How Much Current is Needed & Amp Voltage

[12Many]

I have two more questions that I am having a hard time understanding or finding answers to on the web.

1. I saw the specs on this amp:

Coda S5.5 amplifier has 50 watts into 8 ohms, 100 watts into 4 ohms and 200 watts into 2 ohms all class A. It can output 100 amps.

My speakers dip down to 3 ohms. What is the benefit of having 100 amps over say 29 amps, which my 100 watt amp puts out?

If this is right, W=I*V and W=I squared * R. So at 3 ohm load and 150 watts, yields a current of 7.1 amps. Why would this amp need to output 100 amps if the draw would only be 7.1 amps? Even at 1 ohm load, based on W=I squared * R, then 150/1 = 150 and the sq rt of 150 is about 12.2.

And W-I*V, then if the watts are 100 and current is 100, then the amp output would be 1 volt? that seems too low for real world audio situation.

2. Is the output voltage of an amp based on its wattage ? For example, the AHB2 lists it max voltage output at about 29 volts. Would a 200 watt amp then have a higher max voltage output?

Thanks.

 

[solderdude]

The output voltage of said amp is 20V (40V in bridged mode)
In reality it is only class-A in 8ohm (so 50W in class-A) so would not be in class-A above 25W in 4 ohm and only 12.5W class-A in 2 ohm (would require 1.8A bias) the rest of the power would be class AB.
JA measured 1.26A idle so about 18W class-A in 8ohm
200W in 2ohm would require 10A and as your speaker is 3ohm there would never be more than 6.7A drawn.
When it can deliver 100A it could deliver power in 0.2ohm (2kW when the power supply would not sag or momentarily from the capacitors as the transformer is 1.5kW)
Bridged mono, in 2ohm it can deliver 20A.
I assume the current limiter kicks in at 100A to protect the output devices when the amp is shorted it has no benefits but is a 'side effect' of all the output devices in parallel.

 

[amix]

I have never seen amplifier specs mentioning output voltage. Is there a reason for that?

I ask, because I want to buy a power amplifier soon, but I need to make sure, the amp has high output current (recommendation by speaker manufacturer, which is Dynaudio). Also, how do I calculate the safe max. current my speakers would be fine with?

Thanks.

 

[J.M. Noble]

This is Ohm's Law. For historical reasons, we learn it in engineering classes as:

E = IR

Where E is in Volts, I is current in Amperes, and R is resistance in Ohms

Power is:

P = IE

Where P is power in Watts.

Given any two values, we can compute the others. In the case of audio amplifiers, we are generally given the rated power in Watts and the nominal resistance equivalent (actually a complex "impedance", another topic, but we simplify it) in Ohms.

To figure out the voltage at a given power rating, we need to compute the current first. Algebraic substitution gives us:

P = I(IR) → P = I²R

So, if we have an amplifier rated for 100W RMS (about which more below) into 8Ω, we solve:

100 = I²8Ω → 100/8 = I² → √12.5 = I → I = 3.53 Amperes

Then we solve for voltage:

E = 3.53 * 8Ω → E = 28.24 Volts

Now, we are dealing with AC voltages and RMS power. By convention, we assume sine waves unless otherwise specified. "RMS" means "Root Mean Square", which is a mathematical method for determining the equivalent DC power (and, by extension, voltage) for an AC signal. It greatly simplifies things to do so. In the case of sine waves, we multiply the RMS voltage by 1.414 (square root of 2) to get peak voltage (Vpeak i.e. the positive swing) and 2.828 (Vp-p = 2 * the root of 2) to get peak-to-peak voltage. For this 100W into 8Ω amplifier, we get 39.93V and 79.86V respectively. The amp's actual voltage supply rails will be a few volts past this because most devices cannot swing all the way to the rails. (The math works for the mains current coming out of your wall, by the way: ~339V peak-to-peak sounds a lot scarier than 120VAC!)

So, from a simplified Ohm's Law perspective, we should consider the peak instantaneous current demand based on Vpeak rather than RMS voltage.

E = IR → 39.93Vpeak = I * 8 → 39.93/8 = I → I = 4.99 Amperes

Impedance almost always varies with frequency: our "8 Ohm" speaker could well dip down to 5 Ohms: 39.93 / 5 = 7.99A peak, or more than double what we computed for the continuous steady state current based on RMS figures. Because loudspeakers are dynamic loads, the actual instantaneous current demands could be considerably higher.

