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Stereo and mono out the same jack?

J

Jon

Jan 1, 1970
0
I have two piezo pickups. I can handle building the preamps, but then
I want to send the line level signals through a regular 1/4"
instrument/phone cable. It would be really nice if it would send them
out as stereo tip left/ring right with a TRS plug in the jack, and
then automatically mix the two and put the mix out with a TS plug in
the jack (so it mixes if the ring is grounded). Anyone know of a
circuit offhand to do this? I will think about it myself as well...
I am envisioning something similar to a cross coupled output, like
http://www.rane.com/n124fig5.gif
but without the inverting section at the beginning (I can invert just
by switching polarity), and two different signals into the inputs.

Is this the right approach? Maybe there is something blatantly
obvious that I am not thinking of. And cross coupled outputs are
touchy, anyway, so maybe I should not be basing it off of those if
possible.
 
N

N. Thornton

Jan 1, 1970
0
I have two piezo pickups. I can handle building the preamps, but then
I want to send the line level signals through a regular 1/4"
instrument/phone cable. It would be really nice if it would send them
out as stereo tip left/ring right with a TRS plug in the jack, and
then automatically mix the two and put the mix out with a TS plug in
the jack (so it mixes if the ring is grounded). Anyone know of a
circuit offhand to do this? I will think about it myself as well...
I am envisioning something similar to a cross coupled output, like
http://www.rane.com/n124fig5.gif
but without the inverting section at the beginning (I can invert just
by switching polarity), and two different signals into the inputs.

Is this the right approach? Maybe there is something blatantly
obvious that I am not thinking of. And cross coupled outputs are
touchy, anyway, so maybe I should not be basing it off of those if
possible.

I cant say I followed your cross coupled idea. Care to explain?

One simple approach would be to put dc through the groundable ring so
that when a mono plug is inserted a relay switches the tip output to
mono. A cleaner way would be to monitor its Rout, and when it drops to
near zero switch your opamps to mono.

Regards, NT
 
J

Jon

Jan 1, 1970
0
I cant say I followed your cross coupled idea. Care to explain?

Hmmm... Maybe I'm not thinking about this right. I was seeing the
cross coupled output as a way to output the same signal level if
output to a balanced or unbalanced cable, so each is putting out
opposite and equal signals if the cable is balanced, but if you ground
one of the output terminals the other opamp becomes a summing amp and
puts out two times the signal to the other output. Whether that's how
the cross coupled output behaves or not (I will think about it some
more...) that is the idea I was thinking of using. Grounding one of
the outputs automatically turns the other amp into a summing amp which
outputs the summed mono signal out one pin.
A cleaner way would be to monitor its Rout, and when it drops to
near zero switch your opamps to mono.

Yeah, kind of like that, but where 'monitoring' just consists of the
ground altering the feedbacks of the opamps so that they sum the
signals instead.
One simple approach would be to put dc through the groundable ring so
that when a mono plug is inserted a relay switches the tip output to
mono.

I should have mentioned that this is inside an instrument, so the
smaller the better. Relays probably not a good idea. The best method
of all would be one that only uses the preamp opamps, I guess, but
that doesn't seem very feasible. I am basically just using the
circuit at http://www.oz.net/~walterh/oppiezobuf.htm times 2.
 
G

Glenn Gundlach

Jan 1, 1970
0
I have two piezo pickups. I can handle building the preamps, but then
I want to send the line level signals through a regular 1/4"
instrument/phone cable. It would be really nice if it would send them
out as stereo tip left/ring right with a TRS plug in the jack, and
then automatically mix the two and put the mix out with a TS plug in
the jack (so it mixes if the ring is grounded). Anyone know of a
circuit offhand to do this? I will think about it myself as well...
I am envisioning something similar to a cross coupled output, like
http://www.rane.com/n124fig5.gif
but without the inverting section at the beginning (I can invert just
by switching polarity), and two different signals into the inputs.

