So when an H bridge motor controller is in freewheel, is that all transistors open, or both ends grounded? What is brake mode? It appears they call freewheel opening all transistors, but the current has to go somewhere, so through the body diodes (or protection diodes) with voltage drop and dissipation. Connecting both ends of the load to ground (or both to power) routes the current through a low resistance path dissipating less power, so why is this called "brake" mode?
So a speaker has much less of the "mechanical capacitor" effect and is dominated by the electrical inductance and should be shorted during the other half of the cycle when not driven? Sounds like using a motor controller for a class D amp is doomed unless this can be controlled directly as many chips provide... just not this one.
Diodes Inc, TI, ST have a whole bunch of offerings. Seems to me it's likely that (at most) a bit of stuff around one of those would be a better solution than rolling your own.
If the solution is 20 cent in quantity it might be worth it, unfortunately I experimented with driving class D audio frequency with that device in simulation the other day, and unfortunately even though you can clock PWM very fast It seems the "integrated integrator" gain-bandwidth is just too low for that application, likely to conserve quiescent power, couldn't get more than about 3 kHz baseband frequency out of it in simulation before it couldn't keep up.
The full-power bandwidth isn't dominated by the lone series cap with a symmetrical supply (calculate the Thevenin impedance looking into a capacitive voltage divider), and it avoids asymmetrical clipping if the single supply is unregulated/poorly regulated and droops under load.
I don't think the Thevenin equivalent is appropriate. What I see is the rising voltage on the ground reference cap will raise the reference voltage on the Vin connected cap. This greatly reduces the impact on the power rail drawing all it's current from one cap to the other. Regardless, the Thevenin equivalent is just a 20 uF cap to some voltage. I don't see where that is so useful.
The real problem with this chip is the cost. $6-$7 each in quantity. Not so good, around 5x other solutions.
Not worried about the source supply. Just the total power wasted.
I'm talking to a factory guy from MPS who seems to be saying their MP6519 works like a class D amp while it seems to me to work like a pair of LTC3623 driving only one end or the other at a time eliminating the need for the large caps. I don't know why that schematic in the LT data sheet shows 10 uF caps. They won't do diddly at the low end for any speaker I can find. I only need 250 Hz minimum and that still requires 100's of uF for even a 50 ohm speaker.
It's a virtual ground, once it's in steady-state the small-current-signal impedance looking into that capacitive voltage divider is zero, dude! And it would be if the caps were 1 pf. They're larger than that in practice to keep the virtual ground stiff for large currents. But so long as it stays stiff for larger currents those caps don't create a phase shift.
The single-ended DC-blocking cap adds a pole even for small signals so it has to be large to have negligible impedance at 20 Hz. The voltage-divider caps don't have to be nearly as large as they just have to supply time-averaged current
That the 4.7 u cap for the LC filter is connected to the divider and not the most negative supply should tell you something. It can't connect there for a single-ended circuit! That's not "ground" in that case!
That is to say you size those caps to provide a steady Vcc/2 at the virtual ground node with respect to the maximum output current of the baseband signal at the lowest frequency of interest at the _switching_ frequency, filtering to the DC component just like you were designing a regular buck converter, not like you're calculating the cutoff frequency of a single-ended RC audio filter goddamn it.
If that DC voltage is held approximately constant by the buck converter feedback then those caps don't exist as far as the baseband signal is concerned.
Otherwise you still have to use 1000u blocking caps for even 2 watts into 8 ohms at 20 Hz with a 12 volt supply and your design is "wicked retahded":
Sorry, Thevenin says you are wrong. The equivalent of a voltage divider is the parallel combination of impedances connected to a voltage source defined by the divider of the two impedances. So 20 uF at half the power rail.
It is interesting that in essence the charge on one caps simply shifts to the other cap through the on FET. Very little current would flow from the power supply once the caps are charged. I expect that is the reason two caps are used.
Yes, thinking the two caps are any sort of magical zero ohm impedance ground is retahded. The caps have to be sized for the signal frequency, not the switching frequency. Imagine this, the power converter outputs 1V for a long enough time the cap voltage is 1V and the current through the speaker will virtually stop. Change the output voltage to 2V and the same thing happens. The time for this current to reduce flow is exactly the usual RC time constant with C being the the parallel of the two caps.
As someone said, this approach helps to eliminate the power on pop using a half bridge.
