Dynamic Headroom vs Resistance


I'm in a discussion on another board and a member has asked if he adds 2 more speakers onto his amp (cutting the ohm load in half) will he get a louder, more dynamic system with better headroom?

I was under the impression that
the fact that your powering 2 extra speakers, and drawing more juice outta your amp means you will reach clipping faster as your amp runs outta juice, so you will lose your headroom because your amp will be pushed harder.

to which he responded:
There's more power available. Why would the limit be reached faster?
Half the load means double the power. Wouldn't that increase headroom?
You'd be using the same power as before to reach the same volume level. Only now you're using a smaller percentage of the available power.

Now figure in the extra speakers. The amp needs to deliver even less power than before to reach the same volume because there is twice the cone area. The percentage of power is smaller still.

According to Stereophile, "Dynamic Headroom," is the ratio between the maximum power the amplifier can put out for a burst just 20ms long, and the continuous power. An amplifier that can deliver 200W for 20ms but 50W continuously will have a dynamic headroom of 6dB.

What am I missing here...
is he correct that you'll get more headroom?
but this has no impact on transiet response correct? as thats a seperate function?
sorry if this is confusing...
Geoff
geoffgarcia
I would remind him "doubling power" is in an ideal world. In the real world, most amplifiers fail to double their power because their power supplies run out of juice.
In a parallel circuit, the voltage is the same across each resistor. If the resistors are equal in value, the amperage is equal in value (E=IR) across each resistor. So a parallel circuit of equal resistance value will utilize twice the current since amperage is the sum of both branches of the circuit. Since power is the product of amperage and voltage, the power used doubles in that configuration. So the net "load" might halve, but that drives the amperage to double (since the voltage is constant).

The above is true. Here's my interpretation of how it impacts headroom: (folks, please correct me if I'm wrong)

An amp that puts out 50w continuous into 8 ohms should ideally be able to put out 100w continuous into 4 ohms. If it can, that means that it can deliver twice the current with less load. Many cannot. You'll more often see a 50w into 8 ohm be rated as 80-90 watts into 4 ohms. Basically, those amps borrow the headroom available at 8 ohms -- the amp simply can't generate the current. Headroom, as defined, would go down.

Even an ideal amp that can generate 50 w into 8 ohms and 100 w into 4 ohms would face this. The ability to drive 200w for a brief spurt remains constant regardless of the load because the rail voltage of an amp doesn't vary -- it's the current capacity that is the limiting factor to power, even as measured in short bursts. If the rail voltage is 50 volts and the short burst limit is 200w, then the amp can drive no more than 4 amps for a brief period of time. So 200/50 at 8 ohms is greater than 200/100 at 4 ohms.
thanks ozfly!
interesting!!
this actually is in particular referring to car audio amplifiers which just about all claim to be able to double power when the resistance is halfed.
Some power
400x1 @4ohm
800x1 @2ohm
1600x1 @1ohm

I'm glad I was sorta on the right track with this (or at least I should say Ozfly's thinking follow mine, albeit his is much more factual!)