Is this the right approach? Maybe there is something blatantly
obvious that I am not thinking of. And cross coupled outputs are
touchy, anyway, so maybe I should not be basing it off of those if
possible.

I believe that Rane circuit is a Leunig stage to be a transformerless
line driver. There was a thread last year about that driver having
less than desired balance. I don't believe its what you expect.
GG
 
N

N. Thornton

Jan 1, 1970
0
Hmmm... Maybe I'm not thinking about this right. I was seeing the
cross coupled output as a way to output the same signal level if
output to a balanced or unbalanced cable, so each is putting out
opposite and equal signals if the cable is balanced, but if you ground
one of the output terminals the other opamp becomes a summing amp and
puts out two times the signal to the other output. Whether that's how
the cross coupled output behaves or not (I will think about it some
more...) that is the idea I was thinking of using. Grounding one of
the outputs automatically turns the other amp into a summing amp which
outputs the summed mono signal out one pin.

As far as I can see the cross coupler feeds L into R and R into L, if
in opposite phase, so youd never get straight stereo out, maybe more
like a stereo wide signal out.

I should have mentioned that this is inside an instrument, so the
smaller the better. Relays probably not a good idea. The best method
of all would be one that only uses the preamp opamps, I guess, but
that doesn't seem very feasible. I am basically just using the
circuit at http://www.oz.net/~walterh/oppiezobuf.htm times 2.

A blend switch can use near zero power: the FET that connects the 2
channels has essentially no power consumption, and you can just feed
an extremely small current to one jack ring plus to an opamp input,
plus a Vref to the other opamp input, and the opa will switch state
and turn the fet on or off.

If you used a separate micropower opamp it would eat nothing
noticeable extra. If smallness is the prime issue, maybe use a quad
opamp.


Regards, NT
 
T

Tony

Jan 1, 1970
0
The Rane circuit has been used many times, both in discrete form (eg using two
"B" type SIP resistor packs, so you can vary them to set the gain) and in IC
form (from AD/SSM at least). It does its job well, but you need to watch what
happens with no load (the DC common mode output can easily hit the supply
rails).

In a simple variation on the circuit, the two circuit nodes that are shown as
grounded can be used as a second input, eg for the "-R" signal (where +L is
applied to the main input). This then provides balanced-in, floating
balanced-out - like a transformer.

But not much use for the intended application, which needs something simpler,
eg...

Take the R signal via a buffer and 1k resistor to the Ring terminal.
Feed the Tip terminal with the sum of L+R-Ring. Since R-Ring is zero while Ring
is unloaded, this reverts to the L signal (ie, it's a proper stereo signal), but
if the Ring is grounded, the Tip gets L+R.

I cant say I followed your cross coupled idea. Care to explain?

One simple approach would be to put dc through the groundable ring so
that when a mono plug is inserted a relay switches the tip output to
mono. A cleaner way would be to monitor its Rout, and when it drops to
near zero switch your opamps to mono.

Regards, NT

Tony (remove the "_" to reply by email)
 
T

Tony

Jan 1, 1970
0
The Rane circuit has been used many times, both in discrete form (eg using two
"B" type SIP resistor packs, so you can vary them to set the gain) and in IC
form (from AD/SSM at least). It does its job well, but you need to watch what
happens with no load (the DC common mode output can easily hit the supply
rails).

In a simple variation on the circuit, the two circuit nodes that are shown as
grounded can be used as a second input, eg for the "-R" signal (where +L is
applied to the main input). This then provides balanced-in, floating
balanced-out - like a transformer.

Sorry, edit malfunction; it should have been...
In a simple variation on the circuit, the two circuit nodes that are shown as
grounded can be used as a second input. This then provides balanced-in, floating
balanced-out - like a transformer.

