# Questions on interfacing circuit for an NPN input

I have to use this sinewave outputting sensor with the NPN input of this converter. My previous question was related to the same issue but the question was about a totally different circuit design i.e. a single CE amplifier. Even the circuit was working in practice, from many expert comments I was suggested to make an interface with a comparator instead and use voltage protection.

Input is a sensor which outputs sine waves from 200mVpp to 12Vpp between 1Hz upto 100Hz. The circuit below I aim to convert these sinusoidal inputs to pulses and via comparator's open collector I want to pull down NPN input (Y1) of the F2V converter. Output impedance of Y1 is not given so I measured it with different resistors as around 2k (R1 in the circuit).

In simulation I get the following voltage output at Y1:

It seems this works in simulation for the desired input ranges.

Is my way of driving this sensor to the converter correct? Can this circuit be optimized or improved? Are the locations of TVS and diodes correct? I'm just wondering if there is fundamental error before I start soldering.

edit:

Above is how I'm going to configure the setup. The points A, B, C and D are in question for protection. I was only thinking for the moment MOV or TVS for the point A.

EDIT2:

Corrected comparator circuit:

Suggested self powered transistor circuit:

I apply this noisy signal as an input to both:

With 100mV hysteresis the comparator circuit inputs:

Comparator circuit final output at Y1:

Transistor circuit's final output at Y1:

• What is source impedance of Vin and is there any common mode noise? Another way to clamp the input voltage is two back to front parallel diodes to clamp the Vpp to Vf with say 10K series R. Then amplifying to trigger a non-retriggerable mode one shot ( dual IC) for 10ms will give Vdd out at 100Hz and Vdd/10 at 10 Hz. This can be done in 1 or 2 IC's. Ensure reverse polarity protection to Vdd using 5V automotive LDO. – Tony Stewart Sunnyskyguy EE75 Sep 1 '17 at 13:35
• Is it important to you to have the output @ Y1 in a known state (either high or low) when your sensor is inactive, or disconnected? Your circuit now gives an undetermined output for this condition, depending on resistor tolerances of R21,R22, R23, R24. – glen_geek Sep 1 '17 at 13:55
• Y2 isn't a voltage rail. It's a $20\:\textrm{mA}$ current source. Are you able to see that from the PXF-20 datasheet you linked? – jonk Sep 2 '17 at 1:37
• @jonk Thanks if you say so I didnt get it right it says " Internal sensor supply 12,6V to 14V @ 0-20mA." When I measure Y2 with scope it shows 14.2V thats why I thought it can be used for power. So I have to use a separate power supply right? If I use a separate power supply for the circuit do I need to isolate it via optocoupler or it is ok this way? I would be glad if you have any other inputs for this circuit and values. Especially my concern is for choosing a TVS for overvoltage protection since I will place the sensor in open air. Any of your input is precious thanks. – atmnt Sep 2 '17 at 1:46
• @user134429 If you look at their schematic, it's kind of sloppy but it basically shows you (on the right side) what their schematic is. On the left, they show you three kinds of inputs they can accept and how they would be wired. That part looks like a rats nest, messy. They could have shown three separate diagrams and it would have been clearer. They are supplying you with a $20\:\textrm{mA}$, if you want it. (Worth thinking about.) Or you can just ignore Y2 and give them a clean signal between Y1 and Y0. – jonk Sep 2 '17 at 5:38

To be clear:

1. I don't have your PXF-20 that I might use to validate some assumptions I'm making. And its manual is somewhat "cloudy" to my reading of it.
2. It seems as though your anemometer, from its datasheet, just as a fat, two-axis, four-pole magnet spinning inside a 4100 turn #40 coil to generate the AC.
3. The AC voltages are larger when the anemometer is spinning faster (as would be expected); at the slower rates this can be as little as $80\:\textrm{mV}_\textrm{PP}$; at the higher rates this can be over $12\:\textrm{V}_\textrm{PP}$.
4. There's no indication in your writing (that I recall) which tells me that all you want to do is to hook up your anemometer to your PXF-20 so that it can output a nice, clean signal for you based only upon the frequency.

From these, I conclude that there is no particular reason to complicate what's needed by adding a separate power supply if it is possible to instead power the circuit directly from the PXF-20.

I think that's possible with a fairly simple circuit.

simulate this circuit – Schematic created using CircuitLab

The left side, including $R_1$, $R_2$, $R_3$, $R_4$, $Q_1$, and $C_1$ play two important roles here. Using $Y_2$ and $Y_0$ they provide $\approx 12\:\textrm{V}$ power supply for the rest of the circuit (which requires only a very small amount of current to operate.) The exact voltage of this power supply isn't critical. These parts also provide an output to $Y_1$, which is specified in your datasheet as providing hysteresis (important) with a band sitting between $5\:\textrm{V}$ and $7.5\:\textrm{V}$. Without any input from your anemometer, $Y_1$ will be sitting at about $1\:\textrm{V}$, well below the threshold.

