# Is there a technical reason why you might use a two inverting amplifiers instead of a single non-inverting amplifier?

For example, I found a schematic that uses two inverting opamps to amplify the output of an AD590 with a gain of 1000:

In this case the second opamp appears to be doing nothing but "re-inverting" the amplified voltage as it has a gain of 1. Perhaps this has something specifically to do with the AD590, or the opamp itself (the LT1014), or perhaps it's just to use up an otherwise unneeded fourth opamp in a quad package?

Is it be a safe assumption that it would be possible to replace these two opamps with a single non-inverting opamp with a gain of 1000?

Edit: After probing around (and just looking at the bands on the resistors) it appears that R28 is in fact a 10kΩ resistor.

• One reason would be to get a unity gain. – Eugene Sh. May 16 '16 at 15:09
• There is a distinct difference. The 2nd stage is setup with a 2K resistor to control the source impedance. – Michael Karas May 16 '16 at 15:12
• Due to negative feedback the source impedance will be 2k/0.5*Aol with Aol=open-loop gain. Hence, I ask myself: What is the purpose of the 2k resistor? – LvW May 16 '16 at 15:16
• @MichaelKaras Could this not be achieved with a non-inverting amplifier? – David Freitag May 16 '16 at 15:18
• In the "good old days" when op amps choices were not great (709,741 etc.) building an audio amp with a gain of 100 meant the high frequency response was severely cut off due to poor gain-bandwidth product. To help overcome this two 10X amps were cascaded. Not quite the situation here but a valid technical reason for having the two amps rather than just one. – JIm Dearden May 16 '16 at 16:16

You could simply ground the left end of R30 and connect the AD590 to the non-inverting input of the op-amp. That would actually be better- right now R27 is stupid- it should be 100K to match the bias currents, but with the aforementioned connection scheme it would be the correct value**. The AD590 will 'see' exactly the same voltage across it.

If you want to add R24 in series with the op-amp output it would maintain the current limiting. Note that a capacitive load in either case will tend to destabilize the amplifier, but it's worse in the circuit as shown because the closed-loop gain is lower.

As to why they used this- I can only assume they felt some need to use up the amplifiers in a quad package. A better way to do that would be to connect output to inverting input and ground the non-inverting input.

There are applications where it is better to use two inverting amplifiers rather than one non-inverting, but this isn't one.

** note that R27 may be intended to have a resistor connected between the non-inverting input and a reference voltage to bias the output closer to 0V at 0°C. As shown it will be saturated at room temperature even at the maximum supply voltage of the chip (+/-22V).

• So, something like this? Also, are 100Ω/10kΩ optimal in this situation? Or would 1kΩ/100kΩ or 10kΩ/1MΩ be more reasonable? – David Freitag May 16 '16 at 17:08
• That has a gain of 101 so the output would be about 3V at room temperature. The 100$\Omega$/10K$\Omega$ is good. Note the gain is 1% higher than the original should have been. – Spehro Pefhany May 16 '16 at 17:20

The AD590 has a CURRENT output. For this reason, you must use an opamp wired as an (inverting) current-to-voltage converter - and you cannot (directly) feed a current into the noninverting input terminal of an opamp.

The mentioned gain of 1000 is a voltage gain (but your input signal is a current). Hence, in case of an ideal current source the input current I goes through the feedback resistor and produces a an output voltage of Vout=-I*R28.

If your question is a general one - independent on the shown example circuit, the answer is as follows: There are many cases where an inverting amplifier is required - for example as active devices in active filter circuits, harmonic oscillator circuits or as a controller block in control loops.

• Your math doesn't seem to make sense, at 5V and +25°C the AD590 Iout should be 298µA meaning Vout = -(2.98E-4V) * 100kΩ = -29.8V which would pretty much always leave the output voltage hammered at the -15V rail – David Freitag May 16 '16 at 15:54
• @DavidFreitag Agreed, yet LvW's math is correct. – Spehro Pefhany May 16 '16 at 16:01
• @DavidFreitag, it is not my math - it is just circuit theory. If the circuit will function as desired is another question (I did not comment on). – LvW May 16 '16 at 16:06
• @LvW The issue is that I know the circuit works, but given the available information I cannot reconcile how. I guess I've got some probing to do. – David Freitag May 16 '16 at 16:32
• @DavidFreitag - That's a bad circuit design. R28 is almost certainly a 10k resistor, and it would be a rather old op amp to need either R26 or R27. Is there any chance you could provide a link in the OP to the schematic? – WhatRoughBeast May 16 '16 at 16:42