Resistor Noise in Op Amp Circuits - Page 1
Forum section: electronics / beginners

author: jim_griff
author_rank: Regular Contributor
timestamp: April 10, 2014, 04:28:13 am
content: Hi there. I have a question regarding resistor noise in op amp circuits and how to reduce it, but also about impedance and loading on op amps. Theattached imagesays it all:Example 1: 100k ground resistor / 1M feedback resistor = lots of noiseExample 2: 0.1 ohm ground resistor / 1 ohm feedback resistor = very low noise.How low is too low? Obviously 'Example 2' is too low, since the op-amp I am looking at using (LME49990) would most definitely find it difficult when loaded with 1.1 ohms straight down to ground. The reference LME49990 design for a "balanced mic preamp" shows 10k resistors all around, but I presume these are going to be very high noise for the max 60dB gains I'm looking at implementing.I have searched the internet for days trying to find answers on how different combinations of resistors affect the performance of a basic op amp design. Too much resistance means too much noise, but too little resistance means too much loading straight down to ground. However, I have no idea how to choose the 'sweet spot', or whether the sweet spot is different for different op amps and different situations. i.e. Mic preamp vs high impedance guitar preamp vs line level preamp.Thanks.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 10, 2014, 10:01:28 pm
content: Neither of your examples are particularly noise sensitive. Noise in the microvolts is 60dB down from a quiet mic in the milivolts, which is in turn 60dB down from ~1V line level.Some users may prefer it to be noisy anyway, e.g., phono preamps. So much for analysis.The best (engineering) value to pick is the value that's near the equivalent input noise resistance. For BJT amps, this is usually in the kohms (I think I've seen this spec listed before, though uncommon). FET amps have this more in the 100kohm range.Tim

author: David Hess
author_rank: Super Contributor
timestamp: April 10, 2014, 10:43:26 pm
content: Quote from: T3sl4co1l on April 10, 2014, 10:01:28 pmThat is how I do it. The noise contributed by the resistance is the root of the sum of squares of the current noise through the resistance and the Johnson noise from the resistor itself. The later can usually be ignored so I select the resistance so the current noise through it equals the existing voltage noise. Note that for a feedback or attenuation network, it is the Thevenin (parallel) equivalent resistance that counts.This does indeed work out to about 500 ohms for low noise amplifiers, 1.5 kilohms for slow precision amplifiers, and 10s of kilohms for low input current bipolars with JFET and CMOS input amplifiers being much higher.Linear Technology has a good design note discussing this.

author: Legion
author_rank: Frequent Contributor
timestamp: April 11, 2014, 04:09:46 am
content: Quote from: jim_griff on April 10, 2014, 04:28:13 amAssuming you can see the noise on a scope and that any given resistance on a potentiometer produces the same amount of noise as the same value resistor, you could try replacing the resistors with pots. Then you could watch the noise on the scope as you adjust the pots until you find the sweet spot.Keep in mind I've never tried this, but it seems plausible.

author: David Hess
author_rank: Super Contributor
timestamp: April 11, 2014, 06:34:49 am
content: Quote from: Legion on April 11, 2014, 04:09:46 amI have done this very procedure although not to find the best resistance to optimize the noise because calculating that is trivial. An AC voltmeter will be more sensitive than an oscilloscope. A spectrum analyzer that works down to DC would be even better in some cases.When I did it, I was interested in low frequency noise so I used a high resolution sampling DC voltmeter with the amplifier setup to amplify its own noise by 1000 times. I took a series of samples over ten seconds and calculated the RMS noise from the standard deviation.

author: jim_griff
author_rank: Regular Contributor
timestamp: April 11, 2014, 12:03:16 pm
content: Wow, thanks for all the responses! I'm still a little new at this so it's taking a while to sink in.The project I'm working on at the moment (as a total newbie) is a 48V phantom powered microphone preamp(schematic below). I would like to have a stepped attenuator to control the gain stage, but it would be 24 resistors in series for the lowest gain setting, gradually stepping through all resistors until it hits the 22 ohm input resistor for max gain.Would 24 resistors in series cause 24x the Johnson noise? Or is it the root of the sum of the squares of the Johnson noise per resistor? i.e. sqrt(R1²+R2²+R3²...)Attachments are:(1) Schematic.(2) Calculations for the resistors to be used in the stepped attenuator, corresponding to 2dB increments of gain.(3) Graph showing the actual gain when using best fit resistor vaules.(4) Stepped attenuator.

