Simple current supply for thermistors?

Why constant current? A better way is to have a voltage reference driving a resistor and then the thermistor

A | vref---+-----resistor-----+--------B | | thermistor | | gnd

So, measure Va relative to ground, then Vb, and do the math.

Putting the thermistors in series might get tricky, especially if you use a constant-current source and the temperature swing is wide... you might run out of voltage. It might be better to use a multiplexer to select one thermistor at a time.

John

I agree with John Larkin that the way to go is just to use a resistor divider and some math. The only advantage of the current source would be its linearity, but the thermistor has a very nonlinear response to temperature, so that is out the window. If you select a series resistor that is about equal to the thermistor's resistance in the middle of the temperature range you are interested in, the divider voltage has its largest voltage change per degree in that band of temperature, and is more approximately linear than if the thermistor were driven with a constant current.

All that said, if you still want to use a current source, you can make one with a current reference chip, LM334Z

Of course, the LM334Z is also a temperature sensor all by itself, and a linear one, at that, since it produces a current proportional to absolute temperature, if you just program it with a resistor. So you could replace your thermistor with a resistor, and have the LM334Z sense the temperature. But the signal is smaller than you would get with the thermistor.

Can't get much simpler..

If you want to know how it works....

The base is held at a constant voltage by R1&R2. Therefore the Emitter is held at a constant voltage (=VBASE-VBE).

Therefore the emitter current is constant = (VBASE-VBE)/RE

Therefore the collector current is constant (IC=IE)

You can make a similar circuit using a PNP transistor if you want one lead of the thermistor grounded.

Ghastly analysis.

John

Mine or the one in that paper?

The paper. Assumes a transistor with infinite beta, infinite collector impedance, and simple 0.6 volt Vbe. Plus a couple of silly statements.

John

What about a FET and a source-gate resistor? Ed

Oh that. Yes well it depends what level the pupils are at. They aren't unreasonable assumptions for many applications.

A good way of stimulating a thermistor is with a resistor whose value is equal to the thermistor's resistance at the centre of the req'd temperature range. This gives an automatic linearisation of the thermistor's curve over a moderate range. It's not perfect linearisation, but good enough for most government jobs. --+--Vr | [Rs] = Rth at T-centre. | +---->Vth = Vr*Rth/(Rs+Rth). | [Rth] | --+--0v

eg. For 25C centre-temperature and 0.1mA, Rs = 10k and Vr = 2V.

Plug in various values of Rth and look at the change in Vth per degree. Compare the linearity of DVth/dC with a 0.1mA constant current stimulus.

Usually there is no convenient Vr, so a higher voltage can be tapped down with a pair of resistors.

Vsupply ---+---------------+ | | [R1] [R3] | | +-->V(0C) +--------+-->Vth | | | [R2] [R4] [Rth] | | | 0v----+---------------+--------+

Vr = Vsupply*R4/(R3+R4) and Rs = R3*R4/(R3+R4).

eg. If Vsupply is 5V, then R3= 25k and R4= 16.666k. Nearest preferred values are 24k and 16k, and that could be viable.

R1 and R2 provide the 0C voltage reference, equal to Vth when the thermistor is at 0C.

Thank you to all who replied ... quite a number of good looking suggestions for me to mull over :-)

Mark

It's the usual quality from university academics, most of which have never seen a soldering iron. Maybe there's a reason for outsourcing?

fred

Ed,

Thanks for responding. Could you provide some more details? Perhaps point to a URL with a schematic. Which leg of the FET is used for the current source? Where is the power supply to be connected to?

Mark

Thank you.

view in proportional font (if you are using google, click on "show original")

this circuit gives very tight current regulation: load n-channel (thermistor) mosfet R (see below) _____/\/\/\_________ _______________/\/\/\____________ | _|___|_ _|_ | | | _ / \ | | | |_______/ \____________| | | | c e | | / npn transistor | | \ | | / 1K | |_____________________\ | | ground __|__ _ _____ battery _ or power supply _____ _ | | ground

for R start with a value of .6 divided by desired current and adjust until you get the current you want

This is a sink. If you want a source you could turn everything upside down and do it with a p-channel mosfet and a pnp transistor, and put the thermistor grounded.

The base-emitter voltage of the current-sensing transistor varies with temperature -- the only source of any significant inaccuracy in the circuit, as long as your power supply has stable voltage. Regulation's pretty independent of load. You should look up the b/e temperature coefficient for a bjt, I don't remember offhand.

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

I can't find the original post. What purpose does the OP have in mind for this circuit... maybe he'll find a solution off the shelf, like a temperature sensor chip.