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HAYK
In the early 80's Luxman used Duo-β feedback network in L-58A, L-190A, L-200/205/210/215/220/225/235/, L-410/430/510/530/550 and many more...
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HAYK
The purpose of phase margin is to provide stable operation in high transients. Measuring the phase margin at noise level doesn't say what happens at full power.
This why the square wave test is a better test.
I show on #33 how the transient on simulator is perfect and mentioned that in real world it required 22pF for a model and 50pF for the 2 others to get perfect square wave as without they are perfectly stable but had overshoot.
Using low impedance feedback network reduces the phase shift due to input capacitance, this why it can go bellow minimum 20db gain measured with 1k impedance.
To note that the distortion and noise should be 7 times lower than the spec. at 26db, to be 0.0007%.
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I agree that PM/GM are for small signals and that the circuit needs to work with large transients as well. No argument there. I never suggested otherwise.
Also note that the LM3886 has some interesting quirks as the output signal approaches the supply rails, especially with 4 Ω load. Those aren't covered in the simulation model either, thus, some experimentation in the lab is needed if you want a stable amp.
Tom
Also note that the LM3886 has some interesting quirks as the output signal approaches the supply rails, especially with 4 Ω load. Those aren't covered in the simulation model either, thus, some experimentation in the lab is needed if you want a stable amp.
Tom
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HAYK
Oh, I'm sure you'll notice once you get the temperature up to the smoke point of the oil... 
🙂
Years ago I picked up a heat sink with fans on eBay. Apparently it came from a cell phone base station. It's pretty substantial. I bolted 20 non-inductive 0.8 Ω, 25 W resistors to it. That allows me 2x8 Ω with a tap at 4 Ω for testing stereo amps. With various series/parallel combinations I can get to 4 Ω, 500 W and 2 Ω, 250 W. The whole thing set me back less than $50 plus an evening of drilling and tapping holes in the heat sink.
Tom
🙂 Years ago I picked up a heat sink with fans on eBay. Apparently it came from a cell phone base station. It's pretty substantial. I bolted 20 non-inductive 0.8 Ω, 25 W resistors to it. That allows me 2x8 Ω with a tap at 4 Ω for testing stereo amps. With various series/parallel combinations I can get to 4 Ω, 500 W and 2 Ω, 250 W. The whole thing set me back less than $50 plus an evening of drilling and tapping holes in the heat sink.
Tom
40 years ago, we had a paint can full of resistors and oil. It smelled when it got hot. Later at another job, we used active DC loads for testing power supplies where you could program the current, resistance or voltage. I ran 3 ~KW supplies into 4 N3300 loads and the environmental chamber all driven from an Excel Macro that saved the test log. AC active loads are possible but it requires essentially a big amplifier to dissipate the power. Some DC active loads are also inverters that feed the power back into the mains.
I’m thinking I can just get four 1R 100W resistors and put the m in series. Probably can get cheap heatsink to mount them on.
I don't understand that statement at all.
Resistors in series all 'see' the same current. The current, I, is V/Rtot, where V is the voltage applied across the entire string of resistors and Rtot is the total resistance of the string. For a series combination of resistors, Rtot = R1+R2+R3 .... +Rn. The power dissipated in each resistor can then be calculated as P = I^2*R.
Resistors in parallel all 'see' the same voltage. They then take a share of the total current inversely proportional to their resistance. You can go through and calculate all the currents as I = V/R, add them up, and then determine the total resistance of the parallel combination. You'll find that the resistance of the parallel combination, Rtot, is Rtot = 1/(1/R1 + 1/R2 + 1/R3 .... + 1/Rn). The power dissipated in each resistor can be calculated as P = V^2/R.
If all resistors are of equal resistance the math can be simplified:
Series combination: Rtot = n * R, where n is the number of resistors.
Parallel combination: Rtot = R/n, where n is the number of resistors.
You can then calculate the power dissipated in the entire resistor network as P = V^2/Rtot or P = I^2*Rtot. The power dissipated in each resistor is then P/n, where n is the number of resistors.
Tom
I guess I got confused with voltage vs current.
Looks like it doesn’t matter if it is series or parallel connection, as long as resistance is the same, the power dissipated in each resistor is the same.
Looks like it doesn’t matter if it is series or parallel connection, as long as resistance is the same, the power dissipated in each resistor is the same.
Tom,
I am curious, when you tested for above mentioned high output signal, did you use inverting/non-inverting configuration?
The reason why I’m asking is because I’m wondering if same applies for configuration where LM3886 is used as a difference amp?
I plan on testing for various edge cases in my build in the future, but it is interesting to hear your thoughts on this.
Thanks
I tested with a non-inverting configuration, but I doubt that matters. I'm pretty sure what's going on is that the VAS loses gain as it approaches saturation and that pushes the LM3886 closer to instability. But without access to the internal nodes of the LM3886 I don't know that for sure.
Tom
Tom
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HAYK
You can see in this video he is using 25w resistor in a yellow plastic container filled with water. I have seen the same author using a high power amp where the water was boiling.