Keeping it to the point this time...can anyone shed any light on the required formulae for selecting a resistor to ground from the NI input in a virtual earth summing circuit. I've seen the circuit without this R, but read somewhere that it's inclusion reduces noise and offsets... Anyone?
Ground R??..(short and simple request!)
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pstamler
Well-known member
What you read was right, sort of. In an amplifier with significant input bias current (in other words, bipolar transistors), having the same total resistance to ground on the - and + inputs lowers the DC offset. The way to figure that out is this (assuming the summing resistors are DC-coupled): First, compute the total parallel resistance of the summing resistors; if they're all the same value Rs then the parallel resistance is Rs / n where n is the number of resistors. Call that Rin.
Now use the parallel resistor formula to figure out the parallel resistance of the summing resistors and the feedback resistor Rfb:
1/Rtotal = 1/Rin + 1/Rfb
So let's say you have a 10-input amplifier and the summing resistors (Rs) are all 5k apiece.
Rin = Rs / n = 5000 / 10 = 500 ohms
Let's say your feedback resistor Rfb = 10k; now
1/Rtotal = 1/500 + 1/10,000 = .002 + .0001 = .0021
Rtotal = 1/.0021 = 476 ohms (rounding to the nearest ohm)
That's the total resistance going to DC from the - terminal. Put the same resistance (or the nearest 1% value, 475) from the + terminal to ground.
The down side is that the noise of the amplifier will increase slightly, as there is now twice as much resistance between the amplifier and ground. Of course, you could bypass the resistor on the + terminal with a large cap, eliminating the noise issue.
Note that you don't really need to do this on FET-input amplifiers, as their input bias current is typically small enough to ignore completely.
Peace,
Paul
Now use the parallel resistor formula to figure out the parallel resistance of the summing resistors and the feedback resistor Rfb:
1/Rtotal = 1/Rin + 1/Rfb
So let's say you have a 10-input amplifier and the summing resistors (Rs) are all 5k apiece.
Rin = Rs / n = 5000 / 10 = 500 ohms
Let's say your feedback resistor Rfb = 10k; now
1/Rtotal = 1/500 + 1/10,000 = .002 + .0001 = .0021
Rtotal = 1/.0021 = 476 ohms (rounding to the nearest ohm)
That's the total resistance going to DC from the - terminal. Put the same resistance (or the nearest 1% value, 475) from the + terminal to ground.
The down side is that the noise of the amplifier will increase slightly, as there is now twice as much resistance between the amplifier and ground. Of course, you could bypass the resistor on the + terminal with a large cap, eliminating the noise issue.
Note that you don't really need to do this on FET-input amplifiers, as their input bias current is typically small enough to ignore completely.
Peace,
Paul
bcarso
Well-known member
Great explanation Paul.
I would add (for those who might not know) that the reason this works is that typical bipolar op amps have good matching of input bias currents. The spec that expresses this is called the offset current, the magnitude of the difference between the bias currents. Besides this difference being small or ~zero, it tends to track the bias current magnitude with temperature, hence stays fairly small too.
It's remarkable how many people stick a resistor in series with the n.i. input when it is not needed, just because they have seen it elsewhere and assume it's required. For example, if the amp's output is a.c. coupled there is no need for it unless the d.c. shift due to leaving it out is huge. But that would be a pretty lousy design anyway.
In very rare cases with extremely high impedances one sometimes does the complementary compensation---say a JFET op amp voltage follower looking at 100Mohm at the noninverting input might have a 100Mohm resistor between the output and the inverting input, bypassed with a cap. But in general, for audio signal processing at everyday impedances, the compensating R in the n.i. input with FET amps is pointless, as Paul said.
I would add (for those who might not know) that the reason this works is that typical bipolar op amps have good matching of input bias currents. The spec that expresses this is called the offset current, the magnitude of the difference between the bias currents. Besides this difference being small or ~zero, it tends to track the bias current magnitude with temperature, hence stays fairly small too.
It's remarkable how many people stick a resistor in series with the n.i. input when it is not needed, just because they have seen it elsewhere and assume it's required. For example, if the amp's output is a.c. coupled there is no need for it unless the d.c. shift due to leaving it out is huge. But that would be a pretty lousy design anyway.
