Hi everyone,
my name is Andrea, and I'm an electronic engineer.

I'm trying to develop a Hall Effect sensor interface, to a 0-24V digital electronic device.
I'm using two different Hall Sensors: AH175 and a TLE4935.
The sensor setup is more or less the same, the one with the AH175 has a series resistance on the Output, the TLE hasn't.

I designed an Opamp circuit which brings this square wave with an amplitude of 0-10V, to a signal 0-24V.
This signal, once converted to the 24V, goes into a different line, which is 0-5 V, through a optocoupler.



When I link the AH sensor, everything works fine, but, when I connect the TLE sensor I see a strange behavior on the OpAmp output:


Can you please help me?!
I don't know how it behaves this way.

Thank you very much!

Andrea

 

Hello,

What is the voltage on pin3 of the opamp?
It shows a ground symbol, but there is a text "+Vref motor".
This is confusing.

Bertus

 

You are using OpAmp as comparator, and violating its input common mode range
of the + and - inputs.

The CM range allowed, on a +/- 15V supply is ~ Vpossupply - 4 >= Vin >= Vnegsupply + 4

You would be better off with a comparator, RRIO, and with hysteresis.

Regards, Dana.

 

Hello,

What is the voltage on pin3 of the opamp?
It shows a ground symbol, but there is a text "+Vref motor".
This is confusing.

Bertus

Sorry, the 3rd pin is not Gnd, but is a voltage reference.

 

You are using OpAmp as comparator, and violating its input common mode range
of the + and - inputs.

The CM range allowed, on a +/- 15V supply is ~ Vpossupply - 4 >= Vin >= Vnegsupply + 4

You would be better off with a comparator, RRIO, and with hysteresis.

Regards, Dana.

Thank you for the answer. I’m using it with a single supply of +24V. But how it works with a sensor and with the other it doesn’t?! I cannot understand it

 

You are seriously abusing that op-amp. A series resistor on the input reduces that abuse a bit - add it to the sensor without. Or better yet, change the circuit to use a properly designed comparator with hysteresis.

 

What is the actual voltage on pin 3 of the op am?

Something strange is going on. There should be no reason for the very slow rise of the sensor output signal from 5 V to 10 V - it is supposed to be a digital signal from an open-collector output capable of sinking 100 mA. It simply should not look like it does. The slow rise should not prevent the the op-amp (as a comparator) from switching, but it will make it very susceptible to noise causing unwanted switching without use of hysteresis.

===
D2 serves no useful purpose and only one of R12, R13 and R14 is required.

 

Both Hall sensors have open collector (OC) outputs.

You could eliminate the OpAmp and tie the OCr to D2 and OpTo coupler input,
the internal LED R going to a supply V rather than OpAmp output. That way when OC
turns on the LED in coupler is shut off.

Note AH175 on datasheet states not recommended for new design.

Eliminate R13, not needed. Just R12 to Vsupply and other side to LED OpTo
coupler input pin and OC output of sensor. Also R14 not needed. Just ground the pin.
That way coupler LED has just one R setting its on current, and sensor OC output
when on turns off LED because it shunts all the current away from Opto Coupler
LED to ground.

If OC output sat V causes LED to turn on partially then put a series diode connected
to LED input pin, the other side to OC and LED R the sets LED current. That will make
any current thru LED very small.

Regards, Dana.

 

After looking at the waveform again, my suspicion is that the positive power supply for the amplifier is not connected.

 

First of all, thank you everyone for your answers!

I should give you more details.
The hall sensors and the passives around them comes to me as an “as is”, because they are inside the motor.

I need a power supply of the hall sensors from 5 to 12V, so the signal output of hall sensors, marked as IN on the schematic (pin 2 of the opamp), swallows correctly from 0 to Vmax.

I need 2 different voltages as outputs, 24 and 5 Volts. The waveform I posted before is a screenshot of the signal on the pin 1 of the Opamp.
I put the resistors around the diode of the optocoupler for safety reason, because I didn’t want to broke anything. For me is also the first time dealing with optocouplers.

