Gain in class D amplifier and output power

Thread Starter

leo0001

Joined Sep 30, 2018
79
Hi,

I have a relatively simple question : What is the gain of a class D amplifier ? The voltage output of an push pull ouput stage is probably link to Vdd and to the PWM duty cycle ?

The other question is how I can compute the output power ? The output power depends on Vdd ? If I have a load which is equal to 4 Ohm. The ouput power is Vdd*Vdd/Rload ? (and the dead time, the power is probably less than the previous assertion ?) Assuming I have Vdd equal to 4V, it means i could connect a speaker which has a power (or peak power) of 4W? So the ouput power is directly proportionnal to the voltage that I applied on Vdd and of course to Vin because Vin influence the duty cycle. Actually the instataneous ouput power is something like P = (Vdd*D)^2/Rload and the duty cycle is a function which depends on Vin ...

If you could tell me if it is correct. I m actually trying to design a class D amplifier and the datasheet of my component recommands me to use a VDD which is equal to +-35V which is a lot ... I have on my shelf a 12 V transformer... It will be better if i could use it without doing a boost converter that I don't know to do like this class D amplifier. It will be good if i could do someting simple for the moment x)

Have a nice day ! :D
 

kubeek

Joined Sep 20, 2005
5,733
Can you post the link to your datasheet?
Genereally the supply voltage is dictated by the output power required and teh impedance of the speakers. +/-35V into 8ohms should lead to about 75W continuous.
 

Thread Starter

leo0001

Joined Sep 30, 2018
79
Thanks for your reply,

Here is the datasheet : https://eu.mouser.com/datasheet/2/196/irs2092-1225204.pdf

I do not need all of this power. Actually, 75 W it seems to be enormous to me. But i don't know the relation between power and power sound (decibel). I know that each speaker have this own efficiency to convert electrical power into power sound. I don't know if i can talk about power sound. However i know that i have 20 W speaker with my computer and is large enough for me...

75 W continuous , i.e RMS ?

Do you know if I could go down the VDD voltage to approximately 10 V for having 20 W with a 4Ohm speaker impedance ? Otherwise I will have to do an SMPS...

An other question : What is the difference between 4Ohm 8 Ohm or i don't know may be 16 Ohm ?

If you have any suggestion to do to me, please tell me :D

Thanks a lot :D

Have a nice day !
 

crutschow

Joined Mar 14, 2008
25,660
The theoretical maximum output RMS power from a PWM signal is (Vpp*0.35)²/Rload).

Thus for a ±35V supply, the maximum power into 8Ω would be (70*0.35)²/8 = 75W as kubeek stated.

Fo 10V the power would be (10*0.35)²/4 = 3W with a 4Ω speaker.

The difference in speaker impedance is that the lower impedance will give a higher audio power out for a given output voltage (inversely proportional to impedance).
But you do need to make sure that the amp is rated for the selected speaker impedance.
 

Thread Starter

leo0001

Joined Sep 30, 2018
79
Just an other question if the FB is coming from PWM signal which is the case on the chip that i use (if i correctly understand the principle). Vdd and Vss have to be clearly stable in order to not introduce distortion in the signal ... ? I read that there was problem due to the inductive load which back some energy on the power supply rails during transition between the 2 mosfets. Do you have any idea to limit these problems ? A large capacitor is it the best solution ? A zener diode could clamp the voltage if it exceeds a certain value. It will limit the distorstion to a maximum... I do not understand why the feedback on this chip has been made on the PWM signal ... Actually I 'm not here to do a perfect class d audio amplifier. If i could just do something that works it will be nice !
 

Thread Starter

leo0001

Joined Sep 30, 2018
79
So Vdd to 13 V and Vss to 13 V could be a solution ? Actually I think that I will have to do an SMPS because my transformer gives a 12 V ouput and i have to rectify it so I will have probable something like 10V DC output which is very low... I never made an AC-DC SMPS ... It will nice to learn it :D
 

BobTPH

Joined Jun 5, 2013
2,587
A 12V supply when rectified and smoothed gives you about 15V.

You would need a DC to DC converter to provide a negative supply. They exist, and that could give you + and - 15V.

How much current can the 12V transformer provide? To get 20W it better be 2A or more. If, if you are thinking stereo, twice that.

