## The standby transformer (STDBY)

## Locate the windings of an unknown transformer

*in this case, this is the*)

**STDBY**transformer of my ATX power supply ... E216944 HP CWT 0437 10L 190022-ON VIKING B-2 CLASS (B) |

During this series of manipulations, it is a question of locating the various Primary and Secondary windings, and of determining the direction of the windings, which obviously conditions the phase of the signal present on each of these windings.

I decided to write this article because I have not found anything on the internet that satisfies me ...

The diagram of this transformer is that shown opposite. Certain connections have a "point" which is used to mark the start of the winding, and also beyond the direction of the current passing through this winding.

My spotting method is worth what it's worth, but it's safe and effective. This of course assumes that the transformer is dismantled ...

The material necessary for these operations consists of the following:

- A double trace oscilloscope
- A GBF generator
- An Ohmmeter (
*digital if possible*)

- The first step is to identify the various windings using an ohmmeter in order to determine that of the Primary. This winding must have the highest resistance, and here it is
**P1**with about 5 Ω, the other windings having much lower resistances. - We apply a sinusoidal signal to the
**CH1**input of the oscilloscope as well as to one of the Primary windings, here we will choose the**P1**winding since it has the highest resistance (*the probe on pin 4*,*and its mass on pin 5*), and the**CH2**probe connected to the second Primary winding**P2**(*the probe on pin 7, and its mass on pin 6*).

We set the frequency of the GBF to 1 KHz to start, with a level of 2V_{c/c}, which on a GBF delivering 20V_{c/c}corresponds to an attenuation of 20 dB. If the signal visible on**CH2**is not too readable or noisy, nothing prevents you from increasing the output level of the GBF up to 20V_{c/c}for example, which corresponds to an attenuation of 0 dB. This is also what I had to do for this first manipulation, the digitized signal being too noisy.

With this type of transformer, there is a very good chance that you will not observe anything on**CH2**and for good reason!

These transformers commonly operate at several tens of KHz ...

On the GBF, select higher frequency ranges, in multiples of 10 and sweep the entire frequency range from minimum to maximum until you finally observe a d signal. '**maximum**amplitude (*the frequency of the GBF at this moment will correspond to the resonant frequency of the transformer, as in the case of an RLC circuit ...*).

♦ A tip : In order to better visualize the maximum level of the signal on**CH2**, I use the**ADD**function (*Addition of the 2 channels CH1 and CH2)*, synchronization on**CH1**with the Trigger Mode in**AC**coupling (*Alternative Voltage Coupling*). I thus only obtain one trace, but that will greatly facilitate the adjustment of the frequency of the GBF. I just have to try to get the maximum level, then I go back to**DUAL**mode (*Double trace*).

With the transformer presented here, I obtain the maximum level (4.24V_{c/c}) on**CH2**at the frequency of**312.500 KHz**very precisely.

Here we can see a slight phase difference, which I will measure.

To do this, there are 2 methods, the first of which is that of Lissajous, just to get a first idea ...- not to use too high a measurement frequency. At frequencies above 120 KHz, the phase shift of the two oscilloscope amplifiers (
*HAMEG HM407-2*) may be greater than an angle of 3° in XY function. - that from the screen image alone it is not possible to see whether the test voltage is ahead or behind the reference voltage. An RC element placed in series in front of the oscilloscope's test voltage input can help. The input resistance of 1 MΩ can immediately serve as
**R**, so that only a suitable capacitor**C**is to be connected. If the opening of the ellipse gets bigger (*compared to shorted*) then the test tensin is ahead and vice versa. However, this is only valid in the range of a phase shift of up to 90°. This is why**C****C**must be large enough and only causea relatively small phase shift but sufficient to be noticed.

♦ Note : I tested this method with different values for**C**, and I could not see any difference in the shape of the ellipse ...*Source***:***Instructions for use of the HAMEG HM407 oscilloscope- 2 (pages 28-29)*

Thanks to this Lissajous figure, and if I refer to the indicative figures mentioned in the previous paragraph, the phase angle would be greater than 30 ° (*this is not very precise, just indicative!*).

The second method, I did not invent it on my own, I used the manual for my HAMEG HM407-2 Oscilloscope (*page 29*) for this . It consists in measuring the horizontal difference which exists between the 2 traces, in 2 points which one defines compared to the axis of times (**X**) in order to better be able to locate oneself.

To do this, I will use the so-called "9-square" method which is established as follows:

To easily measure the phase shift, it is useful to finely adjust the time base and to spread the period of the reference signal**CH1**over 9 tiles. A full period is equivalent to 360°. With this setting,1 square corresponds to a phase shift of 40° (*360 / 9*). It only remains then to measure the difference between the 2 traces (**t**) and to multiply it by 40. The phase shift**φ**will be the result of this equation.

In the case that concerns us, a complete period on**CH1**shows us that it measures 6.4 squares. I therefore fine-tune my time base so as to obtain the famous 9 squares for a complete period of**CH1**. The**CH2**trace for the same given point is then offset by 1.1 squares (*this will be our value***t**). If we apply the given formula, we get:**φ = tx 40°**= 1.1 x 40 =**44°**

My estimate greater than 30° with the Lissajous method was good, but it is much more precise!

