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Old 31st October 2005, 07:39 AM   (permalink)
Default impedance matching question

hi there, please help me to answer my question..

what is the priority in the impedance matching in the audio system?
is it we want the maximum power dissipate to the load? or the maximum voltage in the load? when we do impedance matching, i read a book and it said that the input impedance is needed as high as possible, and the output impedance is as lower as possible, when the input impedance is high, the voltage can be absorbed by the input impedance (next stage), but how about the current, the current can't flow very well to the input impedance (next stage), if we need the maximum power, we need the output impedance (previous stage) = input impedance (next stage), but which one is the priority? is it we want the maximum power? or maximum voltage?
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Old 31st October 2005, 08:38 AM   (permalink)
Default Re: impedance matching question

Quote:
Originally Posted by zhi_yi
hi there, please help me to answer my question..

what is the priority in the impedance matching in the audio system?
is it we want the maximum power dissipate to the load? or the maximum voltage in the load? when we do impedance matching, i read a book and it said that the input impedance is needed as high as possible, and the output impedance is as lower as possible, when the input impedance is high, the voltage can be absorbed by the input impedance (next stage), but how about the current, the current can't flow very well to the input impedance (next stage), if we need the maximum power, we need the output impedance (previous stage) = input impedance (next stage), but which one is the priority? is it we want the maximum power? or maximum voltage?
In audio 'matching' you generally want voltage transfer, and NOT power transfer - doing maximum power transfer (impedances equal) reduces your voltage signal by 50%. You don't need the highest possible ratio, but commonly something like five times is enough - so a 600 ohm source would feed into a minimum 3000 ohm load.

It's quite simple to understand if you consider the source impedance as a series resistor, and the load as a resistor from the input to ground. These two 'resistors' form a potential divider, which reduces the signal voltage.
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Nigel Goodwin is offline   Reply With Quote
Old 31st October 2005, 12:45 PM   (permalink)
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okay.. i can understand it now.. thank you very much..
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