# Brainy Memorial Assortment of Miscellaneous Projects

Discussion in 'Tele Home Depot' started by jimdkc, Aug 1, 2016.

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1. ### BarncasterDoctor of Teleocity

Feb 3, 2010
Northern California
jimdkc, come in jimdkc. Over.

2. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO
OK... that is freaking amazing packaging! Not at all what I was planning, but VERY cool!

I've printed out several of your pics and will attempt to trace the circuit and see if I can see anything wrong.

My first inclination was, "Are the jacks isolated?"... but I see you addressed that!

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3. ### Captain NutslotFriend of Leo's

May 3, 2012
Rochester, NY
Hey! What's up?

4. ### BarncasterDoctor of Teleocity

Feb 3, 2010
Northern California
There's no need to fear, Jim is here!

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5. ### BluesBloodedFriend of Leo's

Oct 11, 2012
Thanks Jim, if you need other pictures, let me know

6. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO
I think that the problem is the higher resistance pot you are using in place of the trim pot.

I intentionally designed the circuit to use relatively low values in the zero adjust voltage divider, and then higher values in the output scaling voltage divider, and then the even higher resistance of the volt meter (typically 10 Megohms, or 10 million ohms).

The low resistance of the zero adjust voltage divider has a relatively high current running through it: 9v/(510+200+510) = 9v/1220 ohms = about 7.4 milliamps.

The output scaling voltage divider consists of larger resistance values... a total of 13K ohms... more than 10x the other divider. And the maximum voltage across it will be a theoretical maximum of 2.5 volts. So the maximum current through the output scaling voltage divider is 2.5/13K = about 0.19 milliamps.

Now... if you pull 0.19 milliamps away from the 7.4 milliamps flowing through the zero adjust circuit, it's not going to have much effect.

So, lets put a 2K pot in the zero adjust circuit. current flowing through the circuit is now 9v/(510+2000+510) = 9v/3020 ohms = or about 3 milliamps.

When you draw 0.19 milliamps from the 3 milliamps it will have a more pronounced effect. You are throwing off the zero setting by drawing too much current from it. You are loading down the zero adjust circuit too much.

There are 2 ways to fix this problem:

1. Put a much lower value in for the zero adjust pot. 100 to 200 ohms... maybe 250 ohms tops.

or

2. Increase the resistance values of the output scaling voltage divider. Use 100K instead of 10K and use 30K instead of 3K.

Try one of these 2 fixes and see if it doesn't work better...

If that doesn't help, we may need to add some capacitors to the power supply circuit and the Hall effect sensor circuit. I intentionally designed the circuit to use a battery specifically to avoid having to do that...

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7. ### BluesBloodedFriend of Leo's

Oct 11, 2012
Thank you Jim,

I will try option #2. If it still does not work, I will hook a battery for power. I also have a 200ohm trimpot on the way.

I will report back.

8. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO
For anybody following along, a 250Ω pot is probably easier to find than a 200Ω pot. You might want to also drop the other 2 resistors down a bit from 510Ω to 470Ω to keep the current up in this circuit.

Also, even if you're not having problems, it might be a good idea to increase the 10KΩ-3KΩ resistors to 100KΩ and 30KΩ. This will load down the zero-adjust circuit even less, resulting in slightly better accuracy.

Using an AC power supply rather than a battery, you might want to put a 0.1µF capacitor across the output of the Hall effect sensor (from pin 3 to pin 2). Physically connect this capacitor as close to the sensor as possible. This is called a "decoupling" capacitor, and its purpose is to filter out any high-frequency noise from the power supply. Batteries are inherently "quiet" power supplies, so you can get away without this.

(Anybody catch the error in my previous calcuations? I used 9V instead of 5V. This means that the current in the zero-adjust circuit is even lower, and the output scaling has even more effect than I figured...)

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9. ### BluesBloodedFriend of Leo's

Oct 11, 2012
Jim, I have a 500 omh pot on the way as well. It was the lower pot value I could find.

If at all possible, I would rather use the 500 ohm pot, it goes better with my enclosure design than a trimpot.

Would it work with lower resistors value say around 250ohm?

10. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO
My intention when designing the zero adjust circuit was to provide about 1 volt of adjustment range. To get 1 volt across a 500 ohm pot, you need 2 milliamps of current. This would put the other 2 resistors (at the top and bottom of the pot) at 1000 ohms each.

