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TPA2001D1 total current consumed by IC + speaker

Aditya Pandey
Prodigy 95 points
Other Parts Discussed in Thread: TPA2001D1, TAS5630B



Target is to find the total supply current consumed by TPA2001D1 & Speakers

Calculation 1:

Current consumed by speaker calculation :

cf(db) = 10log ( Pout peak / Pout (Av) )

Crest Factor for sine wave : 3dB

Pout (Av) = 1W

Therefore,

3dB=10log(Pout Peak / 1W)

Pout Peak comes out to be 1.995W consider it to be 2W

Now to find current consumed by speaker we apply

Pout Peak = (Vpeak * vpeak) / Rload

Where Rload is speaker's 8 Ohm

2W = (Vpeak * vpeak) / 8 Ohm

Vpeak comes out to be 4V, this Vpeak is the maximum voltage that can occur across speakers

Now to find current across speakers

Ipeak=Vpeak / Rload

I peak comes out to be 4V/8Ohm = 0.5A

Therefore I total would be current consumed by speakers + current consumed by Amplifier IC (Typical 4mA, Max 6mA . But with filter 7.5mA)

I Total = 500mA + 7.5mA = 507.5mA


Calculation 2 :

Consider this link : http://e2e.ti.com/support/amplifiers/audio_amplifiers/f/6/t/380832.aspx

In above link also they are talking about total current consumed by Class D amplifier.

Please read this part of conversation :

"

Hi, Weiren,

The TAS5630B is a class-D amplifier. This makes your question a little bit complicated.

Unlike a class-AB amplifier where the supply current = the load current, for a class-D amplifier, the supply POWER = the load POWER.

So, you need to determine the maximum power that could be delivered to your load, divide that by the efficiency of the amp, and then convert that power back into a current from your 12V power supply.

-d2

-----

Don Dapkus

Audio Applications Engineering Manager

Dallas, TX USA "

If I apply to my problem :

Efficiency  = load power /Supply power
 
load power max peak possible for worst case = 2W
then Supply power would be = 2W / .78 = 2.564W

Supply voltage = 5v

Supply current = 2.564W / 5V = 512mA

Queries :

1. In calculation 1 Current consumed by IC+Speaker coming out to be 507.5mA
2. In Calculation 2, it comes out to be 512mA
3. Which method is correct?
4. If neither is correct then please guide me & tell me the total current consumed with an example.

  • 0
    TI__Mastermind 37045 points

    Is this issue still open?

    reg,

    Paul.

  • 0 in reply to user5276421
    Prodigy 95 points

    Yes, this issue is still open. I am still waiting for your reply sir.

  • 0 in reply to Aditya Pandey
    TI__Genius 9175 points

    Hello,

    Both methods you discussed are correct and give similar results assuming the peak voltage delivered to the speaker is the same as the voltage rail of the power supply; both methods should be used together for power supply design.

    (Generally the rail voltage should be a bit greater than the peak speaker voltage to account for H-bridge voltage loss across FET Rds-on)

    In Case 1: you found the current necessary to drive 2W into an 8ohm load which is the peak power. Thus you have a peak current demand of 0.5A with 4V across the speaker.

    In Case 2: you ignored the speaker voltage and found the necessary power supply power from a 5V supply to drive the necessary current into the load.

    The difference: Case 1 ignores power supply voltage. Case 2 ignores speaker voltage.

    To adequately design the power supply you must know what the peak voltage is across the speaker so you have enough headroom on the voltage rails of the supply. Case 1 and 2 are interdependent and must be considered together.

    For example: For case 2 I could choose a supply voltage of 2.5V.  To get the desired 2.564W form the supply, it would have to be designed for a current of 2.564W/2.5V = 1.0256A. However this isn’t the full picture. With an 8ohm load the maximum best case peak power across the speaker would be 2.5V(^2) / 8 = 0.78W  (assuming no voltage loss across Rds-on of the h-bridge output FETs). Although the supply is designed for 2.564 W, an 8ohm load cannot draw this much power from a 2.5V supply.

    (note the peak speaker voltage of 4V and a supply voltage of 5V is what gives you your slightly different current requirements between case 1 and case 2)

    If we designed the power supply above without considering the voltage across the 8ohm speaker, we could not actually deliver 2W peak power to the speaker even though the power supply was designed for 2.564W.

     

    A 5V supply at 520mA should be sufficent for your requrements giving 1V of extra headroom.

    Regards,

    Matt