audio switch circuit, when headphones plugged in?

[Gary Funck]

(I know about switched (lifted) jacks, but they won't work
in my case because I don't want to modify the existing jack.
Please read on. <g>)

Recently, I added one of those front panel "hubs" that fits in
the 5.5 in. bay and places USB, firewire, card readers so they're
accessible on the front of the PC. The device I installed is from
Logisys:

http://logisyscomputer.com/viewsku.asp?SKUID=FP808SL

This panel exports headphone audio via a mini-jack. I'd like to plug
my headphones into that jack, and have it cut the output to the speakers.
I have two similar systems equipped with an EPOX 8K9A2+ mother board.
In one system, I'm using the on board AC97 audio, and in the other
case an Audigy 2 sound card. Both sound cards behave similarly.
They have a stereo line out which usually is plugged into amplified
PC speakers, but also allows a nominal 30 ohm headphone.

What I'd like to do is build a widget that will connect the headphones
to the line out when I plug the headphones into the multi-function
front panel, and have the line out go to the amplified speakers
otherwise. I can't easily use a switched mini-jack phone plug
because that would require reworking the front panel.


Here's what I think this switch looks like:

--- sound card line out ---->

swtich <---> headphone --<
panel
<--- speakers amp line in ---<

Note that the sound card line out, and speakers line in would
pass through back of the PC via a PCI card slot retainer that
has a hole in it (this came with the front panel). I'm thinking
I'd put the switch somewhere inside the PC in small metal project
box. It can get power from the PC's supply (either 5v. or 12 v.).
The project box would have three mini-jacks and a power connector
of some sort.

I think I can do this with a DPDT relay which would close if
it sees a path to ground via the head phone jack, when the
jack is plugged into the front panel. But I was thinking that
a digital switch might be also possible. I was looking at the
Analog Devices SSM2404 Quad Audio Switch or equivalent.
What I'm uncertain of is how I'd convert the "path to ground"
that occurs when the headphones are plugged into the front
panel into a logic signal that would drive the Quad Switch
(or the relay for that matter). I'm assuming, but don't know,
that just letting the ground float on the signal input to the
switch is not a good idea as a way of asserting logic zero
(ie, to de-assert the switch enable signal), but I don't know
enough about digital circuits to know one way/other.
The other issue that I'm wondering about, is how the builtin
amp in the PC's sound card behaves when it sees a headphone
load versus the line in load of the PC speakers' amplifier?
I think the headphone load is typically in the 30 ohm range,
but don't know what the input impedence of the speaker amp.
would be. What does the output circuit look like on the PC sound
card? How does it detect whether it is driving headphones
or the speakers? Headphones require some amplification above
the line out level, correct?

Also, I'm wondering is if the 35 ohm load imposed by the
digital switch (as high as 300 ohms in other bilateral switches)
would affect the load seen by the sound card such that it
doesn't drive the headphones properly?

 

[Ban]

Gary Funck wrote:
snip >

Note that the sound card line out, and speakers line in would
pass through back of the PC via a PCI card slot retainer that
has a hole in it (this came with the front panel). I'm thinking
I'd put the switch somewhere inside the PC in small metal project
box. It can get power from the PC's supply (either 5v. or 12 v.).
The project box would have three mini-jacks and a power connector
of some sort.

I think I can do this with a DPDT relay which would close if
it sees a path to ground via the head phone jack, when the
jack is plugged into the front panel. But I was thinking that
a digital switch might be also possible. I was looking at the
Analog Devices SSM2404 Quad Audio Switch or equivalent.
What I'm uncertain of is how I'd convert the "path to ground"
that occurs when the headphones are plugged into the front
panel into a logic signal that would drive the Quad Switch
(or the relay for that matter). I'm assuming, but don't know,
that just letting the ground float on the signal input to the
switch is not a good idea as a way of asserting logic zero
(ie, to de-assert the switch enable signal), but I don't know
enough about digital circuits to know one way/other.
The other issue that I'm wondering about, is how the builtin
amp in the PC's sound card behaves when it sees a headphone
load versus the line in load of the PC speakers' amplifier?
I think the headphone load is typically in the 30 ohm range,
but don't know what the input impedence of the speaker amp.
would be. What does the output circuit look like on the PC sound
card? How does it detect whether it is driving headphones
or the speakers? Headphones require some amplification above
the line out level, correct?

