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| Nicholas Kinar... |
 Posted: Sat Aug 01, 2009 11:21 pm |
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Guest
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Hello--
To preserve bandwidth and slew rate I've selected a high speed op amp
with low wide-band voltage noise and current noise. The 1/f noise is
only 7nV/Hz^(1/2) at 10 Hz, which makes the part suitable for lower
frequency signals extending to DC.
However, the input offset voltage is typically 40 microvolts (300
microvolts max), which may need to be nulled since the op amp is being
used in a non-inverting configuration with high voltage gain.
On my schematic, I have the op amp wired up in a non-inverting
configuration. At the negative input, the feedback resistor is
connected to the output and another resistor is connected to ground.
I've also connected another resistor (with a value of 1 Megaohm) to
this negative input. The terminal of the resistor that is not connected
to the negative input is connected to the output of a 16-bit buffered DAC.
The idea is that the DAC will be used to adjust the voltage at the
negative input, thereby nulling the offset voltage without using a trim
pot. After shunting the positive input of the op amp to GND via a CMOS
switch and load resistor, I'll use a microcontroller to sample an ADC
connected to the op amp output and a minimization algorithm to select
the proper voltage required to reduce the input offset.
However, I'm concerned about whether the 1 Megaohm resistor will inject
significant noise into the feedback loop. Although I know how to
calculate the noise density created by a very large 1 Megaohm resistor,
I am wondering if this noise will be amplified by the op amp feedback.
If so, then is there a way to reduce this noise? Might an RC filter be
an applicable way to go? How do you reduce noise created by a
mechanical trim pot? |
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| John Larkin... |
| Posted: Sat Aug 01, 2009 11:46 pm |
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Guest
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On Sat, 01 Aug 2009 23:21:31 -0600, Nicholas Kinar <n.kinar at (no spam) usask.ca>
wrote:
[quote:b7c1793f8a]Hello--
To preserve bandwidth and slew rate I've selected a high speed op amp
with low wide-band voltage noise and current noise. The 1/f noise is
only 7nV/Hz^(1/2) at 10 Hz, which makes the part suitable for lower
frequency signals extending to DC.
However, the input offset voltage is typically 40 microvolts (300
microvolts max), which may need to be nulled since the op amp is being
used in a non-inverting configuration with high voltage gain.
On my schematic, I have the op amp wired up in a non-inverting
configuration. At the negative input, the feedback resistor is
connected to the output and another resistor is connected to ground.
I've also connected another resistor (with a value of 1 Megaohm) to
this negative input. The terminal of the resistor that is not connected
to the negative input is connected to the output of a 16-bit buffered DAC.
The idea is that the DAC will be used to adjust the voltage at the
negative input, thereby nulling the offset voltage without using a trim
pot. After shunting the positive input of the op amp to GND via a CMOS
switch and load resistor, I'll use a microcontroller to sample an ADC
connected to the op amp output and a minimization algorithm to select
the proper voltage required to reduce the input offset.
However, I'm concerned about whether the 1 Megaohm resistor will inject
significant noise into the feedback loop. Although I know how to
calculate the noise density created by a very large 1 Megaohm resistor,
I am wondering if this noise will be amplified by the op amp feedback.
[/quote:b7c1793f8a]
I assume the resistor from opamp minus to ground (call it R1) is much
lower than 1M... ballpark 100 ohms maybe. In that case, the 1M adds no
significant Johnson noise. It's essentially in parallel with R1 and as
such doesn't change its value much.
Of course, any dac noise gets to the output with a gain of R2/1M,
which should be pretty small. If dac noise is an issue, split the
resistor and bypass the heck out of the junction.
But why not do the zero fix in software? Short the input, digitize the
result, subtract that from future measurements.
John |
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| Eeyore... |
| Posted: Sun Aug 02, 2009 3:40 am |
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Nicholas Kinar wrote:
[quote:1b9c808e3d]Hello--
To preserve bandwidth and slew rate I've selected a high speed op amp
with low wide-band voltage noise and current noise. The 1/f noise is
only 7nV/Hz^(1/2) at 10 Hz, which makes the part suitable for lower
frequency signals extending to DC.
However, the input offset voltage is typically 40 microvolts (300
microvolts max), which may need to be nulled since the op amp is being
used in a non-inverting configuration with high voltage gain.
On my schematic, I have the op amp wired up in a non-inverting
configuration. At the negative input, the feedback resistor is
connected to the output and another resistor is connected to ground.
I've also connected another resistor (with a value of 1 Megaohm) to
this negative input. The terminal of the resistor that is not connected
to the negative input is connected to the output of a 16-bit buffered DAC.
