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Symbolic diagram of circuit being used I built a simple LM741 non-inverting amplifier with a fixed gain of 3.1x and verified it works for a simple +0.3 VDC input on pin 3. It uses a dual power supply consisting of two 9V batteries in series to give me a good 0 V ground reference.

I want to add a Hall effect sensor with operating voltage at 4.5V and added a voltage divider as shown (R1 + R2). I verify that I get ca +4.3v. I connected the A1302 Hall sensor as shown following the pin-out specs. I connected the output lead from the Hall sensor (which my voltmeter says is producing +1.0 to +1.4 V). When I connect it to pin 3, I get a fixed output of -8.05 V.

I know that without the Hall sensor the opamp gives the right amplification and voltage with a variable test input, but it only provides a fixed negative (inverted!) output when the Hall sensor input is added, which should vary as I move the magnet across the sensor. I know the sensor works because I have independently tested it. The expected output range of +3 to +4.2 volts is indide the -9 to +9v range set by the rails. I cant seem to figure out why this doesnt work!

Note: I am an astrophysicist not an EE!

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  • \$\begingroup\$ See my edited version and the included diagram \$\endgroup\$
    – SpaceGuy
    Sep 12 at 14:18
  • \$\begingroup\$ I should also replace R2 with a Zener (5.1 V) and adapt R1 as needed. Be aware also of opamp offset. \$\endgroup\$
    – Antonio51
    Sep 12 at 14:28
  • \$\begingroup\$ The voltage at pin a of A1202 is "too" low, the minimum recommended is 4.5 V min -> 5 V. \$\endgroup\$
    – Antonio51
    Sep 12 at 14:47
  • \$\begingroup\$ Have you looked at the voltages with a scope? It might be oscillating. You should put bypass capacitors (100-500 nF) on the power supply rails and the + input of the op-amp, with a series resistor around 1k. \$\endgroup\$
    – PStechPaul
    Sep 13 at 19:01

1 Answer 1

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I want to add a Hall effect sensor with operating voltage at 4.5V and added a voltage divider as shown (R1 + R2). I verify that I get ca +4.3v.

Voltage ok.

I connected the A1302 Hall sensor as shown following the pin-out specs.

Did you forget that the A1302 current supply is approximately 10-11 mA under 5V?
In this case, the voltage you measure is nomore right. It is something like 2.25 V now.
A1302 is not well supplied, so all can happen.

enter image description here

You should use something like this for supplying A1302.

enter image description here

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  • \$\begingroup\$ Antonio51----Wow! Thanks so much, Ill try this. BTW, my batteries are now at B1=8.65 and B2=8.4 volts so with R1=330 ohms and R2=220 ohms across the first battery my voltage divider produces +5.1volts unloaded, but when I attach the A1302 it drops to 3.8V. The output from Pin6 to ground is now between +5.0 and +5.3V depending on whether the test magnet is off or on the sensor....This confirms what you were saying about the dip in the vvoltage supply! I am at least now getting an output from Pin6 that varies with the magnet and isnt 'pinned' near the + rail voltage like before. Getting close! \$\endgroup\$
    – SpaceGuy
    Sep 14 at 18:02

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