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Schematic:

Schematic

Fritzing BB View:

Fritzing BB View

Total n00b: I am trying to layout my breadboard based this schematic, but I'm not sure if I've even come close. The LM34 in the graphic is actually a SS494B hall effect sensor. The schematic view in Fritzing is all sorts of wrong, and I assume the PCB view is as well.

Basically I wired all of this up, and I still get millivolts at what should be the Pin 7 to Arduino. I don't seem to get any amplification at all at pin 1. My assumption was that i translated the Schematic to the BB completely wrong. How off am I?

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  • \$\begingroup\$ Spaghetti! Mmmm.. that's making me hungry...! \$\endgroup\$
    – Majenko
    Commented Nov 12, 2014 at 15:33
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    \$\begingroup\$ First mistake I can see tracing those wires around is R2. You have a 20kΩ resistor between +5V and GND, and the non-inverting input to the op-amp connected to ground. That's wrong - it should be a 20kΩ potentiometer between +5V and GND, with the wiper connected to the non-inverting input. \$\endgroup\$
    – Majenko
    Commented Nov 12, 2014 at 15:37
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    \$\begingroup\$ Pin 3 is grounded, that will rail the op-amp most likely. \$\endgroup\$ Commented Nov 12, 2014 at 15:38
  • \$\begingroup\$ Make sure your real world breadboard has power busses that connect all the way along. Many are broken in the middle. \$\endgroup\$
    – David
    Commented Nov 12, 2014 at 15:48
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    \$\begingroup\$ For the LM324N the Vcc/GND connections on your breadboard look right, but are wrong on the schematic. Pin 4 is Vcc and Pin 11 is GND. \$\endgroup\$
    – Tut
    Commented Nov 12, 2014 at 15:56

2 Answers 2

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Your R2 is supposed to be a potentiometer connected between +5v and Gnd, and with its wiper connected to pin 3 of your op-amp.
What you have is a single fixed resistor from +5v to the op-amp instead.

The function of R2 is to adjust the DC offset level out of that op-amp (at pin 1) but the way you have it connected that one will never produce anything other than a fixed voltage as close as the op-amp can get to +5v.
Since the second op-amp is configured as an inverting amplifier, this voltage is then 'inverted' to become the millivolts you're seeing (its trying to output 0V and a few millivolts is as close as it can get).

So replace your fixed 20k R2 with a 20k potentiometer and you'll have much better results.

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  • \$\begingroup\$ Thank you. I will replace the 20k resistor with a pot as soon as i can. \$\endgroup\$
    – Rick Bliss
    Commented Nov 12, 2014 at 15:50
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If that image is accurate, then I would suggest simplifying your board layout. When I was a TA for Electronics lab classes I often saw students get confused by a maze of criss-crossed wires on the board. Starting with the IC is definitely a good idea. Some of your resistors can connect directly from the LM324 pin to +5V or ground. R7 is a prime example, and either R4 or R5 could be moved as well. Pins 1 and 6 can be connected together with a short wire right next to the IC.

Also, try to move your components closer to the relevant IC pins. The top left of your board with C1/R1/C2/R3 is compact and easy to follow, but it's off in a corner, and there's no reason for the SS494B to be on the other side. If you wanted to add another circuit, you'd barely have any room! You can also move your IC so that the gap in the power-ground rows isn't right next to pins that need them. Maybe put pins 1 and 14 next to gaps. The power and ground wires can be shorter. If you connect the power and ground rows to the nearest holes possible, it leaves more room for signal wires to go between the pins.

Each of these things is a small change, but if you do them all you'll suddenly find that you have half as many wires and that most of your connections are straight lines. It's much easier to work with.

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  • \$\begingroup\$ Wow, this sounds like excellent advice that I can heed for each project. Consider this printed and posted above my bench! \$\endgroup\$
    – Rick Bliss
    Commented Nov 12, 2014 at 23:02

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