*New Test: Sundown SAZ-1500D*

[tommyk90 5,000+ posts]

Grow up.

Leave?

 

[xluben 10+ year member]

have fun fighting //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

i have to go do my EE assignment. op-amps //content.invisioncic.com/y282845/emoticons/rolleyes.gif.c1fef805e9d1464d377451cd5bc18bfb.gif

 

[tommyk90 5,000+ posts]

One last question immacomputer, so what wattage did you think I'm ACTUALLY getting? In your definition?

 

[CBFryman2 5,000+ posts]

To get a 100% accurate rating he will need to see the O-Scope results. But he can geustimate. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[tommyk90 5,000+ posts]

To get a 100% accurate rating he will need to see the O-Scope results. But he can geustimate. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

Well, i'm just trying to figure out if he thinks i'm getting less or more power than what my test results provided.

 

[ngsm13 5,000+ posts]

As long as all the amplifiers he tested are tested in the same fashion, it is an accurate guage of comparing them.

nG

 

[CBFryman2 5,000+ posts]

Well, i'm just trying to figure out if he thinks i'm getting less or more power than what my test results provided.

Good point. Now that I think of it you would theoreticly be posting MORE power than real power results. Though 2.2ohm is more of a power rating for a 2ohm nominal impeadance. //content.invisioncic.com/y282845/emoticons/laugh.gif.48439b2acf2cfca21620f01e7f77d1e4.gif

 

[xclusiv3 10+ year member]

*laughs*
Yah... Just don't care for doing the stuff. Maybe because my teacher was totally useless and I literally taught myself from a "Dummies" book. Of course, being that I was in the EET major I had to do plenty of it anyway!

i take it you too took it in college?

well im taking it in 11th gr.

and my teacher is totally useless like you say.

maybe if i did hw id do better....?

 

[Immacomputer 10+ year member]

One last question immacomputer, so what wattage did you think I'm ACTUALLY getting? In your definition?

Without knowing the DCR, I calculated about 400rms at 12.3v assuming a 1 ohm DCR. That is just a rough estimate.

To find it, you need the impedance phase angle. We need to go to the impedance triangle to find it. For the hypotenuse, I used your calculated 2.7ohms and for the base, 1 ohm for the DCR value. To find the angle, we take the cosine inverse of the two. The angle comes out to 68.3*.

To find the actual power, we multiply Vrms*Irms*PF. PF is the power factor and it is equal to the cos of the angle we found earlier. Mulitplying that out gives you about 400wrms.

 

[baseballer1100 5,000+ posts]

i take it you too took it in college?well im taking it in 11th gr.

and my teacher is totally useless like you say.

maybe if i did hw id do better....?

yeah?

 

[sundownz Pro Seller]

xclusiv3 ,

Yessir, took it in college. When I moved to North Carolina from Maryland in middle school it totally messed up my course alignment... I got stuck in the "regular" path and only got as high as pre-cal in high school. Ah well, I still learned it one way or another!

As much as I hate to say it... homework helps alot in more advanced subjects like Calculus. Recent studies indicate it helps very little for young children but the effectiveness of HW increases as you go up in age and complexity of material. Maybe try at least glancing at the homework //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif I know I had my share of laziness in high school, but I generally made time first thing in the morning to at least get it done to hand in.

 

[tommyk90 5,000+ posts]

Without knowing the DCR, I calculated about 400rms at 12.3v assuming a 1 ohm DCR. That is just a rough estimate.
To find it, you need the impedance phase angle. We need to go to the impedance triangle to find it. For the hypotenuse, I used your calculated 2.7ohms and for the base, 1 ohm for the DCR value. To find the angle, we take the cosine inverse of the two. The angle comes out to 68.3*.

To find the actual power, we multiply Vrms*Irms*PF. PF is the power factor and it is equal to the cos of the angle we found earlier. Mulitplying that out gives you about 400wrms.

Sub is an actual dual 2 ohm sub, so a 1 ohm nominal load would be correct.

For some reason, I see this amp doing a BIT more than 400 watts, especially with the score it was doing.

 

[sku 10+ year member]

Without knowing the DCR, I calculated about 400rms at 12.3v assuming a 1 ohm DCR. That is just a rough estimate.
To find it, you need the impedance phase angle. We need to go to the impedance triangle to find it. For the hypotenuse, I used your calculated 2.7ohms and for the base, 1 ohm for the DCR value. To find the angle, we take the cosine inverse of the two. The angle comes out to 68.3*.

To find the actual power, we multiply Vrms*Irms*PF. PF is the power factor and it is equal to the cos of the angle we found earlier. Mulitplying that out gives you about 400wrms.

Hmm..very nice...

What about this for layman's terms? 2.8 ohms is about 3 times more then 1 ohm.

Rated power is 1500 watts at 1 ohm. 1/3 of 1500 watts is 500 watts at around ~3 ohms (our impedance is 3 times as much so we have 1/3 the power)...

Just another way of thinking about it...

 

[15nissen 10+ year member]

Leave?

x2

 

[Ford302Redneck 10+ year member]

logically, and real world is 2 different things also //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif