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Off-topic forum - Connect a NTC with a wheatstone bridge to a normal AD-Input

jsiei97 - May 12, 2007 - 08:37 AM
Post subject: Connect a NTC with a wheatstone bridge to a normal AD-Input

Hi

I am thinking about how to connect a wheatstone bridge to a normal analog input. (in a simple cheep way, but a little bit more accurate than just Vdd-R-Rntc-GND)

I am thinking of some sort of electronics that convert the differential measurement to a normal 0-Vdd signal, perhaps with a simple OP (se links).

But I have something in the back off my head that tells me that accuracy is going to be bad anyhow, has any off you any good advice on how to do this.

Thanks
Johan

Note: In this question I am using a NTC, but I am looking for a general solution to measure any resistance in a cheep but good enough way.

Links:
Wheatstone basics:
http://www.omega.com/temperature/Z/TheRTD.html

Wheatstone and OP-amp:
http://www.bbc.co.uk/scotland/education/bitesize/higher/physics/elect/analogue3_rev.shtml

zauberer - May 12, 2007 - 11:17 AM
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Try to see AD62X instrumentation amplifier.

Plons - May 12, 2007 - 12:11 PM
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Instrumentation amplifier is one possibility, another one is the use of the diff amp capability of AVR-ADC itself. See the datasheet of your target AVR, ADC-section.
Unfortunately this feature is "not tested" in PDIP-versions of AVR's (whatever Atmel may mean with that statement Rolling Eyes

)

Nard

ka7ehk - May 12, 2007 - 04:37 PM
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Unless you use a good differential to single-ended coverter, bridge does not help you very much. You CAN use two NTCs in diagonally opposie arms of the bridge to improve sensitivity.

Single-ended is not so bad, especially when the ADC Vref and the source for the divider are the same. It becomes ratiometric, again, almost like half of the bridge is inside the ADC.

Jim

dezellis - May 12, 2007 - 05:51 PM
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Hi,
Depends what you mean by cheap, you may find the following ti.pdf interesting.
http://www.ti.com/litv/pdf/sloa098

Regards

jsiei97 - May 12, 2007 - 06:02 PM
Post subject:

ka7ehk wrote:

Unless you use a good differential to single-ended coverter, bridge does not help you very much.

That makes me wonder what to look for.
What is the difference between a instrumentation amplifier and a good OP-amp?

Or it is possibly easier to explain what to avoid, what is a bad converter Laughing

ka7ehk wrote:

Single-ended is not so bad, especially when the ADC Vref and the source for the divider are the same. It becomes ratiometric, again, almost like half of the bridge is inside the ADC.

If I look on the AT90UsbKey schematics, they have a single-ended solution (attached schematic)?
Is this what you where thinking about? Confused

/Johan

ka7ehk - May 14, 2007 - 03:27 AM
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No, what I am thinking about is just a simple divider between Vcc and ground, one element being a fixed R and one, the NTC. Then set up the ADC to use Vcc as the reference.

Now, the ADC value is proportional to Vcc and so is the divider voltage. Vcc drops out when you write the equation for the ADC output as a function of Rntc. This is the beauty of ratiometic measurement. No op-amp, no instrumentation amp, etc. The one thing you do loose is resolution, because the mid-point of the divider cannot swing all the way from ground to Vcc. You can recover some of this with a differential amplifier, but at the cost of board spacem, power consumption, and expense.

There can be other reasons why an amplifier may be desirable. One happens if the sensor is more than a foot or two from the ADC input or if the environment is especially noisy.

If an amplfier is needed, you can use a regular instrumentation amplifier or the equivalent amp made from several op-amps. It generally takes more than one op-amp to do a good job of converting differential to single-ended.

Another possibility is to use one of the AVRs with a diffential amplifier and gain ahead of the ADC.

Jim

theusch - May 14, 2007 - 05:39 PM
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Quote:

No, what I am thinking about is just a simple divider between Vcc and ground, one element being a fixed R and one, the NTC. Then set up the ADC to use Vcc as the reference.

Ummm--isn't that what the picture shows?

With a 100k bias resistor, Id guess that it is a 30k thermistor--a bias of about 3x the nominal thermistor k works well, depending on the area of interest of the thermistor curve. Also, with a high bias there is a relatively high-impedance signal into the ADC, so slow ADC clock, a small cap on the ADC pin, and/or repeated conversions (tossing the first) may help the sample-and-hold time.

There was a recent extensive thread on this.
http://www.avrfreaks.net/index.php?name ... torder=asc

Lee

ka7ehk - May 14, 2007 - 06:55 PM
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I was hoping that the words would suffice... Here is the idea:

For the (10-=bit) ADC, the following proportion holds, where N is the number produced by the analog-digital conversion:

Vin/Vref = N/1023

For a simple voltage divider with R1 fixed and R2 variable and connected between Vcc and ground:

Vout/Vcc = R2/(R1+R2).

If Vref = Vcc, and Vout (divider) equal to Vin (ADC), then

Vcc*R2/(R1+R2) = Vcc*N/1023

R2/(R1+r2) = N/1023

This is why this circuit arrangment is "ratiometric". Vcc drops out. Given a value for one resistor, you can determine, uniquely, the value of the other from the value of N and never need to know what Vcc is. Simple, no?

This is why, for at least basic cases where the sensor is not too far from the ADC and resolution is sufficient, you don't need an amplifier. Just a resistor and the sensor is all you need. Resolution is the highest, I believe, at the point where the two resistors are about equal.

You DO have to be careful about self heating of the temperature sensor, as well as noise and other such nasties.

Jim

I tried and tried to get a circuit diagram into the message. Sorry, but I could not figure out the magic combo.

Lennart - May 14, 2007 - 07:05 PM
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Quote:

I tried and tried to get a circuit diagram into the message. Sorry, but I could not figure out the magic combo.

If you use "New reply" at top of posts magic combo will be
"Browse" and "Add Attachment"