Simple current supply for thermistors?

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          E      5V
     I = --- = ------ = 5E-8a = 0.05µa
          R     1E8R
Reply to
John Fields
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It's better, not simpler. None of the circuits discussed in this "simpler" subthread will meet the OP's specs. He wants .1 mA, and AFAIK the best he can expect with a gate-source connected FET is about .2 mA. It gets worse.

He's going to have to face some hard realities. At 100 uA, if it can be achieved, physical construction becomes critical. Assuming 5V Vcc, a 100 meg resistance draws 50 uA. His spec was .5% accuracy. 100uA * .005 = .5 uA So if dirt/humiditity/whatever on his pc board yields a 100 meg or lower path for current to follow, he's blown away by two orders of magnitude, or more.

Frankly, I'm clueless as to how to meet his specs. You could use an op amp to get even better current regulation, but I don't know how to accomodate that on a PC board with those specs. Thompson could design it inside a chip, combining the thermistor function with regulation and whatever else, and yielding an output that is not critical. I don't think it can be done with discretes.

Ed

Reply to
ehsjr

Sheesh! Usually, I hate it when I do that. But in this case, I'm glad. I had dismissed the scheme based on the OP's specs and my error. They are *way* less stringent than I figured.

He can buy a chip that will provide the .1 mA current and have at it: REF200AU-ND from Digikey for $5.20.

No more. Thanks John!

Ed

Reply to
ehsjr

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Couldn' t find the original post, but the OP should check out the Dalis DS18B20. It's an addressable 1-wire 12-bit serial device with an accuracy of +/- 0.5 deg. C, and a readability of 0.0625 deg. C. Maxim-Dalis has it for $2.57 one-off, but the lead time is 7 weeks.

Reply to
Charles Jean

Wow. Thanks!

Mark (a.k.a. the OP)

Reply to
redbelly

One trivial way to do this is using a TL431.

The TL431 uses almost no current in through the adj terminal, so by putting a 25k resistor from adj to ground, and putting the thermistor in the same branch, you can measure the thermistor resistance by noting the voltage at the cathode, as below:

5V | .-. | | 1k or so | | '-' | .----o------ output V = 2.5 + Rt*100u .---o-. | | | | Thermistor | | | '---o-' | | | | | | - o--- ^ TL431 | | | | .-. | 25k | | | | | | '-' | | | | | | | '----' GND

(created by AACircuit v1.28.5 beta 02/06/05

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Slightly cheaper than a current source.

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Regards,
  Bob Monsen

If a little knowledge is dangerous, where is the man who has
so much as to be out of danger?
                                  Thomas Henry Huxley, 1877
Reply to
Bob Monsen

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