JordanRLS Posted February 16, 2010 Share Posted February 16, 2010 well, i am aware of the fact that there is a difference between 2, 4 and 8 ohm amps (i believe the lower it is, the higher wattage is passed?) I am just failing to understand why this difference occurs? confused! Quote Link to comment Share on other sites More sharing options...
obbm Posted February 16, 2010 Share Posted February 16, 2010 It's all down to Ohm's Law. Quote Link to comment Share on other sites More sharing options...
AlanP2008 Posted February 16, 2010 Share Posted February 16, 2010 [quote name='JordanRLS' post='747343' date='Feb 16 2010, 03:33 PM']well, i am aware of the fact that there is a difference between 2, 4 and 8 ohm amps (i believe the lower it is, the higher wattage is passed?) I am just failing to understand why this difference occurs? confused! [/quote] <more value than you asked for> Linear solid state amplifiers typically have (inside of them) unregulated +ve and -ve power supplies. The +ve and -ve output devices (bipolar or FET transistors) alternately progressively conduct to supply current to the load (the speakers) for each half of the signal waveform. If a given amplifier produces, say 100W into a 16ohm load, then *if* the power supplies maintained their voltages, it would produce 200W into an 8 ohm load. However, delivering that power actually causes most typical power supplies to "sag" their voltages, so as you halve the load, you actually get less than a doubling in power (ok, there are some other losses involved too - but sag is the main one) - so you might get 16ohm->100W, 8ohm->180W, 4ohm->320W, 2ohm->500W... A better (stiffer) power supply produces less sag, but costs more... But halving the load, in addition to (almost) doubling the power, also doubles the output current into the load. Most amplifiers are designed to supply sufficient current to drive down to a 4 ohm load, but some amplifiers are designed to deliver sufficient current to drive down to 2 ohms - this takes typically twice as many output devices as driving a 4 ohm load, so is more costly to manufacture. In addition, the power supply itself has to be able to deliver that additional current, and there will be more heat to get rid of, which again increases cost... The basic thing is W= V squared / R Watts = volts squared / resistance (in ohms) So as the load resistance is halved, the power output is doubled (provided the maximum output voltage remains the same - which it doesn't quite...) </more value than you asked for> I hope this helps Alan Quote Link to comment Share on other sites More sharing options...
JordanRLS Posted February 16, 2010 Author Share Posted February 16, 2010 [quote name='AlanP2008' post='747381' date='Feb 16 2010, 04:02 PM']<more value than you asked for> Linear solid state amplifiers typically have (inside of them) unregulated +ve and -ve power supplies. The +ve and -ve output devices (bipolar or FET transistors) alternately progressively conduct to supply current to the load (the speakers) for each half of the signal waveform. If a given amplifier produces, say 100W into a 16ohm load, then *if* the power supplies maintained their voltages, it would produce 200W into an 8 ohm load. However, delivering that power actually causes most typical power supplies to "sag" their voltages, so as you halve the load, you actually get less than a doubling in power (ok, there are some other losses involved too - but sag is the main one) - so you might get 16ohm->100W, 8ohm->180W, 4ohm->320W, 2ohm->500W... A better (stiffer) power supply produces less sag, but costs more... But halving the load, in addition to (almost) doubling the power, also doubles the output current into the load. Most amplifiers are designed to supply sufficient current to drive down to a 4 ohm load, but some amplifiers are designed to deliver sufficient current to drive down to 2 ohms - this takes typically twice as many output devices as driving a 4 ohm load, so is more costly to manufacture. In addition, the power supply itself has to be able to deliver that additional current, and there will be more heat to get rid of, which again increases cost... The basic thing is W= V squared / R Watts = volts squared / resistance (in ohms) So as the load resistance is halved, the power output is doubled (provided the maximum output voltage remains the same - which it doesn't quite...) </more value than you asked for> I hope this helps Alan[/quote] Thanks very much, very helpful Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.