Conversion questions

[raveenvijendren]

Good Day,

Can anyone help me with the following,

1) Suppose an amplifier is rated 100 Watts RMS into eight ohms, how do I calculated the power output from the amplifier if it is connected to a four ohm load?

2) Total Harmonic Distortion and Intermodulation Distortion is most often given in a relative value, i.e. a percentage(%). One operational amplifier that I know of says it has THD less
than -120dB. How do I convert THD from a dB value to a percentage value(%)?

Thank you.

[Tweeker]

1) Crank the amp, drop a 4 ohm power resistor in a thermos of water, and see how long it takes it to heat the water? First empirical method I thought of. Not necessarily wise. If the amp is not current limited it will be close to 200 watts.

[Tweeker]

2) dB are logarithmic, -3dB is half, -10 dB is one tenth. So -10dB is 10% and -20dB is 1% and -120dB is not much.

[Stocker]

Originally posted by Tweeker
-120dB is not much.

What he said.

For an amplifier with a stiff power supply and hefty output devices, almost double the output power will be put into a halved load. Note: modern consumer-grade stuff will not necessarily have either of those things. More likely, it will either clip excessively (and sound like poop ) at lower volume setting, or else it will just blow up (really! )

[kelticwizard]

Originally posted by raveenvijendren
One operational amplifier that I know of says it has THD less
than -120dB. How do I convert THD from a dB value to a percentage value(%)?

Thank you.

1% - 1/100

10 Log (1/100) = -20 dB. So 1% THD = -20 dB.

10 Log (1/1,000,000) = -60 dB So .0001% = -60 dB

10 Log (1/1,000,000,000,000) = -120 dB. So .0000000001% = -120 dB.

Put another way, -120 dB is is one trillionth of the total, or one ten billionth of a percent of the total.

[poobah]

C'mon guys..... GEEZE!!!!!!!!!!!!!!!!!!

Back away from the eggnog -

If you're talking about voltage or current, the formula is:
Gain = 10 to the (dB/20) power = 10^(dB/20) = inv log (dB/20)

If you're talking about power, the formula is:
Gain = 10 to the (dB/10) power = 10^(dB/10) = inv log (dB/10)



If you want to find db from a gain, for voltage or current, the formula is:
db = 20 log (Gain)

For power:
db = 10 log (Gain)


So for -120 dB:

Gain = 10 ^ (-120/20) = 0.000001

times 100 to get percent = 0.0001%

All log are base 10... don't use "ln"... that is base e (natural log)

[lineup]

Originally posted by raveenvijendren

1) Suppose an amplifier is rated 100 Watts RMS into eight ohms,
how do I calculated the power output from the amplifier if it is connected to a four ohm load?

Amp rated 100 Watts RMS in 8 Ohms
should teoretically give
200 Watts in 4 Ohms.

If only the voltage supply sets the limit.
It is how high the voltage is over a resistor
that sets how much power.

A voltage supply to an amplifier is fixed to what transformer can give.
now power is BOTH current and voltage.
If the trafo->power supply can not deliver enough current
to reach same voltage over 4 Ohm as over 8 Ohm
than will not be twice as much power,
if we half the Ohm, impedance.

The more current we try to take from a transformer
the lower the voltage will be,
unless the transformer is rated to deliver this current.

Transformers are rated for example 2x30 V, 10 A = 2x30x10= 600 VA
VA. V for Voltage, A for Ampere

But not only transformer and power supply can limit max power.

Also the output transistors have to be able to transfer current.
Transistors have ratings for example:
20 A peak - what can be delivered for a very short time
10 A average - what can be delivered for infinite long time
without transistor burns and destroys.

To know what power is output you need to know
true impedance of you loudspeaker or use a power resistor of 4 Ohm.
Then you measure the AC-voltage across load.
Should be a good AC-meter, that can measure audio frequencies.

Most multimeters are for measuring 50 Hz/60 Hz
and are no good much above 400 Hz.

Say you have connected a 4 Ohm power resistor, RATED 200 Watt!
And AC average voltage ( not peak ) measures 20 Vac across this resistor.
If you use 400 Hz testsignal, you can use most AC voltmeters.

P(Watt) = (UxU) / R
P = 20x20/4 -> gives 100 Watt

If you measure same 20 V across 8 Ohm
P = 20x20/8 -> gives 50 Watt

[kelticwizard]

Okay, it says here:

2] Decibels (dB) = 10 log(P2/P1)

The power ratio need not be greater than unity as shown in the previous examples

So, for -120 dB:
dB = 10 Log (1 / 1,000,000,000,000) =
dB = 10 ( - 12) = -120 dB

So, for power, -120 dB equals one trillionth the power. Multiply by 100 for percentage, and it is one ten billionth of one percent of the power.

[poobah]

Right Keltic,

His question was about voltage though!

[AndrewT]

Hi Keltic,
you would need to be a wizard to come up with those numbers in this celtic land.

10log(p1/p2) for power ratios.

20log(v1/v2) for voltage ratios.

No exceptions.

Poo is spot on with the log base 10 bit as well.