This is a pretty good Ohm's Law calculator that explains the answers when given two values:

Ohms Law Calculator

This free Ohm's Law calculator solves for any of the variables in the Ohm's Law equation using various units of measurement and gives out the solving steps.

www.calculator.net

FWIW, I'm pretty sure based on testing over the past few days that my Dynaudio Evoke 20s will run fine (i.e. mid-80s dB SPL @ 2.5 meters in a 22' X 24' room) on a 10-12Watt/channel Schiit Gjallarhorn. Dynaudio speakers have a reputation for being power/current hogs and at higher volumes that may be true--but I have PA speakers for that kind of horsing around. At more reasonable volumes, I don't think they are any better or worse than other speakers of similar impedance.

 

[amix]

Dynaudio Evoke 20s will run fine (i.e. mid-80s dB SPL @ 2.5 meters in a 22' X 24' room) on a 10-12Watt/channel Schiit Gjallarhorn

I run my Dyns with a 50W/8Ω and they sound quite ok but I believe they will start to shine with a bigger amp. Well, over I'll see in the next week's...

Anyway thank you so much for going through the pain explaining this all in such minute detail and making it so easy to consume. :thumbs up:

 

[RayDunzl]

I have never seen amplifier specs mentioning output voltage. Is there a reason for that?

Maybe just to scare you...

 

[Chrispy]

Maybe just to scare you...

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Prices for each Krell, tho?

 

[RayDunzl]

Prices for each Krell, tho?

Not cheap, but this spec is for 20 year old gear.

 

[Chrispy]

Not cheap, but this spec is for 20 year old gear.

Probably among the few with that sort of specification, tho....

 

[J.M. Noble]

I run my Dyns with a 50W/8Ω and they sound quite ok but I believe they will start to shine with a bigger amp.

It depends on your room, of course, and which speaker model you have. 50W might give you around 17dB over the rated 1W/1 meter rating. 200W only gives you 6dB over *that*, assuming no acoustic compression (which is a bad assumption). Using my own 86dB 1W/1m speakers as an example, I'd expect maybe 99dB peak with 50W/channel and not much more than 102-103dB with 200W/channel at 3 meters/10 feet from the speakers. That's insanely loud for domestic listening!

I do a lot of live sound engineering, and that's significantly louder than I mix rock'n'roll bands when I have 8-10 kilowatts driving efficient PA speakers.

My current Denon AVR-X2100W HT receiver (same amp section as its successor AVR-X2200W reviewed on this site) is good for about 120W/channel. I just did some testing with bass-heavy music and my Dynaudios went into severe and very audible stress at ~35W/channel peak (14.4VRMS/6Ω captured in min/max mode with a Fluke 87 multimeter) even with the ports stuffed with wool socks. If you have a pair of the big floorstanders and/or you don't play music with low bass (i.e. much content under a bass guitar's ~41Hz low E string) you might be able to make use of more power. In my own case I didn't notice any advantage in going from my old NAD C320BEE's 50W/channel to the present 120W/channel. The speakers give up long before the amp does.

I encourage everyone interested in the topic to install a SPL meter app on their phone. While they may not be calibrated when using a phone mic, they give reasonably accurate *relative* measurements and will help you educate your ears. The results can be surprising.

 

[12Many]

Maybe just to scare you...

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Thanks for posting that. I had another thread where i was wondering about amp output voltages. It is interesting to see the values scale as watts increase.

 

[Sokel]

Thanks for posting that. I had another thread where i was wondering about amp output voltages. It is interesting to see the values scale as watts increase.

Here's some more:

 

[Schollaudio]

I have never seen amplifier specs mentioning output voltage. Is there a reason for that?

I ask, because I want to buy a power amplifier soon, but I need to make sure, the amp has high output current (recommendation by speaker manufacturer, which is Dynaudio). Also, how do I calculate the safe max. current my speakers would be fine with?

Thanks.

Some do but they are rare. I like to find a schematic and see the voltage rails to figure peak clipping. Add the rail voltages times .353 square that result and divide by the load. Say the rails are plus minus 50V that would be (100x.353) squared divided by 8 = 155.7watts peak without power supply sag. Some Japanese integrated (Kenwood 5100) had 50 volt rails but were rated at 60 watts per channel.

