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 Chip Amps Amplifiers based on integrated circuits

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 13th July 2007, 09:25 AM #1 jarthel   diyAudio Member   Join Date: Jan 2002 Location: somewhere in Australia current draw of a chipamp I need current draw of an lm3785, lm3886 and lm4780 (as used in PCBs sold by Peter and Brian). I want to use a CLC filter in the supply but I do not know how much current is required by the chip. Thank you ps. I looked at the datasheet and I cannot find it.
 13th July 2007, 10:14 AM #2 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders Hi, the input current is roughly equal to the output current. If your amp is capable of 50W into 8r0 then peak voltage is 28.3Vpk. The peak output current is based on the effective impedance of the load and the shape of the drive signal. Continuous sinewave signals will send Ipk=Vpk/Z for a normal speaker then Z can vary from about 60% of nominal impedance to about 10times nominal impedance. For discontinuous signals with fast stopping or fast starting the effective impedance can drop as low as 35% of nominal impedance. The worst case current is Vpk/nom imp/0.35=28.3/8/0.35=10Apk The worst case input into that 50W amp is about 10Apk from either supply rail. Most of this peak will be met by discharging the main smoothing caps fitted to the supply rails. Any added resistance between the amp and the rectifier will reduce the peak current ability. Your CRC will rely on the last C to supply the peak demand with a little help from any on board decoupling. __________________ regards Andrew T. Sent from my desktop computer using a keyboard
jarthel
diyAudio Member

Join Date: Jan 2002
Location: somewhere in Australia
Quote:
 Originally posted by AndrewT Hi, the input current is roughly equal to the output current. If your amp is capable of 50W into 8r0 then peak voltage is 28.3Vpk. The peak output current is based on the effective impedance of the load and the shape of the drive signal. Continuous sinewave signals will send Ipk=Vpk/Z for a normal speaker then Z can vary from about 60% of nominal impedance to about 10times nominal impedance. For discontinuous signals with fast stopping or fast starting the effective impedance can drop as low as 35% of nominal impedance. The worst case current is Vpk/nom imp/0.35=28.3/8/0.35=10Apk The worst case input into that 50W amp is about 10Apk from either supply rail. Most of this peak will be met by discharging the main smoothing caps fitted to the supply rails. Any added resistance between the amp and the rectifier will reduce the peak current ability. Your CRC will rely on the last C to supply the peak demand with a little help from any on board decoupling.

thank you very much for the extensive reply Andrew. helpful as always

ps. I wouldn't have figured that out myself

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