Bob
 

Thread Starter

leo0001

Joined Sep 30, 2018
79
Thanks Bob for your reply,

How do you know that the output rectified will give a 15V DC ? I know that the output voltage of a transformater depends on the load which is connect to its output. If the load is important, the output voltage will be around 12V and when the load is low, the output will be higher than 12 V, may be 15 V. In my case, the transformer power that i have is 20 VA, so I will have some difficulties to have an output power of 20 W. Besides, the load is not purely resistive ... That seems really bad :D I need to find another one ...

An other question Bob, I don't know how you calculate the current ? According to me, I need Vdd/Rload, so 15/4 = 3A75 ?

My last question, if I have to do an SMPS, why I need a DC to DC converter whereas I have an AC input voltage... Why I can't do an AC-DC converter ? I was thinking about reverse polarity of the DC output to have a negative supply ...

Thanks a lot,

Have a nice day !

Leo
 

BobTPH

Joined Jun 5, 2013
2,587
The voltage of a filtered AC supply is sqrt(2) * Vac - 2 diode drops. 12 * 1.414 - 1.4 = 15.5V.

If a transformer is rated at 12V and 2A it can provide 24W. If you use the filtered DC of 15.5V it can deliver 24W / 15.5 = 1.5A

You can get + and - voltages from the 12V AC input by using what is called a voltage doubler. This uses two half wave rectifiers. Which means each side gets half the wattage max. Essentially, one of the AC wires is common, and the other feeds the positive rectifier on half the cycle and the negative rectifier on the other half.

Generally, if you want a + and - supply from a transformer, use use a center tapped transformer at twice the voltage.

Bob
 

crutschow

Joined Mar 14, 2008
25,660
If a transformer is rated at 12V and 2A it can provide 24W. If you use the filtered DC of 15.5V it can deliver 24W / 15.5 = 1.5A
Actually you need to reduce the maximum DC output by about 50% from the AC RMS output current rating due to the high peak RMS currents that a diode-capacitor supply draws from the transformer.
Thus you should draw no more than about 1Adc total from a 2Arms rated transformer.
 

BobTPH

Joined Jun 5, 2013
2,587
Peak RMS current sounds like an oxymoron to me (except in the case of DC.) Yes, there would be a larger peak current, but it would cause no more heating than drawing a constant current as long as the RMS power is the same.

Bob
 

Thread Starter

leo0001

Joined Sep 30, 2018
79
Thank you for your replies.

First question : 12 V on my transformer mean 12V RMS ? This is not the peak voltage ?

My transformer is rated to be 20 VA ; 1A67 and 12V. 20VA is equal to 20 W if the load is purely resistif. Power is equal to the power factor multiply by VA.

Thank you for the transformer center tapped ;) I forgot this transformer.
 

crutschow

Joined Mar 14, 2008
25,660
Peak RMS current sounds like an oxymoron to me
Yes, I rather mangled my terms.
I should have said peak AC current.
Yes, there would be a larger peak current, but it would cause no more heating than drawing a constant current as long as the RMS power is the same.
If you are talking about the DC average transformer current versus the transformer AC RMS current, no.
Heating is proportional to the square of the current.
Show the math. Depends on the meaning of "a lot".
"a lot" is about twice for a typical diode-capacitor DC supply.

Suppose you have a 1Adc average current, and that is generated by 20% duty-cycle peak current pulses from the transformer into the filter capacitor through the diode(s) .
To simply the calculations, assume the current pulses are square.
This means the peak current will be 5A to get a 1A average output.
That gives an RMS current of √(25A*20%)= 2.24Arms.

Make sense?
 
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BobTPH

Joined Jun 5, 2013
2,587
@crutschow: Okay, I get it now. The average current might be the same, but it is at a higher voltage than the AC RMS voltage, hence more power drawn. I never realized you had to de-rate the current by that much.

Bob
 

crutschow

Joined Mar 14, 2008
25,660
The average current might be the same, but it is at a higher voltage than the AC RMS voltage, hence more power drawn.
It is not directly related to the voltage.
It's the RMS current rating of the transformer (as determined by its VA spec) that's of interest, which is determined by the I²R heating due to the primary and secondary winding resistances.
With the high peak current pulses from a diode-capacitance load, the I²R (RMS) heating current is much higher than calculated using the average current.
 
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