|Class D Switching Power Amplifiers and Power D/A conversion|
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|13th January 2017, 05:16 AM||#12|
Join Date: Jul 2007
this one is well cooked and proffesional
|29th December 2017, 04:22 AM||#13|
Here is a thread on special pricing for the TI TPA 3255 evaluation board.
TI Class D EVM Board 50% Promotion
"The geek shall inherit the earth"
2018 Burning Amp is Sept 30 in S.F. Thread here on diyAudio:
|1st February 2019, 08:44 AM||#14|
Join Date: Jan 2017
how to calculate a PSu for Class D
Here is a very good explanation about calculating a PSuU for Class d amps by Faux French
How to choose the power supply for your class D amplifier module.
Typically less experienced DIYs repeatedly ask about this issue. It is not particularly difficult but requires knowledge of Ohms Law, the functioning of a class D amplifier and a Buck (step-down) power converter and some general electronic overview.
Only one channel (mono) will be considered. If you need two amplifiers (stereo) to operate from the same power supply, you simply double the current capability of the power supply such that also the power rating is doubled.
For deciding the right power supply, the maximum (operational) power supply voltage and the maximum output current (where the over-current protection is activated) of the amplifier need to be known (see the specifications or the chip datasheet). You also need to know if the amplifier operates from a single supply voltage or two symmetrical (+ and -) supply voltages. Next, you have to design for optimum operation with 8 Ohm or 4 Ohm speakers, eventually 2 Ohm speakers for automotive use.
The effective operating voltage
The effective operating voltage (Veff) is the voltage the amplifier can use to generate a sine half-wave.
For class D amplifiers operating in BTL configuration from a single supply voltage it is the single rail-to-ground voltage.
For class D amplifiers operating in SE mode with one speaker terminal connected to ground and a symmetrical supply voltage it is the voltage from one rail to ground.
For class D amplifiers operating in BTL mode with a symmetrical supply voltage it is the voltage from one rail to the other rail, thus, double the voltage from one rail to ground.
This effective operating voltage indicates the maximum peak voltage the amplifier can generate for a sine half-wave. From a sine half-wave peak value to the rms value, the factor 1.41 applies. The rms value is found by dividing the peak value with 1.41.
Though the theoretical conversion factor is 1.41, it is suggested to use the factor of 1.5 when calculating the rms value of a sine half-wave that can be generated from an effective operating voltage (Veff). Then, some output switch drop is taken into account.
As an example, for a BTL-coupled amplifier that can handle a single supply voltage of 36V, the rms value of the sine half-wave that can be generated by the amplifier is 36Veff/1.5=24Vrms.
With the rms voltage of the sine wave signal, the power in a load can be calculated using the expression Vrms*Vrms/Rload. Using the values from the example above: 24Vrms x 24Vrms / 8 Ohm = 72W. Or, for a 4 Ohm load: 24Vrms x 24Vrms / 4 Ohm = 144W.
This way we can calculate how much AC signal power it is possible for the amplifier to generate in the load PROVIDED the amplifier can stand the resulting current.
The maximum current for an amplifier output
This value is more difficult to find in a chip datasheet. Sometimes it is specified as the (peak) current the amplifier can handle at an output, sometimes as the output current where the (over-) current limiter is invoked. Some datasheets are not clear about if the chip can stand the maximum (peak) current on more outputs at the same time.
When we have calculated the maximum output power the supply voltage will allow, we have to compare the resulting current against the maximum current for the amplifier chip. To calculate the maximum peak output current, we simply divide Veff with the load impedance. Veff / 8 ohm for 8 Ohm load and Veff / 4 Ohm for 4 Ohm load.
Using the values from the example above, the peak current value in 8 Ohm will be 36Veff / 8 Ohm = 4.5 Apeak. In 4 Ohm it will be 36Veff / 4 Ohm = 9 Apeak. Let’s assume the amplifier can handle 6 Apeak at the output(s), operation with 8 Ohm load is fine (max. 4.5Apeak) but operation with 4 Ohm load (max. 9Apeak) will result in over-current at a certain output level.
If we know we will use the amplifier from the example above with 4 Ohm speakers, we need to lower the supply voltage (Veff) such that the maximum current at an amplifier output is not exceeded. Reducing the supply voltage (Veff) to 24V means the peak current in 4 Ohm will be 6A and the amplifier maximum current is not exceeded.
Calculating the power supply capacity
When the maximum output power that is possible with a certain supply voltage has been calculated and the maximum current at the amplifier output has been successfully checked, we need to calculate the power capacity of the power supply.
Here, we use the knowledge that a class D amplifier has an efficiency of typically 90%. To be prudent we use 80%.
The classical way assuming test with a constant amplitude sine-wave: The maximum output power from the amplifier is divided by 0.8 (80%) to take amplifier losses into account. Using the values from the example above, 72W in 8 Ohm requires a power supply of 72W / 0.8 = 90W. In 4 Ohm, 144W / 0.8 = 180W.
The realistic way assuming use with music: The power supply capacity (Watt) calculated above is divided by two. Why? Because music has got a “crest-factor” meaning that no music is demanding full power for long. Actually, most music has long passages with rather quiet sound and short passages with powerful sound. Therefore, it is seen as unrealistic to design the power supply for maximum sound level all the time. Clever and competent people estimate the steady-state power supply need to be half of what is normally needed for the maximum power the amplifier can output. The power line decoupling capacitors are handling the need for short but powerful surge currents.
Using the example above with 72W in 8 Ohm will then require only a 45W power supply (half of 90W).
Can we swap 4 and 8 Ohm loads?
Yes, but with two different results. If the amplifier and power supply are optimized (designed) for a 4 Ohm load, you can always use an 8 Ohm load instead – but, you will only get half the output power. No overload. If the amplifier and power supply are optimized for an 8 Ohm load, you can use a 4 Ohm load instead at lower output levels. But at a certain output level, the higher current demanded by the 4 Ohm load will exceed the current capability of the amplifier and perhaps also power supply such that you have an over-current situation.
Therefore, 8 Ohm speakers are the most versatile if you do not need the extra power.
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