Tony (remove the "_" to reply by email)
 
J

Jon

Jan 1, 1970
0
I believe that Rane circuit is a Leunig stage to be a transformerless
line driver. There was a thread last year about that driver having
less than desired balance. I don't believe its what you expect.
GG

Yes that is what that circuit is. It is not very near to what I was
thinking after all. I thought about it a bit and here is a much
better description of the concept I originally had:


___ ___
Vleft-------------|___|---o----|___|----.
100k | 100k |
| |
___ | |\ |
.-------|___|---o----|-\ | ___
| 100k | | >-----o----|___|----Vtip
| | .-|+/ 1k
| ___ | | |/
| .--|___|---' |
| | 100k ===
| | GND
| |
| '------------------------------------.
| |
| ___ |
| .----|___|----. |
| | 10k | |
| | | |
| ___ | |\ | |
Vright----o-------|___|---o----|-\ | ___ |
10k | >-----o----|___|--o-Vring
.-|+/ 1k
| |/
|
===
GND

Note that this is just an IDEA. Brainstorming. But if I figure
correctly (and I certainly don't sometimes), connect instrument amps
or something to both outputs, and Vleft from the left preamp, Vright
from the right preamp. Vtip will be -(Vleft+Vright-Vright) => -Vleft,
and Vring will just be -Vright. (Overall inversion doesn't matter.
Only bad if one is opposite the other.) If the ring is then grounded,
by a TS plug, the right channel op amp is still driving 1k (is that
big enough?) to ground and Vtip will now be -(Vleft+Vright+0) =>
-(Vleft+Vright), a 50/50 mix of the two, although twice as loud.

Does this look right? What problems would it have? Is there a
simpler way? Could this be merged with the preamp circuit itself, so
there are only two opamps total?

There is a lot I don't know about practical op-amp design. (Which I
am making a list of, to be asked in another post...)

Also, ideally, the mixed output would be (Vleft+Vright)/2, so a mono
or stereo output is at the same overall level, but it's not life or
death.

As far as I can see the cross coupler feeds L into R and R into L, if
in opposite phase, so youd never get straight stereo out, maybe more
like a stereo wide signal out.

Yeah, I was trying to get the basic idea across, of one channel
feeding back into the other and being grounded out. See above.
A blend switch can use near zero power: the FET that connects the 2
channels has essentially no power consumption, and you can just feed
an extremely small current to one jack ring plus to an opamp input,
plus a Vref to the other opamp input, and the opa will switch state
and turn the fet on or off.

That sounds like an even better idea.
If you used a separate micropower opamp it would eat nothing
noticeable extra. If smallness is the prime issue, maybe use a quad
opamp.

Yes, smallness and low power consumption are most important. I don't
understand your blend circuit though. Is the quad op-amp an
alternative to the FET? Or both are being used together?

So two op-amps for the preamps, and one to mix? And one as a
comparator to turn on or off the FET, which is used to switch the
mixed or unmixed signal? Wouldn't that need 2 FETs? I am imagining
an FET transmission gate. This is all very fuzzy...
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Jon <[email protected]>
Note that this is just an IDEA.

No, it's the way, or at least a good way, to do what you want. It's
virtually the same as the solution proposed by Tony Roe Whether 1 k to
ground is OK or not depends on the op-amp. If you use NE5534 (or any one
of several more modern types) it's OK, but 741 and 324 aren't so keen on
1 k (for different reasons). I wouldn't use a 324 for this job anyway.
 
