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TPA6120A2 : VCC needed to achieve 150mW channel

Mathieu Lesperance
Prodigy 50 points
Other Parts Discussed in Thread: TPA6120A2, TPA6020A2

Hi everyone,

I am trying to do a DAC device powered by the USB bus only. I am worried about the power consumption of this IC. I know this chip can draw up to 700mA to reach it's maximum power, but it seem also quite reasonable when it's powered with a lower VCC like the Hi-Fi Smartphone application (±5V). I need a 150mW/channel for a load between 16 to 64 Ohms. I need to know the value of VCC needed and the current drawn from the power supply. I have done some math and i have found:

VCC = ±7V (with a Vp=4.5V, I suppose that a dropout of 2.5V from power supply to output voltage occur.)

and

Currentmax= 115mA (Rload=16Ohms)

If someone could look at this, i would be very grateful.

Thanks, Mathieu.

  • 0
    Prodigy 50 points
    I've redone my math because i have forgotten the output resistance of 10Ohm, so :

    VCC = ±7.5V

    and

    MaxCurrent = 100mA (Rload=16Ohms + Ro=10Ohm)
  • 0 in reply to Mathieu Lesperance
    TI__Guru** 103055 points
    Hi Mathieu,

    How are you approximating these current levels?
    I've tested TPA6120A2 EVM at Vcc = ±6V, Rload = 16Ohm. At ~150mW power supply was at ~60mA.
    If you need I could try to test it at Vcc = ±7.5V

    Best regards,
    Ivan Salazar
    Texas Instruments
  • 0 in reply to Ivan Salazar
    Prodigy 50 points

    I've used the following formulas from the datasheet:

    where:

    Pl=0.150W

    Rl=[16Ohms(headphone)+10Ohms(output resistor)] or [64Ohms(headphone)+10Ohms(output resistor)]

     then:

    Vp=2.8V (Rl=26Ohms)

    Vp=4.7V (Rl=74Ohms)

    Since the amplifiers in the TPA6020A2 aren't rail-to-rail i've guessed a ±2.5V difference between the Vcc and the Maximum output voltage. From there i've added my highest value of Vp + 2.5V + some headroom, which give me a Vcc of ±7.5V.

    Then i calculated the power needed for a 16Ohms load, since it's my biggest load in term of power dissipation:

    Psup= [(15V*2.8V)/(3.1416*26)]+30*0.015

    Psup= 0.74W

    Since I used the chip in stereo i double this value so the power drawn from my power supply is 1.5W approximatively and from the ohm law i found the current which is 100mA. I know this value of current is average and spike of current will occur, but i will insert suffisent decoupling capacitor to minise this effect.

    I would be very pleased if you could try this situation with the EVM. 

    Thanks for your quick answer, Mathieu.

  • 0 in reply to Mathieu Lesperance
    TI__Guru** 103055 points
    Hi Mathieu,

    Sorry for the late response. I've tested again the EVM at Vcc = +-7.5V, RL = 16 Ohm. With Vi = 1Vrms I got Po ~ 155 mW with the power supply at
    Icc ~ 43mA.

    Best regards,
    Ivan Salazar
    Texas Instruments