Tacho Circuit - Transformer question

[Dave B]

 This circuit was posted by Hugh. I think several here are using this tach circuit with the 2917 frequency to voltage chip. My question is : What type of step down transformer should be used and also what value and wattage should the current limiting resistor be just before the clamping diodes there ? Let's say it's a 3 phase 48 volt system and so maybe any single phase may see 100 - 120 volts AC tops.

 What I am thinking is that if this transformer is just parralleled off one phase of a 3 phase alternator then won't this imbalance the load to say the 3 phase rectifier ? A higher resistance for the primary would be less of a load and therefore lower current and although very small but wouldn't this be less work also for the alternator and therefore easier turning ? I would think that maybe a doorbell transformer would have a low primary coil resistance and not be a good choice for the step down transformer ? I just see that tap off the one phase as being a short and therefore quite a load just to knock down the voltage. Maybe I'm over thinking this. Thank you in advance for anyone helping me understand this circuit.  Dave B.

[joestue]

Any small transformer is will work, provided you can get at least 50mv ac into the chip.

If you use a nominal 10:1 transformer, a 1 kohm 1/4th watt resistor is fine.

What is the volts/hz? you may need to find a 240volt Tx, or, put the resistor on the line side of the transformer.

btw, a small transformer should present an almost undetectable load on your system.

[Dave B]

Thanks for the reply. When you say "small" transformer I assume you mean the physical size ? Right now I'm using a 120 to 8 VAC door bell transformer. It works but I can't help but think that if I want to lessen the load by using a "smaller" transformer of similar ratio that it would burn up ? I have trouble not seeing the primary as a short and with fine wire why woudn't it fry ? The primary of this transformer measures 44 ohms and the secondary 8 vac measures .8 ohms. Wouldn't this mean that at say 44 vac input from the one phase of the alternator that this would be 1 Amp and approx. 44 watts produced ? This is not too far above my 48 volt dc cut in, very significant I would say if I need to produce an additional 44 watts just to drop down the AC voltage of one phase to protect the tacho circuit from frying. Am I all wet here with my thinking ? Why wouldn't a much smaller transformer burn up ? Too many years since days in the lab.  Thanks for anyone's help. Dave B

Any small transformer is will work, provided you can get at least 50mv ac into the chip.

If you use a nominal 10:1 transformer, a 1 kohm 1/4th watt resistor is fine.

What is the volts/hz? you may need to find a 240volt Tx, or, put the resistor on the line side of the transformer.

btw, a small transformer should present an almost undetectable load on your system.

[SparWeb]

No you're fine.  That's a good choice of transformer.

120V / 44 ohms = ~3 amps (and that's ignoring reactance, which is the more important part of the equation, here)
The transformer coil has a high reactance, so expect much less current than that on the primary winding of the Tx.  Like, 100's of mA.
120VAC is what's it's designed to get from the wall power, after all.

Your stator's output is actually more like 60VAC line-to-line isn't it?  Depends on if you're Wye/Delta or IRP connected, and which way to pick up your AC.  If you have 3-phase Wye and you measure from a line to a center tap, then you have nearly half the AC voltage.  An option if you're concerned about high voltage on the transformer, but I wouldn't be.  I'm skipping some details but I don't think they matter.  (If I'm wrong someone will surely point it out)

[oztules]

The resistance of the primary is only of interest with DC ......or very low frequency.

Once you get bubbling along, your are only interested in the frequency and the inductance.
Ohms law does not have any bearing once the frequency starts to get going.... think impedance not resistance. The frequency through the inductor will give you a impedance to current flow..... nothing like resistance which would increase with the voltage....with no regard to frequency

Unloaded the tranny will have almost no current draw. As you are only lightly loading it, it will not change much as voltage rises, as the frequency rises in lock step.... so your reactance  will be proportional to your frequency (XL (reactance)=2phi F(frequency) x L (inductance in henries).... L is constant, phi is constant.... so reactance=constant x frequency).

If you pick your tranny to have 110v@60hz with your mill, you can speed up as much as you want, the current draw will change very little unless you load it differently. Your mill will not see it I suspect at all. If you hit 110vac at higher frequency ( your mill ), then your home and hosed anyway... if your voltage hits 110vac at less than 60 hz, then a resistor as Joe suggests will fix it.