N

N. Thornton

Jan 1, 1970
0
Yes that is what that circuit is. It is not very near to what I was
thinking after all. I thought about it a bit and here is a much
better description of the concept I originally had:


___ ___
Vleft-------------|___|---o----|___|----.
100k | 100k |
| |
___ | |\ |
.-------|___|---o----|-\ | ___
| 100k | | >-----o----|___|----Vtip
| | .-|+/ 1k
| ___ | | |/
| .--|___|---' |
| | 100k ===
| | GND
| |
| '------------------------------------.
| |
| ___ |
| .----|___|----. |
| | 10k | |
| | | |
| ___ | |\ | |
Vright----o-------|___|---o----|-\ | ___ |
10k | >-----o----|___|--o-Vring
.-|+/ 1k
| |/
|
===
GND

Note that this is just an IDEA. Brainstorming. But if I figure
correctly (and I certainly don't sometimes), connect instrument amps
or something to both outputs, and Vleft from the left preamp, Vright
from the right preamp. Vtip will be -(Vleft+Vright-Vright) => -Vleft,
and Vring will just be -Vright. (Overall inversion doesn't matter.
Only bad if one is opposite the other.) If the ring is then grounded,
by a TS plug, the right channel op amp is still driving 1k (is that
big enough?) to ground and Vtip will now be -(Vleft+Vright+0) =>
-(Vleft+Vright), a 50/50 mix of the two, although twice as loud.

Does this look right? What problems would it have? Is there a
simpler way? Could this be merged with the preamp circuit itself, so
there are only two opamps total?

There is a lot I don't know about practical op-amp design. (Which I
am making a list of, to be asked in another post...)

Also, ideally, the mixed output would be (Vleft+Vright)/2, so a mono
or stereo output is at the same overall level, but it's not life or
death.
[email protected] (N. Thornton) wrote in message news:<[email protected]>...
the idea I was thinking of using. Grounding one of


Yeah, I was trying to get the basic idea across, of one channel
feeding back into the other and being grounded out. See above.


That sounds like an even better idea.


Yes, smallness and low power consumption are most important. I don't
understand your blend circuit though. Is the quad op-amp an
alternative to the FET? Or both are being used together?

So two op-amps for the preamps, and one to mix? And one as a
comparator to turn on or off the FET, which is used to switch the
mixed or unmixed signal? Wouldn't that need 2 FETs? I am imagining
an FET transmission gate. This is all very fuzzy...


OK, I like your circuit. The only slight disadvantage is that it will
double its distortion level when running in mono, but thats very small
point if you use a reasonable quality opamp. The other thing is if you
have a patchy connection on the shorted ring this would feed into your
mono sound channel.

What I had in mind, lets see...


|\
left --|-\ ___
| >---|___|----+------- Vtip
--|+/ 5k |
|/ |_|
-------|_
|\ | |
right -|-\ ___ |
| >---|___|----+------- Vring
-|+/ 5k
|/


There we go. With the fet off it works in stereo, with the fet on it
outputs mono. To switch the fet youd need another opamp which detects
the dc level on the jack ring, plus an RZR etc to put a small dc
offset onto the ring. When a mono plug is inserted, this dc offset is
shorted and the opamp switches the fet on, resulting in mono ouput.

The advantage is avoiding distortion, and avoiding possible problems
with dirty ring contacts. But it takes an opamp and a fet. If you
already have 2 opamps in place though this would be less parts than
another 2 opamp cross coupled circuit. If youre designing the whole
lot I'd probably prefer your circuit for lower parts count.


Regards, NT
 
J

Jon

Jan 1, 1970
0
OK, I like your circuit. The only slight disadvantage is that it will
double its distortion level when running in mono, but thats very small
point if you use a reasonable quality opamp. The other thing is if you
have a patchy connection on the shorted ring this would feed into your
mono sound channel.

Ok. When you say distortion you are talking about the overall signal
level and clipping? I don't think that is a problem. I intend to
drop the signal down a bit anyway so that it is more like a guitar
level signal, and so it can handle the battery dropping in voltage
somewhat. (That is how it will behave, right? As the 9V battery dies
it will just reduce the possible output swing of the op-amp (to a
point, of course)?) The shoddy ring connection is a valid concern,
but wouldn't it do the same thing with the FET circuit?

|\
left --|-\ ___
| >---|___|----+------- Vtip
--|+/ 5k |
|/ |_|
-------|_
|\ | |
right -|-\ ___ |
| >---|___|----+------- Vring
-|+/ 5k
|/


There we go. With the fet off it works in stereo, with the fet on it
outputs mono. To switch the fet youd need another opamp which detects
the dc level on the jack ring, plus an RZR etc to put a small dc
offset onto the ring. When a mono plug is inserted, this dc offset is
shorted and the opamp switches the fet on, resulting in mono ouput.

