But then what would he use for the voltage source, + and - 4.5V? It appears he wants the + and - 9V available for the amplifier. But my EET is extremely rusty.
Leave the opamp with it's ground connected to the joint between the two resistors, choose the bias so it drives the output from the -2.5 V the opamp sees (which really is the 0 from the computer) to the +2.5V (which is the +5 of the computer).
I'm only giving the circuit a cursory look, but I thought he wanted a 9V reference voltage added to the sensor voltage going into R7. But I guess a lower reference voltage may be adequate too.
The midpoint voltage of the voltage divider used for the reference is 0.45V, well within the -2.5 to +2.5 swing.
The reason why I think it matters is that batteries are a nuisance, they run empty and will cause a problem with long running experiments, so it's worth the extra time to engineer them out of the circuit.
From what I can see they're only used to power the op-amp.
The 'floating ground' trick is s.o.p. when designing op-amp circuitry that needs to be fed from a single supply.
I said reference voltage, but I meant the voltage for the summing amp would need to be 9V if that's the step up range he's going for. I totally agree about eliminating the batteries if possible; I just wasn't sure how to get adequate voltage available to the amp without the 9V batteries. But again, it's been many years since I've even looked at a schematic. The floating ground idea looks extremely cool though.
Reading the post again he might be able to completely get rid of the whole circuit by placing two Ge diodes in series with the 0 terminal of the LM35, that will raise the 0 by 0.4 V, just enough to get over the threshold.
Have you tried simply raising the ground of the LM 35 by placing a diode in series with the 0 terminal?
That way you might just add a single 10ct component and get rid of the problem completely.
Just enough to raise the ground of the LM35 above the input sensitivity threshold.
One Si diode would probably work, two Ge in series would be slightly better (because you get more range).
So that would be:
+
|
|
+---+--+
| lm35 +---- out
+---+--+
|
---
\ /
v
---
|
|
---+-- gnd
The diode is 'up side down', you simply use the voltage drop to raise the 0 of the lm35 (like a zener diode), it doesn't know any better and will add the output voltage to the ~0.6 V the diode provides.
What happens to the voltage across the diode when the temperature changes? Looks to me like you've just replaced a calibrated temp sensor with two sensors in series!
The temperature of the diode will change right along with the temperature of all the other hardware, it's not as though you'll go out of your way to get the diode heated up.
The way it's drawn is not the way it is built, the diode would normally sit near the computer end of things, not right on top of the sensor.
The thermal drift of the other circuit is at least as large as that of a diode at room temperature, probably a whole lot larger.
