# mute de-amplify sound volume

I want to de-amplify the sound volume of my television. My native language is not english, and I do not know the correct word, if any, for the opposite of amplify. When my TV is turned from volume zero to volume one, the heard sound jumps to something that is too loud.

TV volumen scale              :  0, 1, 2, 3, 4, 5,...,49,50,51,52,...100
Corresponding subjective scale:  0,11,12,13,14,15,...,59,60,60,60,...60


What I would really like is a sound volume on this subjective scale from e.g. 3 to 20. So I would like to decrease the volume by a factor of app. three. Can it be done by connecting a suitable resistor in series or in parallel with the internal speakers?

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The correct word is "Attenuate" –  Majenko Dec 17 '11 at 22:55
Thank you - now I know:o) –  Hans-Peter E. Kristiansen Dec 17 '11 at 23:07
Put a pillow over the speaker. –  Olin Lathrop Dec 18 '11 at 2:28

Absolutely. Your internal speakers will have some nominal impedance, perhaps 8 ohms (you would need to do some research to know for sure). If you were to hook an 8 ohm resistor in parallel (across) the speaker, then half of the "volume" (power actually) would come out the speaker and the other half would simply be wasted as heat in the resistor.

Assuming the same 8ohm speaker, if the resistor were greater than 10 ohms, then less power would go to the resistor and the volume you hear would be louder.

As a practical matter you could try a few different resistors to see what works best for you. One thing to be aware of is that you need to use a resistor with an adequate power rating, perhaps 5 watts.

While it would be possible to compute the value of an "ideal" resistor, the mathematics is not as simple as you may think due to the way the human ear perceives sound. The perceived volume is not a linear relationship but rather a logarithmic function measured on the decibel scale.

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Thank you - my only concern with this solution is that the television sound amplifier would see e.g 4 Ohm instead of 8 Ohm. Would I not risk to draw too much current, and burn the amplifier? –  Hans-Peter E. Kristiansen Dec 17 '11 at 23:26
Without knowing the specific TV it is impossible to give a definitive answer. For most modern TVs operated at less than full volume continously it is unlikely. –  JonnyBoats Dec 18 '11 at 2:05

The simplest way would be with a voltage divider

The voltage $V_{in}$ is reduced at a ratio of $R_1:R_2$.

Specifically:

$V_{out} = V_{in} \times \frac{R_2}{R_1+R_2}$

So if you have $R_1$ at twice the resistance of $R_2$ you will end up with a third of the voltage, and hence a third of the volume.

Typical example resistances would be $R_1=20K\Omega$ and $R_2=10K\Omega$.

So, feeding those into our formula, and an instantaneous voltage of 1.5V for $V_{in}$, we get:

$V_{out} = 1.5 \times \frac{10,000}{10,000+20,000} = 1.5 \times \frac{1}{3} = 0.5V$

This is how most attenuating volume controls operate - the potentiometer acts as a voltage divider with the wiper being the centre tap.

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... and connect the speakers to V_out? I do not think it would work. If the voltage divider is to be stiff, the resistors needs to be much smaller than the resistance of the speakers and I do not know anything about the build in sound amplifier. –  Hans-Peter E. Kristiansen Dec 17 '11 at 23:04
with Z_2 in the kOhm region it would be pulled down by the speaker. –  Hans-Peter E. Kristiansen Dec 17 '11 at 23:06
Sorry, didn't see the bit about being in line with the internal speakers. You might get away with very small resistances then, in the order of 1 to 10 ohms. –  Majenko Dec 17 '11 at 23:11