MisterBill2
- Joined Jan 23, 2018
- 18,176
Pulse counts are inconvenient to use in linear systems, Max.What is the reason for using this method for RPM detection? Cannot pulse count be used?
Max.
Pulse counts are inconvenient to use in linear systems, Max.What is the reason for using this method for RPM detection? Cannot pulse count be used?
Max.
The OP's transducer is digital.Pulse counts are inconvenient to use in linear systems, Max.
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.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.
Max.
A
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.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 {/QUOTE]
It is not going to work without the pull up.
The one ringed IS 8.2k!
Max.
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
Here is the waveform: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.
I do have a load resistor of 100K OhmsDid you include the load resistor, Rl in the datasheet schematic?
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!You appear to have a wiring error. Is that a cap (blue thing) between pin 2 and Gnd? It should be a resistor there.
The Sensor datasheet lists Vsignal as <1.8V. I am powering both the sensor & the LM331 @ 24V.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.!!
Signal voltage | NPN HIGH ⁄ LOW, Approx. VS ⁄ < 1.8 V |
OK, then the signal is 410 millivolts with a 300 millivolt offset from zero. Probably that is the problem, although I have not looked at the specs for that LM331. You may need some amplification. and possibly an offset adjustment. Possibly a small stepup transformer would work.Here is the waveform:
View attachment 196347
Don't think so. It's required for differentiation of the input pulse when the IC is used as a F-to-V converter.You need to remove the 470pF capacitor
Agreed. The latest pic clarifies things.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!