Bass power

[nag1_uk]

Hi all,
Right I know you need to have more power for a bass signal then a high frequency siganl but why???
I am in the middle of my uni project writeup and I am getting a but confused!

The reason I ask is that I have done a graph showing the power consumed by the amp against the input amplitude with a sinewave input at 550Hz with, then the same with a 15KHz wave and the low frequency test consumed a lot more power!

Any technical reasons???

Thanks

Mike

 

[tiroth]

This is false: obviously, P=V^2/R which is independent of frequency.

Where you might be getting confused is that if you look at a FFT analysis of acoustic power of music signals, you will find that there tends to be more power at lower registers. Once you drop below 40-50Hz though the power again drops off.

It is also possible that the efficiencies of the drivers were simply different if you are comparing two discrete data sets instead of a general case.

 

[nag1_uk]

This was done with resistive loads so no speakers!

 

[Bill Fitzpatrick]

We need to know WHAT you are measuring, WHERE you are measuring and HOW you are measuring.

 

[nag1_uk]

I was applying a sine wave and monitoring the input amplitude (voltage) and monitoring the voltage and current drawn by the power supply, thus the power supplied.
The output of the amp was running into a resistive load.
The voltage was kept constant as normal and and current was monitored on the power supply scale.

Mike

 

[tiroth]

Either there is measurement error or the load wasn't a resistor. I lean towards the former. Are you sure your oscillator's voltage is stable as frequency is swept?

There is no getting around the fundemental physics.

 

[nag1_uk]

I was not sweeping the frequency, I was "sweeping" the input amplitude. I simply ran the test twice, one at 550Hz and the other at 15KHz.
The Load was 2 18ohm wire wound resistors in parrellel.

I am getting even more confused now. Might just miss this bit out of the experiment writeup!

Mike

 

[Cal Weldon]

In very simple terms, it seems reasonable that since the driver moves a lot further to produce a low note than a high note, the amount of power required will be more. But that only deals with a driver in the equation, not a resistive load.

Cal

 

[johnnyx]

Maybe the amp was clipping at 550Hz and not at 15kHz. Did you monitor the output with a scope? Waveform distortion could be the cause of the error.

 

[nag1_uk]

The inputs and outputs were on a scope, there was no sign off clipping. I am guessing I am going to have to re-do these reading to check now!

Thanks

Mike

 

[macboy]

In very simple terms, it seems reasonable that since the driver moves a lot further to produce a low note than a high note, the amount of power required will be more.

I'd argue with that. Consider the forces required to accelerate something faster (at higher frequency). If you wanted the cone to move as far at 10 kHz as a 1 kHz, then you would need 10 x the acceleration forces on the cone, and 10 x the power.

To get back on-topic though, it is entirely possible (and likely) that the wirewound resistors being used for the original test are not non-inductive types. Their inductance would cause them to draw less current at the higher frequency, and that could explain the test results.

nag1_uk, try monitoring the load current in addition to the voltage. Do this by placing a small (0.1 to 0.5 ohm; the value doesn't really matter) reisistor in series with the load resistor and measure the voltage across it (with the scope). Please make sure this resistor is non-inductive. If your load is truly non-inductive, then the current will always be exactly in phase with the voltage at all frequencies. If you see the two waveforms drift out of phase (and/or change in relative amplitude) as you change the frequency, then your load is inductive.

p.s. be careful about how you connect the resistors and how you connect the scope probes to them... don't create a short with the scope probe grounds.