Reply by yvancastilloux April 12, 20052005-04-12
Hi, thank you for your answer.

However, our hardware consultant made simulations with the complete
model.  He was able to approximate the following formulas:

C2 =  Lowpass cacitor
C1 = ADC capacitor (40 pF)
R1 = Sensor output resistance (32kOhm)
R2 = ADC capacitor (2 Kohm)

The initial sampled value when the sensor is first sampled is equal to
:C2/(C1+C2) 
 
 The first order settling time is:
R2 * C1 *  10.  It represents the minimum time to charge C1.
 
The second order settling time is
R1 * C2 *  3  It represents the minimum time to charge C2.

For C2 = 10nF :
                The initial sampled value will be at 99.6% of the
final sampled value ~ 8 LSB with a settling time of 1ms
 
For C2 = 100nF
                The initial sampled value will be at 99.96% of the
final sampled value - 0.8LSB with a settling time of 10ms.

These values were observed experimentally as well.  The 8LSB noise
value is much larger than the theorical one.

Yvan


--- In msp430@msp4..., "Karl Adler" <specific2@m...> wrote:
> Your 32Kohm+2kohm tell me that your sensor is
an ADXL accelero,
which should 
> be considered a current source.If you don't buffer
it then you
should allow 
> enough time (SHT0_10 for example) for the ADC12
input caps to
charge. The 
> noise that you see  while sampling is normal
according to data sheet.
> Karl.
> 
> >From: "yvancastilloux" <kayoux@h...>
> >Reply-To: msp430@msp4...
> >To: msp430@msp4...
> >Subject: [msp430] ADC sampling time problem
> >Date: Sun, 03 Apr 2005 22:05:04 -0000
> >
> >
> >
> >Hi,
> >
> >We're using the 12-bit ADC for 5 sensors.
> >
> >Reading the spec sheet of the ADC, it says that it can sample at 8kHz
> >and more.  The model specifies an impedance of 2000kohm and 40pF.
> >
> >Our sensors have first-order low-pass filters.  The output is not
> >buffered by an amp.  That is, the output of low-pass is fed directly
> >to the ADC input.
> >
> >I calculated the time constant.  when the internal switch of the ADC
> >closes, I assume that the low-pass capacitor(10nF) is an open-circuit.
> >The 32kOhm low-pass resistor is then in series with the 2kOhm.
> >
> >The time constant is then: R*C=(32k+2k)*40pF=0.00000136 sec.
> >Setting the sampling time to 10*R*C=0.0000136 sec.   If a drive the
> >ADC with a calibrated DCO at 4 MHz, this amounts to 54.4 clock cycles.
> >  However, the signal is still very noisy at 64 cc.  It takes 50 ms of
> >sampling before the signal stabilises.  If I increase it to 1024 cc,
> >the signal stabilises faster but is still very noisy.  If I put a 1uF
> >capacitor(low bandwidth) and sampling time of 1024 cc, the signal
> >becomes much clearer but not as clean as it should be.
> >
> >I believe that this is due to 2 things:
> >
> >1)the low-pass capacitor exchanges charges with the ADC capacitor.
> >The lower the capacitor, the more a loss of charge affects its
> >voltage.  However, I just can't figure out the equation that shows
> >that. Anyone does?
> >
> >2)cross-talk between sensors is amplifed by not having a buffered
> >output. Is this true?
> >
> >Thank you for your answers.
> >
> >Yvan
> >Newtrax Technologies
> >www.newtraxtech.com
> >
> >
> >
> >
> >
> >.
> >
> >
> >Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> >
> 
> _________________________________________________________________
> Don't just search. Find. Check out the new MSN Search! 
> http://search.msn.click-url.com/go/onm00200636ave/direct/01/




Beginning Microcontrollers with the MSP430

Reply by Karl Adler April 3, 20052005-04-03
Your 32Kohm+2kohm tell me that your sensor is an ADXL accelero, which
should 
be considered a current source.If you don't buffer it then you should allow 
enough time (SHT0_10 for example) for the ADC12 input caps to charge. The 
noise that you see  while sampling is normal according to data sheet.
Karl.