That's a valid way of mixing? Whatever happened to "use a summing amp
to sum voltages! never connect two outputs together!"? clearly
something else I need to learn...

So there are some different possibilities as I see it:

1. Build my type of circuit with the preamp included so it is just two
opamps total. But the current preamp circuit requires high input
impedances (see below), to match the piezo, and my above switching
circuit has 100k. If it is possible to use 10M or so resistors for
the switching circuit, I guess that would work. (but I always see
op-amp circuits with resistors from 1k to 100k and never more or less,
so it must be bad to use those values, right?)

2. Use the preamps and switching circuit separately, which requires 4
op-amps and either two duals or one quad chip.

3. Use the FET switching circuit, which requires 3 op-amps, and hence,
two duals or one quad chip, same as 2.

4. Can the FET be triggered with the dc voltage directly, without the
op-amp? It seems like it could be, with the right voltages in the
right places and the right FET.

Here is my preamp circuit, exactly. I intend to drive either guitar
amps, around 50k inputs, i believe, or sound cards, about 600ohm
inputs (and certainly feel free to tell me if I did something wrong or
need something else to make it better.):


___ |\
.---|___|----o------------|+\ #| ___
| 6.6M | | >---o-----#|----|___|----o
| | .--|-/ | #| 1k
--- Piezo .-. | |/ |
[===]6 Vpp | |2.2M | | 4.7 uF
--- at max | | '---------' Vout
| '-'
| |
| |
'------------o----------------------------------------o
|
| ---(Top section x2)---
|
9V |
+ |
| |
o------------|-------o-------.
| | | |
.-. | | |
| |4.7k | | |
| | | | |
'-' | |\| |\|
| | |-\ |-\
o---------o--' | > | >
| | |+/ |+/
.-. | |/| |/|
| |4.7k --- | |
| | --- | |
'-' | 103 | |
| | | |
o---------o----------o-------'
|
===
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

John Woodgate said:
No, it's the way, or at least a good way, to do what you want. It's
virtually the same as the solution proposed by Tony Roe Whether 1 k to
ground is OK or not depends on the op-amp. If you use NE5534 (or any one
of several more modern types) it's OK, but 741 and 324 aren't so keen on
1 k (for different reasons). I wouldn't use a 324 for this job anyway.

I am using a tl062 right now because that is what i had in my parts
box. I need to order an unrelated chip (and probably this circuit's
output jack) from a catalog, so i should get some other things while
I'm at it. ne5532s are pretty good quality? I see them used in audio
a lot. Are the more expensive opa134 burr brown type chips really
worth it?

Oh wait... *Looks through box* I guess i do have some 5532s, and
072s, 741s, and another 062, and one each lmc60 and lf412. I remember
having problems with the lf412 and lmc60 in the older version of the
circuit. Which would you use?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Jon <[email protected]>
wrote (in <[email protected]>) about
'Stereo and mono out the same jack?', on Mon, 16 Feb 2004:

[big snip]
That's a valid way of mixing? Whatever happened to "use a summing amp
to sum voltages! never connect two outputs together!"? clearly
something else I need to learn...

Look at the 5 kohm resistors: they prevent the two outputs 'seeing' each
other.
So there are some different possibilities as I see it:

1. Build my type of circuit with the preamp included so it is just two
opamps total.

Yes, that's what I recommend.
But the current preamp circuit requires high input
impedances (see below), to match the piezo, and my above switching
circuit has 100k. If it is possible to use 10M or so resistors for
the switching circuit, I guess that would work. (but I always see
op-amp circuits with resistors from 1k to 100k and never more or less,
so it must be bad to use those values, right?)

