How to use the ratiometric Hall effect sensor(UGN3503) in a levitron

[Shahriar]

Hi
I want to build a levitron by using a PWM IC (UC3843). I explain the
PWM Controller behavior as below very simple to explain my problem.

PWM=1 if VFB < 2.5V (Controller Vref)
PWM=0 if VFB > 2.5V

In most of Levitrons on the web, The sensor is placed near the ElectroMagnet.
the ratiometric sensors output Voltage (Vout) is 2.5V when there is no magnetic field and Vout will Increase upto 4.2V or Decrease downto 1.2V depending on the magnetic field direction.
and this is what I can't understand:
When PWM is ON, the Vout (i.e.) will be 4.2V and absorb the levitron and Vout will Increase or decrease a little depending on the levitron magnet direction. if we add a Negative Offset voltage to Vout to bring it less down the controller Vref (2.5) to oblige the controller turn off, then when the electroMagnet turns off the Vout becomes a very low voltage and keep the Controller Off. (Vfb = Vout - Voffset)

maybe I couln't explain it very good, to understand my problem try to imagin the Sensor output voltage in different figures when it is close to electroMagnet and PWM turns ON and OFF .
Or In the link below, some one has opened a bought levitron device, can anyone explaine what happens to Sensors output when PWM is On and OFF

http://hi.ece.olin.edu/icb/spring_04/ttgb/student/Ukiyo/gallery.htm

ThanX in avance
Shahriar

 

[ante1]

Hi Shahriar,

Why not buy one, they are not very expensive?
http://shop.conrad.se/cgi/websale6.cgi?Ctx=%7bver%2f6%2fver%7d%7bst%2f3eb%2fst%7d%7bcmd%2f0%2fcmd%7d%7bm%2fwebsale%2fm%7d%7bs%2fconrad%2dschweden%2fs%7d%7bl%2f01%2daa%2fl%7d%7bctx%2fffbcd3b2%2fctx%7d%7bmi%2f000355%2fmi%7d%7bpi%2f%2e%2e671157%2e%2e%2fpi%7d%7bpo%2f1%2fpo%7d%7btpl%2fhead%2ehtm%2ftpl%7d%7bsf%2f%3cs1%3esv%e4vande%20jordglob%3c%2fs1%3e%3cs2%3e%24%241%3c%2fs2%3e%2fsf%7d%7bmd5%2fb92489bd2b990446ddddfb1095826ede%2fmd5%7d

It’s less than 14€.

 

[Shahriar]

Yeah, maybe buiding one is more cheaper.
But I am just curioud about how this man have done this.
The only part that I can't understand is: when the Electromagnet switches, will affect the Hall sensor. and everything goes wrong...
can you explain.

Here the file, I think you have seen it before a year ago

Shahriar

Levitation.pdf

 

[ante1]

It

 

[audioguru2]

Hi Shahriar,
I didn't read the whole article but maybe it pauses the drive to the electromagnet then measures the magnetic field from the permanent magnet during each pause. It would be easy to use the pauses during PWM.

 

[Shahriar]

Dear Ante
Would you mind just telling me the sensor output voltage in two condition (PWM=1 & PWM=0) if we assume the Distance between The sensor and Electromagnet is constant (just approximate numbers).
i.e If the distance is 2Cm the

Vout = ??? when PWM =1
Vout = ??? When PWM =0

Dear Audioguru
I don't think it is using such trick, the controller is just a Fan control.

ThanX
Shahriar

 

[audioguru2]

The sensor is at a right angle to the electromagnet and picks up both its North pole and its South pole that would cancel. Then all it detects is the strength of the permanent magnet in the levitating object which is determined by its distance. Maybe???

 

[ante1]

Dear Ante
Would you mind just telling me the sensor output voltage in two condition (PWM=1 & PWM=0) if we assume the Distance between The sensor and Electromagnet is constant (just approximate numbers).
i.e If the distance is 2Cm the

Vout = ??? when PWM =1
Vout = ??? When PWM =0

That is impossible to answer, it depends on a lot of factors.