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Wakum Mata!
Politcally Incorrect Musings
Time for Toy Mods 
3rd-Dec-2007 11:27 am
science
(Note: This is a project I started back on May 30, 2006. Due to my moving to another town, it fell off the map for a while. I had most of this post already finished... but it just sat in the bit-bucket until I could return to it.)

I am surprised that there have not been a huge number of lawsuits against the toy manufacturers for damaging kid's hearing. Today I am modifying one of my daughter's toys to make it more quiet. That means soldering and... math. Logarithms, to be exact. I am talking dB.

This is electronics DIY and math is involved.


This is not some simple task like adding in a resistor. No, no, no. The electronics in this toy is what's known as a "glop-top". This is where a silicon chip is bonded directly to the circuit board and sealed with some sort of epoxy. The manufacturer saved money by putting the speaker driver directly into the chip. There are no driver transistors that I can play with. The speaker is wired directly to the chip and the driver is tuned for an eight ohm load. I know this because when I measured the speaker resistance, I measure eight ohms (while in the circuit). What this means is that I need to halve the volume of the toy ( or thereabouts ) but still make the circuit think it is an eight ohm load so the speaker driver circuit does not burn out. If the load on the circuit is too small, then the driver might burn out because it is trying to pump out too much current.

If I remove the speaker, I measure that it is 16 ohms and not eight. I deduce that there is a 16 ohm resistor in parallel with the speaker and that halves the resistance to about eight ohms. The math is pretty simple:

R/2 = 1/(1/R+1/R), this is the basic formula for resistors in parallel where there are two equal resistors.

A shorthand way of writing this is: R/2 = R || R

So, 8 = 1/(1/16+1/16). But I want to, at most, halve the volume to this toy. It is really loud and there is no volume control. This is where dB and logarithms come in.

First off, a dB is a deciBel. It is a ratio between two similar things. These things can be power, voltage, sound pressure, photons, and so on. Many times you may see some brochure for a stereo or a radio where it says is has a gain of XYZ dB. But, dB what? The unit, dB, on it's own is meaningless without knowing a dB of what. Is it dB current? dB voltage? dB power? It makes as much sense as saying that an object is more colorful. What does "more colorful" mean? It is just marketing speak.

A good primer on the decibel can be found here.

What I want to do is reduce power to the speaker. That is accomplished by reducing the amount of current it receives. I can reduce current to it in two ways: by putting a resistor in series, or by putting a resistor in parallel. If you clicked on the links to view the images, then you can see that both of these will change the resistance of the circuit. I want to keep it the same, so I will need a mix of serial AND parallel. This is not as bad as it sounds.

I am not going to delve deep here. Logarithm's make me feel lightheaded, too. I take it as a challenge to actually (re)learn them. Funny thing is that they are not so bad if you remember this: a logarithm is a formula to find an exponent.

For example, let's say that you have the equation:

1000 = 10x

You don't know what x is, but need to find out. A logarithm is what is used to find x. It looks like this:

x = log10 1000

Note that 10 is called the "base". So, you would say "x equals the log to the base 10 of 1000". Base ten is what we count in... like counting fingers. Computers use a base of 2 (binary).

Logarithms are the single most important concept in mathematics. It is how the real world works and it would be impossible to model or understand the world without logarithms. Learn more about logarithms here. Now, I request that we give a moment of silence out of respect for this globally important and world changing mathematical device.......

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...
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...

Back to the decibel.

A dB is a ratio, or comparison, of two things. Kinda of a "before and after" comparison. The formula for power in dB is:

dB difference = 10 LOG (Power2/Power1)

Here, I am comparing the difference between two power levels. I want half the volume so the ratio "Power2/Power1" needs to be equal to at least 0.5, or -3dB. So, -3dB = 10 LOG (1/2), this is half power.

Half the power means half volume, right? WRONG!

The human ear also works on a logarithmic scale as well as does the transducer device that produces the sound (the speaker). Lots of ugly math later and we find that half the wattage to a speaker does not mean half the volume. Because of the way we humans perceive loudness, I have to cut the power by a lot more. It turns out that we perceive "half-as-loud" between 1/4 to 1/10 power (-6dB to -10dB).

So, half audio power is -6db = 10 LOG (1/4)

A speaker is a transducer. It changes electrical energy to mechanical energy. That is, electricity in the speaker coil causes a movement in the speaker cone. This is the source of sound.

The voltage is fixed. I cannot alter the silicon or batteries. It doesn't matter anyhow, since cone-type speakers are current driven devices. That means I need to change the amount of current going to the speaker. Putting a resistor in series with the speaker will indeed reduce the current. However, it will have the effect of raising the voltage output the driver is tuned for and could cause the audio to clip. The thing is already low-fidelity.

What I mean about the voltage is that the driver is putting out a signal that fluctuates between two voltages. The center point between the high and low points (peaks) is such that it will not hit the maximum or minimum voltage the driver is capable of producing. That is, the signal will not hit the voltage rails. If I change the resistance on the speaker by making a higher resistance, then the center point of the signal will shift higher as well. The signal might then bounce into the rails and clip the audio signal.

But, because I want the signal's voltage levels to remain fixed where they are, I can take advantage of that to change the current without ever having to measure the voltage or the current.

P = IV, meaning power equals current times voltage.

V = IR, meaning voltage equals current times resistance.

Using a bit of math-fu :

P = V2/R

-6 = 10 LOG (V2/R2) / (V2/R1)


Keep in mind that these voltages are the same value. They cancel. What we really have is a ratio of two resistances: R1 and R2

Simplifying the terms, we have -6dB = 10 LOG R1/R2

We know that -6dB corresponds to a ratio of 1/4. Therefore, the new resistance, looking into the speaker, needs to be 4 times greater in order to achieve a reduction in power of 1/4.

The speaker is 16 ohms. 4 x 16 = 64 ohms total. We need to add 48 ohms in series with the speaker.

Now I need to fiddle around a bit so that the resistance seen by the driver looks like 16 ohms.

Rtotal = R1 || R2

16 = 64 || R2

16 = 1 / (1/64 + 1/R2)

Solving for R2:

R2 = (16*64)/(64-16) = 21.33 ohms.

So I place a ~22 ohm resistor in parallel with the speaker and 48 ohm resistor (in series):

             _____
             \   /
    +--/\/\---|_|--+
    |   48   SPKR  |
    |              |
    |              |
    +-----/\/\-----+
    |      22      |
    |              |
    |              |
    O              O


Solder it all together and you are done!

Isn't this nice? I didn't even need an oscilloscope or meter!

The toy's volume is reduced and the audio fidelity is maintained.

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