It can be. Even your 2.2 Mohm is pushing your luck on offset current.
[snip]
Here is my preamp circuit, exactly. I intend to drive either guitar
amps, around 50k inputs, i believe, or sound cards, about 600ohm
inputs (and certainly feel free to tell me if I did something wrong or
need something else to make it better.):

Well, now, you've got 6.6 Mohm, which isn't a standard value (6.8 Mohm
is). In that position, you can have as many megohms as you like. But the
2.2 Mohm is too big. The way to overcome that is to use a 'bootstrap'
connection. You apply the output signal of the op-amp, or a fraction of
it, to the ground end of the resistor, which greatly increases its
effective value at signal frequencies.
___ |\ 4.7 uF
.---|___|----o------------|+\ #| ___
| 6.8M | | >---o---o-#|----|___|----o
| | .--|-/ | | #| 1k
--- Piezo .-. | |/ | = 470 nF
[===]6 Vpp | |220k | | |
--- at max | | '--220k --' | Vout
| '-' 1k
| +-----------------------+
| 10k
'------------o-----------------------o----------------o
[rest snipped]

The 220 k in series with the - input is to keep the impedances at the
two inputs nearly equal, minimising current offset. The 220 k at the +
input is bootstrapped with 0.9 times the output voltage, which is also
0.9 times the input voltage, so appears at signal frequencies to be 2.2
Mohm.
[snip]

I am using a tl062 right now because that is what i had in my parts
box.

It's a rather low-power device, not capable of driving loads below
several kohms.
I need to order an unrelated chip (and probably this circuit's
output jack) from a catalog, so i should get some other things while
I'm at it. ne5532s are pretty good quality? I see them used in audio
a lot.

Yes, but they are not unity-gain stable; you can't use them as
followers.
Are the more expensive opa134 burr brown type chips really
worth it?

IMHO, not for your application.
Oh wait... *Looks through box* I guess i do have some 5532s, and
072s, 741s, and another 062, and one each lmc60 and lf412. I remember
having problems with the lf412 and lmc60 in the older version of the
circuit. Which would you use?

TL072s are quite good. I don't know the LMC60 at all. The LF 412 is
good, and, although the data sheet doesn't seem to say so explicitly, is
unity-gain stable, as shown by its use as a follower in the 'sample and
hold' application circuit.
 
J

Jon

Jan 1, 1970
0
John Woodgate said:
It can be. Even your 2.2 Mohm is pushing your luck on offset current.

Ok. First, what problems would the offset current cause?

Second, if i am combining the two circuits they will need to be
inverting amplifiers so i can sum them and also to get -Vright.
That's what i meant by input impedance. The inverting amps as i drew
them had 100k inputs, which is not high enough for the piezos.

as described here: http://www.amp.com/products/technology/articles/interface.stm

so i need an inverting (summing) amp configuration that will interface
with piezo. does such a thing exist? it looks like the inverting amp
can be configured as in figure 9 on that site, but can it sum?

here is the theoretical combined circuit with super big resistors,
which i was talking about when i said that resistors this big are not
"allowed":


___ ___
.-----o---------|___|--o----|___|--.
| | 10M | 10M |
| .-.10M | |
--- | | ___ | |\ | 4.7uF
|===| | | .----|___|--o----|-\ | ___ #| Tip
--- '-' | 10M | | >---o--|___|---#|-----.
| | | | .-|+/ 1k #| |
| | | ___ | | |/ |
=== === | .-|___|--' | |
GND GND | | 10M === |__ .-.
| | GND V | |
| '-----------------------------------. | |
| ___ | ____A | |
| .----|___|--. | | '-'
| | 10M | | | |
| | | | | ===
| ___ | |\ | 4.7uF | | GND
.----------o----|___|--o----|-\ | ___ #| | |
| 10M | >---o--|___|---#|--o--'
| .-|+/ 1k #| Ring
--- | |/
|===| |
--- ===
| GND
|
===
GND


maybe this can be bootstrapped, too? i don't see how...
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Jon <[email protected]>
Ok. First, what problems would the offset current cause?

