Hello,
I purchased the BIAMP 6-24 CROSSOVER pcb at the DIYaudio store.
The left graph is the output of the on pcb CRCRC DC filter. The wall wart feeds the filter.
The analyzer input is taken at the output of the pcb onboard DC filter.
Thanks DT
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Indeed, often the feedback network is crucial and benefits from physically larger resistors than necessary to reduce the self-heating.
There's a clever trick (though I do wonder if its ever done) of using a string of identical resistors as the feedback network, thus cancelling out non-linearity in resistance (for instance a 20-fold divider made from 20 1k resistors in series to ground, with the feedback tap from the first one.
As the currents are the same (assuming input stage is very high impedance) it doesn't matter if R(I,t) is non-linear in I or depends on t, all resistors have the same R(I,t), same current, and same time (relativity can be ignored!). Each resistor is in an identical state so the voltage division ratio is simply 1:20, the integer ratio of the resistor counts. Basically the current through the network can be distorted without this appearing as a voltage.
Before going to such lengths though you might want to pay careful attention to feedback take-off points first as this can be a source of distortion if care isn't taken, and no amount of careful feedback network design/component choice can mitigate this.
[ thinking about it the 20 resistors in a row idea probably solves the self-heating issue just by spreading the power over 20 components! Its a brilliant idea apart from the large BoM - perhaps more suited to SMT where its hard to get thin film resistors as performant as through-hole versions, but 20 cheap thick-film resistors might do surprizingly well - some research for those with good audio analysers perhaps? ]
Mark,
If you want to see resistor distortion test different load resistors at the output of some of the best power amplifiers and or headphone amplifiers.
Some of the Topping headphone amplifiers A30 PRO or power amplifiers Benchmark AHB2. Some of these amplifiers push the limits of the audio analyzers on the market.
The quality of the test load resistors is quickly apparent.
Thanks DT
Check the Resistors non magnetic leads those are the one to get. We tossed thousand of r
at TI when we were design seismic instrument, resisters with magnetic leads. 🙁 They metal film resistor are made by cutting a spiral into the body and they can cause a small magnetic field at higher frequency. sometimes they can cause problems but not often. depending on the app you’re better off with a film resistor it just depends.
at TI when we were design seismic instrument, resisters with magnetic leads. 🙁 They metal film resistor are made by cutting a spiral into the body and they can cause a small magnetic field at higher frequency. sometimes they can cause problems but not often. depending on the app you’re better off with a film resistor it just depends.
The point is that they all self heat the same amount, so they all change value the same amount so the division ration does not change with temperature. The same story with voltage coefficient. The idea came originally from Bruce Hofer, one of AP's founders and now retired.
But. If the current through the resistors is distorted, so is the voltage at the pick-off point. That doesn't change.
Jan
Yes resistors can vary with and they have tempo and most signal you have nothing to be concerned about. I personal feel that higher wattage resistor sound a little better like RN60 and 65. I have no proof except listening. Not often do you run resistors in audio at high power. If high power is necessary change to use a wire wound. If you need ultimate go for Vishal bulk metal film for accuracy $$$ they are as good as you can get.
This used to be a fairly common technique for integrated circuit designers, when we still had to use quite nonlinear diffusion resistors (in separate wells, you also had to connect the wells properly). On an IC, it also improves resistor matching and hence gain accuracy.
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As long as they all change by the same amount, you should get distortion compensation in the voltage at the pick-off point. The non-linear voltage-to-current transfer of the resistor string distorts the current, but the opposite current-to-voltage non-linearity of the bottom resistor compensates for that when you look at the voltage across it.
The momentary power varies a lot more with signal when there is a large DC bias: (VDC + v)2 = VDC2 + 2 VDC v + v2. The second term causes extra temperature variations that cause even-order distortion. I think that's Nelson's point.
Sorry Marcel, we misunderstand each other. The point was that if the current through the string is distorted (which was the stated case), the voltage at the pick-up point is also distorted, this is the case even with perfectly linear resistors.
The other point was the thermal and voltage modulation of the resistor values by the signal, those changes are nulled out by the string.
Jan
I see I have used the term take-off point incorrectly. Mark meant the point in the output stage that the feedback network is connected to, I meant the tap of the feedback network that goes to the input stage.
Anyway, suppose you have two very non-linear resistors with i = K v3. Connect them in series and put a voltage A sin(omega t) across them. The voltage across each of them will then be a perfectly undistorted A/2 sin(omega t), while the current will be a grossly distorted K A3/8 sin3(omega t). Replace one of them, but not both, with a linear resistor and the voltages will also be distorted.
Anyway, suppose you have two very non-linear resistors with i = K v3. Connect them in series and put a voltage A sin(omega t) across them. The voltage across each of them will then be a perfectly undistorted A/2 sin(omega t), while the current will be a grossly distorted K A3/8 sin3(omega t). Replace one of them, but not both, with a linear resistor and the voltages will also be distorted.
Fully agree with that Marcel. My take was that a current was flowing which in itself was distorted to begin with. In that scenario the tap voltage would also be distorted even with perfectly linear resistors. Sorry for the confusion.
Jan
Jan