..............oztules


Edit: As an example, if we look at a welding transformer (big one 220v), it has a resistance of about 1 ohm or so on the primary.... very thick wire. By your reckoning, it would draw 220amps just idling, and blow every fuse in the area.... but that does not happen. It draws very little if the transformer steel is good steel, until you actually weld... and then probably less than 20A for a decent size welder..................... so something else is at work... not resistance... but impedance.

[Dave B]

Sparweb,

  My stator is wired 3 phase wye and runs about 37 vac line to line open voltage @ 80 rpm. I'm picking up the ac for the transformer parrallel off one phase, this of course before the 3 phase rectifier. Rare if ever that the input to the transformer will ever see even 110 vac. I need to keep my AC and DC thinking separate, I've been more in the DC mode not thinking about the reactance. What about the frequency though ? At 80 rpm if my math is correct this will be 13.3 hz and 37 vac. (20x2 magnets 15 coils) Lets say 240 rpm for the top end which would be 40 hz and 111 vac. Again, it's been a long time since lab but something tells me I might not be too far out of line to think the transformer will load the alternator considerably at lower rpm right when I would like the least amount of load (free wheeling) up to speed until at least cut in. Please help me if I'm going off track. I want to use this circuit but I don't want to load the alternator and sacrifice some performance just to use it. A few watts is one thing but 10,20,30 or so watts especially when the machine should be free wheeling up to cut in is considerable. Any more thoughts or comments on this is much appreciated. Thank you,  Dave B.
 

No you're fine.  That's a good choice of transformer.

120V / 44 ohms = ~3 amps (and that's ignoring reactance, which is the more important part of the equation, here)
The transformer coil has a high reactance, so expect much less current than that on the primary winding of the Tx.  Like, 100's of mA.
120VAC is what's it's designed to get from the wall power, after all.

Your stator's output is actually more like 60VAC line-to-line isn't it?  Depends on if you're Wye/Delta or IRP connected, and which way to pick up your AC.  If you have 3-phase Wye and you measure from a line to a center tap, then you have nearly half the AC voltage.  An option if you're concerned about high voltage on the transformer, but I wouldn't be.  I'm skipping some details but I don't think they matter.  (If I'm wrong someone will surely point it out)

[OperaHouse]

As to the question of vert small transformer, even the small transformer out of an old modem will worl.  The lmiting resistor canadditionally be placed on the primary and be 20-68K.  The core just saturates at higher voltages.  Using a LED opto isolator may be an even better choice.

[SparWeb]

Hi Dave,

You're asking for the details I was trying to skip!  Okay...   As Oztules pointed out, the impedance is the thing.  In a transformer it depends on the input frequency somewhat, but 13-40 Hz is not so far off that it will matter much.  The impedance is still high.   Use impedance just like resistance with Ohm's law.

A little (fist-sized) transformer may draw about 10 watts, max load, on the primary with 120VAC input.  If you're not sure, go try it:  connected it to a power cord and plug it into the wall!  If it came out of something that was plugged into the wall, then you're not doing anything that it wasn't designed to do.

10W / 120VAC = 0.083 Amps  (normal operation with grid power)

Then

120 VAC / 0.083 A = 1446 Ohms Impedance     < property of the transformer itself.  A lot more than your DC 44 ohms, eh?

Take this to your situation.  Here are the calculations (back of the envelope style) for the transformer primary on your power lines:

37 VAC / 1446 Ohm = 0.026  Amps =  26 milliAmps.    <Low speed turbine

111 VAC / 1446 Ohm = 0.077  Amps =  77 milliAmps.    <High speed turbine

Nothing to worry about!

(Footnote for picky people:  yeah I know that the impedance will be lower at lower frequency.  I'm not working it out but my gut tells me the result will still be <100 mA at cut-in.  Prove me wrong.)

[Dave B]

Spar,

  OK, I posted before Otzules (Thanks for that too) reply ealier and have since read that over as well. I'm getting it and it's scarey to think I'm starting to remember it from way back. My only thought on this now is that in the very low RPM range (before cut in) and just needing to get the machine moving that as you say the impedence will be lower than for example the 1446 ohms that you state. Then again the voltage is lower also so the current will be less as well and therefore so will the load on the alternator to get it moving. Bottom line is I guess I'm over thinking it but I was not off track with my thoughts but rather the anount of effect it will have which is minimal. This is a 16' machine also. I sure appreciate the help from everyone, the circuit works like a champ no matter. Thanks Hugh for posting that way back when as well.  Dave B.

