Filtering 50hz out

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charge

Joined Mar 13, 2010
4
Hi

Does anyone know the formula to work out the value of the capacitor to block the 50Hz signal passing through the transformer but allowing other higher frquencies through (around 10KHz).


-------||--------|---------------|-------
..........c? ..........|.....................|
50Hz .................|...Transformer..|..No 50Hz on this side
........................|......... 1:1...... |
-----------------|---------------|-------

transformer: L = 1.4mH
 

Thread Starter

charge

Joined Mar 13, 2010
4
yes i would like to use the capacitor and the inductance of the transformer as a filter as shown in the image of this pdf file.

www.avnet.co.za/Designers_Choice/issues/DC2006-17.pdf

i am using that exact transformer and it says a capacitor can be used with that transformer to create a filter. was just wondering if anyone knew the formula related to that setup.

Also it will be at high voltage so i want to use as few components as possible
 

t_n_k

Joined Mar 6, 2009
5,455
It depends by how much you want to attenuate the 50Hz component and where the signal component frequency band starts.

If Xc is the capacitive reactance then you will make Xc >> XL where XL is the transformer reactance at 50Hz.

If Xc=1/(ωC) and XL=ωL

then you will set Xc=K*XL where K is the desired ratio of the reactances.

If K=100 (say) and L=1.4mH (from your data sheet)

then 1/(ωC)=K*ωL

or C=1/(K*ω^2*L)

then C=1/(100*314^2*1.4e-3)=0.72uF

If you want K=1000

then C=0.072uF [say .068uF as a preferred value]

But this C value [0.068uF] will have a significant reactance at 10,000Hz - 234Ω, and this is significant in comparison to the inductive reactance of 88Ω at that same frequency.

So there is a trade-off in terms of the overlap between the frequencies you want to attenuate and those you want to pass relatively unattenuated.

Of course I forgot that the reactances have sign relative to one another so it's not quite that bad.

The transfer ratio G, at 10kHz with .068uf

|G|=|XL/(XL-Xc)|=88/146=0.6

With C=0.68uF at 10kHz

|G|=88/(88-23.4)=1.36 - meaning the resonance effect (@~5.2KHz) is becoming a factor.

In the long run it's probably better to do a series of frequency-magnitude plots as a function of C to determine what's the best compromise value.
 
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t_n_k

Joined Mar 6, 2009
5,455
You also need to keep in mind that there will be a limiting condition for the choice of C based on the 50Hz voltage. You used the term high voltage - exactly how high?

As the value of C increases so does the current drawn from the 50Hz side.

At 50Hz the 0.68uF has a reactance of 4.7KΩ. While the current would not be excessive if the voltage were 240V (say), you still have to choose a suitably rated capacitor type that can withstand direct 240V AC continuously without any degradation.
 

t_n_k

Joined Mar 6, 2009
5,455
If K=100 (say) and L=1.4mH (from your data sheet)

then 1/(ωC)=K*ωL

or C=1/(K*ω^2*L)

then C=1/(100*314^2*1.4e-3)=0.72uF

If you want K=1000

then C=0.072uF [say .068uF as a preferred value]
Bit of a brain seizure on this one unfortunately ....

if K=100 C=72uF

if K=1000 C=7.2uF

if K=10,000 C=0.72uF

So 1uF would probably be fine, giving better than 1000 x attenuation at 50Hz.

This value puts resonance at ~4.3kHz. Whether that's a "problem" for the signal transmission task I'm not sure.

The other major issue one would have to watch out for is that the transformer core doesn't become saturated during the 50Hz AC cycle. I haven't checked whether the data sheet gives any indication as to what primary 50Hz voltage level would lead to the onset of saturation.
 

Thread Starter

charge

Joined Mar 13, 2010
4
thank you for all the information, i will read it and try to understand all of it. it will be used at 240v blocking out mains frequency.
 
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