If the d.c. impedances between the inputs and ground are not equal, both
bias and offset currents will cause d.c. offset voltage at the output,
perhaps affecting coupling to the next stage and restricting output
voltage swing.
Second, if i am combining the two circuits they will need to be
inverting amplifiers so i can sum them and also to get -Vright. That's
what i meant by input impedance. The inverting amps as i drew them had
100k inputs, which is not high enough for the piezos.

as described here: http://www.amp.com/products/technology/articles/inter
face.stm

Ah, well, you've now changed the question.
so i need an inverting (summing) amp configuration that will interface
with piezo. does such a thing exist? it looks like the inverting amp
can be configured as in figure 9 on that site, but can it sum?

Yes: the normal summing amp IS inverting. Indeed, the virtual earth
point (the - input) is also called 'the summing junction'.

Figure 9 shows a 'capacitive feedback' inverter, or 'charge amplifier'.
It is particularly useful for capacitive sources, and has much better
noise performance than a solution using high-value resistors.
here is the theoretical combined circuit with super big resistors, which
i was talking about when i said that resistors this big are not
"allowed":
[snip]


maybe this can be bootstrapped, too? i don't see how...

You wouldn't NEED to bootstrap 10 Mohm resistors. But you would have big
problems with d.c. offset.
 
N

N. Thornton

Jan 1, 1970
0
[email protected] (N. Thornton) wrote in message

Ok. When you say distortion you are talking about the overall signal
level and clipping?

no, distortion, THD. No amp is perfect and the cross coupled circuit
will increase it. But I dont think you need worry about it, it will
still behave well.

I don't think that is a problem. I intend to
drop the signal down a bit anyway so that it is more like a guitar
level signal, and so it can handle the battery dropping in voltage
somewhat. (That is how it will behave, right? As the 9V battery dies
it will just reduce the possible output swing of the op-amp (to a
point, of course)?)

yes, down to a point. As long as the opamps still in spec, and the
battery can deliver the required i_out.

The shoddy ring connection is a valid concern,
but wouldn't it do the same thing with the FET circuit?

no, a small cap would be used to prevent switching at a.f.


That's a valid way of mixing? Whatever happened to "use a summing amp
to sum voltages! never connect two outputs together!"? clearly
something else I need to learn...

So there are some different possibilities as I see it:

1. Build my type of circuit with the preamp included so it is just two
opamps total.

easiest option to build.
3. Use the FET switching circuit, which requires 3 op-amps, and hence,
two duals or one quad chip, same as 2.

easy conceptually
4. Can the FET be triggered with the dc voltage directly, without the
op-amp? It seems like it could be, with the right voltages in the
right places and the right FET.

yes, the only downside is going to be a large voltage thump when you
plug stuff in, so that wouldnt be great practice, but it should work.
You can do better though.

Here is my preamp circuit, exactly. I intend to drive either guitar
amps, around 50k inputs, i believe, or sound cards, about 600ohm
inputs (and certainly feel free to tell me if I did something wrong or
need something else to make it better.):


___ |\
.---|___|----o------------|+\ #| ___
| 6.6M | | >---o-----#|----|___|----o
| | .--|-/ | #| 1k
--- Piezo .-. | |/ |
[===]6 Vpp | |2.2M | | 4.7 uF
--- at max | | '---------' Vout
| '-'
| |
| |
'------------o----------------------------------------o

John has covered the impedances. I'd add something else: a 1M resistor
from output terminal to ground. It bleeds the dc off your output cap,
so you dont get an ear bashing woofer bending thump when you plug
things in.


