Creating line-level output from speaker output

I have this electronic talking/musical educational toy that only has a
built-in speaker for audio output, but I'd like a line or headphone
output for it as well.

So I did the logical thing: disconnected one of the wires to the
speaker, connected it through a mini-jack's "switch", then connected the
other output wire to the 3rd pin of the jack. In other words: when a
jack plug is connected to the device, the speaker is cut off. If the
jack plug is disconnected, audio is heard through the speaker.

So far so good, but the signal is way to loud for a headphone and
line-levels, so I experimented with a trimpot to see which value I would
need to get a decent level and ended up with approx 30K Ohms which I
found as a regular resistor and connected this between one of the signal
wires and the jack-plug.

OK, the line level has been taken care of, but the sound is crap!!!
It sounds like almost all high-level audio signals are filtered out. The
toy isn't exactly hi-fi I'm sure, but it's got to be better than this!
So am I doing something wrong? If so, what should I do to get a decent
line or headphone level output when the device only has a speaker?

 

Two possibilities spring to mind:
1. By using such a large series-resistance, you've inadvertently formed a
low-pass RC filter
2. The audio output stage can't handle such a large output resistance

The amplifier is designed to work into a low-impedance: typically 8 ohms

Try a potential divider like this [View in a fixed-pitch font]:


In o-----,
|
.-.
| | R1
| |
'-'
|
+----o Out
|
.-.
| | R2
| |
'-'
|
GND o----+----o GND


Check the loudspeaker impedance Z.

Aim for R1+R2 in the same ballpark as Z (a bit higher is probably OK).

R2 / (R1+R2) sets the attenuation.

 

You have probably connected your trimpot in series with the signal,
that is the reason why it doesn't work as it should.

Use a 100 Ohm pot instead, connect the input signal between the ends,
and the output signal between one end and the wiper. That is how you
connect a volume control.

If you dont have exactly a 100 Ohm pot you can use values between 50
and 1000 Ohm.

 

Just put two 1/4W metal oxide resistors of 100r and 16r in series across amp
o/p, 16r being the one to common terminal, and arrange to send to line
across the 16r.
This will be at 1/7 of amp level, about -19dB of amp o/p voltage.
If now too low, double that 16r to 33r - with about 12dB voltage loss.
Bandwidth will be "flat" at receiving end when there is *no* l/s in use.
But the loudspeaker /network load behaviour will back-influence the amp's
damping properties with the non-linear shunt - with some introduced spectrum
colouration.

 

Yes, I measured the speaker to 8 Ohms.

So the two resistors should (ideally) be 4 Ohms then, but since there
are no 4 Ohm resistors easily available, could I use something more
common like two 3.3 Ohm or 4.7 Ohm resistors? Of the two, it sounds like
you're saying 4.7 Ohms would be better.

Your diagram helps, but I'm still a little confused.
Are you saying that I should leave one of the existing output pins
unused, while the other one is connected to R1 etc., creating one of the
new line-level output pins, while ground (- of the battery connector)
should be connected to the other line-level output pin?

I assumed that one of the speaker pins would already be connected
directly to - of the batteries, but that's not the case.

Here's what I'm talking about as a picture explains things so much
better (as for the speaker, I think it'll make things a whole lot easier
if I disconnect it altogether for now).

o-----,
existing |
audio o .-.
output | | R1 (4.7 Ohms)
| |
'-'
|
+------------+
| |
.-. |
| | R2 (4.7 |
| | Ohms) +----o
'-' New line-level output
|
GND o------------+-----------------o

 

Oh. I also assumed that. It's a little trickier, if the speaker isn't
grounded. You might try connecting a resistor (4.7, 5.6, 8.2 or even
higher) between the output terminals, and then use a capacitvely
coupled voltage divider with somewhat larger resistors (an order of
magnitude).

If that doesn't work, you'll have to use a transformer.

 

Oops. That should be: 8.2, 10, 12 ohms or higher.

 

I haven't quite managed to work it out, but there's this other talking
toy *with* a headphone socket, only that it's way too loud for a line
output (the computer used for sampling it has the input level way down
and it's still too loud), so I figured out by opening up the toy that
there was a resistor along the ground line of the headphone socket.

I can't remember the value, but it was pretty low. Under 1 K Ohm or
something. I put a 30 K Ohm resistor in place of it and audio output is
more in the acceptable level.
I found out something very interesting regarding audio quality...
The voice doesn't have much of an high-end. It's mostly low frequencies,
but playing around I figured that by placing a capacitor alongside (in
parallel with the resistor) I suddenly got a crispier, more high-end
voice, just as if I was to turn up the treble!

Now, I don't know much about electronics theory, but like to build stuff
and so on, so I don't know what's "right" or "wrong" here, but to me,
the audio quality suddenly became a lot better and easier to comprehend,
so can someone tell me why the producers of the (educational) toy didn't
bother to add a capacitor like this in the design?
Is there a downside to this that I haven't noticed yet?
Could I end up damaging the device by modifying it like this?

 

It's doubtful that you'll ruin anything playing around like this. What
you've
done with the cap is to allow the resistance to appear smaller as the
frequency
goes up. The reactance of the cap is 1/(2*PI*F*C)...that reactance (in
Ohms)
is in parallel with the resistor. The effective resistance will be less as
the
frequency goes up so more amplitude at higher frequencies. You have given
a slope to the frequency response.

This was not done originally because it costs! Cheap is the action word.

 

Thanks for explaining!
Does this mean that the more amplitude higher frequencies get, the more
lower frequencies have their amplitude lowered? Or are the lower
frequencies unaffected while the higher ones are just raised? (higher
capacitance means more amplitude to higher frequencies I think is what I
figured out).

What companies do to save a few cents....
And what little I have to spend in order to improve the toy! ;-)

 

As the frequency goes down, the caps impedance goes up. Looking at the
Xc equation, you will see that there is an inverse relationship. Well,
anyway,
the cap finally appears to be an open circuit and that would leave only the
resistor to modify the amplitude rather than the resistor in parallel with
Xc.

You can get the reverse effect by placing the cap between the signal source
and ground. This would be called a low pass filter. The low frequencies
would be passed more so than high frequencies....(high frequencies would
be shunted to ground)

 

That's a good explanation! Might come in handy some time I need to lower
the high frequencies.

 

Wow! Not really knowing what I was doing and not making sense of the
circuitry (not having a common grounding for the speaker output among
other things) I experimented with a lot of things, among others finding
an 8 Ohm resistor and placing it across the speaker output terminals
-and it helped!!! The sound actually became a little better.
But I can see now that the sound source in itself wasn't very good, and
you can't improve on what's bad to start with ;-)