Author Topic: 9V ac = 13V dc.  (Read 2075 times)

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Jerry

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9V ac = 13V dc.
« on: December 28, 2008, 03:35:17 AM »
I had connected a small pair of wires directly to one of the 11 coils on the 14-11 ceramic alt for frequency measurements.


Just for fun I took a few ac voltage readings from the coil while the alt was runing at 13.93 volts dc output no load. The ac voltage from one coil is just a littel over 9vac while on the dc side of the fullwave bridge the voltage is 13.93vdc


Then I loaded the alt down to 3 amps.


The AC side measured 13.5 volts while the DC side was measureing 13.54 volts.


I'm just thinking where have the diode voltage loses gone? I know they are there at higher amperage from the alt but why not at no load or low load.


Flux HELP!


Here is the wave form under load.


                       JK TAS Jerry

« Last Edit: December 28, 2008, 03:35:17 AM by (unknown) »

esc

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Re: 9V ac = 13V dc.
« Reply #1 on: December 27, 2008, 10:28:50 PM »
That's pretty interesting!


My $0.02? (If it is worth that much)


It has to be a difference in how quickly the two meters respond to the difference in voltage.  The AC meter is probably slower to react (mass of needle?) and is missing the voltage "peak" while the DC meter (no mass) responds more quickly and locks in on the "Peak" voltage.


The oscilloscope looks pretty interesting (don't have any experience reading one).  What is the scale?  Can you get a voltage off of it?

« Last Edit: December 27, 2008, 10:28:50 PM by esc »

Flux

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Re: 9V ac = 13V dc.
« Reply #2 on: December 28, 2008, 02:40:43 AM »
Jerry this is a bit complicated, don't worry too much about it, you haven't devised a means of eliminating the diode loss.


Probably on no load your coil waveform is roughly sinusoidal . Your ac meter looks to be a rectified moving coil meter so it will measure average volts but will be scaled to rms ( that is what the world normally considers ac volts). 9vrms will be 12.7v peak, which is near enough to your 13.93 ( average of the peaks of all the coils).


As you start to load it the ac volts will increase as the diode conduction angle increases. The coil resistance comes into play and the effect is seen on the scope, the flat top bit is battery volts plus diode drop. You can see the diode drop as the notch after the flat bit. The waveform is much nearer a square wave and for the square wave, peak mean and rms are the same. this would imply the ac meter reading the same as the dc. The things will not be exact as the ac meter is still reading mean but scaled rms  so it will have a strange factor. The waveform is not a true square wave either so in reality there is no real significance in the readings.


If your coils had much lower resistance the ac voltage would stay below battery voltage up to higher currents.


Diode loss is far from easy to sort out except in smooth dc circuits such as solar. The loss from diodes in alternators is difficult to determine as it affects the waveform and makes conventional measurements very doubtful.


What you are looking at is very similar to a dc power supply using a transformer and full wave rectifier with a very large smoothing capacitor, voltage waveforms with conventional meters tell you very little. The scope shows what is going on but does nothing to give you real answers as the maths for the set up is beyond most of us with complex waveforms.


Hope this helps but don't take the measurements seriously just use it as a comparison.


Flux

« Last Edit: December 28, 2008, 02:40:43 AM by Flux »

esc

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Re: 9V ac = 13V dc.
« Reply #3 on: December 28, 2008, 06:38:50 AM »
So, if I understand your explanation, I wasn't TOTALLY lost.


The digital meter is reading peak voltage and the analog is reading "mean but scaled rms".


Thanks Flux, your explanations are always helpful, even if frequently over my head.

« Last Edit: December 28, 2008, 06:38:50 AM by esc »

Jerry

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Re: 9V ac = 13V dc.
« Reply #4 on: December 28, 2008, 10:06:36 AM »
Yes Flux you are correct. The wave form was much smooter and more rounded no load.


I tryed to get a piture but didn't have any luck. Timeing I'd guse. I just got lucky on the loaded picture.