 

[J.M. Noble]

Some do but they are rare. I like to find a schematic and see the voltage rails to figure peak clipping. Add the rail voltages times .353 square that result and divide by the load. Say the rails are plus minus 50V that would be (100x.353) squared divided by 8 = 155.7watts peak without power supply sag. Some Japanese integrated (Kenwood 5100) had 50 volt rails but were rated at 60 watts per channel.

Directly using the rail voltages will almost always overstate the available power because there are voltage drops across the output devices (at a minimum), and unless the power supply is pretty stiffly regulated the rails will sag significantly (strictly speaking, you will see increased ripple) under "full power" conditions. Also, I would use the .707 (half the square root of 2) times a single rail for RMS for better accuracy--that's how I learned it in circuit analysis class, anyway.

 

[Schollaudio]

Directly using the rail voltages will almost always overstate the available power because there are voltage drops across the output devices (at a minimum), and unless the power supply is pretty stiffly regulated the rails will sag significantly (strictly speaking, you will see increased ripple) under "full power" conditions. Also, I would use the .707 (half the square root of 2) times a single rail for RMS for better accuracy--that's how I learned it in circuit analysis class, anyway.

True but that's small and unknown sag will have more impact.

 

[J.M. Noble]

True but that's small and unknown sag will have more impact.

They are in the same neighborhood.

2V drop across the output transistor isn't a bad ballpark figure, which gives us about 144W RMS into 8Ω.

Ripple in a bridge rectifier-type supply is given by Vripple = Idc/(2FC), where Idc is load current in Amperes, F is supply frequency in Hertz, and C is capacitance in Farads. Assuming a North American supply and 10,000μF for our now 144W amp, we get V = 4.24/(2*60*.01) or 3.53V. We want to stay below that ripple, so our peak swing can't be over about 44V, or 31.1VAC RMS. This gives us about 121 Watts RMS best case with an ideal power supply that has no voltage drops of its own.

With 4Ω speakers, ripple gets a lot worse: around 7V! Figure something under 40V peak swing, and maybe 180-200W RMS in ideal circumstances. Going straight off the +/- 50VDC rail voltage would seem to indicate ~312W RMS ...

If you're keeping score you will note that I glossed over some details that will likely make the actual output before clipping lower. I would guess that 90-100W continuous is the most you'd reliably get from that +/- 50VDC supply with a class AB amplifier. Dynamic power could plausibly be the 144W we just computed without ripple.

 

[valerianf]

"I would guess that 90-100W continuous is the most you'd reliably get from that +/- 50VDC supply with a class AB amplifier. Dynamic power could plausibly be the 144W we just computed without ripple."
To go above this limit a Arcam Class G topology with voltage rail switching may be the solution.
It can provide 200W @8 Ohms and 350W @4 Ohms for the FMJ P49 amplifier.

 

[amix]

What could the reason be for Dynaudio customer support telling me, that power is less of a concern with their speakers but current is? Better: why could a higher amount of current matter more for a speaker than power?

To be honest, these were not exactly the words they used, they said "quality of power", but we all know this term is on the edge of marketing, prolly trying to keep the customer away from asking too many questions. However, around forums, people tend to say, it's current rather than power, so I guess, that's what support meant with "quality".

 

[J.M. Noble]

Power = Current X Voltage ... so it's odd to say current matters more than power. They are intertwined.

According to my recollection, Dynaudio was somewhat unusual back in the 80s and 90s in producing drivers with less than 8Ω impedance for the home audio market. Lower impedance means a higher current draw from a given amplifier since amplifiers are voltage sources--see Ohm's Law in previous posts and again here:

Volts = Amperes X Ohms ... with simple algebra, we get Volts (divided by) Ohms = Amperes

Thus, the lower the impedance, the higher the current. Home audio amplifiers weren't commonly designed with less than 8Ω loads in mind back then, so Dynaudio picked up a reputation for being difficult to drive. It has kind of become part of the lore 30 years later, in my opinion. Sensitivity was an issue, too, but it seems like their modern designs are pretty average at ~86dB/W/m.

I just took delivery of a Schiit Gjallarhorn. It probably produces about 12 Watts per channel into my 6Ω Dynaudio Evoke 20s. That's about 10% of what they were getting from my Denon AVR-X2100W, yet it sounds great at any volume I actually would listen to (low/mid-80dB range)--and I'm over 10 feet from the speakers in a 22' X 24' room. An active highpass filter would help since the little Gjallarhorn is fully capable of making the Evokes fart with low bass--more power would not help!

Perhaps owners of the Confidence, Contour, etc. lines have a different experience.