The power supply is connected to the Opamp.
I would like to remark the fact that the circuit, as is, works fine with one sensor and it behaves that way only with the TLE.
I thought that I wasn’t stressing that much the opamp, because the frequencies are quite low, the highest speed is around 500 Hz, so the SR and the voltages of the opamp should be in-line with the specs.

Previous tests reported a strange behavior of the TLE sensor corresponding to the variation of the R10. If I lower down the R value the ouput waves of the opamp get better, but the opencollector will drain a lot of current when the output is low.

I hope that my further explanation will help.

Thank you again!

 

You still haven't said what the reference voltage on pin 3 is.

What does the signal from the TLE sensor look like with just R10 and R11 in the circuit and no connection to the amplifier?
[EDIT] section deleted
[EDIT 2 - replacement]: I previously thought the waveform was the output from the sensor. I still think there is something actually wrong in the op amp circuit, and it still looks like a power supply problem. The voltage at the output should not be that shape. The low level is OK, but the high level is not - it should go to at least 18 volts or so without that step followed by the slow rise. There is no capacitance on the output and the DC load is not excessively high. Nothing in the circuit as shown explains why that ramp should be there. It shape should not be changed at all by the values or R10 or R11. If it is improved by lowering the value of R10 it suggests that the amplifier is getting its power via R10 and the input protection diode.


I recommend a resistor of 10k to 50k between pin 2 of the amp and the junction of R10 and R11. This will help protect the amplifier input against static discharge and also possible damage if the 12 volt supply to R10 is connected when the 24 volt supply for the amplifier is off. It should not affect the normal operation of the circuit, though it might make some difference with an op amp with lower input impedance. If you add this, please check the waveforms at both ends of the resistor and post photos.

 

Oh, sorry, I lost it.
The circuit is made by two channels, absolutely identic to each other.
The reference is made by a voltage divider, attached to a non inverting buffer. the reference voltage is 8 Volts. It is 8 V because of other circuits specifications.( when 12V sensor power supply)
I will attach more screenshots the next days.

I also do not know why that waveform could appear! I was thinking about a supply current issue of the opamp. It’s a JFET input amp.
I also designed the output stage in order to guarantee a correct current flow inside the zener and the optocoupler diode. Could it be possible that the opamp could not supply enough current to polarize correctly the diode? When putting the probe on the 5V output at the optocoup end, I see a 0-5V square wave!

If you watch in detail the waveform, the ramp starts around 5V, which is the zener voltage.
Do you think that it could be linked to that? I also tried to attach a resistor in series between the IN pin and the TLE sensor, without any results.

 

A square wave at the zener makes sense since it will "clip" all of the ramping part off.

If you look at the Texas Instruments datasheet for the TL084 you will find a curve for output voltage versus current, but at ±15 V supply. You must do a bit of arithmetic and estimation. If we assume that the zener in your circuit is actually a short circuit to ground, which would mean the current would be higher than normal, the maximum possible current from the output would be 24 V/3600 Ω = 6.7 mA. Using the datasheet curve, the output current would be at about that level with ±15 V supply and a bit less than 2 k load (to "ground"), with the output at about 12 volts - so the output would be about 3 volts less than the positive supply. For your circuit, this would be about 21 volts. I said about 18 volts to allow for a bit larger voltage drop. In reality, with the zener conducting at 5 V, the maximum current from the amplifier in your circuit, even with the output right at 24 volts would be 5.3 mA, so our estimate of 18 volts or more should be very safe.

The zener breakdown voltage probably has something to do with the waveform, though IRED in the optocoupler will start to conduct well before the zener does - about half a milliamp at 2 volts, about 2 mA at 4.5 V at the output of the amp. The current will not rise sharply when the zener begins to conduct.

With the amplifier power supply at 24 volts, the current into the input at pin 2 should be only nanoamperes, so changing R10 or R11 should have no effect you would ever see on an oscilloscope.

When you are making measurements, be sure to also scope the power supply pin of the amplifier right on the pin - not nearby on the circuit board or breadboard. Also check the ground pin of the amp the same way. If the power supply is OK, then it is almost certain that the amplifier is defective.