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Hall-effect sensor, need to know resistance

by user_name
Tags: halleffect, resistance, sensor
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user_name
#1
Mar9-10, 04:18 PM
P: 13
Im trying to use a hall effect linear sensor for position sensing. the sensor im looking at getting is here.

The application I will be using it in is an electronic drums hi hat pedal. To achieve variable control of the virtual opening/closing of the hi hat, the drum module I am using requires the pedal "must have a hat down resistance of <100 ohms and hat up of >10k ohms."

I know that V=IR but not sure how to find the resistance here. Based on this products specs I would like to know the resistance in ohms of this sensor in order to figure out if its feasible to use and whether or not ill need to add a pull up resistor or trim pot, etc in line to make it compatible, etc. Thanks in advance!!
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berkeman
#2
Mar9-10, 05:05 PM
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Quote Quote by user_name View Post
Im trying to use a hall effect linear sensor for position sensing. the sensor im looking at getting is here.

The application I will be using it in is an electronic drums hi hat pedal. To achieve variable control of the virtual opening/closing of the hi hat, the drum module I am using requires the pedal "must have a hat down resistance of <100 ohms and hat up of >10k ohms."

I know that V=IR but not sure how to find the resistance here. Based on this products specs I would like to know the resistance in ohms of this sensor in order to figure out if its feasible to use and whether or not ill need to add a pull up resistor or trim pot, etc in line to make it compatible, etc. Thanks in advance!!
Welcome to the PF. How are you planning on making a magnetic field that varies linearly with position? The Hall sensor says it's outout is linear with B-field, but proximity sensing a magnet wouldn't seem to be real linear itself. How linear do you want to make the output with respect to the high hat opening size?
sophiecentaur
#3
Mar9-10, 05:08 PM
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I see your problem. This device appears to give a 'Voltage' output swing of 0.4V into a 4.7kΩ load yet the drum module is stating its required input in a different way. Why not feed a simple transistor 'driver' circuit with this output to pull the drum module 'up' or 'down', whichever you need? Can you look at the spec of the drum module to find what's actually at its input? I should imagine that the required resistances are to ground.
Is this just an on/off requirement or is it somehow proportional control? If it's just on/off then a reed switch would provide the necessary 'logic level' translation.

user_name
#4
Mar9-10, 05:45 PM
P: 13
Hall-effect sensor, need to know resistance

@berkeman:thanks nice to be here. this has been accomplished by others using a traditional acoustic type hi hat stand by fixing the hall effect sensor inside the bottom hi hat cymbal then attaching a neodymium ring magnet to the hi hat clutch just underneath the top hi hat cymbal then adjusting their proximity to achieve desired response. The magnet is moved towards and away from sensor via hi hat pedal mechanism.

What I'm trying to do is a little bit different. In my electronic drum kit there is a hi hat pedal which controls the virtual opening/closing of the hi hat and theres a fixed cymbal (rubber pad with a piezo) for the drummer to hit..there is no stand.
Currently, my pedal operates as a momentary switch; either fully open or fully closed. I have upgraded drum modules and the new one allows for variable control of the hi hat provided the pedal is capable of sending positional signals.
The sensor would be fixed to the base of the controller and the magnet attached to underside of foot pedal just above the sensor. Pushing the pedal down brings magnet almost in contact with the sensor which alters the resistance and Vout. I can fine tune the linearity of the circuit by adjusting placement and strength of the magnet and fine tune via parameters in the module.
user_name
#5
Mar9-10, 06:28 PM
P: 13
Quote Quote by sophiecentaur View Post
I see your problem. This device appears to give a 'Voltage' output swing of 0.4V into a 4.7kΩ load yet the drum module is stating its required input in a different way. Why not feed a simple transistor 'driver' circuit with this output to pull the drum module 'up' or 'down', whichever you need? Can you look at the spec of the drum module to find what's actually at its input? I should imagine that the required resistances are to ground.
Is this just an on/off requirement or is it somehow proportional control? If it's just on/off then a reed switch would provide the necessary 'logic level' translation.
Typically, the module reads the voltage coming in from the foot pedal and varies its response accordingly. The hall effect sensor would be wired to a TRS output jack and 1/4" stereo cable routes the signal to the module.