>From: "yvancastilloux"
<kayoux@kayo...>
>Reply-To: msp430@msp4...
>To: msp430@msp4...
>Subject: [msp430] ADC sampling time problem
>Date: Sun, 03 Apr 2005 22:05:04 -0000
>
>
>
>Hi,
>
>We're using the 12-bit ADC for 5 sensors.
>
>Reading the spec sheet of the ADC, it says that it can sample at 8kHz
>and more.  The model specifies an impedance of 2000kohm and 40pF.
>
>Our sensors have first-order low-pass filters.  The output is not
>buffered by an amp.  That is, the output of low-pass is fed directly
>to the ADC input.
>
>I calculated the time constant.  when the internal switch of the ADC
>closes, I assume that the low-pass capacitor(10nF) is an open-circuit.
>The 32kOhm low-pass resistor is then in series with the 2kOhm.
>
>The time constant is then: R*C=(32k+2k)*40pF=0.00000136 sec.
>Setting the sampling time to 10*R*C=0.0000136 sec.   If a drive the
>ADC with a calibrated DCO at 4 MHz, this amounts to 54.4 clock cycles.
>  However, the signal is still very noisy at 64 cc.  It takes 50 ms of
>sampling before the signal stabilises.  If I increase it to 1024 cc,
>the signal stabilises faster but is still very noisy.  If I put a 1uF
>capacitor(low bandwidth) and sampling time of 1024 cc, the signal
>becomes much clearer but not as clean as it should be.
>
>I believe that this is due to 2 things:
>
>1)the low-pass capacitor exchanges charges with the ADC capacitor.
>The lower the capacitor, the more a loss of charge affects its
>voltage.  However, I just can't figure out the equation that shows
>that. Anyone does?
>
>2)cross-talk between sensors is amplifed by not having a buffered
>output. Is this true?
>
>Thank you for your answers.
>
>Yvan
>Newtrax Technologies
>www.newtraxtech.com
>
>
>
>
>
>.
>
>
>Yahoo! Groups Links
>
>
>
>
>
>
>

_________________________________________________________________
Dont just search. Find. Check out the new MSN Search! 
http://search.msn.click-url.com/go/onm00200636ave/direct/01/


Reply by yvancastilloux April 3, 20052005-04-03
Hi,

We're using the 12-bit ADC for 5 sensors.

Reading the spec sheet of the ADC, it says that it can sample at 8kHz
and more.  The model specifies an impedance of 2000kohm and 40pF.

Our sensors have first-order low-pass filters.  The output is not
buffered by an amp.  That is, the output of low-pass is fed directly
to the ADC input.

I calculated the time constant.  when the internal switch of the ADC
closes, I assume that the low-pass capacitor(10nF) is an open-circuit.
The 32kOhm low-pass resistor is then in series with the 2kOhm.

The time constant is then: R*C=(32k+2k)*40pF=0.00000136 sec.
Setting the sampling time to 10*R*C=0.0000136 sec.   If a drive the
ADC with a calibrated DCO at 4 MHz, this amounts to 54.4 clock cycles.
 However, the signal is still very noisy at 64 cc.  It takes 50 ms of
sampling before the signal stabilises.  If I increase it to 1024 cc,
the signal stabilises faster but is still very noisy.  If I put a 1uF
capacitor(low bandwidth) and sampling time of 1024 cc, the signal
becomes much clearer but not as clean as it should be.

I believe that this is due to 2 things:

1)the low-pass capacitor exchanges charges with the ADC capacitor. 
The lower the capacitor, the more a loss of charge affects its
voltage.  However, I just can't figure out the equation that shows
that. Anyone does?

2)cross-talk between sensors is amplifed by not having a buffered
output. Is this true?

Thank you for your answers.

Yvan
Newtrax Technologies
www.newtraxtech.com