i would like to do an amp which has around 600rms at 2 ohm with supply around +/-56V!
Is it posible to get around 800 rms with this amp using a 4 ohm load(need to push up supply voltage to...................how mutch)?
amp has 2x6 darlington tranzistors!
i think this could be done but will the amp heat up too much?
in my oppinion the amp would heat up more when using 2hm load witth supply +/-56V(more current) than using a 4 ohm load at 800rms
Give some answer!!!!!
Is it posible to get around 800 rms with this amp using a 4 ohm load(need to push up supply voltage to...................how mutch)?
amp has 2x6 darlington tranzistors!
i think this could be done but will the amp heat up too much?
in my oppinion the amp would heat up more when using 2hm load witth supply +/-56V(more current) than using a 4 ohm load at 800rms
Give some answer!!!!!
Hi,
Ocool there is something wrong with your calculator. 56.6^2/2=1600Wpk =800W into 2 ohm, more than he wants.
49^2/2=1200Wpk =600W into 2ohm.
Taking losses into account then +-55Vdc when loaded may be a target to aim for. 2 times 40Vac to 42Vac 1kVA.
I would go for MJ21193/4. Neither darlingtons nor Lfets can match at 2 ohm loading.
I'm told that Vfets might be able to compete and probably cheaper if you go quasi Nchannel.
Ocool there is something wrong with your calculator. 56.6^2/2=1600Wpk =800W into 2 ohm, more than he wants.
49^2/2=1200Wpk =600W into 2ohm.
Taking losses into account then +-55Vdc when loaded may be a target to aim for. 2 times 40Vac to 42Vac 1kVA.
I would go for MJ21193/4. Neither darlingtons nor Lfets can match at 2 ohm loading.
I'm told that Vfets might be able to compete and probably cheaper if you go quasi Nchannel.
Smokingmachine,
As said this is indeed a difficult question with no simple unique answer.
But certain topologies are better than others. In an analysis by Prof. Otala a long time ago, a double emitter follower was rated 10 times better than quasi-complementary, and full complimentary topology (the best) 20 times better. The use of "better" is of course very unspecific; it refers to relative distortion but also to other characteristics. The details are somewhat long to post here, and it also depends on the type of semiconductor, etc. as others suggested. (These tests were for transistors but would broadly be valid also for MOSFETs).
This as a very general indication. It would appear that MOSFETs may have the edge over transistors, but I have no experience of them. When I last heard there was a very large spread in characteristics, which could make matching difficult. I wonder if this has improved.
Regards.
As said this is indeed a difficult question with no simple unique answer.
But certain topologies are better than others. In an analysis by Prof. Otala a long time ago, a double emitter follower was rated 10 times better than quasi-complementary, and full complimentary topology (the best) 20 times better. The use of "better" is of course very unspecific; it refers to relative distortion but also to other characteristics. The details are somewhat long to post here, and it also depends on the type of semiconductor, etc. as others suggested. (These tests were for transistors but would broadly be valid also for MOSFETs).
This as a very general indication. It would appear that MOSFETs may have the edge over transistors, but I have no experience of them. When I last heard there was a very large spread in characteristics, which could make matching difficult. I wonder if this has improved.
Regards.
well the amp i am trying to do has mj11015(6x) and mj11016(6x)!!this are bipolars!!!
i wonder if this amp will do if I raise voltage to +/-80 or 85 volts and than it could deliver around 800 --- 900 rms at 4 ohm load or the amp will be heating too much(i will use a vent)
i think if amp could deliver 600 rms at 2 ohm load then the amp could deliver around 800 rms at 4ohm load(with proper supply)and still won`t be heating as much as if the amp would run at 2 ohm load with supply around +/- 56 V per rail!!!!
As I calculated the amp with +/-56V at 2 ohm load woult took much ampers than with +/-80........85 V(4 ohm load)!!
Am i right or no?
Ant thanks for all your answers!!!!!
i wonder if this amp will do if I raise voltage to +/-80 or 85 volts and than it could deliver around 800 --- 900 rms at 4 ohm load or the amp will be heating too much(i will use a vent)
i think if amp could deliver 600 rms at 2 ohm load then the amp could deliver around 800 rms at 4ohm load(with proper supply)and still won`t be heating as much as if the amp would run at 2 ohm load with supply around +/- 56 V per rail!!!!
As I calculated the amp with +/-56V at 2 ohm load woult took much ampers than with +/-80........85 V(4 ohm load)!!
Am i right or no?
Ant thanks for all your answers!!!!!
"how much power?"
Hi there is a very interesting article at the musical fidelity site
---->www.musicalfidelity.com about power levels in domestic
hifi powr amps (thats very loud) the article is also showing no
mercy on small hifi amps rendering them low-end (kinda rude)
marketing thing or not these guys are banking on it
good luck to them
regards
Mastertech
Hi there is a very interesting article at the musical fidelity site
---->www.musicalfidelity.com about power levels in domestic
hifi powr amps (thats very loud) the article is also showing no
mercy on small hifi amps rendering them low-end (kinda rude)
marketing thing or not these guys are banking on it
good luck to them
regards
Mastertech
Hi Smoking,
I think you are there but this just to confirm.
Ppk=V*V/Rload
P=V*V/Rload/2
i.e. peak power = 2 * P based on sinewave.
Add about 4V to 8V to obtain the +-Vrail voltage at full power into the load.
Then add the droop in rail voltage when testing at full power.
100W into 8ohms needs 40Vpk at the load.
At full power into that same load the Vrail will be about +-45Vdc.
At quiescent Vrail will be about +-50Vdc.
Hope this clarifies.
I think you are there but this just to confirm.
Ppk=V*V/Rload
P=V*V/Rload/2
i.e. peak power = 2 * P based on sinewave.
Add about 4V to 8V to obtain the +-Vrail voltage at full power into the load.
Then add the droop in rail voltage when testing at full power.
100W into 8ohms needs 40Vpk at the load.
At full power into that same load the Vrail will be about +-45Vdc.
At quiescent Vrail will be about +-50Vdc.
Hope this clarifies.
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