Please help me understand this amp circuit

The amplifier is a speaker dock for MP3 players. It's just a cheap single-chip amplifier with two cheesey speakers from Ebay, and is of no real consequence, but I would like to understand why the designer may have made certain changes from the example schematic shown in the datasheet, and what the consequences would be of changing back to the datasheet version. The as-built amp produces very little output volume, and appears to be capable of much more, and I'm curious why they elected not to let it do that.

The datasheet and a circuit drawing of the amp as received are here:

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It appears the designer went way out of his way to reduce the input signal and the amp gain. I don't really know why. The actual circuit differs from the drawing in the datasheet in three ways:

  1. The input signal first goes through a 50K pot to ground, and the wiper off that then goes through another divider circuit of 39K/4.7K to ground, with the chip's input pin connected at the junction. I really don't understand this at all. I even wonder if the 39K resistor was intended to be 3.9K. (It is indeed 39K, on both channels.) In any event, I don't see why either of these resistors need to be there. It appears to be the Homoepathic version of the input signal - diluted down so much that there's hardly anything left. I tried jumpering around the 39K, and got nice loud volume out of the speakers.

  1. The resistor labeled Rf in my drawing is the feedback resistor. It's referred to in the datasheet as reducing amp gain. Again, it's not clear this is really needed, but I assume the intent is to reduce distortion.

  2. Capacitor C8 in the datasheet is not present in the actual circuit. I don't know what this does. I don't know what "bootstrap" does here at all.

Obviously I'm not an EE, and analog isn't exactly my strong point. Well, at least it's not RF. But I would like to understand this if I can. Any help would be appreciated.

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** Your schem and the datasheet are both too small to be readable.
** The datasheet tells you - try reading it.
** Bootstrapping in an audio amplifier is a way of improving the available voltage swing from the driver stage and hence improves overall linearity. Usually, the load resistor for the class A driver stage is split into two and the mid point connected to the speaker output via an electro capacitor.

..... Phil

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Phil Allison

What is the source that gives the input signal ? In some cases it may be that the input amplitude voltage is high enough that it should be reduced to avoid distortion.

This resistor fixes the gain : G = 10K / (50 + Rf) Have a look at the data sheet, there are equations for the gain that includes Rf.

C8 at the output is a first-order low-pass (high-cut) to avoid oscillations of the AOP at high frequencies when the gain is high.

I Hope This Helps.

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This amp just plugs into the earphone jack of an MP3 player such as an iPod, Sansa, or whatever. So it's not really a typical line input. I wouldn't think the voltage would be all that high, but in any case, the pot is there to take care of that if it starts to clip. I'm testing it with a Sandisk Clip player, and at maximum volume on both the player and the amp, it's still not very loud at all. So I think I'm going to try playing with the 39K and 4.7K resistors and see what effect they have.

Yes, I found it, but didn't understand what "JWC1" is. Here's the formula:


R1 _____________________________

1 Rf + R2 + __________


Anyway, if R2 is 50 ohms, then a Rf of 100 ohms will cut the voltage gain by as much as two-thirds, depending on what JWC1 is.

Ok, so maybe they left it out because the gain is low. And I probably out to add it back if I increase the gain.

Yes it does. Thanks very much.

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Yes : adjust these resistors with the pot to maximum, to get the max volume you need for your purpose.

"j" is the symbol for imaginary, i.e sqrt(-1) "w" is omega (e.i 2*PI*F) and F is the frequency "C1" is the value of the condensator C1 in Farad.

Then the expression (1 / j*w*C1) is the value of the resistor that is equivalent to C1 at frequency F. The value of C1 should be such that its equivalent resistance is low in regard to the serial resistor at minimum frequency.

Two-thirds of what ? Just use: G = 10K / (50 + Rf)

Sure, maybe you won't even need it, but keep in mind that for a high gain, if the AOP oscillates you may not notice it at all, because the frequency could be above 20 KHz.

Don't mention it.

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Well, I just meant that with Rf = zero as shown in the datasheet circuit, the gain is 200, but if Rf is 100 ohms as in my amp, the gain is 67, ignoring in both cases the effect of C1. I'm just saying that the 100 ohms makes a big difference. I assume this was done because the amp produces too much distortion when operated at maximum possible gain. If that's the case, then I'm better off if I can get where I need to be solely by adjusting the input resistors and leaving Rf alone.

Thanks again for your help.

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