output LR filter

Helix · 8th February 2002, 09:28 AM

Most of the class AB amps i have built seem to have a parallel L R output filter. Say a inductor of 10 turns and a 1R resistor. i was wondering if any one could tell me :
a) Why is it there?
b) Can one recalculate the turnover freq, and make the inductor smaller?

Geoff · 8th February 2002, 09:58 AM

Helix

The output inductor is fitted to ensure that amplifier stability is maintained into capacitive loads. Common values are 2 to 6mH (10 to 20mohm dcr) with a 1 to 10ohm damping resistor in parallel to reduce ringing. Air cored inductors are preferred.

If you are using normal speakers, and the speaker cable is not one of the high capacitance types, it should be possible to remove the inductor/resistor completely. Alternatively, the inductor/resistror could be replaced with a suitably rated 0R1 resistor which will ensure stability but will reduce the maximum power a little, particularly into a 4ohm load.

Geoff

Helix · 8th February 2002, 10:34 AM

my speaker cable isn't one of the strange types, but the speakers are 3 way and may introduce a bit of capacitance from the crossover networks, i think?

the amp in question has a
"inductor of 2mH and you make it by winding 15 turns of 18 SWG enamelled wire on a 3W wire wound resistor"
where the resistor is 1R.

i was wondering if you could not eleminate it (just incase) but just reduce the values a bit.

Geoff · 8th February 2002, 11:05 AM

Helix

The smallest inductance value I can remember seeing was about 1mH (10 to 12 turns of 22# round a 2W 10ohm resistor). There is no way that I know of for easily calculating the required inductance. The normal method of determining it would be to measure the amp into various capacitive loads and to observe the overshoot and ringing on a square wave.

Assuming that you do not have the test equipment to do this, I see no reason why you shouldn't reduce your inductor to say 10 turns and listen to the results. There may or may not be an audible difference between the two inductance values. My guess is that there won't be.

Geoff

Helix · 8th February 2002, 12:36 PM

ok, i will just leave it be then, i just thought that a 1R resistor in to 4 ohm load would loose a little power at 100 or so watts.

isn't L for a air core solenoid coil :
L=(u*N^2*A)/l
where:
u = permeabilty of free space
N=number of turns
A=cross section area
l = inductor length

Helix · 8th February 2002, 12:39 PM

having said that, the inductor resistance is far lower and hence the resistor value is meaningless

Geoff · 8th February 2002, 01:06 PM

Helix

Sorry, I obviously hadn't woken up properly when I made the previous posts. All references to mH should have read uH.

At audio frequencies the resistor has virtually no effect since the low impedance of the inductor effectively bypasses it. The resistor comes in to play at higher frequencies when the impedance of the inductor begins to rise towards a similar value to that of the resistor. In your case, with a 1ohm resistor and 2uH inductor (I assume your 2mH was in error as well), this will be somewhere above 200kHz.

Geoff

hifi · 8th February 2002, 01:08 PM

BTW: i once read an article on magnetic coupling between output coils in an amplifier....but i dont remember the conclusions or where i read it, Electronics World? ....

anybody?

/micke

kokoriantz · 29th November 2020, 01:16 PM

How to calculate the L & R ? any known method?

OldDIY · 29th November 2020, 04:30 PM

The loads are different. Amplifiers are different.

8th February 2002, 09:28 AM	#1
Helix diyAudio Member Join Date: Jul 2001 Location: UK	output LR filter Most of the class AB amps i have built seem to have a parallel L R output filter. Say a inductor of 10 turns and a 1R resistor. i was wondering if any one could tell me : a) Why is it there? b) Can one recalculate the turnover freq, and make the inductor smaller?

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8th February 2002, 09:58 AM	#2
Geoff diyAudio Member Join Date: Apr 2001 Location: UK	Helix The output inductor is fitted to ensure that amplifier stability is maintained into capacitive loads. Common values are 2 to 6mH (10 to 20mohm dcr) with a 1 to 10ohm damping resistor in parallel to reduce ringing. Air cored inductors are preferred. If you are using normal speakers, and the speaker cable is not one of the high capacitance types, it should be possible to remove the inductor/resistor completely. Alternatively, the inductor/resistror could be replaced with a suitably rated 0R1 resistor which will ensure stability but will reduce the maximum power a little, particularly into a 4ohm load. Geoff

8th February 2002, 10:34 AM	#3
Helix diyAudio Member Join Date: Jul 2001 Location: UK	my speaker cable isn't one of the strange types, but the speakers are 3 way and may introduce a bit of capacitance from the crossover networks, i think? the amp in question has a "inductor of 2mH and you make it by winding 15 turns of 18 SWG enamelled wire on a 3W wire wound resistor" where the resistor is 1R. i was wondering if you could not eleminate it (just incase) but just reduce the values a bit.

8th February 2002, 11:05 AM	#4
Geoff diyAudio Member Join Date: Apr 2001 Location: UK	Helix The smallest inductance value I can remember seeing was about 1mH (10 to 12 turns of 22# round a 2W 10ohm resistor). There is no way that I know of for easily calculating the required inductance. The normal method of determining it would be to measure the amp into various capacitive loads and to observe the overshoot and ringing on a square wave. Assuming that you do not have the test equipment to do this, I see no reason why you shouldn't reduce your inductor to say 10 turns and listen to the results. There may or may not be an audible difference between the two inductance values. My guess is that there won't be. Geoff

8th February 2002, 12:36 PM	#5
Helix diyAudio Member Join Date: Jul 2001 Location: UK	ok, i will just leave it be then, i just thought that a 1R resistor in to 4 ohm load would loose a little power at 100 or so watts. isn't L for a air core solenoid coil : L=(uN^2A)/l where: u = permeabilty of free space N=number of turns A=cross section area l = inductor length

8th February 2002, 12:39 PM	#6
Helix diyAudio Member Join Date: Jul 2001 Location: UK	having said that, the inductor resistance is far lower and hence the resistor value is meaningless

8th February 2002, 01:06 PM	#7
Geoff diyAudio Member Join Date: Apr 2001 Location: UK	Helix Sorry, I obviously hadn't woken up properly when I made the previous posts. All references to mH should have read uH. At audio frequencies the resistor has virtually no effect since the low impedance of the inductor effectively bypasses it. The resistor comes in to play at higher frequencies when the impedance of the inductor begins to rise towards a similar value to that of the resistor. In your case, with a 1ohm resistor and 2uH inductor (I assume your 2mH was in error as well), this will be somewhere above 200kHz. Geoff

8th February 2002, 01:08 PM	#8
hifi diyAudio Member Join Date: Mar 2001 Location: sweden	BTW: i once read an article on magnetic coupling between output coils in an amplifier....but i dont remember the conclusions or where i read it, Electronics World? .... anybody? /micke

29th November 2020, 01:16 PM	#9
kokoriantz diyAudio Member Join Date: Dec 2015 Location: south east asia	How to calculate the L & R ? any known method?

29th November 2020, 04:30 PM	#10
OldDIY diyAudio Member Join Date: Dec 2009	The loads are different. Amplifiers are different.