Also, I'm wondering is if the 35 ohm load imposed by the
digital switch (as high as 300 ohms in other bilateral switches)
would affect the load seen by the sound card such that it
doesn't drive the headphones properly?

As you have already stated, how does the relais know if headphones are
connected? Some phones have 600R others only 25R. They do not need
amplification either. And the amp usually doesn't mind if there is any load
connected.
The best would be to check out if the built-in mini-jacks do have switches
built-in. Most of them do, even if they are not used. You could then make a
bridge from the switch O/P to the shielded cable to the speakers.
What you want to do is unfortunately impossible. If there are no built in
switches, forget the whole thing and do it manually with a DPDT switch.

 

[Gary Funck]

[...]

As you have already stated, how does the relais know if headphones are
connected? Some phones have 600R others only 25R. They do not need
amplification either. And the amp usually doesn't mind if there is any load
connected.

Well ... when the headphones are plugged into the jack at the front panel,
there will be a closed circuit between both the L and R circuits and the
common. As so:

+----------+
| |
V V

L C

where L = Left and C = Common (Ground) for example. If we can impose a
detector circuit, that impose a very small DC current through L and C above,
we can detect that the phones have been plugged in. If this DC current can
be converted into a logic level voltage, we can than trigger the CMOS switch
(or a relay). I'd have some questions about how we'd isolate our detector
circuit, which would go wiggy once the 1 volt at 100 mA of A/C audio starts
heading back towards the earphones.

Maybe we could work it the other way ... we could always allow a
signal to flow to the headphone circuit. If the headphones are plugged in,
a current will flow. We (somehow) convert that A/C current into a logic
level, which switches out the speakers. Briefly (50 ms. or less) we'd drive
both the speakers and the headphones until our switch disabled the
connection
to the speakers. Hopefully the amplifier circuit in the sound card won't
care
about briefly seeing both loads in parallel?

I'd be a little concerned that there might be a higher current/voltage sent
to speakers
when we remove the headphones before the sound card amplifier has had time
to
adjust to speakers only ...

The best would be to check out if the built-in mini-jacks do have switches
built-in. Most of them do, even if they are not used.

Thanks. Will do that. It certaionly would be easier. Is some sort
"de-popper"
circuit needed to keep the insertion/removal of the headphone jack from
causing
a loud noise in the loud speakers?

 

[Ban]

[...]

As you have already stated, how does the relais know if headphones
are connected? Some phones have 600R others only 25R. They do not
need amplification either. And the amp usually doesn't mind if there
is any load connected.

Well ... when the headphones are plugged into the jack at the front
panel, there will be a closed circuit between both the L and R
circuits and the common. As so:

+----------+

|

V V

L C

where L = Left and C = Common (Ground) for example. If we can impose
a detector circuit, that impose a very small DC current through L and
C above, we can detect that the phones have been plugged in. If this
DC current can be converted into a logic level voltage, we can than
trigger the CMOS switch (or a relay). I'd have some questions about
how we'd isolate our detector circuit, which would go wiggy once the
1 volt at 100 mA of A/C audio starts heading back towards the
earphones.

Too complicated, a dc current will also degrade headphones performance.

Maybe we could work it the other way ... we could always allow a
signal to flow to the headphone circuit. If the headphones are
plugged in, a current will flow. We (somehow) convert that A/C
current into a logic level, which switches out the speakers. Briefly
(50 ms. or less) we'd drive both the speakers and the headphones
until our switch disabled the connection
to the speakers. Hopefully the amplifier circuit in the sound card
won't care
about briefly seeing both loads in parallel?


I'd be a little concerned that there might be a higher
current/voltage sent to speakers
when we remove the headphones before the sound card amplifier has had
time to
adjust to speakers only ...

No need to be concerned, there will be neither a "plop" when the phones are
inserted. As I told you already, the soundcard output doesn't care too much
what is plugged in. Maybe the overall level drops a bit, depending on the
phones, but when the soundcard is able to drive the phones alone loud
enough, the speakers can also be operated in parallel without any problem.