Noise of a Non-inverting Operational Amplifier Circuit

[@analogdevicesinc]

@Pianoman0488 Hi Paul, we are glad that you liked it and feel free to give us some topic ideas.

[@RobertsAudioTech]

Your tutorials are great! Probably the most succinct and to-the-point explanations I've come across on the topic(s) (especially with your videos regarding the topic) of noise analysis regarding inst-amps et al. I greatly appreciate people that do this to share their enlightenment and teach others. Thanks!

[@analogdevicesinc]

Hi Gabriel, Thanks for the comment... we are glad that you liked it.

[@Jackandthebeanstock]

People dont realise that this is GOLD! for ultra low noise design these calculations are soo useful.

[@antoleyo3857]

I'm an audio-DIY, this video quiet enrich my knowledge.

[@Kabodanki]

I like what analog devices is doing, their site is full with a lot of usefull things,like schematics, eval board even for the simple thing

[@analogdevicesinc]

@knightmoto Excellent catch. You are exactly right. My mistake in the video. When calculating the noise of this configuration referred to output, as in the video, the current noise should not multiply by the parallel combination of R1 and R2 as I stated, but only by R1, which would make the current noise contribution in this example significantly larger. Your formula is nice and clean, but I think the opamp_input_voltage_noise and opamp_current_input_noise should be squared?

[@MantisRay861]

Nice video! I have not learned about this in the slightest in class yet, so thank you for explaining it.

[@iblesbosuok]

Great crystal-clear explanation. I suggest to use learning glass method by prof. Matt Anderson and prof. David Ruzic. Thank you.

[@analogdevicesinc]

Thanks for the tip!

[@ChipUeltschey]

Thanks so much, this was a very clear explanation that leads to more intelligent selection of resistor values.

[@samruthkrishna2947]

Thank you, very brief and useful calculations

[@alessiocaligiuri]

Very good video, many thanks for this explanation!

[@brown50714]

Great resource, will use it again this year.

[@analogdevicesinc]

+Trevor Brown Thank You!

[@rosssilverwood6276]

Now this is information I can use!

[@mmaranta785]

This is great to know

[@ferraritsai]

I have one china equalizer, but I found the OPAmp gain is 2, but the output voltage is too small, so I change from 10k ohms to 2.4k ohm, and SE output can provide 2 Vrms, the original gain noise floor I cannot hear, but rise the gain to 5.16, I can hear some noise when idle. Should I lower the resistor from 10K to how low?

[@rizkiputraprastio7553]

Thanks for your videos. It helps me a lot.

[@manissri]

grt presentation Matt keep it up

[@rogeronslow1498]

Great presentation. Thank you.

[@ggmb1032]

Thank you so much sir. But where does 100 that you multiply by R2 come from?

[@stnhndg]

The opamp is configured as amplifier. From the inverting pin point of view - it's just an inverting amplifier with the gain of R1/R2 (which is 100). R2 noise signal essentialy just a signal that goes on an inverting pin. So it would be multiplied by 100 (just as if it were a usual signal source).

[@kuldeepreddy]

Excellent video, Please do noise calculation for a charge amplifier aattached to a piezo electric sensor. Thank you.

[@fjord2141]

Good video but conclusion at 7:12 about sizing resistors isn't correct. To minimize noise it isn't accurate to say reduce the value of R1. This reduces the output noise contribution of R1, yes, but also reduces the signal gain and hence the RTI noise contribution from R1 goes up as R1 is made smaller (under the assumption that R1>>R2). To see this, let Av = 1+(R1/R2) = (R1/R2) (when R1>>R2), then you can work out that the input-referred noise contributions of the resistors is RTI = 4kT(R2) + 4kT(R2)^2/R1)] [V^2/Hz] = contribution from R2 + Contribution from R1. This equation suggests that one should make R1 larger (not smaller) to reduce its noise contribution. You can re-write the above equation as Noise RTI = 4kT(R2)(1+1/G) where G = R1/R2. Then, you can see if the gain is held constant, you should reduce R2 to improve noise performance. (R1 will reduce as well since gain is being held fixed). So more precise statement = "For a fixed gain, use the smallest resistors possible". Also as another poster seems to have pointed out (based on your reply -- I don't see the original post), the current noise contribution to voltage noise RTO is: in^2*(R1)^2 [V^2/Hz]. You can see this by noting that the virtual ground enforces 0V across R2 and therefore no signal current through R2, so all the current must flow through R1.

[@bambinodeskaralhes]

Incorrect, better signal to noise ratio is achieved by reducing the value of the sqrt( R1//R2)

[@hugovillegas6277]

Thanks for this excellent video; very easy to follow. Just one thing: the assumption here is that the used bandwidth is just 1Hz? If this is the case, if I want to use this device to build an audio pre-amp, I would have to use a BW of 20kHz, so the Johnson noise in R1 would be 575nv instead of 4nv. ¿Am I understanding this right? Thanks again. :)

[@jp_engineer]

Wait at 4:42 the current noise is not passing by the 1k resistor. The positive input is grounded so the negative input will tend to ground. So the current noise should flow over 100k back to the amplifier, because over 1k there is no potential difference. What am I missing ? Can someone help me with this?

[@stnhndg]

First, real amplifers have finite input impedance. That's why it consumes some current (and that's why it has input current noise). Next, since it's a noninverting amplifier it supposedely has some signal source at the + input. It's that just for the sake of noise analysis we got rid of the source (we consider it noiseless). But what it means is that the voltage on the + pin is not always zero.

[@jp_engineer]

@@stnhndg I get you in the matter of the real world. But we simplify with models to get an understanding of specific things, so it lacks that the model is explained?

[@stnhndg]

@@jp_engineer Well, any simplification has its cost. Nevertheless, even in this case it's valid. Since we're dealing with the current source (not the voltage source) it has to draw the current from other points. If it were a voltage source in the same location, your assumption would be correct (considering zero impedance signal source). ... I know, it's a bit complicated in the beginning. Just think in terms of superimposing different noise sources one at a time. It just requires some time to get used to.

[@tramontz2998]

thanks a lot

[@vtrandal]

4.52 what? Volts?

[@e_vt0_pl]

nV/sqrt(Hz).

[@weinihao3632]

It's the output voltage noise referred to the input, thus 457 nV/sqrt(Hz) / 101, where 101 is the gain of the opamp