# LM4871 for audio amplication

Discussion in 'The Projects Forum' started by nailz7, Jan 14, 2009.

1. ### nailz7 Thread Starter New Member

Jan 5, 2009
6
0
I m using LM4871 to power up a 8 ohm, 1W speaker. LM4871 will have a 5V Vccc to power it. According to the data sheet, the diagram should be correct, but how do i check if the current provided is sufficiant to allow the speaker to be functionable?

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2. ### mik3 Senior Member

Feb 4, 2008
4,846
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The power delivered to the speaker is:

P=(Vout RMS)^2/R

P=power on the speaker
R=impedance of the speaker

If you assume the peak output (Vo) voltage is 5V (equal to the power supply voltage which is not in practise but its close) then the RMS value is:

Vrms=Vo*0.707=5*0.707=3.54Vrms

Then the power is:

P=Vo^2/R=3.54^2/8=1.5W

If the speaker can handle it will be ok otherwise reduce the gain accordingly as the peak output voltage to be 2.83V to reduce the output power to 1W.

3. ### nailz7 Thread Starter New Member

Jan 5, 2009
6
0
ok thanx, but what about the current i should provide to the op amp? cause i am unsure about the current output of the op amp to the speaker? Also i will actually be stepping down the op amp's VCC of 5V from a 9V source, which is being used to power up other device, but the only current info i can get from the data sheet is Quiescent Power Supply Current IDD. Is this the current i need to supply to the op amp for it to function?

4. ### jj_alukkas Distinguished Member

Jan 8, 2009
753
8
That chip is not an opamp, its a 3W power amplifier. There is an easy way but not a safe idea during the testing process.To easily limit the power, connect a 47k pot to the input of the amp in series set to minimum position. Increase the volume till you feel the sound is clear and loud and not distorted. Find the max position till which its clear. It will be near the mid point. Shutdown the amp, disconnect the pot. Find the resistance till the current position from the side you connected the input. Substract (47k - that value ) and replace a fixed resistor for that value. This will limit the maximum output however input you feed in.

5. ### nailz7 Thread Starter New Member

Jan 5, 2009
6
0
By input of the amp, do u mean pin 4? cause i need to know the current to pin 6 Vdd.

6. ### SgtWookie Expert

Jul 17, 2007
22,202
1,793
Current = Power(Watts) / Voltage
If your output is 1W, then you'll need 200mA current, plus the 10mA quiescent current.

Keep in mind that the speaker is rated for 8 Ohms at 1kHz. At higher frequencies, it will have a higher impedance; at lower frequencies, a lower impedance.

If you're thinking of powering it from a 9v "transistor" battery, that won't work very well. Use three "C" or "D" alkaline cells in series instead. If you are using rechargeable batteries, you can use four in series.

7. ### nailz7 Thread Starter New Member

Jan 5, 2009
6
0
sorry for the trouble but as it isn't specified in the data sheet, does that mean that it will be ok to just pump 200mA current and the amplifier will still function?

8. ### jj_alukkas Distinguished Member

Jan 8, 2009
753
8
There are 2 ways to Limit an Amplifier Output
1.By Limiting Power Supply Current
2.By Limiting Input Signal

An Amplifier can work in low currents as well as higher than it is required to.
But in low current when the sound output reaches the peak,it will distort.But In your Situation the amp supplies more output than necessary. So it is ok to try limiting supply current first. It must work unless you feel the output is abnormal. Don't worry abt the chip, Nothing will happen unless you don't blow it up with out-of-range voltages and speaker impedance.

This chip is a mono Amp. Max Input Voltage is 6 Volts Only. Don't supply more than that. Chip will blow out.Your audio input is given at pin 4.Thats the pin I was talking abt before.

9. ### nailz7 Thread Starter New Member

Jan 5, 2009
6
0
ok, will give it a try, thanx guys