masly Posted September 2, 2011 Report Share Posted September 2, 2011 Hi there, I am, new to the forum so I apologise if this question is out there before. I am looking to build a circuit that can power two cabinets that are 60W 8Ohm (2 speakers in each at 30W). In the project section there is a 20W amplifier using a LM1875 IC, is their any way of modifiying this circuit to make it 12v and using a different IC to make it more powerful.Thanks for the help. Quote Link to comment Share on other sites More sharing options...

audioguru Posted September 2, 2011 Report Share Posted September 2, 2011 I am looking to build a circuit that can power two cabinets that are 60W 8Ohm (2 speakers in each at 30W). In the project section there is a 20W amplifier using a LM1875 IC, is their any way of modifiying this circuit to make it 12v and using a different IC to make it more powerful.Two 8 ohm speakers in parallel have a total of 4 ohms so the current is doubled and the power is also doubled. But a 12V supply results in an output swing of only about 8V peak-to-peak which produces a power of only 2W into 4 ohms.For many years, car radios use two amplifiers in a bridge for each channel so that the output voltage swing is almost doubled. Then the current is also almost doubled and they produce 3.5 times the power which is 7W into 4 ohms with a 12V supply. When the supply is 14.4V then the output is about 15W. When the volume control is turned up too high producing a lot of clipping distortion then the output is 20W.A couple of years ago there were more than 100 car radio bridged amplifier ICs. One had a built-in supply voltage doubler. Today most are not made anymore. Quote Link to comment Share on other sites More sharing options...

Hero999 Posted September 3, 2011 Report Share Posted September 3, 2011 Yes, you need a peak voltage of 21.9V to deliver 60W into 4 Ohms.It can be calculated using Ohm's law:P = VII = V/RSo:P = V*V/R = V^{2}/RNote, the above formulae relate to DC. The RMS power is half the peak power:P_{RMS} = Vp^{2}/(2R)Rearrange for VpVp = sqrt(P_{RMS}2R) Vp = sqrt(60*2*4) = sqrt(60*8) = sqrt(480) = 21.9VYou need a full bridge with a voltage doubler to stand any chance of getting 60W with a 12V supply. Quote Link to comment Share on other sites More sharing options...

audioguru Posted September 3, 2011 Report Share Posted September 3, 2011 But many car radio amplifiers produce 60 Whats when the supply is only 12V.60 Whats is the true power of 1W multiplied by the age of the school kid designer's grandmother. Just a bunch of lies.Many car radios have four channels. Then they produce the true power of 60 Watts (15W per channel into 4 ohms). Quote Link to comment Share on other sites More sharing options...

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