current draw of a chipamp

jarthel · 13th July 2007, 09:25 AM

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.

AndrewT · 13th July 2007, 10:14 AM

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.

jarthel · 13th July 2007, 11:45 AM

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

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.

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
calculating total current draw	woodturner-fran	Tubes / Valves	3	9th March 2008 11:11 AM
Current rating and tube current draw	cbutterworth	Tubes / Valves	1	13th October 2006 04:42 AM
Do CLC filters work well at low current draw?	leadbelly	Solid State	2	28th March 2004 12:51 PM
current draw on a gainclone?	cowanrg	Chip Amps	9	9th August 2003 06:58 AM
Opamp current draw	5th element	Solid State	2	21st July 2003 12:11 AM

New To Site?	Need Help?
Register to Participate Search Privacy Statement Contact Us	Frequently Asked Questions Did you forget your password? Mark Forums Read


	Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving

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.