The idea is that the DAC will be used to adjust the voltage at the
negative input, thereby nulling the offset voltage without using a trim
pot. After shunting the positive input of the op amp to GND via a CMOS
switch and load resistor, I'll use a microcontroller to sample an ADC
connected to the op amp output and a minimization algorithm to select
the proper voltage required to reduce the input offset.
However, I'm concerned about whether the 1 Megaohm resistor will inject
significant noise into the feedback loop. Although I know how to
calculate the noise density created by a very large 1 Megaohm resistor,
I am wondering if this noise will be amplified by the op amp feedback.
If so, then is there a way to reduce this noise? Might an RC filter be
an applicable way to go? How do you reduce noise created by a
mechanical trim pot?
[/quote:1b9c808e3d]
I don't quite follow your description but if you have a series R to help
cancel input bias current effects at DC, simply bypass it with a C or R-C
series network to reduce the noise.
Graham
--
due to the hugely increased level of spam please make the obvious adjustment
to my email address |
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| Winfield Hill... |
| Posted: Sun Aug 02, 2009 4:33 am |
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On Aug 2, 1:46 am, John Larkin wrote:
[quote:a6efbf431d]On Sat, 01 Aug 2009 23:21:31 -0600, wrote:
To preserve bandwidth and slew rate I've selected a high speed op amp
with low wide-band voltage noise and current noise. The 1/f noise is
only 7nV/Hz^(1/2) at 10 Hz, which makes the part suitable for lower
frequency signals extending to DC.
However, the input offset voltage is typically 40 microvolts (300
microvolts max), which may need to be nulled since the op amp is being
used in a non-inverting configuration with high voltage gain.
On my schematic, I have the op amp wired up in a non-inverting
configuration. At the negative input, the feedback resistor is
connected to the output and another resistor is connected to ground.
I've also connected another resistor (with a value of 1 Megaohm) to
this negative input. The terminal of the resistor that is not connected
to the negative input is connected to the output of a 16-bit buffered DAC.
The idea is that the DAC will be used to adjust the voltage at the
negative input, thereby nulling the offset voltage without using a trim
pot. After shunting the positive input of the op amp to GND via a CMOS
switch and load resistor, I'll use a microcontroller to sample an ADC
connected to the op amp output and a minimization algorithm to select
the proper voltage required to reduce the input offset.
However, I'm concerned about whether the 1 Megaohm resistor will inject
significant noise into the feedback loop. Although I know how to
calculate the noise density created by a very large 1 Megaohm resistor,
I am wondering if this noise will be amplified by the op amp feedback.
I assume the resistor from opamp minus to ground (call it R1) is much
lower than 1M... ballpark 100 ohms maybe. In that case, the 1M adds
no significant Johnson noise. It's essentially in parallel with R1 and as
such doesn't change its value much.
[/quote:a6efbf431d]
Nick, try this. Instead of resistor voltage-noise, it's often useful
to think of the noise current generated by a resistor. The familiar
e_n = sqrt (4kTR) for voltage-noise density is modified by moving
the R downstairs, giving us its current noise i_n = sqrt (4kT / R).
Also, memorize 4kT = 1.6 x 10^-20 at room temp.
Taking your case, but now with R in the denominator, it's clear that
higher R means less current noise into your summing junction. |
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| Nicholas Kinar... |
| Posted: Sun Aug 02, 2009 9:37 am |
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Guest
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[quote:23fadcbe6e]I assume the resistor from opamp minus to ground (call it R1) is much
lower than 1M... ballpark 100 ohms maybe. In that case, the 1M adds no
significant Johnson noise. It's essentially in parallel with R1 and as
such doesn't change its value much.
[/quote:23fadcbe6e]
Yes, the resistor is close to 100 ohms. It's a good observation that
since the resistor is in parallel, the change is minimal.
[quote:23fadcbe6e]
Of course, any dac noise gets to the output with a gain of R2/1M,
which should be pretty small. If dac noise is an issue, split the
resistor and bypass the heck out of the junction.
[/quote:23fadcbe6e]
I was thinking of using cascaded RC or LC filters to get rid of noise,
but this might prevent the DAC from changing the voltage quickly.
[quote:23fadcbe6e]But why not do the zero fix in software? Short the input, digitize the
result, subtract that from future measurements.
[/quote:23fadcbe6e]
That's a great idea, John! Thank you for suggesting this. Would this
also be effective for cascaded op amps (i.e. when the output of one op
amp is fed into another op amp input)? I would think that it would be. |
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| Nicholas Kinar... |
| Posted: Sun Aug 02, 2009 9:58 am |
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Guest
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[quote:3069913be7]Nick, try this. Instead of resistor voltage-noise, it's often useful
to think of the noise current generated by a resistor. The familiar
e_n = sqrt (4kTR) for voltage-noise density is modified by moving
the R downstairs, giving us its current noise i_n = sqrt (4kT / R).