Now, the right side is powered and when the anemometer generates even a small signal the remaining circuit will alternately release and then drive the Darlington pair, $Q_4$ and $Q_5$, allowing $Y_1$ to rise well above the $7.5\:\textrm{V}$ threshold and fall well below the $5\:\textrm{V}$ as the small AC signal continues. This will work with even a small AC input, but it will also work with the higher generated voltages, too.

I've picked high-$\beta$ NPN transistors here and I think it's worth the trouble to use them. (I didn't include it, but you might also include a $680\:\textrm{k}\Omega$ resistor from the base of $Q_4$ to $Y_0$.)

There is no lightning protection here! Just be aware.

I've made some assumptions that are possibly unwarranted about your anemometer. If I had it here, I could do some testing on it. But I don't. I'm guessing that the wire resistance alone is at least $300\:\Omega$ in the wound coil. I honestly don't actually know, though. But I think the circuit is worth a shot.

If you do decide to try it, first just build up the portion that includes $R_1$, $R_2$, $R_3$, $R_4$, $Q_1$, and $C_1$. Then verify that there is something from about $10\:\textrm{V}$ to about $12\:\textrm{V}$ on the emitter of $Q_1$. You should also see about $1\:\textrm{V}$ less than that at $Y_1$. If that's working out about right, then you can consider adding the rest and seeing if that works okay.

Mostly, this is to help you consider the idea of a self-powered circuit and show you one of the simpler (but less accurate) ways to approach solving that problem when interfacing an anemometer like this. It takes advantage of the provided current source and provides signal conditioning sufficient perhaps to drive the $Y_1$ input, as well.

I'm a little more worried about the slower anemometer speeds because of the tiny AC voltages generated. If that's still a problem more can be done, of course. You might try and change the value of $R_6$ a bit, one way or another, and see how that adjusts things better. But you'd probably need an oscilloscope tied to $Y_1$ and grounded at $Y_0$ to know better, then.

By the way, if you want to try and simulate your permanent magnet anemometer you might look over "Permanent Magnet Synchronous Machine Model for Real-Time Simulation" to see what I had to play with in trying to make sure the circuit I provided had a shot at working. I was able to get a behavior that relatively closely mimics the specifications provided by the anemometer datasheet, where merely changing the rate also changes the voltage approximately in the way they said it did. But it was annoying to develop and it's non-trivial. But it gave me a little better confidence that the circuit might work well, too. Worth a moment's trouble.

• Many thanks for the nice answer. You are right Y2 is a current source I verified it by connecting 220 Ohm resistor I read around 5V. But I dont know why they write 0-20mA instead of just 20mA is its a current source(?). Also what's the purpose of Y2 as current source, I would think better to provide a voltage source for one's gadget. I opened the sensor and it is like 4 pole magnet rotating and inducing voltage across the coil. So the coil resistance hence the sensor's resistance I measured as around 680 Ohm. – atmnt Sep 2 '17 at 11:45
• There will be 12V or 24V power supply anyway for PFX, so I was planning to use the same supply for my comparator circuit's power rails. Is your transistor circuit have enough hysteresis? Im a bit worried about overvoltage caused by weather lightning ect. Can I add TVS between sensor output and the circuit in your transistor circuit like in my circuit? – atmnt Sep 2 '17 at 11:50
• @user134429 My circuit doesn't have any hysteresis. Your PFX-20 has the hysteresis at the input $Y_1$. I think it would be wise to worry about lightning. But I'm not qualified to comment on providing good protection, either. I do know that it can be complicated, involving design elements of your structure and nearby areas ("rolling circles" design, for example) and also specific elements at the entry point into the home. This is why I said I might go with fiber optic cable at the home entry point. The resistance you measured seems reasonable, about twice the absolute minimum I estimated. – jonk Sep 2 '17 at 17:00
• @user134429 The circuit can be simplified a little if you have your own power supply rail. But you'd need to make a FINAL decision about exactly how you want to arrange things. My own feeling is that you should concentrate on supplying the PFX-20 properly and just self-power the circuit itself from the 20 mA current source they provide. This isolates the two issues. And since the PFX-20 supplies the 20 mA whether you like it or not, might as well just use it. – jonk Sep 2 '17 at 17:10
• With comparator circuit I can set hysteresis for noisy sinusoids like 100mV hysteresis. If this is not done at the beginning Y1 will not know that it was noise. I dont know if I could explain what in my mind. But one more question to you if you dont mind, in their datasheet for the wind sensor nrgsystems.com/assets/resources/an40C-IF3-interface.pdf I look at their suggested block diagram. The over voltage protection component's one leg is tied to "protection earth" not to the circuit's own ground. Is that really what is supposed to be Im really confused why it is like that. Thanks – atmnt Sep 2 '17 at 17:29