In very rare cases with extremely high impedances one sometimes does the complementary compensation---say a JFET op amp voltage follower looking at 100Mohm at the noninverting input might have a 100Mohm resistor between the output and the inverting input, bypassed with a cap. But in general, for audio signal processing at everyday impedances, the compensating R in the n.i. input with FET amps is pointless, as Paul said.
Thanks folks, been trying to search out that answer for a while..You've solved it overnight. I'm using FET input amps, TL072's due to poor availability of parts in my current location, so I think I'll forget that calculation. Not only an answer but one that makes life easier!
Afraid I've come back for more on this one..Just for the sake of learning really.
Say I decide to go with a bipolar input IC, like the Ne5532, which has a significant bias current figure. My mix stages are to sum just 4 inputs. 3 of these have just a 10k mix R from the source, the other needs pot control- so I use a 10k pot in front of the 10k R for that input.
The R for the pot controlled input now varies between 10k and 12.5k. My parallel R for all 4 inputs is now between 2.5k and just over 2.6k. These values paralleled with the feedback R at 10k give me between 2k and 2.08k. What i'm wondering is, does this tiny fluctuation matter with regard to the R to ground from the NI input? Obviously a 2k resistor would be enough to correct the offset to within 5%. How spot on do I need to be?
Perhaps my point would have been better illustrated using higher values of pot/mix R... 100k would give me up to an 800 ohm difference depending on pot position..
My final question is this... As my summing amp is mixing few inputs, the bus source impedance will be relatively large..2k as above with 10k mix R's. In this case would a fet input like the TL072 actually give better noise performance than the 5532..which I believe would prefer a lower source impedance at it's bipolar inputs?
Say I decide to go with a bipolar input IC, like the Ne5532, which has a significant bias current figure. My mix stages are to sum just 4 inputs. 3 of these have just a 10k mix R from the source, the other needs pot control- so I use a 10k pot in front of the 10k R for that input.
The R for the pot controlled input now varies between 10k and 12.5k. My parallel R for all 4 inputs is now between 2.5k and just over 2.6k. These values paralleled with the feedback R at 10k give me between 2k and 2.08k. What i'm wondering is, does this tiny fluctuation matter with regard to the R to ground from the NI input? Obviously a 2k resistor would be enough to correct the offset to within 5%. How spot on do I need to be?
Perhaps my point would have been better illustrated using higher values of pot/mix R... 100k would give me up to an 800 ohm difference depending on pot position..
My final question is this... As my summing amp is mixing few inputs, the bus source impedance will be relatively large..2k as above with 10k mix R's. In this case would a fet input like the TL072 actually give better noise performance than the 5532..which I believe would prefer a lower source impedance at it's bipolar inputs?
> Ne5532, which has a significant bias current figure
Data needed. How high is this current? Bias and offset?
And while you have the datasheet open: what is the offset voltage? Not a heck of a lot of point in nulling the current errors if the voltage error is high.
How much DC is coming off the sources?
How much final offset voltage can you stand? 2 microVolts? 2 Volts?
You have to have a DC blocking cap (or servo) somewhere between input and output. The mixing stage is not a bad place to lose all the DC that accumulates in the inputs.
> would a fet input like the TL072 actually give better noise performance than the 5532
Data needed. What is the noise voltage and current for each device? (You may not find noise current on a GP FET device.)
Data needed. How high is this current? Bias and offset?
And while you have the datasheet open: what is the offset voltage? Not a heck of a lot of point in nulling the current errors if the voltage error is high.
How much DC is coming off the sources?
How much final offset voltage can you stand? 2 microVolts? 2 Volts?
You have to have a DC blocking cap (or servo) somewhere between input and output. The mixing stage is not a bad place to lose all the DC that accumulates in the inputs.
> would a fet input like the TL072 actually give better noise performance than the 5532
Data needed. What is the noise voltage and current for each device? (You may not find noise current on a GP FET device.)
Data needed. How high is this current? Bias and offset?
Input Offset: 10na Typ. 150na Max.
Input Bias : 200na Typ. 800na Max.
