as the other posters remarked.

When the power is switched off however, the second variant is less likely to have the output bleed back into the input and will probably make the input load independent from the output connections. For some applications (audio?), that can be a desirable property. Whether this is indeed the case here depends on the internal circuitry of the opamp in question. Since a specific type is given, this may indeed have been part of the design.


In the schematic you have drawn, as others have answered, there isn't so much benefit from this layout.

If however there are two different model op-amps and the resistor values are different, then there can be good reasons for using such a layout. I created a similar circuit, which needed to amplify a relatively high frequency signal, and then drive the output in to a 50 ohm load. These two functions require op-amps with different characteristics. For the first op-amp, it needs to have a higher bandwidth to allow it to amplify a high frequency without any loss of gain at high frequencies. For the second op-amp, it had to have a higher rated output current to be able to drive a 50 ohm load at the maximum output voltage, but didn't need such a high bandwidth as it only had a gain of 1.


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