Hi Dave,

You're asking for the details I was trying to skip!  Okay...   As Oztules pointed out, the impedance is the thing.  In a transformer it depends on the input frequency somewhat, but 13-40 Hz is not so far off that it will matter much.  The impedance is still high.   Use impedance just like resistance with Ohm's law.

A little (fist-sized) transformer may draw about 10 watts, max load, on the primary with 120VAC input.  If you're not sure, go try it:  connected it to a power cord and plug it into the wall!  If it came out of something that was plugged into the wall, then you're not doing anything that it wasn't designed to do.

10W / 120VAC = 0.083 Amps  (normal operation with grid power)

Then

120 VAC / 0.083 A = 1446 Ohms Impedance     < property of the transformer itself.  A lot more than your DC 44 ohms, eh?

Take this to your situation.  Here are the calculations (back of the envelope style) for the transformer primary on your power lines:

37 VAC / 1446 Ohm = 0.026  Amps =  26 milliAmps.    <Low speed turbine

111 VAC / 1446 Ohm = 0.077  Amps =  77 milliAmps.    <High speed turbine

Nothing to worry about!

(Footnote for picky people:  yeah I know that the impedance will be lower at lower frequency.  I'm not working it out but my gut tells me the result will still be <100 mA at cut-in.  Prove me wrong.)

[oztules]

Oh Sparweb, what have you done???

I don't wish to be pickky...... but your creating the wrong ideas with this..... Let me explain.

Sparweb says:
"A little (fist-sized) transformer may draw about 10 watts, max load, on the primary with 120VAC input.  If you're not sure, go try it:  connected it to a power cord and plug it into the wall!  If it came out of something that was plugged into the wall, then you're not doing anything that it wasn't designed to do.

10W / 120VAC = 0.083 Amps  (normal operation with grid power)

Then

120 VAC / 0.083 A = 1446 Ohms Impedance     < property of the transformer itself.  A lot more than your DC 44 ohms, eh?

Take this to your situation.  Here are the calculations (back of the envelope style) for the transformer primary on your power lines:

37 VAC / 1446 Ohm = 0.026  Amps =  26 milliAmps.    <Low speed turbine

111 VAC / 1446 Ohm = 0.077  Amps =  77 milliAmps.    <High speed turbine


This is leading folks astray. If we ignore the resistance of the windings for the moment, and just look at the inductance and reactance it imparts to the changing emf we get this (I'm assuming 60hz for your afore mentioned fist sized transformer)

If we say  your right, and we have Xl=ohms, and 120/0.083= 1446 ohms..... then Henries=3.82 (L)

At 37v our mill Hz drops to 18hz and so our henries from above are still 3.82 our reactance in ohms is now about 433ohms, and current is 37/433= about .085A ... 85ma

At 111vac our mill  hz is now 55hz, and reactance is now 1323 ohms..... current is 111/1323 for about 84 ma

At start up of 1hz, our mill is making 2v..... so we get 24 ohms Xl.... for ... you guessed it 84 ma..... the relationship stays the same in this (ignore the coil R) context......


In fact even though Dave has  different volts/hertz relationship, the same rules apply. (40hz@111v or 166v@60hz for equivalence sake)

Whats wrong with this??
Well the coil resistance is of little consequence at 120v where the reactance plus resistance is 1446 ohms, but at 1hz, the reactance was 24ohms, plus the original resistance in the windings (ac needs to get through this to create the field in the first place.....) so our start up current will look more like 24+44=68 ohms for 2volts.... or at 1hz is only 2/68=29ma.

Now Dave needs to change the hertz/voltage ratio to suit his particular mill for his calculations..... but this is more like what he will see.  If he finds a 240v tranny, he should be good to go with even less current, and no saturation problems..... from memory saturation occurs or is most prevalent at no load..... and his load is going to be virtually nil. His frequency/volts ratio makes a resistor likely for a 120v 60hz transformer..... but will probably still be safe.... maybe just warm up a bit more than expected....and I suppose the regulation will be better with a saturated core




...................oztules

[Flux]

Yes the simplest thing is to use a 240v transformer. To prevent core saturation it would then handle 110v at 25/30 Hz. . The smaller the transformer the higher the primary resistance so it could stand a fair bit of saturation and still be happy. Once you have one point right, it will follow in volts/cycle so the lower frequency at cut in will still not cause saturation.