Offset voltage is: 0.5mv Typ, 4mv Max.
How much DC is coming off the sources?
Sorry, but How do I know?
How much final offset voltage can you stand? 2 microVolts? 2 Volts?
Sorry, but I don't know how significant this is either...thats part of what I'm trying to work out!
-The mix stage has a large value blocking cap.
would a fet input like the TL072 actually give better noise performance than the 5532 ...Data needed.
Ne5532.. en= 5.0-8.0 nv/ Sq.Rt Hz dependant on frequency.
Input Noise Current= 0.7-2.7 pa/Sq.Rt Hz dependant on frequency.
TL072.. en=15 nv/ Sq.Rt Hz @1Khz. You were right, no data on input noise current.
There's the data, now I hope you're gonna give me a clue what to do with it...
Input Offset: 10na Typ. 150na Max.
Input Bias : 200na Typ. 800na Max.
Offset voltage is: 0.5mv Typ, 4mv Max.
How much DC is coming off the sources?
Sorry, but How do I know?
How much final offset voltage can you stand? 2 microVolts? 2 Volts?
Sorry, but I don't know how significant this is either...thats part of what I'm trying to work out!
-The mix stage has a large value blocking cap.
would a fet input like the TL072 actually give better noise performance than the 5532 ...Data needed.
Ne5532.. en= 5.0-8.0 nv/ Sq.Rt Hz dependant on frequency.
Input Noise Current= 0.7-2.7 pa/Sq.Rt Hz dependant on frequency.
TL072.. en=15 nv/ Sq.Rt Hz @1Khz. You were right, no data on input noise current.
There's the data, now I hope you're gonna give me a clue what to do with it...
> I don't know how significant this is either
There are things that are known, and things that are unknown; in between are The Doors.
Build it as it is; nothing looks like a mistake.
Use a socket; buy both TLO72 and 5532 chips. Also buy a $13 DVM.
Wire it up and poke around. You will have a better sense of proportion. You will find that many sources have many milliVolts of DC on them, that even if you perfectly match the DC resistances around an opamp there is a couple mV of built-in offset, and small unbalances are lost in the overall.
> The mix stage has a large value blocking cap.
Wise move. Then the main effect of accumulated amplified DC on the mix is a loss of headroom. If you run +/-15V supplies and the chip can swing +13 and -13V when output DC is 0.0000V, then when output DC is +2V it can swing (relative to +2V) only +11V and -15V. Assuming audio is symmetrical, you can only get 11V peaks. This 1.5dB lower than with 0.000V DC output. Audio covers a 60dB, 90dB, or more range: losing a couple dB is not a big problem. Anyway your next stage probably doesn't need 13V or even 11V peak: mix stages usually run 1 Volt max at nominal level because real-life mixing is so unpredictable.
There is an additional reason to match DC and AC impedances at inputs: common-mode distortion. However a summer is not too prone to this. And if you were designing at that level of cleanliness, TL072 and 5532 would not be on your palette. They are fine chips; but as ears and meters have improved people have found fault with them. Really funny: they obsess about matching resistances and use DSL chips to listen to commercial recordings that have been through a dozen 5532s (and probably a TL072 or ten) on the way to the master disk.
There are things that are known, and things that are unknown; in between are The Doors.
Build it as it is; nothing looks like a mistake.
Use a socket; buy both TLO72 and 5532 chips. Also buy a $13 DVM.
Wire it up and poke around. You will have a better sense of proportion. You will find that many sources have many milliVolts of DC on them, that even if you perfectly match the DC resistances around an opamp there is a couple mV of built-in offset, and small unbalances are lost in the overall.
> The mix stage has a large value blocking cap.
Wise move. Then the main effect of accumulated amplified DC on the mix is a loss of headroom. If you run +/-15V supplies and the chip can swing +13 and -13V when output DC is 0.0000V, then when output DC is +2V it can swing (relative to +2V) only +11V and -15V. Assuming audio is symmetrical, you can only get 11V peaks. This 1.5dB lower than with 0.000V DC output. Audio covers a 60dB, 90dB, or more range: losing a couple dB is not a big problem. Anyway your next stage probably doesn't need 13V or even 11V peak: mix stages usually run 1 Volt max at nominal level because real-life mixing is so unpredictable.
