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Author Topic: GE 800Hz 150VAC meters questions  (Read 8574 times)
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Opcom
Patrick J. / KD5OEI
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« on: August 30, 2014, 11:44:28 PM »

These are high quality GE meters. I don't know the construction of them.  Before realizing they were 800Hz meters, I tried them on 60Hz. They work well and do not consume a lot of current, just a few mA. They will rise and fall with the AC voltage as expected. but when the voltage is changed by just a few volts the needles do not move. When up in the 80+ volts range, they move for gross changes of 10V or more. They are very free when there is no juice just like any meter. I have not been using them for fear of ruining them. On an 800Hz voltage made by driving a transformer with a tube amp, they behave 'properly'.

So why would they appear to stick at 60Hz? Just too much current/too much magnetism? Unlike other 400Hz stuff like selsyns and fans, using a low voltage like 12-24V won't work here as the meters just indicate the lower voltage.

1. If they are 800Hz meters, then shouldn't they have less windings , therefore passing more current, therefore indicating higher than they should?

2. Why don't they indicate higher voltage? And what makes them seem to stick when on 60Hz?

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« Reply #1 on: August 31, 2014, 02:00:00 PM »

A lot would depend on the type of movement;  Iron vane, rectifier, eTc.  Coil reactance would only matter if not a rectifier type.

A quick check of just how much I they pull at 60 & 800 Hz would indicate how much is just turning into heat vs. motor energy.

73DG
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« Reply #2 on: August 31, 2014, 02:37:02 PM »

I may be able to measure the current with a series resistor, it is very small. Maybe a 100 Ohm one would do. Last time, I didn't have a sensitive AC voltmeter handy.

I can not see down into the motor area as there is a gray plastic part there. It looks similar to one of these pictures.
Nice big meters about 5" size. I have two and want to use them to monitor the incoming voltage to the building.


* 70043482.jpg (10.88 KB, 200x200 - viewed 645 times.)

* mxpUvPEF69OKUqWAspLoSOQ.jpg (9.89 KB, 225x213 - viewed 654 times.)
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« Reply #3 on: August 31, 2014, 02:42:12 PM »

Opcom

If the current is very small, you may want to use a 1000 ohm series resistor to sense the current. Although the larger series resistor may affect the reading slightly... you can still see if there is any difference between the 800Hz and 60Hz results. If the meter is drawing less than 1mA... then a 1000 ohm series resistor will drop the voltage by less than 1V.

Is there a part number visible on your meter?

As an experiment, if you increase the 60Hz AC voltage at a place where the meter appears to be stuck... will gently taping on the meter's front surface cause the needle to readjust its position to indicate the new/higher value of the voltage? [Just to see if the movement or the needle gets mechanically "stuck" on anything (for some reason TBD) when the applied AC waveform is at 60Hz]

Stu
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« Reply #4 on: August 31, 2014, 06:14:17 PM »

I'll look for the part number on it.

"As an experiment, if you increase the 60Hz AC voltage at a place where the meter appears to be stuck... will gently taping on the meter's front surface cause the needle to readjust its position to indicate the new/higher value of the voltage? [Just to see if the movement or the needle gets mechanically "stuck" on anything (for some reason TBD) when the applied AC waveform is at 60Hz]"

Yes, tapping will make it re-adjust.  If the power is off, the needle is free to swing the full range.
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« Reply #5 on: August 31, 2014, 07:37:27 PM »

Opcom

Okay...

Another experiment... to see if we can figure out what kind of meter movement you are dealing with:

1. If you apply DC to the meter (maybe 10V in series with a 1000 ohm resistor to be on the safe side):

Does the meter move (i.e. no capacitor in series and/or no capacitive voltage divider)?
Does it read 10V?
Does it read 10V regardless of the polarity with which the DC is applied to the meter terminals (i.e. no rectifier)?

Stu
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« Reply #6 on: August 31, 2014, 09:09:20 PM »

Every rectifier type AC Voltmeter I've ever had used a full-wave bridge, and didn't care what polarity DC went into it, it just reads low.

73DG
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« Reply #7 on: September 01, 2014, 12:55:10 AM »

Tomorrow looks like a good time to check this. I have been working on the Tucker TX this week end, changing the grid circuit according to the discussions. It's a lot of fiddly bits but coming along!

PJ
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« Reply #8 on: September 04, 2014, 11:26:59 AM »

Opcom

I suspect that the magnetic field that produces rotational torque (relative to the axis) on the meter movement is also producing an orthogonal rotational torque that is tilting (very slightly) the movement in such a way as to increase the friction between the axial pin and the jeweled sockets that hold it.

Just as an ultrasonic cleaner can break loose particles stuck to a surface...  800Hz input current (and the harmonics associated with a full wave or half wave rectifier) may be doing a better job of breaking the static friction than 60Hz current.


Stu
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« Reply #9 on: September 04, 2014, 07:21:25 PM »

dI/dt isn't sufficient at 60cps vs. 800cps to apply to "L" (at a given winding inductance) enough torque to smoothly overcome bearing friction.
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« Reply #10 on: September 04, 2014, 10:18:02 PM »

I got too busy with the transmitter last weekend. Make myself a note to get the meters out and check all those suggestions.
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« Reply #11 on: September 05, 2014, 12:00:29 PM »

Rick

The torque is proportional to i (not di/dt).

The time varying component of the rectified current (whose rate of change is, by definition, di/dt) will produce increases and decreases in the torque... between its maximum and minimum values. A larger value of di/dt will not produce more maximum torque. To the extent that increasing and decreasing the torque more rapidly (between the same maximum and minimum values) does a better job of breaking the static friction... a larger value of di/dt would be beneficial... and  that would explain what Opcom is observing. 

Stu

dI/dt isn't sufficient at 60cps vs. 800cps to apply to "L" (at a given winding inductance) enough torque to smoothly overcome bearing friction.
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« Reply #12 on: September 05, 2014, 01:13:50 PM »

Quote
dI/dt isn't sufficient at 60cps vs. 800cps to apply to "L" (at a given winding inductance) enough torque Crossings  Grin to smoothly overcome bearing friction.

Ok. Dependent on freq., those crossings, heh, heh.

Sounded professional though.
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« Reply #13 on: September 05, 2014, 07:17:27 PM »

I like it when you guys confuse me. It's an unusual pleasure.
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« Reply #14 on: September 06, 2014, 08:39:53 AM »

Ok, we try.  Grin

Faraday's  law of induction and because of energy storage by the inductance;
vl = L (di/dt). -Defines what "L" is via the voltage across the inductance assuming a hypothetical zero resistance of course.

Now when you look at with a frequency added component you are transformed into the world of Reactances and Impedances.

First let's assume a steady state sinusoidal voltage and current.
i = Iejwt. ( that "w" is omega or 2pi*f)

And di/dt = jwIejwt
Then VLejwt = jwLIejwt

Already you can see that di/dt is frequency dependent.

Moving right along, canceling the exponential,

VL = jwLI. Which expression resembles ohms law where wL looks like R but is now called reactance and is directed in quadrature by "j".

So you'll correct me if I'm wrong but the relationship to me states that if "w" or 2 pi * F,  or just freq. is greater across a given winding, then VL is greater, that is more energy across that coil at 800cps vs. 60cps which certainly will help trigger that lil' baby. ..transferred into the mechanical domain, just plain more scrote. 

Well I digressed somewhat. The question was varying torque, not the definition of inductive reactance. Grin
But you can see where the thought process, the original statement initially railed.

And do you know how long it took to write this on an iPad? --Never again.
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