Tachometer IC that does not require zero-crossings?

Question

I recently built the basic tachometer circuit in this LM2917/LM2907 Datasheet shown below using the 8-Pin version of the device. I realize this circuit requires a VR pickup sensor, and so the LM2907/LM2917 must rely on triggering off of zero-crossings. But what if I have a square, 0-5V TTL input waveform? Can this IC even trigger off of a non-VR input frequency, or do I need to find another tachometer?

I can only think to use an op-amp to convert the 0 to 5 V TTL square wave to a -5 to 5 V. But that requires a negative voltage supply to the -Vcc rail of the op-amp, which I will not have access to.

Upon further reading of the datasheet, it talks about triggering off of a differential input using the 14-pin version of the device, and further explains this in section 10.1 paragraph 2 (page 13). Perhaps this is what I need? If so, how do you adjust the threshold to which it will trigger off of?

Answer 1

I realize this circuit requires a VR pickup sensor, and so the LM2907/LM2917 must rely on triggering off of zero-crossings.

The data sheet tells you that a zero cross does not trigger anything: -

You need to exceed a positive threshold and then return back below a negative threshold or it won't register anything - this is to avoid noise triggering causing a false count on the tacho circuit.

But what if I have a square, 0-5V TTL input waveform?

AC couple to the input with an RC network is the usual ploy and note that the 8 pin version can have a negative input as low as -28 volts with respect to ground: -

Upon further reading of the datasheet, it talks about triggering off of a differential input using the 14-pin version of the device, and further explains this in section 10.1 paragraph 2 (page 13).

If you want to put a 0 volt to 5 volt square wave in then set the "freed-up" and spare input to a DC voltage of 2.5 volts or use something like this: -

Basically the circuit above is called a data slicer and converts the average signal into a DC signal for use on one of the comparator inputs.

Answer 2

The data sheet for the TACH chip that you linked makes it very clear how the inputs operate. See this:

9.3 Feature Description 9.3.1 Differential Input This device features a Schmitt-Trigger comparator that is the first stage in converting the input signal. Every time the output of the comparator flips between high and low correlates to a half cycle elapsing on the input signal. On the LM29x7-8 devices, one terminal of this comparator is internally connected to ground. This requires that the input signal cross zero volts in order for device to detect the frequency. On the LM29x7 devices, the input terminals to the Schmitt-Trigger comparator are both available for use. This open terminal allows the potential at which the comparator’s output is flipped to be applied externally. This allows the device to accept signals with DC offset or compare differential inputs.

What this means that if your input signal does not cross through zero volts you want to use the device that has both the + and - inputs pinned out. Then you bias one of the inputs to a reference voltage level that matches the midpoint level of the sensor input voltage. The sensor then connects to the other input.

Answer 3

Use this kind of RC circuit. The op amp in this circuit is the internal op amp of your tachometer IC. The RC network will change your 0-5V signal to a -2.5V to 2.5V signal. R1 should be around 10k, C1 should be selected such that the RC time constant is much (at least 5 times) greater than the time period of the square wave which you are expecting.

Answer 4

This Tachometer IC uses any repetitive to make edge-triggered fixed pulse width using a Schmitt comparator and a one-shot.

If you have a unipolar logic level pulse, then the comparator needs a reference Vdc with a resistor divider to the Vin midpoint voltage or 2.5V in your case using 2 equal large R values with a small filter cap.

Alternatively Vin- = gnd , AC couple the pulse to a load R to gnd.

Input current = 1uA max for +/-50mV means input impedance ~ >= 50k ohms. Thus choosing 1k Ohm induces V=IR= 1mV offset which is insignificant. Choosing C=0.1uF gives a 100us time constant pulses +/5V which is more than adequate for 30 mV hysteresis. (Nom) and 1k give relatively low impedance and decay time is slow enough to be detected as a zero crossing on the positive edge.