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Old 9th January 2012, 08:29 PM   #11
RJM1 is offline RJM1  United States
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@ KatieandDad
This is true, but before the switching power supplies were introduced, which apparently he has, they sometimes had a separate winding instead of the inverter . I'm talking late 70's here.

Last edited by RJM1; 9th January 2012 at 08:35 PM.
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Old 9th January 2012, 08:36 PM   #12
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Mid 70's VCR ???

My comment is still valid above but obviously you have to scale down the expected load available from that particular winding. A display would only need microamps or one or two milliamps at HV. The filament current would be the higher load.

It's a valve in disguise.
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Old 9th January 2012, 08:41 PM   #13
RJM1 is offline RJM1  United States
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AS I said "florescent display at a few milli-amps."
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Old 9th January 2012, 08:46 PM   #14
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Lets wait for him to reply to see if he understands what we are talking about.
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Old 9th January 2012, 10:57 PM   #15
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I was just thinking about buying the switching power supply if i can't use the transformer from the vcr. I would like to build a lm1875 powered stereo amp and I was wondering if I could use the transformer I scavanged from the VCR.
I'm a newbie to diyaudio, and it would be my first project. I have relatively no experience in schematics, I've done a lot of soldering though. I'll post the results of the transformers output tommorow.
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Old 10th January 2012, 11:56 AM   #16
AndrewT is offline AndrewT  Scotland
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Build a bulb tester.
Insulate every transformer tapping by inserting each into a separate terminal of an insulated terminal strip.

Find which tappings are on the same winding. Label them.

Power up the presumed primary via the bulb tester.
If the bulb goes OFF, then you do not have a short circuit in the wiring.

Very carefully measure the mains input voltage at the terminal strip. Note reading
Measure the voltage at the ends of each winding. note readings.
Note mains voltage again.
Report your findings.
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Old 10th January 2012, 06:08 PM   #17
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So I Took the measurments on Acv setting with my multimeter from the secondaries. the results were this
(pin)8 - 9 = 15V
8 - 10 = 18V
8 - 11 = 10V
9 - 10 = 5V
9 - 11 = 0V
10 - 11 = 9V

14 - 15 = 32V
14 - 16 = 10V
14 - 17 = 12V
15 - 16 = 0V
15 - 17 = 2,5V
16 - 17 = 3V

The transformer had 8 pins on the secondaries, by groups of four. That is the reason I think there are two secondaries. Now for the second part, How do I measure the ampers of this thing? If the VA rating is 40, then do i divide 40 with the volts and get ampers? and would this suite the needs of a lm1875 stereo amp?

PS. Are the voltage results strange and I'm doing something wrong? cause they seem wrong to me, so many different results.
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Old 10th January 2012, 06:20 PM   #18
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This is not a very user friendly transformer but dont give up yet.

Look at #10.

You can load up each pair of terminals maybe 100mA at a time and see what happens.

Be careful though, as already intimated, there may be some very low current HT windings that are only rated at a few mA each.

If you load each winding at say 50mA and it is significantly different to your original readings, turn OFF immediately and report your results.
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Old 10th January 2012, 06:25 PM   #19
AndrewT is offline AndrewT  Scotland
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You can't measure the amperes.
You can estimate the relative current capabilities of the windings by comparing the wire diameters used for each (and every) part of the windings.

One winding, let's say it is 12-5-0-5-12Vac and all the wire in the winding is 0.9mm diameter. then you can estimate that each part of that winding can carry the same maximum current.

If the other winding were say, 10-6-0-6-10Vac and the wire were 0.8mm diameter from 6-0-6 but only 0.6mm diameter from 10-6 and from 6-10 then you can see that the winding has a lower current capability in the middle sections and a much lower current capability in the two outer sections.

Proportioning the VA ratings between all these windings can be done, but there is quite a bit of arithmetic involved.
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Old 10th January 2012, 06:27 PM   #20
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Yes, but he cant see the wire diameter.

This is going back to the Black Box approach of the unknown quantity.
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