Power Supply for the Audio Amplifier LM4880
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 30th May 2014, 04:42 PM #1 amendiola diyAudio Member   Join Date: Sep 2010 Power Supply for the Audio Amplifier LM4880 I need to calculate the current that consumes my LM4880 audio amplifier, the datasheet indicates that this chip can deliver 250mWrms by channel. The LM4880 will work with 5 VDC. I don't know how to calculate the DC current that the amplifier will use. Is this calculation correct: I = P/V = 0.250/5 = 0.05 = 50 mA (by channel)? The output power is the sinusoidal wave rms power. Datasheet: http://www.ti.com/lit/ds/symlink/lm4880.pdf I want to know the current for selecting the correct power supply. I'm using an 7805 power supply. Regards. Alfredo Mendiola Loyola Lima, Peru
 30th May 2014, 04:56 PM #2 Arty diyAudio Member   Join Date: Feb 2011 allmost. besides the output power, chips are not 100% efficient, and so consume a bit more than the output power. a rule of thumb , that is no way bad to follow, is to choose a powersupply that is 2 times bigger than the sum of the outputs. in Your case, it would mean 0.5 watt / channel. supposedly the rails will not collapse if You calculate it like this. You use a regulator, that is verry nice. The regulator needs at least 7 volt input, and to be safe i would suggest something around 8-9 volts. So, the powersupply would need to supply 8-9 volt DC, and least 100mA / channel. I would not think You would find any traffo that would be so small, so if i where you, i would go for a 9 volt wallwart type, rated at 9 Vdc, that is a common thing to come by, and will work nicely with the circuit. Given this, for 0.5 watt power @ 5 volt You would need 100mA / channel.
amendiola
diyAudio Member

Join Date: Sep 2010
Quote:
 Originally Posted by Arty allmost. besides the output power, chips are not 100% efficient, and so consume a bit more than the output power. a rule of thumb , that is no way bad to follow, is to choose a powersupply that is 2 times bigger than the sum of the outputs. in Your case, it would mean 0.5 watt / channel. supposedly the rails will not collapse if You calculate it like this. You use a regulator, that is verry nice. The regulator needs at least 7 volt input, and to be safe i would suggest something around 8-9 volts. So, the powersupply would need to supply 8-9 volt DC, and least 100mA / channel. I would not think You would find any traffo that would be so small, so if i where you, i would go for a 9 volt wallwart type, rated at 9 Vdc, that is a common thing to come by, and will work nicely with the circuit. Given this, for 0.5 watt power @ 5 volt You would need 100mA / channel.

 31st May 2014, 10:01 AM #4 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders the absolute max Voltage is 6Vdc The recommended maximum voltage is 5.5Vdc If you run the chip from a regulated 5Vdc, then I suggest you add a ~5.5V trip to protect the chip from overvoltage. The quiescent current is quoted as 6mA max. I don't know if this is per channel or per chip. Does the datasheet clarify? 250mW into 8r0 is equivalent to 2Vpk and 250mApk into 8r0. A 16ohms headphone or similar dynamic driver will draw peak currents approaching that same 250mApk per channel. That means the capacitance after the regulator should be sized to provide at least a 500mApk transient capability. A 470uF 16V electro does not take up much space. Or maybe a pair of 470uF? __________________ regards Andrew T.
amendiola
diyAudio Member

Join Date: Sep 2010
Quote:
 Originally Posted by AndrewT the absolute max Voltage is 6Vdc The recommended maximum voltage is 5.5Vdc If you run the chip from a regulated 5Vdc, then I suggest you add a ~5.5V trip to protect the chip from overvoltage. The quiescent current is quoted as 6mA max. I don't know if this is per channel or per chip. Does the datasheet clarify? 250mW into 8r0 is equivalent to 2Vpk and 250mApk into 8r0. A 16ohms headphone or similar dynamic driver will draw peak currents approaching that same 250mApk per channel. That means the capacitance after the regulator should be sized to provide at least a 500mApk transient capability. A 470uF 16V electro does not take up much space. Or maybe a pair of 470uF?
Dear Andrew:

What formula did you use to convert 250mW to 250mApk ? (with 8 ohm load.)

I'm using a 100uF/25v electrolytic at the LM7805 IC.

Regards
Alfredo Mendiola Loyola

 31st May 2014, 05:15 PM #6 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders P=VČ/R=IČR=IV __________________ regards Andrew T.
amendiola
diyAudio Member

Join Date: Sep 2010
Quote:
 Originally Posted by AndrewT the absolute max Voltage is 6Vdc The recommended maximum voltage is 5.5Vdc If you run the chip from a regulated 5Vdc, then I suggest you add a ~5.5V trip to protect the chip from overvoltage. The quiescent current is quoted as 6mA max. I don't know if this is per channel or per chip. Does the datasheet clarify? 250mW into 8r0 is equivalent to 2Vpk and 250mApk into 8r0. A 16ohms headphone or similar dynamic driver will draw peak currents approaching that same 250mApk per channel. That means the capacitance after the regulator should be sized to provide at least a 500mApk transient capability. A 470uF 16V electro does not take up much space. Or maybe a pair of 470uF?
Can a 100uF capacitor provide 500mApk transient capability?

Is there a fomula to calculate the transient capability?

I'm using a Lm7805 regulator.

Regards.
Alfredo Mendiola Loyola
Lima, Peru

 31st May 2014, 06:38 PM #8 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders Yes, but not for long. The transient will draw down the capacitor voltage at a rate determined by the current and the capacitance. 1 Farad will supply 1 Ampere and the voltage will drop by 1 volt every 1 second. The rate of volts drop depends entirely on capacitance and current. No other complications. __________________ regards Andrew T.

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