Unfortunately, I dont know the voltages the module is looking for and the manual only mentions the 'ohms' specs I mentioned above. I guess I am assuming that if I can create a circuit within those parameters that I'll be able to fine tune the functionality via control params within the module menu.
I am attempting to make a on/off pedal into a continuously variable pedal by using a hall effect sensor. It has been accomplished before with other drum modules so I'm sure it will work here I just need to find out how. blazing trails...

Im sorry sophiecentaur, I didn't follow you when you stated the 'Voltage' output swing of 0.4V into a 4.7kΩ load.... I'm learning all of this on the fly...would you please tell me what you meant and where/how you got the 4.7 kohm load?
Bob S
#6
Mar10-10, 11:52 AM
P: 4,663
I suspect that your drum module input is a current source (or perhaps a voltage source with a series resistor), perhaps 10k to match the range specified in the OP. The best thing you can do is a) measure the voltage of the module input open circuit (with no load connected) and b) measure the module input current with the input shorted. This will allow designing a matching circuit from the Hall Effect sensor.

Bob S
user_name
#7
Mar10-10, 12:13 PM
P: 13
Ok I can do that but I need to know how I go about getting those measurements.. is it possible to get the measurements by reading them from the tip/ring/sleeve of the cable plugged into that input? If so, can you guide me as to how? Thank you for your help.

Rob
Bob S
#8
Mar10-10, 02:05 PM
P: 4,663
Quote Quote by user_name View Post
Ok I can do that but I need to know how I go about getting those measurements.. is it possible to get the measurements by reading them from the tip/ring/sleeve of the cable plugged into that input? If so, can you guide me as to how?
Yes, you read the voltage and current by measuring them at the tip/ring/sleeve of the input cable. Use a simple voltmeter with a built-in milli-amp-meter. The dc voltage is probably less than 10 volts (open circuit), and the dc current probably less than 10 milliamps (shorted circuit).

Bob S
user_name
#9
Mar14-10, 09:20 PM
P: 13
Quote Quote by Bob S View Post
Yes, you read the voltage and current by measuring them at the tip/ring/sleeve of the input cable. Use a simple voltmeter with a built-in milli-amp-meter. The dc voltage is probably less than 10 volts (open circuit), and the dc current probably less than 10 milliamps (shorted circuit).

Bob S
Alright, so the voltage is 5 volts and I'm not sure I measured the current correctly but I believe it is 6.5-6.6mA? How I measured current is set the multimeter to 20mA, with the module on, I turned it off then put the red lead on the sleeve and black lead on the tip and the reading jumped around for a bit then settled down around those numbers. Does that sound about right?
I really want to get this hall effect sensor hat pedal set up so I hope this info does the trick! Whats the next step? thx
Bob S
#10
Mar14-10, 10:03 PM
P: 4,663
These measurements are very informative. It implies that the input impedance is a ~760-ohm resistor connected to a 5-volt source. I will guess that a suitable Hall Effect output would be a common-emitter open-collector NPN transistor with the output collector current being 0 to 6.5 milliamps. A voltage output 0 to 5 volts should also work, as long as it can sink up to 6.5 milliamps. The hat-down resistance < 100 ohms corresponds to an input voltage less than about 100*5/(760+100) = 0.58 volts, and a hat-up resistance > 10 kohms corresponds to ~ 5 volts.

Bob S
user_name
#11
Mar15-10, 01:30 AM
P: 13
Thanks so much Bob for this great information. Based on the above information, will this product be suitable for the intended application?

Also, will this require any additional electronics (resistors, battery, etc) in the circuit to make it compatible with the drum module? If so what are they?

