HELP - Frequency to Voltage Convertor using a Retro-Reflective Sensor to Measure RPM

MisterBill2

Joined Jan 23, 2018
5,174
The OP's transducer is digital.
The absolute requirement for analogue was not made clear until Post #6!!
Either way a conversion is required for this (digital) device. :rolleyes:
Max.
A
Pulse rate to frequency at a speed adequate for most physical systems is conveniently handled in the analog mode by any of a number of devices, including my favorite, the LM2917. That particular device is able to work with a wide range of pulse amplitudes and wave-forms. Any counter requires also a time base, and any result is delayed by the counting time. And doing it in software demands programng hardware and software, plus the function withing the processor is hidden from normal observation. In addition, the TS obviously wanted an analog solution.
 

Thread Starter

JustinG_KNF

Joined Jan 6, 2020
16
I am still getting 0V as my output. I've switched out the 270k Resistors with 8.2K Resistors.
I've confirmed none of the metal is touching across pins.
I've added my updated circuit below, does anything appear odd?

Does the 10K Resistor (Changed out to 8.2K resistor) in the circuit diagram act as a pullup resistor? Would this mean that the second pullup resistor that I added is not necessary? For what its worth, I tested this without the second pullup resistor and still did not achieve a reading.
1578401751260.png




1578401447749.png
 
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ericgibbs

Joined Jan 29, 2010
9,532
hi KNF,
Sharpened up your image.
Have you confirmed that you have square wave signal of approx 24V coming from the Sensor output, ie: the NPN output junction with the 8k2 to +24V.??
Post #5 circuit.
E
 

MaxHeadRoom

Joined Jul 18, 2013
19,981
If that 8.2k is meant to be the sensor load resistor, it is open circuit to the sensor via the 470pf.
You need a pull up on the sensor output itself.
Max.
 

Alec_t

Joined Sep 17, 2013
10,901
You appear to have a wiring error. Is that a cap (blue thing) between pin 2 and Gnd? It should be a resistor there.
 

MisterBill2

Joined Jan 23, 2018
5,174
I am still getting 0V as my output. I've switched out the 270k Resistors with 8.2K Resistors.
I've confirmed none of the metal is touching across pins.
I've added my updated circuit below, does anything appear odd?

Does the 10K Resistor (Changed out to 8.2K resistor) in the circuit diagram act as a pullup resistor? Would this mean that the second pullup resistor that I added is not necessary? For what its worth, I tested this without the second pullup resistor and still did not achieve a reading.
View attachment 196321




View attachment 196322
Those 82000 resistors look like 8200 on my screen. I suggest a meter check.That would have the input voltage held towards the V= rather solidly. Also, I would say that 24 volts is higher than I would be comfortable with using for IC power, no matter the ratings. Can you use the scope to examine the waveform and voltage at pin 6? That will tell us a whole lot.
 

Dodgydave

Joined Jun 22, 2012
8,805
I am still getting 0V as my output. I've switched out the 270k Resistors with 8.2K Resistors.
I've confirmed none of the metal is touching across pins.
I've added my updated circuit below, does anything appear odd?

Does the 10K Resistor (Changed out to 8.2K resistor) in the circuit diagram act as a pullup resistor? Would this mean that the second pullup resistor that I added is not necessary? For what its worth, I tested this without the second pullup resistor and still did not achieve a reading.
View attachment 196321




View attachment 196322

You need to remove the 470pF capacitor, because at 200Hz it will have a reactance of 1.6Meg ohms . Just input your sensor directly into pin 6 with a 8K2 pullup resistor.
 

schmitt trigger

Joined Jul 12, 2010
135
Did you include the load resistor, Rl in the datasheet schematic?

The output of the LM331 is a current, and it requires a resistor to develop a voltage.

I built an identical circuit for a small steam engine I have. (see attached photo). I placed white stripes on the black flywheel which I then use a photoreflective coupler to read. Pointed by white arrow.

The pulse output goes to the LM331, which outputs a current. That current is directly measured with a 100 microamp analog meter, which on the photo you can see I re-labeled the scale with the revolutions per minute. (U/min)

Because I employed a current meter I did not use a load resistor, but if you plan to measure voltage, you do require it.

IMG_1597.PNG
 

Thread Starter

JustinG_KNF

Joined Jan 6, 2020
16
Those 82000 resistors look like 8200 on my screen. I suggest a meter check.That would have the input voltage held towards the V= rather solidly. Also, I would say that 24 volts is higher than I would be comfortable with using for IC power, no matter the ratings. Can you use the scope to examine the waveform and voltage at pin 6? That will tell us a whole lot.
Here is the waveform:
1578419868493.png
 

ericgibbs

Joined Jan 29, 2010
9,532
hi K,
If those scope images are of the sensor output, there is a problem, they should be approx +5v in amplitude.

E

Edited the approx high level as I see you are now using a 5V supply to the 8k2, not the 24V as originally stated.!!
 

Thread Starter

JustinG_KNF

Joined Jan 6, 2020
16
You appear to have a wiring error. Is that a cap (blue thing) between pin 2 and Gnd? It should be a resistor there.
I think it was just an optical illusion! The resistor goes from pin 2 to gnd. The capacitor goes from pin 1 to ground. Sorry!


Here is my update circuit. I've removed the 470pf capacitor & connected the signal directly to pin 6. No luck so far, still reading 0V.

1578420796055.png
 

Attachments

Thread Starter

JustinG_KNF

Joined Jan 6, 2020
16
hi K,
If those scope images are of the sensor output, there is a problem, they should be approx +5v in amplitude.

E

Edited the approx high level as I see you are now using a 5V supply to the 8k2, not the 24V as originally stated.!!
The Sensor datasheet lists Vsignal as <1.8V. I am powering both the sensor & the LM331 @ 24V.
1578421020263.png
 

ericgibbs

Joined Jan 29, 2010
9,532
hi,
You are reading the d/s incorrectly
High is Vsupply - 1.8v, so if VS is say 24V then Vout high would be 24V-1.8V = 22.2V.

E

Update: from d/s
Signal voltageNPN HIGH ⁄ LOW, Approx. VS ⁄ < 1.8 V
This could be also read as High = Vs and Low = < 1.8v

Either way the output pulse high should be close to VS.
 
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Alec_t

Joined Sep 17, 2013
10,901
You need to remove the 470pF capacitor
Don't think so. It's required for differentiation of the input pulse when the IC is used as a F-to-V converter.
I think it was just an optical illusion! The resistor goes from pin 2 to gnd. The capacitor goes from pin 1 to ground. Sorry!
Agreed. The latest pic clarifies things.
Can you post a scope shot of the signal on pin 6 for the updated circuit?
 
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