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Author Topic: Converting amps needed based on ohm change  (Read 572 times)

David Allred

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Converting amps needed based on ohm change
« on: August 03, 2017, 10:18:07 am »

If a power amp draws 9.5A @ 1/8th power @ 4 ohms, is there a formula to convert the draw @ 8 ohms.  Is it basically 1/2 or 8 and x2 for 2?  Seems a little too easy.

thanks.
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John Roberts {JR}

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Re: Converting amps needed based on ohm change
« Reply #1 on: August 03, 2017, 11:37:02 am »

I like easy....  take it when you can.

JR
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Tim McCulloch

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Re: Converting amps needed based on ohm change
« Reply #2 on: August 03, 2017, 12:20:31 pm »

Georg Ohm laid it out for you.
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Bill Koonce

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Re: Converting amps needed based on ohm change
« Reply #3 on: August 03, 2017, 12:22:40 pm »

I'd think that a lot of that would depend on the amplifier class and power supply. A Class A amp will draw nearly full power regardless of output power, for example. If your amp draws 9.5A @ 1/8 power and that draw was linear, then it would draw 76A @ full power @ 4 Ohms, and over 150A @2. That's one mighty big AC circuit that your amp would need...if input power was linear to output power.
« Last Edit: August 03, 2017, 01:30:10 pm by Bill Koonce »
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David Allred

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Re: Converting amps needed based on ohm change
« Reply #4 on: August 03, 2017, 12:55:51 pm »

I like easy....  take it when you can.

JR

Thanks, JR, for confirming that I was correct.  (Or at least being the least cryptic.) ;)
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John Roberts {JR}

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Re: Converting amps needed based on ohm change
« Reply #5 on: August 03, 2017, 01:22:27 pm »

I'd think that a lot of that would depend on the amplifier class and power supply. A Class A amp will draw nearly full power regardless of output power, for example.
What do you think the odds are that the OP has a class A amp?  ::)
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If your amp draws 9.5A @ 1/8 power and that draw was linear, then it would draw 76A @ full power @ 8 Ohms, and over 300A @2. That's one mighty big AC circuit that your amp would need...if input power was linear to output power.
That was not the OP's question.

The 9.5A was for 4 ohm 1/8th power.   8 ohm 1/8th power would be half that current, all else equal. (not exactly 1/2 because power supply sag would be different but sag at 1/8th power should not be significant for either load..

A current draw @ 1/8th (speaker?) power for a class A amp is not a meaningful spec.  Try not to confuse the issue with superfluous information.

JR
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Jeff Bankston

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Re: Converting amps needed based on ohm change
« Reply #6 on: August 03, 2017, 02:43:26 pm »

How would you figure a multi step amp ? I would use an amp meter. But what would be the point of this ?
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John Roberts {JR}

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Re: Converting amps needed based on ohm change
« Reply #7 on: August 03, 2017, 02:59:51 pm »

How would you figure a multi step amp ? I would use an amp meter. But what would be the point of this ?
You are over thinking this... if 1/8 power for 4 ohm is X...  1/8 power for 8 ohm is half X...  finis.

The point of 1/8th power is for sizing mains ampacity since that is closer to typical consumption than full power.

JR
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David Allred

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Re: Converting amps needed based on ohm change
« Reply #8 on: August 03, 2017, 03:58:32 pm »

You are over thinking this... if 1/8 power for 4 ohm is X...  1/8 power for 8 ohm is half X...  finis.

The point of 1/8th power is for sizing mains ampacity since that is closer to typical consumption than full power.

JR

I found another document on the QSC site.  The PLX3102 is listed 8.8A (not 9.5) @ 4 ohms @1/8th power and 5.7A @ 8 ohms @ 1/8th power.  So not 1/2, but 35% less.
Per QSC... 1/8th power is light clipping, 1/3rd power is very heavy clipping, all using pink noise to simulate heavily compressed program material.
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John Roberts {JR}

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Re: Converting amps needed based on ohm change
« Reply #9 on: August 03, 2017, 05:49:55 pm »

I found another document on the QSC site.  The PLX3102 is listed 8.8A (not 9.5) @ 4 ohms @1/8th power and 5.7A @ 8 ohms @ 1/8th power.  So not 1/2, but 35% less.
Per QSC... 1/8th power is light clipping, 1/3rd power is very heavy clipping, all using pink noise to simulate heavily compressed program material.
OK sorry not that simple...

for probably more info than you want the 3102 is a  "2 tier class H technology" amp. What class H means is that they use different voltage power supply rails inside the amp, so for small voltages they can pull from a low voltage rail and not waste all the heat dissipation from a single higher voltage rail. The 3102 is 2 tier so has 3 different power supply voltages inside.

1/8 power @ 4 ohm is a smaller voltage than 1/8 power at 8 ohm. The higher voltage 8 ohm 1/8 power is probably drawing from the bottom and middle rail voltage so less efficient than the 4 ohm just pulling from the smallest bottom rail. So not exactly 1/2 the current.

A class D amp without the multiple PS rails would be more like the linear 2:1 ratio, but the 3102 is not class D or class A.  ::) Sorry.

JR

PS: Since 1/8th power is an important metric for install amps it is likely that the internal power supply voltages were selected to be most efficient at 1/8 power for typical load impedance. Note: these internal voltages are probably different for most efficient 1/8 sinewave power and 1/8 pink noise power. I'll bet they are tweaked for best noise power.
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