I plan to wire the HES to a 1/4" output jack in a project box with a TS cable connecting to the drum module hat controller input jack. The project box will reside at the base of the hi hat pedal. The magnet will be mounted to the underside of the pedal and distance from the sensor proportional to pedal open/close position. Hopefully, this will accomplish with <$10 what Yamaha, Roland, etc attempt to do while charging >$150. Plus the solution is more elegant.. no pots, no opticals, no moving parts other than the affixed magnet. Much thanks again to you Bob for your generous help.
user_name
#12
Mar15-10, 12:09 PM
P: 13
Quote Quote by user_name View Post
Alright, so the voltage is 5 volts and I'm not sure I measured the current correctly but I believe it is 6.5-6.6mA? How I measured current is set the multimeter to 20mA, with the module on, I turned it off then put the red lead on the sleeve and black lead on the tip and the reading jumped around for a bit then settled down around those numbers. Does that sound about right?
I really want to get this hall effect sensor hat pedal set up so I hope this info does the trick! Whats the next step? thx
Hi
Could you please verify for me that I indeed measured the DC current using the correct technique? I am just uncertain as to whether I did or not and if this value is actually valid. Thanks!
Bob S
#13
Mar15-10, 03:39 PM
P: 4,663
Quote Quote by user_name View Post
Could you please verify for me that I indeed measured the DC current using the correct technique? I am just uncertain as to whether I did or not and if this value is actually valid. Thanks!
Your measurements are correct. You need to do three things.
1) You need a source of dc volts between 9 and 15 volts. You need to purchase a 7806 3-pin regulator (at local electronics store) to generate a regulated 6 volts for the Honeywell Hall Effect sensor and an op-amp.

2) The Honeywell sensor will not sink 6 milliamps, so you need to buffer the output with an opamp. Buy a LM324 quad op amp or preferably a LM358 dual op amp (these work on a single rail 6 volt supply).You need only one op amp. Initially this should be wired as a noninverting gain of one (voltage follower)**. Put a ~50-ohm resistor in series with the op-amp output to the drum module.

3) Connext the ground of the sensor and the op-amp to the neg terminal of the 7806. connect the positive termial of the sensor and the op-amp to the +6 volt output of the 7806

** connect the Hall Effect sensor output to the (+) input of the op-amp. Connect the output of the op-amp back to the (-) input of the op amp. Connect the output of the op-amp to a ~50 ohm resistor in series with the drum module.

Bob S
user_name
#14
Mar15-10, 08:09 PM
P: 13
Quote Quote by Bob S View Post
Your measurements are correct. You need to do three things.
1) You need a source of dc volts between 9 and 15 volts. You need to purchase a 7806 3-pin regulator (at local electronics store) to generate a regulated 6 volts for the Honeywell Hall Effect sensor and an op-amp.

2) The Honeywell sensor will not sink 6 milliamps, so you need to buffer the output with an opamp. Buy a LM324 quad op amp or preferably a LM358 dual op amp (these work on a single rail 6 volt supply).You need only one op amp. Initially this should be wired as a noninverting gain of one (voltage follower)**. Put a ~50-ohm resistor in series with the op-amp output to the drum module.

3) Connext the ground of the sensor and the op-amp to the neg terminal of the 7806. connect the positive termial of the sensor and the op-amp to the +6 volt output of the 7806

** connect the Hall Effect sensor output to the (+) input of the op-amp. Connect the output of the op-amp back to the (-) input of the op amp. Connect the output of the op-amp to a ~50 ohm resistor in series with the drum module.

Bob S

For the 50-ohm resistor, which wattage do I need (1/8, 1/4, 1/2, 1, 2, 5, 10, 50, etc)?

Are any capacitors and/or diodes necessary?

Project parts list
- SS495A Hall Effect Sensor linear
- LM7806T 6 VOLTS @ 1 AMP POSITIVE VOLTAGE REGULATOR-3 PIN TO-220 PACKAGE
- LM358N DUAL OPERATIONAL AMPLIFIER (LOW POWER)-8 PIN DIP PACKAGE
- 51 OHM 1/2 WATT 5% CARBON FILM RESISTOR
- Magnets
- Breadboard
- Project Box
- 1/4" TS Phone Jack

