From: John Woodgate
Subject: Re: Output Impedence
Date: Sat, 26 Oct 2002 07:19:25 +0100
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Sat, 26 Oct 2002 10:49:36 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that James Meyer
wrote (in ) about 'Output Impedence', on Fri, 25 Oct 2002:
>On Fri, 25 Oct 2002 20:16:06 +0100, John Woodgate
>>I read in sci.electronics.design that Michael R. Kesti
>>wrote (in <3DB97CF0.1682A5C0@gv.net>) about 'Output Impedence', on Fri,
>>25 Oct 2002:
>>> Then load the output with
>>>about 1000 ohms (Rl) and again measure the output voltage (Vl). Maximum
>>>accuracy will occur when Vl is about half of Vo. If Vl and Vo are almost
>>>equal, then the value of Rl is too high and you should try again with a
>>>smaller value. Conversely, if Vl is small compared to Vo, try again
>>>using a larger value for Rl. Doubling or halving the value will quickly
>>>get you to an appropriate value.
>>>Another approach is to use a potentiometer for Rl and adjust it until
>>>Vl is exactly half of Vo. Rl and Ro are equal under this condition
>>>and May be determined by measuring the resistance across the potentiometer.
>>DON'T do this with the outputs for loudspeakers! You will zap the output
> The output devices will zap about 0.001% of the time.
> If you solder the connections to the output, DON'T pick up the soldering
>iron with the pointy end.
James, the output source impedance of an output for a loudspeaker is
typically less than one-tenth of the minimum rated load impedance. So
searching for the output voltage to drop to half involves applying loads
far less than that minimum. While many audio amplifier ICs are short-
circuit protected, they are not necessarily protected against loads of
intermediate impedance. This is particularly the case if you put a pot
across the output and wind the pot down.
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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