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bharat

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inverting and non-inverting op-amp
Sep 13^th, 2007, 3:13pm

Hi All,
To achieve the gain of mod 2 (+2 or -2) considering the inverting and non-inverting op-amp configuration, which one I should prefer? The gain of inverting amplifier is (-R2/R1) and non-inverting amplifier is (1+R2/R1). By replacing the R1 and R2 appropriately, in both the places one can get the gain of mod 2. Now, my question is what are the advantages/disadvantages one configuration has over other?

Thanks
-Bharat

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monte78

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Re: inverting and non-inverting op-amp
Reply #1 - Sep 14^th, 2007, 1:22am

bharat wrote on Sep 13^th, 2007, 3:13pm:

Hi All,
To achieve the gain of mod 2 (+2 or -2) considering the inverting and non-inverting op-amp configuration, which one I should prefer? The gain of inverting amplifier is (-R2/R1) and non-inverting amplifier is (1+R2/R1). By replacing the R1 and R2 appropriately, in both the places one can get the gain of mod 2. Now, my question is what are the advantages/disadvantages one configuration has over other?

Thanks
-Bharat

Hi Bharat,

Inverting amplifier has an input impedance given by R1, while non-inverting amplifier has an input impedance given by the operational amplifier positive input (usually greater than R1). It depends on your input impedance requirements.

Bye!
Monte

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HdrChopper

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Re: inverting and non-inverting op-amp
Reply #2 - Sep 14^th, 2007, 7:37pm

Hi Bharat,

Among other things you might want to consider apart from the input impedance is the signal swing. While the non-inverting configuration is limited by the input diff pair common mode range, in the inverting one the input signal CM can be independent from the one at the input differential pair and also the input signal swing can be rail 2 rail (for the non-inverting amplifier you would need a r2r diff pair, which is more complicated to implement).

Bandwidth is also a difference between the two options. The non-inverter option has a GBW equal to the unity gain frequency, while the inverting configuration has a GBW lower than the fu by R1/(R1+R2). Therefore the inverting one is slower and your choice will depend on the desired speed.

Regards
tosei

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bharat Community Member Offline Posts: 57	Re: inverting and non-inverting op-amp Reply #3 - Sep 16^th, 2007, 7:40am Thanks tosei I got the first point of ICMR in both the configuration. Regarding BW, can you please explain how the unity gain freq is reduced by R1/(R1+R2) for inverting configuration? The Vout of inverting configuration will be loaded by feedback resistance R2 (applying miller here). -Bharat
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HdrChopper

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Re: inverting and non-inverting op-amp
Reply #4 - Sep 16^th, 2007, 8:13pm

Hi Bharat,

For both topologies you have GBW~ f-3dB(CL)�|A(CL)| ~ [1] --> "CL" means closed-loop.
Now, from [1] and for the NON-inverting topology you get

GBW=(R1+R2)/R1�fu(loop gain) = (R1+R2)/R1�( fu(opamp) �R1/(R1+R2) )

Thererfore, GBW = fu (opamp)

For the inverting case we have,

GBW= R2/R1�( fu(opamp)�R1/(R1+R2) ) = R2/(R1+R2)� fu(opamp)

*** actually this is the correct reduction factor and not the one in my previous post where I switched the resistors ratio ****

Consequently the GBW is lower for the inverting case assuming the same amplifier.

Regards
tosei

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ywguo

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Re: inverting and non-inverting op-amp
Reply #5 - Sep 17^th, 2007, 12:56am

Hi tosei,

For inverting case fu(loop gain)=fu(opamp)�R1/(R1+R2) is NOT correct. It is shunt-shunt feedback, which is described very well in Chapter 8, Analysis and Design of Analog Integrated Circuits, by Gray, Hurst, Lewis, and Meyer. The feedback facor is fb=if/vo=1/R2.

Furthermore, the two topologies have the same GBW if the basic amplifiers have the same fu (unity gain frequency) and the closed loop gains are equal.

However, the above is theoretially correct only. Simulation proves the two topologies have different GBW where a LM7372 opamp model is used. LM7372 is an opamp produced by National Semiconductor. I thought the parasitic caps and practical loading affect the Gain Band Width.

Best regards,
Yawei

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HdrChopper

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Re: inverting and non-inverting op-amp
Reply #6 - Sep 17^th, 2007, 8:30am

Hi Yawei

I agree: the feedback factor you stated is correct. However, the input current can be modeled by its norton equivalent of Ii with a resistance R1, in which case, the closed loop voltage gain is what I stated in my previous post.The GBW would be the same for both if the inverting case, the transfer function is only VO/Ii

In the "CMOS Mixed Signal circuit design" from Jacob Bakker, page 335-336 it is clearly stated the GBW of both topologies is different.

tosei

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