Having a series primary resistor will absolutely prevent any damage even if the core does saturate, the differentiated third harmonic waveform will still trigger the 2917 tacho chip with no problem.

Using a bell transformer is fine and the loading will be negligible, only a Watt or two,composed partly of iron loss and primary copper loss, the secondary loss will be infinitesimal.

There will be negligible in phase primary current with no load on the secondary, the current will be almost entirely magnetising current and inductive in nature. The dc resistance is something you can't avoid, but is irrelevant for ac operation, it's normally a nuisance but in this case it does give you a lot more margin to protect against core saturation, it just means you need less external series resistance if you have too many input volts for the chosen primary at nominal operating frequency.

As someone else said, you can also use an opto isolator but at 110v the resistor loss will be much more than the magnetising current of a small transformer , in either case it is small enough not to worry about.

Flux

[SparWeb]

Oztules,
You're right, of course!

I was waving the red flag in front of the bull wasn't I?

The relationship of impedance (variable) to inductance (constant) was in the back of my mind, so I was pretty sure that the current wouldn't go up at low frequency/speeds.  My judgement was to not complicate the explanation.  I decided to try it "oversimplified".  Then I wondered if it would ruffle some feathers. 

I was afraid of doing the math, though.  It would rapidly become "complex" math, if you see what I mean....  It was a conscious decision not to delve into the rest of it.

[Flux]

There are various ways to look at this but essentially we are looking at an iron cored inductor with no air gap.

At low flux levels the magnetising current is negligible and is mainly supplying the core loss. As you take the iron into saturation the magnetising current shoots up and when the core is fully saturated it is only the winding resistance that limits the current.

The emf equation is the simplest way to look at this   E=4.4 ( phi max) f N  where phi max is peak core flux

N is number of turns and fixed. So to limit the peak core flux to a safe value then E ( volts) is proportional to frequency.

This gives the condition that doubling frequency will double the voltage rating without saturating. If you choose the volts for safe peak flux at one frequency it will be safe for any frequency when supplied with a pm alternator as the alternator follows exactly the same emf equation. Double the speed ( frequency ) and you double the volts.

It's ok thinking about impedance below saturation but in effect the impedance is non linear and a function of the core flux and the shape of the BH loop of the core.

Flux

[Dave B]

Wow, this is all going nuts but I sure appreciate the information. So for my 120 / 8 vac doorbell transformer of 44 ohms for the primary and .8 ohms for the secondary it appears that the load seen from the alternator's one phase (of 3) is small even in the lower rpm before cut in all the way down to start up, is this correct ? After cut in and cruising along it's even less significant ?Therefore the machine should be running near free wheeling before cut in correct ? Those were my concerns, I forgot just how much fun AC circuits can be.   Dave B.

[Flux]

Basically if it works and doesn't get hot then all is well. you are a bit close to the limit for real comfort and as I said previously it would be safer to use a 240v transformer.

Taking your worst case figures of 110v in at 40 Hz you may be a bit close for a nominal 60Hz transformer, if it is designed for 50/60 Hz use then it will be less stressed.

Added to this the waveform on the input of the rectifier will be a square wave and that makes things a bit worse.

What will happen in reality is that the transformer may start to saturate at something under 100v. You then have the difference between your input ac and the saturation voltage of the transformer being current limited by thew 44ohm transformer dc resistance.

It would be a bit more comforting to add some series Resistance in series with the primary to add to the 44 ohms. In fact you have so much more secondary volts than needed for the tacho chip that it wouldn't hurt to add a fair bit , say 1k.

These small transformers are quite forgiving and you probably have no need to worry at all. The only reason to do anything would be if you found the thing running rather warm, it certainly will have no effect on the start up of the machine as at start up frequency the 44 ohms will swamp the magnetising current. and take near nothing.

Flux

[oztules]

Yes Dave...... just bang it in and run it.

You will find your mill  cannot tell if it was there or not.