I was completely wrong about the current flow in the two caps. To change the voltage at the junction twice the amount of current flowing into either cap has to flow through the speaker which exactly matches the idea of the Thevenin equivalent being the parallel combination of the two caps. The advantage is the circuit powers up with equal charge on both caps with no current flowing through the speaker.
The DC bias at the mid-point of the capacitive divider is fixed, it's not going anywhere over a long period of time or otherwise.
You'll have to explain then why they connect the LC filter cap to the capacitive divider midpoint point because it makes zero sense to connect it there if that's not a virtual ground.
They chose the input resistor for the audio signal high-pass to be a very specific value, 121k, for a reason, 50uA of DC current flows thru it and the feedback network to the error amp is DC coupled all the way from the capacitive divider mid point.
And the input high-pass network has a low-frequency corner of about 100 Hz but two 10 uF in parallel has an impedance of about 75 ohms, why would they choose an input low-frequency corner of 100 Hz but make the impedance the output sees at that frequency ten times that of the 8 ohm speaker by itself AAHHHHHHH! No values in this circuit are picked arbitrarily. If the input low frequency corner is 100 Hz then its output bandwidth extends at least that far.
Of course it is not fixed. It is set by the current flowing into the capacitors which change voltage depending on that current. The only way to have zero net current when the cap voltage is half the power supply is to wiggle the SW pin at a 50/50 duty. That's a pretty boring circuit. Change that ratio to get current through the speaker and the caps will charge up to a new voltage defined by the duty cycle, D * Vcc. This is not rocket science. The RC is the speaker impedance and the parallel capacitance. To provide a stable reference for the waveform being produced the RC value needs define a corner frequency well below the frequency of the signal being reproduced.
It's not trying to filter wrt ground. It is filtering the signal on the speaker. So across the speaker is the perfect place for it.
That is an extremely weak technical argument, relying on the value of a 10 nF AC pass cap. So you think that cap should be 1.322 nF? Any other value is bogus?
Try simulating the caps, the speaker, the inductor and the switches. Change the duty cycle from 50/50 to 10/90 and measure the response time. I can assure you it will reflect a 20 uF capacitance blocking the DC in that path.
Why can't you actually analyze the circuit instead of waving your hands all around? Is there any math to show a capacitor divider does what you say it does? If one end of the caps were driven in some manner, then I would be inclined to have some confidence you were talking about something real. But two capacitors connected in this way are just in parallel given stability in the Vcc and ground rails.
What they might do, is to provide some benefit as decoupling the Vcc rail to ground. When the SW drives the speaker at a higher level the increase in speaker current will flow into the midpoint junction of the caps raising that voltage of which half will come from the PSU and half will come *from* the top cap. So it gives the DC coupling benefit of a 20 uF cap and at the same time offers capacitance to stabilize the Vcc rail as well as reducing the draw from the PSU.
The DC voltage on one terminal of the error amp is 6 volts due to the
50uA current thru the 121k resistor. Therefore the DC voltage on the other terminal of the error amp must be 6 volts. The other terminal is DC coupled to the mid-point of the divider. So what else could it be.
Caps are there for the same reason they're there in a half-bridge converter, to create a virtual ground:
The two points are only DC coupled in the sense that the capacitors will follow. It's not like an op amp where the impact is nearly instantaneous. Once you start running the voltage on the 121k resistor up and down the error amp will force the output of the regulator up and down and the caps will follow that too, but only as much as results from the current through the speaker, et. al.
What the heck are you arguing about this for. You keep talking in absurd terms. I take it you don't do this for a living?
Yes, a *virtual* ground. It works well as long as you understand the principle of RC filtering.
Is this done yet? Did you do your simulation to show the RC effect? If you do, it will be very educational.
One cap will charge up on one half-cycle and the other will discharge, and vice versa. Energy goes in, energy comes back. Energy goes in, energy comes back. It's very efficient
If the DC midpoint voltage is held approximately constant they don't contribute to the small-signal frequency response at all so they're sized like any other supply rail bypass capacitance, appropriate to the maximum output power and efficiency of the amp. Why do you need 1000uF of supply bypass capacitance for a puny 2 watts of output power at 100 Hz? Even a class AB amp doesn't need that! This f**ka is supposed to be high-efficiency, remember?
Just have to be large enough to keep all the DC potentials stiff under load and DC feedback is helping, sure don't need 1000u of capacitance to do it.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.