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Old 30th May 2014, 04:42 PM   #1
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Default 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
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Old 30th May 2014, 04:56 PM   #2
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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.
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Old 30th May 2014, 05:12 PM   #3
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Join Date: Sep 2010
Quote:
Originally Posted by Arty View Post
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.
Thank you for your help.
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Old 31st May 2014, 10:01 AM   #4
AndrewT is offline AndrewT  Scotland
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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?
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Old 31st May 2014, 03:46 PM   #5
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Join Date: Sep 2010
Quote:
Originally Posted by AndrewT View Post
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
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Old 31st May 2014, 05:15 PM   #6
AndrewT is offline AndrewT  Scotland
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Join Date: Jul 2004
Location: Scottish Borders
P=VČ/R=IČR=IV
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regards Andrew T.
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Old 31st May 2014, 06:01 PM   #7
diyAudio Member
 
Join Date: Sep 2010
Quote:
Originally Posted by AndrewT View Post
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
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Old 31st May 2014, 06:38 PM   #8
AndrewT is offline AndrewT  Scotland
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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.
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