4 ms·

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.

• jacquesm 17y ago
  The 5 V from the USB. 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).
  • jeromec 17y ago
    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.
    • jacquesm 17y ago
      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.
      • jeromec 17y ago
        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.
        • jacquesm 17y ago
          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.
          • jeromec 17y ago
            Wow, now I know where to look the next time I have a hardware project. Nice.