In this figure, which is nothing but a different representation of the oscillogram "**Primary winding P2**" presented in a previous paragraph, I aligned the 2 curves and balanced their levels so as to be able to distinguish the phase difference. Is it positive or negative?

To find out, the timewhich separates 2 rising edges (*t 0**or 2 falling edges*) is called the**time difference**between the 2 signals:*t 0 = t 1 - t 2*

→ Ifis*t 0***positive**thenis**CH2****ahead**of**CH1**

→ Ifis*t 0***negative**thenis**CH2****delay**on**CH1**

This induces that the output signal (**CH2**) is in advance with respect to that present at the input (**CH1**), at an angle of 44°.**T**and**T**constitute the duration of a complete period for each trace concerned, ie 360°.*Source:**http://physique.vije.net/1STI/electricite.php?page=sinusoidaux2* - not to use too high a measurement frequency. At frequencies above 120 KHz, the phase shift of the two oscilloscope amplifiers (
- As in the previous measurement, I connect the
**CH1**probe to the Primary winding**P1**. This will remain unchanged during all of my measurements.

This time, I will take the measurement on Primary winding**P3**(*the*). First observation, the output level is much lower than previously ... normal, all the windings are not made the same way. The number of turns and the diameter of the enamelled wire used to wind them may differ ...**CH2**probe on pin 8, and its mass on pin 6

A quick glance would tend to tell me that the phase shift is less than 180° ... but it is better to verify this more surely!

It remains to measure the phase shift since it exists, and I will again use the Lissajous method, then I will then confirm by calculation the real value of the phase angle.

I am going to apply the "9 tiles" method again, in order to obtain a more convincing result:

I carry out the fine adjustment of my time base so as to obtain the famous 9 tiles for a complete period of**CH1**. The**CH2**trace for the same given point is then offset by 3.4 squares (*this will be our value***t**). If we apply the given formula, we get:**φ = (t / T) x 360°**= (3.4 / 9) x 360 =**136°**

The**CH2**curve is 136° behind the curve of reference**CH1**.

It is all the same more precise like that! If the Lissajous figure is a good visual indicator of the phase shift between 2 signals, this calculation remains essential. - For the Secondary part, I will apply the same principle as before, immutable. With the
**CH1**probe still on its Primary winding (*on pin 5 and mass on pin 4*), the**CH2**probe will naturally find its place in turn on each of the secondary windings, with the mass of the probe on the common pin which does not is here other than the midpoint (*pin 1, which is not always the case ...*).

I connect the**CH2**probe to pin 2 to start.

On this oscillogram, a slight phase difference is perceptible, which I will measure. Same procedure as for the other measurements, I first use the Lissajous method, just to get a first idea ... and there, according to the figure below, we see a phase shift of about 180° (*I say approximately, because it is not very precise, just indicative!*). - Always the same "9 squares" method, really easy to use and accessible to everyone:

I fine-tune my time base so as to obtain the famous 9 squares for a complete period of**CH1**. The**CH2**trace for the same given point is then offset by 0.9 squares (*this will be the value***t**). If we apply the given formula, we get:**φ = (t / T) x 360°**= (0.9 / 9) x 360 =**36°**

The**CH2**curve is 36 ° ahead of the curve of reference**CH1**. - Processing of the second Secondary winding, always the same method, I leave the GBF on
**CH1**and on the Primary winding (*pin 2*), and I move the**CH2**probe to pin 3.

As for the previous winding, a small figure of Lissajous to see what this weak phase shift looks like ... At first glance, the phase shift seems weak ... I immediately check as for the previous oscillograms:

I fine-tune my time base so as to get all 9 tiles for a full period of

**CH1**. The trace

**CH2**for the same given point is then offset by 0.9 squares (

*it will be the value*

**t**). If we apply the indicated formula, we get:

**φ = (t / T) x 360°**= (0.9 / 9) x 360 =

**36°**

The curve

**CH2**is 36 ° ahead of the reference curve

**CH1**. Exactly as for winding

**S1**.Here is finished with the auscultation of this particular transformer, which is none other than the one which transmits the PWM implusions coming from the integrated circuit TL494 to the 2 transistors mounted in a "Push-Pull" circuit and which "cut" the positive supply of nearly + 300V in the Primary circuit for this ATX type power supply.I got there because I have this power supply for PC, new because it has never been used, and which displays brâvemant 400 W max while it unfortunately can hardly provide more than 230W max (

*see the article concerning the "deviations" of ATX power supplies*) ... diagram not found on the Internet, that's why I undertook to dismantle it to recreate its diagram as faithfully as possible, in order to perhaps modify it, but that is already another story ...I want to thank you for having had the time and the courage to read me until the end of this article which I had the great pleasure of writing.

I am like I like to say it from time to time for the sharing of information and knowledge ...Philippe

♦ Here is the link to the blog of my friend Félix, which presents the transformation of an ATX type power supply for PC into a small laboratory power supply:

http://xaaander.blogspot.fr/2014/04/concevez-votre-alimentation -de.html? showComment = 1396618334812 # c5713801989805613594

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