Please help me understand this amp circuit

[Peabody]

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:

http://drop.io/wnkf425dd

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.

2. 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.

3. 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.

 

[Phil Allison]

"Peabody"

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

http://drop.io/wnkf425dd

** Your schem and the datasheet are both too small to be readable.

3. Capacitor C8 in the datasheet is not present in the
actual circuit. I don't know what this does.

** The datasheet tells you - try reading it.

I don't know what "bootstrap" does here at all.

** 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

 

[[email protected]]

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.

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.

2. 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.

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.

3. 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.

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.

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.

I Hope This Helps.

 

[Peabody]

[email protected] says...

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 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.

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.

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

VOUT/VIN =

R1
_____________________________

1
Rf + R2 + __________

JWC1



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.

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.

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.

I Hope This Helps.

Yes it does. Thanks very much.

 

[Jean-Christophe]

[email protected] says...


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

Yes, I found it, but didn't understand what "JWC1" is.

"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.

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.

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

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.

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.

Yes it does. Thanks very much.

Don't mention it.

 

[Peabody]

Jean-Christophe says...

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

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.