There is an additional reason to match DC and AC impedances at inputs: common-mode distortion. However a summer is not too prone to this. And if you were designing at that level of cleanliness, TL072 and 5532 would not be on your palette. They are fine chips; but as ears and meters have improved people have found fault with them. Really funny: they obsess about matching resistances and use DSL chips to listen to commercial recordings that have been through a dozen 5532s (and probably a TL072 or ten) on the way to the master disk.
> Input Bias : 200na Typ. 800na Max.
Smoke some really good stuff so we can pretend the "typical" really is a likely number.
Assume the other input is just grounded.
200nA times a 2K resistor is 0.0004V= 0.4mV.
200nA times a 2.08K resistor is 0.000416V= 0.416mV.
0.400mV-0.416mV= 0.016mV drift as you move the pot.
Everything is 4 times worse for the 800nA worst-case. 1.6mV input offset voltage due to bias current and resistors; 0.064mV wobble as you fiddle the pot.
> Input Offset: 10na Typ. 150na Max.
The smoke wore off: numbers that represent cancellation of two numbers, like Offset Current, are never typical; bet on nearly worst-case. (Actually your design should always tolerate a worst-case chip without getting a Warranty Return or requiring you to sort-out marginal chips.)
First: take a wild round number and see if we are in trouble. If the chip in hand has 100nA offset current, then if we "match" DC resistances we will still have half the error due to the uncancelled 200nA bias current. Not a big improvement. True, a lot of chips will be less than 100nA, and maybe even dead-nuts-zero (at specific temperature and moon-phase), but could be 150nA too.
> Offset voltage is: 0.5mv Typ, 4mv Max.
So for ~2K resistances, all this current-error is small compared to the voltage error. If milliVolt errors bug you, you are going to have to trim the offset voltage, and in the process you can also semi-trim the offset current errors. If you are building DC-accurate amplifiers for analog computing, or thermocouple or strain-gauge meters, you have to do that (or buy spiffier chips). For most musical applications, small DC is not a big problem, and big DC can be blocked with a cap.
Remember that until this op-amp concept invaded audio, all our stuff had 6V to 200V output DC offset, and we didn't know there was a problem with that. Low offset +/- power amplifiers can give some circuit simplification, but don't let it go to your head.
Smoke some really good stuff so we can pretend the "typical" really is a likely number.
Assume the other input is just grounded.
200nA times a 2K resistor is 0.0004V= 0.4mV.
200nA times a 2.08K resistor is 0.000416V= 0.416mV.
0.400mV-0.416mV= 0.016mV drift as you move the pot.
Everything is 4 times worse for the 800nA worst-case. 1.6mV input offset voltage due to bias current and resistors; 0.064mV wobble as you fiddle the pot.
> Input Offset: 10na Typ. 150na Max.
The smoke wore off: numbers that represent cancellation of two numbers, like Offset Current, are never typical; bet on nearly worst-case. (Actually your design should always tolerate a worst-case chip without getting a Warranty Return or requiring you to sort-out marginal chips.)
First: take a wild round number and see if we are in trouble. If the chip in hand has 100nA offset current, then if we "match" DC resistances we will still have half the error due to the uncancelled 200nA bias current. Not a big improvement. True, a lot of chips will be less than 100nA, and maybe even dead-nuts-zero (at specific temperature and moon-phase), but could be 150nA too.
> Offset voltage is: 0.5mv Typ, 4mv Max.
So for ~2K resistances, all this current-error is small compared to the voltage error. If milliVolt errors bug you, you are going to have to trim the offset voltage, and in the process you can also semi-trim the offset current errors. If you are building DC-accurate amplifiers for analog computing, or thermocouple or strain-gauge meters, you have to do that (or buy spiffier chips). For most musical applications, small DC is not a big problem, and big DC can be blocked with a cap.
Remember that until this op-amp concept invaded audio, all our stuff had 6V to 200V output DC offset, and we didn't know there was a problem with that. Low offset +/- power amplifiers can give some circuit simplification, but don't let it go to your head.