Since this would be even lower current, you would want to go with the higher values in the output scaling circuit (100K + 30K instead of 10K + 3K).

This is why I chose the low values for the zero-adjust circuit in the first place.

The advantage of using higher resistances throughout would be longer battery life. The disadvantage would be noise susceptibility due to the tiny currents being measured.

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11. ### BarncasterDoctor of Teleocity

Feb 3, 2010
Northern California
André,
Let me know if you need me to send you a calibrated rod magnet for testing.
Rob

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12. ### BluesBloodedFriend of Leo's

Oct 11, 2012
That would be great, PM sent.

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13. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO
Very generous offer Rob. The calibration magnet you sent me really helped to verify and legitimize this whole concept!

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14. ### BarncasterDoctor of Teleocity

Feb 3, 2010
Northern California
Always glad to aid in the effort!

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15. ### BluesBloodedFriend of Leo's

Oct 11, 2012
I'm reporting back

Rob sent me a magnet that reads on his gravitas
South 1465 gauss
North 1570 gauss

I redid my board with the new values

1k resistor on each side of the 500ohm pot to make the the first voltage divider

100k and 33k resistor (I did not have a 30k resistor on hand)

I added a 0.1uf cap on pin 2 and 3 of the hall sensor.

My results are
South 1136
North -1235

I'm off by a 300 gauss. Can this be due to the higher 33k resistor in the last voltage divider?

16. ### BarncasterDoctor of Teleocity

Feb 3, 2010
Northern California
Hi André,
Looks like it's getting closer! Also, my readings are peak readings across the face of the magnet poles. Across the face of a given charged Alnico rod, my meters can show a difference of perhaps +\- 75 gauss as you move the sensor about. Their little magnetic fields would seem to have minor irregularities. I would say if you got within 75 gauss of my values, you could call it done.
Rob

17. ### BluesBloodedFriend of Leo's

Oct 11, 2012
Hey Rob,

Yep it's getting there. The good thing is that the 500ohm pot works great to adjust the dmm to zero before probing.

I also noticed that the hall sensor effect needs to be flat out on the end of the rod to give the best reading. All in all although it does vary, it's quite consistent reaching the peak value. It just takes a couple of readings and fooling around until I get the highest reading.

Since this is not a commercial endeavor but a personal measuring tool, I now have a comparison tool. I can use it to compare what gauss magnets have and what impact they might have on sound and study different gauss level. Accurate or not I'm better equipped with the meter than without.

The fact that you sent me a reference magnet is priceless because I now have a comparison point. I will build another one on the prototype board and see if playing with different resistor value impact the reading. If only I did not to sleep so much

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18. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO

The 33K resistor will make the readings a bit low... but not that low.

I calculated what the readings would be with a 30K resistor, and they came out:

South 1162
North -1364

I took a look at the data sheet of the Allegro A1302 Hall Effect Sensor. The specification for the sensitivity of the sensor can actually vary from 1.0 to 1.6 millivolts per gauss. My circuit was designed for the TYPICAL sensor sensitivity of 1.3 millivolts per gauss.

I lucked out and got a sensor that is very close to typical. It looks like you may have gotten one on the lower end of the spec.

Since you have a reference magnet, here's what you can do to calibrate the circuit:

1. Temporarily replace the 33K resistor with a pot. Maybe 50K.

2. Turn that pot all the way up, then zero your meter.

3. While measuring your reference magnet, and turn the pot down until you read the correct reading for the reference magnet.

4. CAREFULLY (without changing the value) disconnect the pot and measure its value.

5. Replace the temporary pot in the circuit with a resistor that is as close as you can find to the value you read on the pot.

6. Verify your new circuit by measuring the reference magnet again in both directions.

I've uploaded a copy of the sensor datasheet if you want to take a look at it.

#### Attached Files:

• ###### A1301-2-Datasheet-2.pdf
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757.1 KB
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19. ### jimdkcFriend of Leo'sSilver Supporter

Age:
61
Mar 12, 2009
Independence, MO
Also, based on some calculations... you could just try a 3.0K or 3.3K resistor in place of the 33K... should be pretty close! (Actual value I calculated was 3.13K)

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Oct 11, 2012