Need help understanding the output of a Transformer

[sgkini]

Hi,
    I was trying to learn about transformers. So I purchased one from ebay. I received  it yesterday . The detail of the transformer is attached along with the message.  I connected the primary (blue + brown) to the mains (230 V 50 Hz) and connected the secondary to my Siglent DSO and captured the output .Then I connected a 6.8 K power resistor across the secondary and then captured the image again.

The output looks like an output from a bridge rectifier. I thought transformer converts from AC to AC.  Is this transformer custom made and has the bridge rectifier inside in the center somewhere or is my understanding totally  wrong ?


Regards
Satish

[xygor]

Yes, it is rectified according to the label. After SEC it says ... DC.
According to the line after that, it also appears to be fused.
I'm not sure what the "T" means.  Thermal protection maybe.

[JOERGG]

I'm not sure what the "T" means.  Thermal protection maybe.

The T behind the 6,3 A Fuse stands for "träge" witch means delayed action or slow fuse.
The thermal protection switch at 125 degrees C is insertet in the primary winding, according to the label.

[tautech]

Now you know it has an integral full wave rectifier, and the giveaway is the waveform in your second image.
Also note the frequency of 100 Hz is your line frequency x 2

[sgkini]

Thanx for helping me understand.  I did not read the DC part till  xygor pointed it out. 

I assumed that 25.5 V was RMS voltage so the Vpp would be around 72 V and the output from the bridge rectifier would be around 36 V  - 2 * Vf  . I connected a 2200 µF cap across the 6.8 K resistor and found the dc voltage to be arounf 51V (2*25.5 ) . Can you please explain how that is possible or is there some thing I understood wrong? 

ThanX for taking the time to explain it to me.

Satish 

[codeboy2k]

25.5 V RMS is  (25.5  / 0.707 ) peak = 36.07 peak AC Volts. When full wave rectified it will be 36.07 - (2*Vf) = 34.87, assuming Vf=0.600 .

After the 2200 uF it will be 34.87 * sqrt(2) = 49.31V

You are seeing almost 51V most likely because the transformer is unloaded so the voltage is actually higher than 25.5V.  The 25.5V is probably the output voltage at the rated 5.4A load.

Working backwards from your scope output, which shows 51.2 volts ==> ( ( 51.2 / sqrt(2) ) + 2*Vf)   = 26.44 37.4 VAC RMS

[sgkini]

I understood the first part, that the peak voltage of the AC wave would be 25.5 * sqrt(2) = 36.06 and the peak after full wave rectifications will be around 34.6 V. But you lost me after that

Why are you mutliplying the peak value again with sqrt(2) ?  I thought that the smoothing cap will charge to the peak value of the full-wave rectified voltage (34.86 v) and discharge depending on the load and forming a saw-tooth like ripple.

Would be great if you could explain why sqrt(2) is multiplied to the  peak value ?


Satish

[megajocke]

sgkini: You are right about that. It should only appear once in the calculation.

However, that transformer is marked "ED=20%" which means that it is not rated for continuous operation but for 20% intermittency at 190 VA. (20% of the time on, 80% off) I'd guess it's the transformer for powering the motor of a desk with adjustable height. The intended application would have the mains connected to the transformer all the time, but it would only be loaded for short periods quite seldom.

If you only load it to 40% of its rated current it should be okay for continuous operation because this gives 20% of the heating compared to the rated current which it is only rated for to carry 20% of the time.

The reason the voltage is quite high is to compensate for the voltage drop at rated load. The voltage drop at rated load can be quite large for a low-intermittency tranformer.

25.5V might also be the average rectified voltage (rather than RMS), including the diode drops. The average rectified value is about 0.9 times the RMS value for a sine, so it adds about 10% to the expected peak value.

[N2IXK]

Are transformers with built-in rectifiers common in Europe? The only ones I have seen here in the states are the "wall wart" power pack types, which usually have a filter cap as well.

Have never come across a discrete transformer that had a built-in bridge. Seems like a pretty stupid idea, actually. If the bridge fails, you need to replace the entire thing, which is going to cost a LOT more than a simple rectifier would...

[sgkini]

Are transformers with built-in rectifiers common in Europe? The only ones I have seen here in the states are the "wall wart" power pack types, which usually have a filter cap as well.

I never knew one existed before I got this one. At least learned something new

Have never come across a discrete transformer that had a built-in bridge. Seems like a pretty stupid idea, actually. If the bridge fails, you need to replace the entire thing, which is going to cost a LOT more than a simple rectifier would...

That's exactly what came to my mind, what if the bridge fails? 

[codeboy2k]

I understood the first part, that the peak voltage of the AC wave would be 25.5 * sqrt(2) = 36.06 and the peak after full wave rectifications will be around 34.6 V. But you lost me after that

Why are you mutliplying the peak value again with sqrt(2) ?  I thought that the smoothing cap will charge to the peak value of the full-wave rectified voltage (34.86 v) and discharge depending on the load and forming a saw-tooth like ripple.

Would be great if you could explain why sqrt(2) is multiplied to the  peak value ?

Satish

Doh! my sleepy mistake... I've been awake all night.. I screwed that calculation.. , you're right ... There is only 1 multiplication by sqrt(2), which I already did when I divided by .707

[megajocke]

That's exactly what came to my mind, what if the bridge fails?

I'd guess you throw it away, just as if the thermal fuse had failed. If this really is from an adjustable height desk I doubt you'd do any component-level repair.

Still, it seems highly unusal to pot the bridge rectifier inside the transformer.

[sgkini]

@codeboy2k no worries . Every one make mistakes and I know how its after pulling an all nighter .

@megajocke Yes seems very unusual as per the very little knowledge of transformer I have. I was thinking that something was wrong with my scope when I saw the no load waveform. But things got a little clear after adding a 6.8K load and a lot clearer here. Got it on ebay for 7 euro + shipping. While buying I thought the output of the transformer was 25.5 RMS VAC . Next time should read more carefully 

[xygor]

That's exactly what came to my mind, what if the bridge fails? 

Why do you think it's a bridge?

Measure the output with your ohmmeter on the diode setting (with the primary not energized) and see how many diode forward drops you measure.

It could be a center-tapped secondary with two diodes.

[tszaboo]

That's exactly what came to my mind, what if the bridge fails? 

Why do you think it's a bridge?

Measure the output with your ohmmeter on the diode setting (with the primary not energized) and see how many diode forward drops you measure.

It could be a center-tapped secondary with two diodes.

Usually silicon is cheaper than copper. This is interesting, I dont think I've seen transformer with built in rectifier potted in the middle. But it actually makes sense, less cabling, automatically good size rectifier. I kinda like it.