author: tszaboo
author_rank: Super Contributor
timestamp: April 11, 2014, 12:14:58 pm
content: The sweet spot is opamp and application dependent. In your example, the opamp has to drive 1 Ohm load. Normal opamps will not do that, that is way too much current. I assume, you want this for audio. Most audio opamps are designed for 600Ohm load or 10Kohm load, headphone amplifiers can be lower than that. You will have some load on the output, but if you connect this to the input of an other opamp, than the load is basically 0. Than depending on your taste, you can make your feedback 10K, or even 600 Ohm. Also, look into the THD+N/load graphs! The higher the load, the bigger the THD. So you dont want to decrease the resistor noise when it will significantly increase the THD.If you are looking for the least thermal noise possible, I've seen suggestions to parallel circuits. Not opamps, the whole circuits have to be paralleled, and the output is summed. I suspect cheap 5532 circuits outperform more expensive opamps in terms of noise, but I've jet to meet real data (Douglas Self doesnt count).

author: Kevin.D
author_rank: Frequent Contributor
timestamp: April 11, 2014, 01:08:06 pm
content: Quote from: jim_griff on April 11, 2014, 12:03:16 pmYes or a little shortcut sqrt (R1+R2+R3....) x 4 nV/ rtHz .where R is value's of resistors expressed in K Ohm.A nice little vid here :-(he has a few videos on noise in amps ,all Matt Duffs short videos are usefull , concise and to the point. )

author: macboy
author_rank: Super Contributor
timestamp: April 11, 2014, 01:27:54 pm
content: Quote from: Kevin.D on April 11, 2014, 01:08:06 pmMore to the point, youcouldcalculate the thermal noise of each resistor, then add them as sum of squares, but the result is the same if you just sum resistor values and then calculate the thermal noise once.For parallel resistors, calculate the parallel resistance then find the thermal noise of that. e.g. the input and feedback resistors of the op-amp are effectively in parallel (from AC/noise perspective) so the effective resistance seen at the op-amp input is the parallel value. This means that the total noise is lower than the noise from either resistor alone.

author: jim_griff
author_rank: Regular Contributor
timestamp: April 11, 2014, 01:28:37 pm
content: Quote from: Kevin.D on April 11, 2014, 01:08:06 pmOh nice, thanks. I found another video with the same guy a couple hours ago. Nice trick too. I was hoping for something simple to work out the noise of resistors. Some of the formulas out there are really mind boggling. I can stare at them for hours and they still look like hieroglyphics.

author: jim_griff
author_rank: Regular Contributor
timestamp: April 11, 2014, 01:31:11 pm
content: Quote from: macboy on April 11, 2014, 01:27:54 pmThanks. I was wondering about parallel resistors and how to calculate the noise from those.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 11, 2014, 06:10:30 pm
content: Noise doesn't care how many parts it's coming from... it all adds in the same way.You don't even need a resistor component to generate noise. A 50 ohm transmission line has... you guessed it, a 50 ohm equivalent noise resistance. Assuming it's correctly matched, of course. (Which might ultimately be a dumb resistor terminating one end, but it might also be an antenna* -- not that free space is any less noisy -- or an amplifier with no resistors at all in the signal path.)*Also, note that the antenna is resonant, so it's only matched over a narrow range of frequencies. The noise is filtered accordingly, of course.Tim

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: April 19, 2014, 02:41:29 pm
content: Quote from: T3sl4co1l on April 11, 2014, 06:10:30 pmIts my understanding that when talking about noise levels as a single value, a set of phase-randomized signals will add as the root of the sum of the squares.The sum of several uncorrelated sources will obviously be a simply linear sum of their levels (e.g. voltage). But as a function of probability, at any given time one source might be 'high' and another might be 'low'. So they tend to sum to a much lower value.

author: djsb
author_rank: Super Contributor
timestamp: April 19, 2014, 03:02:10 pm
content: Very informative topic. I've often wondered how the noise figures quoted on data sheets actually relate to a real circuit. I've just subscribed to Analogs youtube channel.David.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: April 19, 2014, 03:38:12 pm
content: It will cost you a few bucks, but I highly recommend Doug Self's book on Small Signal Audio Design. He covers some interesting ways to cheat mother nature and achieve lower noise levels than the theoretical levels of a physical resistor.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 19, 2014, 06:50:53 pm
content: Quote from: Conrad Hoffman on April 19, 2014, 03:38:12 pmAn op-amp around a feedback resistor is a refrigerator.The thermal power coupled through that resistor is considerably miniscule, so it doesn't actually do anything noticeable, not at room temperature anyway, but it is thermodynamically true.Tim