I've noticed what your saying about ac power supplies. Hook a fullwave bridge rectifier to  a 10 vac transformer and get 10 volt dc out. Add a large value cap to the dc side and get 14 volts dc no load.


A battery has the same effect. And depending on the amperage/resistance of the transformer and the load from none to heavy changes the #s both ac and dc.


                           JK TAS Jerry

« Last Edit: December 28, 2008, 10:06:36 AM by Jerry »

Jerry

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Re: 9V ac = 13V dc.
« Reply #5 on: December 28, 2008, 01:07:31 PM »
Hi Flux.


As you have stated its very dificult to measure frequency from one coil with any stability.


The scope I'm using also has a frequency meter but the #s are jumping all over the place on the FRQ meter.


The #s are going up and down radicaly +- 10 to 50 HZ under load. I've just spent an HR Googleing the board looking for an apropriate filter to place between to coil and the FRQ meter and no luck.


I suspect it would be a PY filter or maybe an LC network? Do you have any sugjestions on CRCT configurations and componet valuse?


I have many speaker crossover componetes that may smooth things out? Or is it a simple as placing a cap accross the coil?


The coil measures 6 Ohms DC. 14 magnetic poles.


                         JK TAS Jerry

« Last Edit: December 28, 2008, 01:07:31 PM by Jerry »

FuddyDuddy

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Re: 9V ac = 13V dc.
« Reply #6 on: December 28, 2008, 07:54:43 PM »
Converting from AC to DC usually uses the standard conversion ratio: DC x 1.414, or AC/ 0.707. This is standard throoughout the industry, so your values look good if you are doing a single wave conversion. Standard losses for a silicon diode are 0.7 volts, and that doesn't depend on their size in amperage.

If you do add a capacitor, it goes across the DC side, not the AC. This will raise your apparent votlage, but not your current. It will, however, help filter out any noise.

I must add, your wave form looks very good, but the sharp edges will give you a bit of harmonics. The capacitor will help that. You will find that the apparent voltage will "sag" somewhat under load.

Since you are only at about 10 volts, a 25 volt capacitor at, say, 50mfd, would do the job, but you might want to add a smaller one, say at 0.1mfd, just to help the noise filtering. I would recommend all capacitors at a 100 volt level just to take care of any "surges". That's safer and they aren't that spendy. The different size capacitors do different jobs. One is low frequency, the second, higher.

Good luck, looks like you're going in the right direction.

Very good question.

« Last Edit: December 28, 2008, 07:54:43 PM by FuddyDuddy »

dnix71

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Re: 9V ac = 13V dc.
« Reply #7 on: December 28, 2008, 09:48:03 PM »
That's a $400 scope. I'm envious. The meter looks like a Harbor Freight special, though, that's more my budget (they're $2.99 again, no coupon needed)
« Last Edit: December 28, 2008, 09:48:03 PM by dnix71 »

Flux

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Re: 9V ac = 13V dc.
« Reply #8 on: December 29, 2008, 01:35:17 AM »
There may be two issues with the stability of the scope trace when you try to measure frequency.


If the scope frequency measuring bit is a conventional frequency meter it will be affected by fast things that you probably won't see on the waveform. That vertical notch after the peak of the waveform is caused by the diodes and there will be very fast oscillations in this region if you trigger the scope on them and expand the timebase.


You should be able to use a simple filter with 100k in series and about 1uf across the scope input to remove the fast edges. For frequency the waveform doesn't matter and you should add enough filtering to get it to a near sine or triangle waveform.


Looking at waveforms of this type on a scope also has the issue that with imperfect magnet spacing the wave shape of each pole may be slightly different. If you trigger on a complete group of pole waves the waveform will be stable but the individual shapes may not be the same. If you trigger on any non integral number of waves then the trace may jitter as the slightly varying images will not fall on each other. If you try to get frequency from the graticule it will add confusion so alter the trigger and timebase speed to get a stable trace.


Flux

« Last Edit: December 29, 2008, 01:35:17 AM by Flux »