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How to amplify Hall Effect Sensor Output

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cesimkaol

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Hi Dears,

I have a hall effect sensor 506 you can check it here http://pdf.datasheetcatalog.com/datasheet/MicronasIntermetall/mXuxyww.pdf

I made basic connection . it has 3 pins .

1-> VDD
2-> GND
3->Output

İt is working it gives 0.3 V normally . When I keep magnet close to sensor outpot voltage drops to 0 V.

I want to detect this 0.3 with microcontroller , but this meanless for microcontroler TTL.

So give me a way to me for following .

Thank Your for your attention.
 

Couldn't you use a simple non-inverting op-amp setup with a gain of 10 to bring the output from 0.3V to 3.0V?
 

In the datasheet, page 4, figure 1-1, it clearly shows it is an open drain output. It requires a pullup resistor to Vdd in order to work. I would use a 4k7 or 10k resistor.
 

schmitt trigger said:
In the datasheet, page 4, figure 1-1, it clearly shows it is an open drain output. It requires a pullup resistor to Vdd in order to work. I would use a 4k7 or 10k resistor.
Click to expand...

Friend as you see in the datasheet page 8 , in the table output Voltage you cant see 3 volt or 4 volt on sensor output ;

VOL Output Voltage 130 280 mV

As I showed above , I read ~300 mV when there is no magnet close to sensor, when you put magnet close to sensor , output become 0V .

So as I defined above , I want to amplify this 300 mV with opamp to 3-4 Volts in order to read with microcontroller.

I tried it with LM741 non-inverting type....I change a lot of resistor for different values ( Vout = 1+ R2/R1) for example 10k-1k , 10k-5k , 10k-330ohm but it always give 4.2V on output , I cant handle it.... :((((((

Give me solution....Thank You...
 

I can read datasheets...As an applications engineer, for several years I used to review or write datasheets myself.

Vol means the low state output voltage when the output is sinking 20 mA, and that current which comes from a a pullup load to Vdd.

Why do you think there is no Voh (high state output voltage) specification? Because in an open drain circuit like this one, the output voltage will be equal to the Vdd value (minus a small drop caused by the high state leakage current Ioh).

This case is identical to a LM339 comparator...if you don't have a pullup resistor, you cannot have a high output voltage. The 0.3 volt that appears at the output is the Ioh leakage current which, as read with a high impedance DMM, will appear as a small voltage.

You want a solution? I already gave you one. Use a pullup resistor to Vdd.
 
schmitt trigger said:
I can read datasheets...As an applications engineer, for several years I used to review or write datasheets myself.

Vol means the low state output voltage when the output is sinking 20 mA, and that current which comes from a a pullup load to Vdd.

Why do you think there is no Voh (high state output voltage) specification? Because in an open drain circuit like this one, the output voltage will be equal to the Vdd value (minus a small drop caused by the high state leakage current Ioh).

This case is identical to a LM339 comparator...if you don't have a pullup resistor, you cannot have a high output voltage. The 0.3 volt that appears at the output is the Ioh leakage current which, as read with a high impedance DMM, will appear as a small voltage.

You want a solution? I already gave you one. Use a pullup resistor to Vdd.
Click to expand...

I see that you can read datasheets . Thank you very much :) it helped me , I used pull up on output....And it works...
 

A new question , This project is battery supplied and battery is 20000 mAh.
According to datasheet in Electrical Characteristics table that supply current Idd = 3.2 mA .
Is that 3.2mA that sensor consume while it works? If it is , then battery can supply for 20000/(3.2*24) ~= 260 days .

Is my calculation right ? Can and I am not sure about that 3.2mAh and how many current is sensor consume while it works?
 

Hey cesimkaol,
Supply current(Idd) corresponds to the current drain on the Vs terminal. The supply current is dependent on the supply voltage.And yeah this is the current,sensor consumes while its working.
Your calculations seems right according to me,but there will be other losses too in circuit,due to magnetic field.So it won't work for 260 days.
 

I saw a system which was working with hall effect sensor and was supplied by battery and system seem to me long lasting ( +5 years ) , that means their system consumes at small currents .May be at uA levels instead of mA.

Guess about it , I cant figure out...Thank you!
 
Last edited:

Depending on the battery chemistry, self discharge could make it go flat far earlier than the load calculations predict.
 
I've red now your post, and may be you have already solved all the problems you were facing, but anyway, will you find useful a simple project for the readout of an Hall magnetic sensor?
 
Last edited:

Yes , I solved problems which were we faced.The problem is battery life and ultra low power sensor.... :) I will try to contact distributors for ultra low power sensor.
 

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