This is going to be awesome if it works
Bob S
#15
Mar16-10, 07:35 PM
P: 4,663
Please confirm that you want a variable analog hi-hat control ("variable control of the virtual opening/closing of the hi hat") and not an up/down open-close switch, because the circuits are different (although both use the same opamp). For the circuit I described above, change the connection from the output of the opamp back to the (-) input with two 10k resistors in series. This is so that if you need gain, you can attach a third 10k resistor from the junction between the two 10k's to ground. This will give x2 gain.
Bob S
user_name
#16
Mar17-10, 02:05 AM
P: 13
Quote Quote by Bob S View Post
Please confirm that you want a variable analog hi-hat control ("variable control of the virtual opening/closing of the hi hat") and not an up/down open-close switch, because the circuits are different (although both use the same opamp).
Yes, variable analog hi hat control is what I'm after. Drum modules capable of this functionality typically allow a range for hi hat pedal 'openness' and are able to generate random minute changes to the samples played based on whether the pedal is closed, 1/4 open, 1/2 open, 3/4 open or fully open. The pedal I am currently using can only send on/off signals but is perfect for use in a DIY conversion like this.

Quote Quote by Bob S View Post
For the circuit I described above, change the connection from the output of the opamp back to the (-) input with two 10k resistors in series. This is so that if you need gain, you can attach a third 10k resistor from the junction between the two 10k's to ground. This will give x2 gain.
Bob S
Given that I am trying to achieve continuously variable hi-hat control, is this change still necessary? I am unclear as to whether you meant for this change to be implemented regardless of analog or on/off hi-hat control or if I should make the change to the circuit only IF I'm trying to achieve on/off control.
user_name
#17
Mar17-10, 11:22 AM
P: 13
Quote Quote by user_name View Post
Alright, so the voltage is 5 volts and I'm not sure I measured the current correctly but I believe it is 6.5-6.6mA? How I measured current is set the multimeter to 20mA, with the module on, I turned it off then put the red lead on the sleeve and black lead on the tip and the reading jumped around for a bit then settled down around those numbers. Does that sound about right?
I really want to get this hall effect sensor hat pedal set up so I hope this info does the trick! Whats the next step? thx
Ok I think I made a mistake above. I rechecked the module input after doing some reading and what I quoted above isn't for the current, it is the resistance. When the selector dial on my multimeter is pointing to '20k' for [tex]\Omega[/tex], the other end points to '20m' for ~A.

So with the module turned off I get 6.5k [tex]\Omega[/tex] and with the module turned on, if I reverse the leads, I get -4.96 with the selector at 20k [tex]\Omega[/tex] and -38.3 with the selector at 200k [tex]\Omega[/tex]. Not sure if those last 2 numbers would tell you anything, I just thought it was odd that they were so different..

As for the current, I retesed and with the module on I get 1.03mA and 0.00mA when it is off.

The voltage is 4.95.

I apologize for the error, I just misread where the dial was pointed. I assume this will change the design of the circuit?
Bob S
#18
Mar17-10, 02:11 PM
P: 4,663
Hi-
What I present here (in thumbnail) is an amplifier circuit that interfaces between the Hall Effect sensor and the module input. The amplifier can be run with a gain of one, or a gain of 9.5.

The amplifier shown is the Linear LT1413, which is the same as the National LM358. The power supply is the 7806 (6 volts out).

V1 represents (simulates) the Hall Effect sensor output with a 2.5 volt quiescent dc offset and a 0.1 volt peak, 1000Hz sine wave. For a gain of one, all you need is the 10k (R1) from the LM358 output back to the (-) input. The simulation shows two more resistors, R2 and R3 (2k and 2.8k) with the ratio chosen such that the midpoint voltage (between R2 and R3) out is also 2.5 volts. If the (-) input is also tied the the R2 R3 midpoint via the gain switch (as shown), the amplifier gain is about 9.5, which is shown in the simulation. Do NOT put a capacitor from the midpoint to ground, as R2 and R3 are part of the feedback gain circuit.

If you need the gain, you may have to use a potentiometer to tweak the R2 R3 midpoint voltage to get a 2.5 volt quiescent voltage out.

Bob S
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