Also, memorize 4kT = 1.6 x 10^-20 at room temp.
Taking your case, but now with R in the denominator, it's clear that
higher R means less current noise into your summing junction.
[/quote:3069913be7]
Hi Winfield,
Thank you so much for your response! That's an interesting way of
looking at noise in the system.
Memorizing the value 4kt is useful for calculations at room temperature.
It's interesting to note that with higher R, current noise drops. Since
I work with electronics which must operate at lower temperatures (-40
deg C), lowering the temperature will reduce the current noise.
Your reply prompted me to take a look in that book that you've written.
There's a really good section on noise and noise sources! |
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| John Larkin... |
| Posted: Sun Aug 02, 2009 10:46 am |
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Guest
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On Sun, 02 Aug 2009 09:58:22 -0600, Nicholas Kinar <n.kinar at (no spam) usask.ca>
wrote:
[quote:31aff61ec8]Nick, try this. Instead of resistor voltage-noise, it's often useful
to think of the noise current generated by a resistor. The familiar
e_n = sqrt (4kTR) for voltage-noise density is modified by moving
the R downstairs, giving us its current noise i_n = sqrt (4kT / R).
Also, memorize 4kT = 1.6 x 10^-20 at room temp.
Taking your case, but now with R in the denominator, it's clear that
higher R means less current noise into your summing junction.
Hi Winfield,
Thank you so much for your response! That's an interesting way of
looking at noise in the system.
Memorizing the value 4kt is useful for calculations at room temperature.
[/quote:31aff61ec8]
You can also just remember that a 60 ohm resistor has 1 nv/rthz noise,
and figure everything else from that.
John |
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| John Larkin... |
| Posted: Sun Aug 02, 2009 11:01 am |
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Guest
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On Sun, 02 Aug 2009 09:37:31 -0600, Nicholas Kinar <n.kinar at (no spam) usask.ca>
wrote:
[quote:21ba9a1f49]
I assume the resistor from opamp minus to ground (call it R1) is much
lower than 1M... ballpark 100 ohms maybe. In that case, the 1M adds no
significant Johnson noise. It's essentially in parallel with R1 and as
such doesn't change its value much.
Yes, the resistor is close to 100 ohms. It's a good observation that
since the resistor is in parallel, the change is minimal.
Of course, any dac noise gets to the output with a gain of R2/1M,
which should be pretty small. If dac noise is an issue, split the
resistor and bypass the heck out of the junction.
I was thinking of using cascaded RC or LC filters to get rid of noise,
but this might prevent the DAC from changing the voltage quickly.
But why not do the zero fix in software? Short the input, digitize the
result, subtract that from future measurements.
That's a great idea, John! Thank you for suggesting this. Would this
also be effective for cascaded op amps (i.e. when the output of one op
amp is fed into another op amp input)? I would think that it would be.
[/quote:21ba9a1f49]
Sure. Short the input with your cmos switch and digitize the output of
the entire chain. We call this "software autozero."
In some of our products, like thermcocouple scanners, we have, say, an
8-input mux, and one of the inputs is a high-quality ground. All 8
inputs are scanned on a rotating basis. The "ground" value is
regularly digitized in its turn and software lowpass filtered to make
the internal variable ZOFF that's subtracted from all other channel
measurements. The filtering removes most of the noise from the
ground-measurement data.
The filter is just
ZOFF = ZOFF + (GNDSAMPLE - ZOFF)/2^N
where the divide by 2^N is a right-shift. If you have the luxury of
working in floats, you can do
ZOFF = ZOFF + F * (GNDSAMPLE - ZOFF)
where F is a small number, 0.02 or some such.
This just simulates a 1st order RC lowpass filter.
The ZOFF value is also useful as a gross error check.
If you have multiple gain ranges, it may be prudent to have a zero
factor for each.
John |
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| Nicholas Kinar... |
| Posted: Sun Aug 02, 2009 3:02 pm |
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[quote:62bfa407f2]I auto-correct offset at _every_ SAR step. (I'm using the old flying
capacitor stunt ;-)
[/quote:62bfa407f2]
Ah yes, the capacitor is connected at the input of the SAR ADC.
[quote:62bfa407f2]The real stunt is how I get a 10-bit monotonic DAC with a crap
process, but that has to remain a trade secret for the moment 
[/quote:62bfa407f2]
Trade secrets could be traded. |
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