Also, the theory is just theory, in practice your tied to a battery bank, and even though you have emf generated in the stator of over 100 volts...... the terminal volts will only still be less than 60 due to the batteries holding the stator terminal volts down. If the bank is low, that figure will drop to 53 or so volts..... so saturation may not be a bother either.


...............oztules

[boB]

It would be a bit more comforting to add some series Resistance in series with the primary to add to the 44 ohms. In fact you have so much more secondary volts than needed for the tacho chip that it wouldn't hurt to add a fair bit , say 1k.

These small transformers are quite forgiving and you probably have no need to worry at all. The only reason to do anything would be if you found the thing running rather warm, it certainly will have no effect on the start up of the machine as at start up frequency the 44 ohms will swamp the magnetising current. and take near nothing.

Flux

How about adding a non-polarized capacitor in series as well ??  (of the proper value of course).

This could help reduce saturation at very  low RPM frequencies if needed and might make a lower voltage transformer usable.

boB

[BrianSmith]

Just put a 100K resistor in series with your phase and some 4148s to gnd and chip power on the IC side to keep it from going past gnd - 0.7 and chip VCC + 0.7.   Don't think you need a transformer here.  I've done this with PIC inputs and 120VAC before no problem.  The 100K limits it to less than a milliamp at upto 100V.   

[boB]

Just put a 100K resistor in series with your phase and some 4148s to gnd and chip power on the IC side to keep it from going past gnd - 0.7 and chip VCC + 0.7.   Don't think you need a transformer here.  I've done this with PIC inputs and 120VAC before no problem.  The 100K limits it to less than a milliamp at upto 100V.   

They could certainly do that if the turbine phases were referenced to ground, but it probably isn't

But, expanding on that idea, he could use a differential amplifier (op-amp) and resistively divide
that down to make a ground referenced output to drive the IC.

Or, just go the opto isolator route.

boB

[Flux]

I don't like the idea of a capacitor in series with the primary of the transformer. Although it will act as a divider to limit the primary voltage it will be frequency sensitive in the wrong way to be of use for the low frequencies, it will restrict the volts at cut in a great deal, when it is already low and doesn't bother the transformer. It may still give enough to trigger the tacho chip though.

Normally you need some form of isolation to be sure of things and an opto isolator fed from a resistor or the transformer is the easy solution.

You can in some cases do it without isolation if the common for the tacho chip is rectifier negative. One rectifier line to rectifier negative will still give a pulsating component that you can use with a divider. You only need the positive clamp diode as it can't reverse in this case. Although this may feed directly into a microprocessor, the 2917 tacho chip needs a trigger that goes positive and negative so you need a differentiator ( series capacitor and following resistor from the trigger to ground to act as a dc restorer and let the ac component through. If you need additional filtering at the trigger to remove noise then it gets quite messy.

The transformer is simplest in the end, but the opto isolator can give the cleanest signal with minimum filtering if there is high frequency noise ( not a problem unless you have a converter in the circuit.

You can of course use a differential amplifier with series resistors and diode clamps but unless you use a dual supply you may still have to capacitor couple the output to take the 2917 input below zero.

Flux

[boB]

I don't like the idea of a capacitor in series with the primary of the transformer. Although it will act as a divider to limit the primary voltage it will be frequency sensitive in the wrong way to be of use for the low frequencies, it will restrict the volts at cut in a great deal, when it is already low and doesn't bother the transformer. It may still give enough to trigger the tacho chip though.

Flux

The capacitor idea was more or less to reduce the volt-seconds on the core at VERY low RPMs and be able to reduce the resistor value.
The resistor should do the job by itself  though and probably better.

We use Iso-Optilators for tachometer RPM readings from 3-phase turbines.

boB

[joestue]

Its not the volt seconds at the low end that matter.
The 44 ohm resistance of the transformer equals the ac reactance as oztules pointed out at 2 hz, within reasonable margins of error.
Furthermore the core saturating at low voltages and frequencies really doesn't even matter for several reasons, first is that on average the frequency is 1/3rd normal. this also means the eddy current portion of the iron loss is 1/9th normal losses and thus we can assume that with one third the frequency the iron losses are something like 1/4th normal operating losses. Now say we double the amps needed to magnetize the core because its running 50% higher than normal, and the loss goes from 'negligible' to 'might not be negligible'.

