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Calculating LP filter values for TDA1541A
Calculating LP filter values for TDA1541A
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Old 18th July 2013, 01:45 AM   #1
Ceglar is offline Ceglar  Australia
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Default Calculating LP filter values for TDA1541A


I'm wanting to know how to calculate the low pass filter frequency, for an ultrasonic filter, for use with TDA1541A.

I've seen various values of capacitance placed across (typically) an I/V resistor, and also some mention of scaling the caps value depending on the value of that resistor - although, it seems to me, the filter itself would be formed by the source impedance of the 1541A and the value of that shunt capacitance - but then again I dont know that, or the source impedance of TDA1541A (not in the data sheet??).

Can any one shed any light on this?.

Shane Ceglar

Last edited by Ceglar; 18th July 2013 at 01:46 AM. Reason: .
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Old 23rd July 2013, 12:38 PM   #2
Ceglar is offline Ceglar  Australia
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Yeah, well - does no one really know???

What I've gleaned is that the relationship between the cap shunting the I/R resistor forms the corner freq. of LP filter. Not in series, but perhaps with a current source they appear in parallel. I dont know, but it appears that way. For example 33R I/V // 47nF forms a filter that is quite similar to the 2.2nF//1.5k . The numbers all come out the exact same. To which end a couple of russian military teflon film and Al. foil were removed from their casings and potted in wax, sounds great - and up ya bum!

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Old 23rd July 2013, 02:05 PM   #3
lcsaszar is offline lcsaszar  Hungary
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You can set the cutoff frequency to effectively attenuate the sampling frequency (176.4 kHz in most cases) and the aliasing sidebands. So you set it as low as possible, but so that it should not affect the top end of the audio range. For lower sampling frequency you need a higher order filter, that probably will have some unwanted phase shift in the audio band.
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Old 24th July 2013, 12:03 AM   #4
PlasticIsGood is offline PlasticIsGood  United Kingdom
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The feedback resistor determines the I/V ratio, so for example a 1k resistor will result in 1V per mA, more or less. A cap in parallel with the resistor will reduce the impedance of the combination as frequency increases, resulting in less V per mA. When f is such that the impedance of the cap is the same as the resistor, the combination will be halved, so the V out will be halved, etc.

If you think in that direction you can answer all your questions and direct yourself to the maths of simple L-C-R combinations. Then there are a few useful filter calculation programs for free. Sallen Key filter design in particular is worth a search. Illustrated tinkering with complete filters will show the I/V stage in context. Graphs of both group delay and phase angle are useful.

Input and output impedance of opamps are assumed to be infinite and zero, respectively, for the purpose of calculation.

You will find debate on several issues. Whether to roll off slowly and early, or sharply and late; whether to use Butterworth or Bessel, or something more drastic; whether to include the I/V stage in the filter; and whether active or passive.

There appears to be no single, simple and comprehensive survey of issues.
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Old 24th July 2013, 01:55 AM   #5
abraxalito is offline abraxalito  United Kingdom
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If your aim is ultrasonic filtering, then you don't get much of that with a simple C across the I/V resistor.

If you're using your TDA in NOS, the images start after 22.05kHz. Depending on how they produced the recording, there might be absence of content until 24.1kHz but its probably best not to assume that.

Bottom line - if you're serious about chopping out ultrasonics, then a multi-pole filter's a must and that means inductors or going active.
'The total potential here must be nothing less than astronomical.'
'Nothing less. The number 10 raised almost literally to the power of infinity.'
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