T&L Publications

Bit of a problem:<p>I need to measure a resistance from 1000 to 8000 ohms and have the output be in volts. i.e. 1 V = 8000 ohms and 8 V = 8000 ohms.<p>I need to read the voltage into a computer port. Thats the wasy part, i just cannot come up with the circuit. I just need the voltage to have some sort of factor associated with it. Example 0.5 V per ohm.<p>thanks

Possibly a constant current source? Then I can measure the voltage drop accross that?<p>Can Anybody design that?

I'm assuming you meant 1V = 1000 ohms.
-Rick

What supply voltages do you have available? Do you have an A/D converter to convert volts to digital data? What is the maximum voltage you can tolerate?

This shouldn't be too difficult.<p>Since 1V = 1000 ohms, 8V = 8000ohms, you need a current source of 1V/1000 = 1mA.<p>That can be done minimally with a pnp transistor and two resistors. One goes to emitter and Vcc, the other goes to base and ground.<p>Then your current source equation would be
I ~=Vcc/2/R if R are the same.<p>Of course, this doesn't take into account the beta of the transistor, nor does it take into account fluctuations in supply voltages, but it is the simplest current source. Also, I'm assuming that 8k is the highest resistance you want to measure, so use a Vcc that's greater than 12V to avoid saturating the current source.<p>If you want beta/supply independence, then it will require an additional zener diode for a reference voltage at the base of the transistor.

Yes, this is not hard you need two transistors, prefferably matched, and two maybe three resistors counting the resistor under test. I'd rather draw but, will try with words to describe the circuit.
1. Two NPNs (Q1 and Q2) wired base to base and emitter to emitter. THis is ground
2. Q1:c to Q1:b is wired together (diode connected).
3. R1 to VCC and Q1:c.
4. R2 to Rtest and VCC.
5. Rtest to Q2:c
6. R1 = (Vcc - Vbe1)/I; set I to 1mA cause it's easy. You might need to make this higher to get a more solid current source, can't remember exactly. This might be in Widlar Current Source territory. You could make one of those, too.
7. R2 = (Vce1(max)/I) - 1 kohm (smallest Rtest) if less than zero, you don't need it.
8. The current going through both transistors should be mirrored, AHA! I finally remembered what this circuit is called a "current mirror". Eureka! I've discovered something I already knew, . . . again.
9. 1 mA through a 1k ohm resistance is 1 volt.
Or 1 mA through an 8k ohm resistance is 8 volts.
10. Google current mirror and see if I'm really senile.
-Rick<p>

Thanks guys!<p>I am taking the data into an A/D.<p>Also...9V WallWart or Battery<p>
Thanks...will try..<p>[ August 26, 2003: Message edited by: Greg ]</p>

Another way to do it is with a OP-AMP. If you use two 9V wall warts. Connect the negative of one to the positive of the other one. This junction will be the ground. Then you will have plus and minus 9V with respect to ground.
Take a opamp, 741, TL081 etc. Take a a 10k or greater pot(potentiometer) and conect the two outside leads to the -9V supply. Connect a 1k ohm resistor from the wiper of the pot to the inverting input(-) of the opamp. Connect a 1k ohm resistor from the non inverting input of the opamp
(+) to ground. Connect a 1k ohm resistor from the inverting input (-) to the output of the opamp. Connect the +9V of the wall wart to the V+ of the opamp and the -9V of the wall wart to the V- of the wall wart. Adjust the pot untill you have exactly 1 volt between the output of the opamp and ground. Replace the resistor that is connected between the inverting input and the output with your variable resistance. The output will vary from 1V to 8V with 1k to 8k of resistance.
It just so happens that the inverting input of the op amp becomes a virtual ground and hence a constant current source of 1ma.
Ned

rosborne,<p>just what I am looking for.<p>When I take the output off the Collector I have 8 Volts for 1 Kohm and 1 Volt for 8 Kohm.<p>Must be measuring wrong.

That's what it should do. Your actually measuring the drop across the transistor if your reference is ground, which is the complement of hte voltage across the resistor under test. If you can use VCC as a reference it should work perfectly. Or you can do an easy software fix which is what I'd do. I hate messing with something that works.
-Rick

what i suspected.<p>done and working.<p>thanks<p>greg

Now i need the source with 500 uA.<p>So I calculated a resistance of 16.7 k Ohm.<p>Is this right? It does not simulate right?

16.7k is correct. If you make Rick's circuit with PNP transistors, the output will be ground referenced.

Russ,
with the same positive supply?

Russ is correct, I should have thought of that and the same supply is fine, but you must change the topology of the circuit. The pnp's emitters go to 9V and the collectors go to the current control resistor and the resistor under test.
-Rick