If he were actually going to run it at 110vac and 37hz for more than a half hour at a time then perhaps it might overheat.
but since the ac volts is clamped to 50-60 or so square wave (peak voltages), at 37hz its actually less flux than design. Somewhere in the middle at 40vrms and 15hz will be peak iron loss, but since the frequency is 1/4th design, it might only be something like 1/2 as much iron loss as it would be at 120vac/60hz.

Capacitors would be needed to limit the startup current for larger transformers if the turbine doesn't have enough torque to generate the magnetising current, which would in theory remain constant.

[boB]

Its not the volt seconds at the low end that matter.

 I'm sure you're right.  Most likely not a problem at all for a small transformer like this with a high  resistance.
A larger transformer with low resistance might keep a turbine from starting up though if
it dragged the turbine down enough IF it saturates.  I've seen where an auxiliary supply connected to a 1KW
48V turbine trying to turn on at 10 Volts would keep it from starting up by trying to draw too
much current.  I could see this happening with the wrong transformer too.

boB

[Flux]

Yes it's a small transformer with a lot of resistance and very little magnetising current. DaveB's originall worry was about hindering start up, but this won't be an issue here.

With large transformers directly connected to a mill there can be a start up problem unless they are chosen to be run at very low core flux compared with normal operation. Even at full flux the magnetising current on the high voltage side is normally very modest, but run it as a step up and refer the magnetising current to the input side and it becomes something that can easily delay or prevent start up. It's the same thing that delays start up with iron cored pm alternators and it is a bigger factor than cogging, although cogging invariably gets the blame.

.Flux

[SparWeb]

Hi,
Dredging up an old thread, here.  I put this circuit together again today, on the "breadboard", and had it out at the mill to test my work before soldering it permanently.  It was windy and a perfect day to test with the WT spinning full-tilt.  Batteries full and dump load going; it was a nice day for wind energy.  That's when I had a "whoops" moment (which is a sure sign I'm about to learn something).

I had already satisfied myself that it was working fine, and just starting to take it out, when it suddenly began to give erratic readings.  What I'd done was to un-hook the primary of the transformer from the wire with the mill's AC.  With no AC from the turbine I expected the output voltage reading to drop to zero (or more exactly 0.025mA as I'd seen on the desk before).  Instead, the voltage rose up to over 10V and stayed there, with only occasional dips to about 6V, then rising back up to 10V.  Not really related to the speed of the WT, so now I'm worried that I broke the IC or something.  Hook the transformer back up and everything goes back to normal.  Huh.  Why would the output voltage go up (indicating higher frequency), when the transformer's primary is disconnected (frequency should be zero)?  I tried shorting the wires of the primary, and then the output voltage did go to zero...  Started to clue into what was going on, but I still didn't believe it!  I took the board and the transformer into the separate box where I have the rectifiers, and hooked them up to the turbine's AC there.  No more odd behaviour.

Long story short,  the LM2907 chip is SOOO sensitive, that the disconnected leads of the transformer could detect the radio noise from the Tristar charge controller, because it was dumping energy.   Like an antenna, just waving the wires near the TS-60 was enough to trick the IC into thinking the signal frequency was umpteen kHz.  Wow.

I'm going to need some shielded wires for this.

[Flux]

The 2917 can be a bitch to tame. Normally a decent filter close to its input is all that is needed but noise can be coupled into it through the inter winding capacitance of the transformer if the actual 2917 ground is noisy.

I had trouble with my boost converter circuits, using a 2917 close to the converter running in that case at about 5khz. I found an otpo isolator was far easier to use than a transformer. At the time I had some BASEEFA approved isolators where the separation between led and photodiode was several mm. These worked brilliantly but i have no reason to suspect a standard DIL isolator wouldn't have worked fine. In the later mppt buck converter i went back to a transformer with the pwm running at 30k it was easy enough to filter out.

Screening things may help but unless you do it properly you may have even more trouble from the stray capacitances. All depends on your grounding.

If your transformer has an inter winding screen then ground this to the ground of the 2917 and it might clear the problem.

Flux

[SparWeb]

Interesting point about the ground.  The tach circuit was running on its own little battery, thus "floating" relative to the main system voltage.  When permanently installed the circuit will be driven from the main system battery.  This may introduce new problems that I ought to go check for before I have another whoops moment.  I have some common opto-isolators (in a box somewhere...), if I do have trouble.  Thanks again!