Drive speakers with motor bridge

Rick C · 2021-02-22T13:25:00+00:00

Trying to drive a speaker at some power level without burning a bunch of power in an A or AB amp is hard to do with discretes. I guess I shouldn't say "hard", but messy. There are conditions to be dealt with and the number of components starts to add up. Class D audio drivers work much more efficiently, but they don't provide some features I require unless you go with the big package, high power, more expensive drivers. I found a motor controller that should work, even if it is not fully specified for switching speed for example. LV8548MC is low cost, readily available and pretty basic, providing drive for two units. The feature that is hard to find in other units is measuring the current. I need to detect shorts and opens. This device has nothing built in, but the switches are high enough resistance that a high enough voltage will be dropped on the low side that it can be directly compared to a reference using LVDS inputs on the FPGA. It takes a few pins, but that's one thing FPGAs are good at, providing I/O pins. With 200 mA of speaker current the voltage drop should be between 70 mV and 125 mV (they actually spec both min and max FET resistance even if it is for the combination of high and low side). The peak current may be more than that even. So thresholds at say 55 mV and 200 mV should distinguish a short and an open condition. I think it was a switching voltage regulator I saw they internally measured the voltage on the FET to determine when the current was reaching a level of interest. I thought it might work here and it seems like it can, mostly because the thresholds can be very non critical. Any comments?

Feb 22, 2021 Last reply: 5 years ago 45 Replies

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Rick C 5 years ago

Yes, capacitors are efficient. I don't get your point. A single capacitor in series with the speaker is just as efficient. Capacitors are not dissipative other than parasitic effects.

I'm getting tired if you spouting claims that you never support. The capacitance requirement is not changed when two capacitors are used across the power rails. Nothing you have shown refutes that statement. The two 10 caps uF in the AC sense behave exactly as if they were in parallel with one another and both connected to ground.

Yes, and keeping the DC potential "stiff" under load is defined by f = 1/ 2pi R C

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bitrex 5 years ago

Look at me, I can adjust the output voltages of this "DC" half-bridge converter up and down by 5 volts at 0.2 Hz with a screwdriver on a pot in the feedback network, therefore C5/C6 have to be like, 20,000uF to "pass" such a low frequency:

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bitrex 5 years ago

No idea who drew the arrows btw, wasn't me

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bitrex 5 years ago

No the supply bypass capacitance for a power amp isn't determined by the lowest frequency you have to amplify, that's nuts.

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Rick C 5 years ago

The two series capacitors connected to the speaker are not supply bypass.

It is abundantly clear that you have no way to support your suppositions and it is equally clear you will continue beating around the bush with this issue.

Continue without me. :)

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bitrex 5 years ago

On the topic of half-bridge PWM sine inverters, which have to output 60 Hz, obviously:

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"If no power loss occurs in the inverter, the power supplied from the DC link to the inverter is similar to the effective power of the inverter"

averaged over a full cycle of the baseband waveform. That is to say assuming the inverter is very efficient the link capacitance is sized proportional to the inverter max power output.

Note the equations for the link capacitor's delta V over a full cycle of the baseband waveform at the bottom. If the DC power delivered to the load integrated over the first half cycle of the baseband waveform, Dc1, is the same as the DC power delivered over the second half cycle, then Dc1 - Dc2 = 0 and you'll see those equations then have no time dependence and therefore no frequency dependence, and if the supply rails are split exactly evenly the AC ripple at the midpoint of the DC link divider is in fact zero.

A class D audio amp connected this way is functioning like a half-bridge converter so why it would behave differently I dunno.

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