How can we build an efficient microcontroller-based amplifier?

Kragen Javier Sitaker, 2016-07-13 (5 minutes)

Some folks at the Fabricicleta were asking me a few years back about making high-efficiency amplifiers for audio from bicycles, so that you can really blast music driven by a five-watt hub generator I guess.

Suppose you use, say, a 180MHz Cortex M3 LPC1830 as your processor, and suppose that you can only output one bit every two CPU cycles, or 90MHz. A simple pulse-density modulation scheme that swings from 100010001000... to 111011101110... should limit the artifactual waveform from the PDM modulation to one fourth of the bit rate, or 22½MHz; it can approximate waveforms of up to -6dB. Then you can feed this bitstream into a Darlington or a power FET or whatever, then filter its output reactively to avoid power dissipation.

(There may actually some lower frequencies that show up from dithering: 10101101010110..., for example, has an oscillation whose period is seven bit times.)

This 22½MHz is three decades or ten octaves above anything we need to reproduce for audio purposes. Even with just a single-pole 6dB/octave LR or LC filter, you get 60dB of attenuation as long as the cutoff frequency is at or below 20kHz; I think that with two stages of LC filter, you’ll get 24dB/octave, which would give you 240dB of attenuation. You should be able to use fairly small-value inductors and capacitors for this, because the total amount of energy stored for half a cycle of a 22MHz wave is very small.

Supposing we’re driving an 8Ω speaker, we’d probably like the output impedance of the final-stage filter to be a lot smaller than that at 20kHz. This actually means that the entire signal path from the amplifier through the two inductors to the speaker needs to have a lot less than 8Ω impedance, say, 0.8Ω.

Capacitive reactance is 1/ωC, while inductive reactance is ωL. This suggests that our maximum total inductance is when ωL = 0.8Ω, which happens when the inductors total 6.4μH. And we’d like the second one to have most of that, so, say, 0.6μH in the first inductor, and 5.8μH in the second one, thus 75 milliohms of reactance in the first and 730 in the second.

Now we want our capacitors to ground to have the same reactance as the inductors at our 20kHz cutoff frequency. So, for the first one, 75 mΩ = 1/(2π 20kHz)C, ∴ 2π 20kHz C = 1/75mΩ, ∴ C = 1/(2π 20kHz 75mΩ) = 106 μF. That’s a reasonably attainable value, though it has to be electrolytic. The second one need only be 11μF.

Digi-Key’s most popular inductor, with 1,077,238 in stock at the moment, is a 10¢ TDK 82nH 150mA inductor, the MLK1005S82NJT000. Sadly, its DC resistance is 1.8Ω, because it’s in an 0402 package (1mm × 0.5mm), so it can’t serve as a high-efficiency reactive filter for an 8Ω speaker unless we put a bunch of them together. Which is actually a reasonable thing to do.

Digi-Key's most popular inductor in the right range of inductance and DC resistance is Murata’s LQM2HPN1R0MG0L, a 34¢ 1μH (±20%) 1.6 amp 55mΩ ferrite-core inductor in a 2520 (2.5mm × 2.0mm) package. Its minimum self-resonant frequency is 60MHz, which suggests that it may actually not be a great choice here; it might capacitively couple through the exact high-frequency energy we want to stop.

If I restrict to only inductors with a self-resonant frequency over 180MHz, the most popular is the TDK MLZ1005MR47WT000, a 28¢ 470nH 500mA 200mΩ multilayer ferrite-core inductor with a 260MHz self-resonant frequency, in the same 0402 package as the most popular one. It's marketed as an “inductor for decoupling circuits”. Its saturation current is 120mA, which I guess means that above 120mA it doesn’t induct any more. 120mA on an 8Ω load would be 115 milliwatts, but I guess you could use several of these inductors.

I don’t know, I guess I have a lot more to learn about reactive filters.

Could you just wind the inductor yourself?

L = μ n² A / l, where μ is the permeability, n is the number of turns, A is the cross-sectional area, and l is the length. So if I want an inductor of 6 μH or so, 12 turns with a radius of 1 cm and a length of 1 cm and no core would be about right; or 39 turns with a radius of 1 cm and a length of 10 cm and no core. I feel like you could probably keep the parasitic capacitance and resistance down to trivial levels with that approach.

Topics