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?
a) Why is it there?
b) Can one recalculate the turnover freq, and make the inductor smaller?
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
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
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.
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.
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
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
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
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
having said that, the inductor resistance is far lower and hence the resistor value is meaningless
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
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
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
anybody?
/micke
If you want the amplifier to be small-signal stable with any passive load:
1. Determine the amplifier's output impedance without the L and R
2. Look for frequency regions where the real part of the output impedance goes negative. I'll assume there is a negative peak at frequency f_neg.
3. Choose an R that is greater than two times the opposite of the most negative real part of the output impedance, for example four times the opposite of the most negative real part of the output impedance
4. Connect an L in parallel with a reactance equal to R at f_neg
5. Determine the impedance of the amplifier with the L and R included and fine tune their values as needed to make the real part of the impedance positive at all frequencies
I took a more pragmatic approach when I designed my amplifier: I came up with a target for the magnitude of the output impedance at 20 kHz, divided that by 2 pi * 20 kHz to find the inductance, put a resistor in parallel with a value close to the nominal load impedance (8.2 ohm) and shunted the output with a capacitor with value C ~= L/R^2 to turn the whole thing into a first-order series filter. (Making it a first-order series filter was recommended in an article of A. N. Thiele, as it helps to keep RF signals out of the amplifier.) I have no idea if it is stable with any passive load, but it certainly is with all loudspeakers and cables I ever connected to it.
1. Determine the amplifier's output impedance without the L and R
2. Look for frequency regions where the real part of the output impedance goes negative. I'll assume there is a negative peak at frequency f_neg.
3. Choose an R that is greater than two times the opposite of the most negative real part of the output impedance, for example four times the opposite of the most negative real part of the output impedance
4. Connect an L in parallel with a reactance equal to R at f_neg
5. Determine the impedance of the amplifier with the L and R included and fine tune their values as needed to make the real part of the impedance positive at all frequencies
I took a more pragmatic approach when I designed my amplifier: I came up with a target for the magnitude of the output impedance at 20 kHz, divided that by 2 pi * 20 kHz to find the inductance, put a resistor in parallel with a value close to the nominal load impedance (8.2 ohm) and shunted the output with a capacitor with value C ~= L/R^2 to turn the whole thing into a first-order series filter. (Making it a first-order series filter was recommended in an article of A. N. Thiele, as it helps to keep RF signals out of the amplifier.) I have no idea if it is stable with any passive load, but it certainly is with all loudspeakers and cables I ever connected to it.
Thank you for details. A standard amplifier exhibits resistive output impedance at low frequencies up to OLBW, than it becomes inductive. If a capacitive reactance comes on the output, the two opposite reactances in series provoke resonance. The serie LR will come in series with the output inductance to lower the resonance and damp it. The question is if I know my internal LR how to estimate the external one. Also to note that there is the Zobel which comes into account. If I understand, the Zobel, internal LR and the external LR must be link to each other, but how?
As a result, get a compensated active circuit. At least at critical frequencies.
All together is called Zobel 🙂
All together is called Zobel 🙂
True, but at yet higher frequencies, the real part of the output impedance will usually go negative, which may cause oscillations with the wrong load impedance. That can be solved with either an RC series circuit across the output or an LR parallel circuit in series with the output. I haven't a clue why designers often do both.
Most Japanese, McIntosh, put the Zobel after the LR. The Zobel is mandatory for high impedance current outputs as the OLG depends on the load. If the load is inductive, the amp oscillates. All explanations I found about Zobel by google, relate to speaker impedance. An amp is universal impedance.
Not sure there is any mutual coupling between these coils.
Too small, too few turns, too far from each other.