Regards, NT
 
J

Jon

Jan 1, 1970
0
no, distortion, THD. No amp is perfect and the cross coupled circuit
will increase it. But I dont think you need worry about it, it will
still behave well.

yeah. that is a negligible problem for this circuit.
yes, down to a point. As long as the opamps still in spec, and the
battery can deliver the required i_out.

ok. that's what i thought.
no, a small cap would be used to prevent switching at a.f.
ahh.

John has covered the impedances. I'd add something else: a 1M resistor
from output terminal to ground. It bleeds the dc off your output cap,
so you dont get an ear bashing woofer bending thump when you plug
things in.

ok. i will add that. in fact, i will use that for other appropriate
circuits from now on...


John Woodgate wrote in message
If the d.c. impedances between the inputs and ground are not equal, both
bias and offset currents will cause d.c. offset voltage at the output,
perhaps affecting coupling to the next stage and restricting output
voltage swing.

ah. it's not going to be a lot, though, is it? and putting an equal
resistor in the feedback loop (of the voltage follower) would make it
somewhat better? i don't think a small dc offset is going to affect
this type of audio signal in an important way. It gets a little bit
of slowly changing DC offsets from the piezo already.
Ah, well, you've now changed the question.

of course. :)
Figure 9 shows a 'capacitive feedback' inverter, or 'charge amplifier'.
It is particularly useful for capacitive sources, and has much better
noise performance than a solution using high-value resistors.

well, piezo sensors are modeled as purely capacitive sources, so that
would theoretically be the best, however, i don't see a summing
function being built with one. especially one that will sum from two
piezos *and* an op-amp output.
You wouldn't NEED to bootstrap 10 Mohm resistors. But you would have big
problems with d.c. offset.

Is there a better way to get the same functionality and the same input
impedances?

what's that you say? a toggle switch? hmm... that would solve the
dirty ring problem too. and could also have a 'panic' setting to
output the piezos directly into whatever, in case the battery dies. i
used to play it like that. it doesn't sound great, but it would be
useful in a tight spot. hmm...
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Jon <[email protected]>
ah. it's not going to be a lot, though, is it?


It can be: it depends on a number of factors, so the simplest thing is
not to let it happen.
and putting an equal
resistor in the feedback loop (of the voltage follower) would make it
somewhat better?

A lot better, if the resistor is hundreds of kohms.
i don't think a small dc offset is going to affect
this type of audio signal in an important way.

It might NOT be small.
It gets a little bit of
slowly changing DC offsets from the piezo already.

I don't see how: the piezo is a capacitor.
of course. :)


well, piezo sensors are modeled as purely capacitive sources, so that
would theoretically be the best, however, i don't see a summing function
being built with one. especially one that will sum from two piezos
*and* an op-amp output.

No problem whatsoever! A conventional inverting summing amplifier has a
resistor in series with each input and a feedback resistor. A summing
amplifier for capacitive sources has capacitors at the inputs and a
feedback capacitor. You make the op-amp output look like a capacitor by
simply putting a capacitor in series with it (you may need 100 ohms as
well, to improve op-amp stability). For the same sensitivity, use the
same value capacitance as that of your piezos. You can adjust the
sensitivity by varying the capacitor value.
 
J

Jon

Jan 1, 1970
0
John Woodgate said:
I don't see how: the piezo is a capacitor.

"slowly changing" being the key words. if i plug the piezo directly
into the op amp input, noninverting, it will create a dc offset when a
constant pressure is applied. sure, it leaks through the input
impedance slowly, and drops back to zero, so it is technically just
very low frequency, but it stays around for quite a while, and since
this is under a violin bridge, which has different constant pressures
at different times, it was initially causing problems because the
signal would float up and down past the opamp's rails, without the 10M
resistor to ground.
No problem whatsoever! A conventional inverting summing amplifier has a
resistor in series with each input and a feedback resistor. A summing
amplifier for capacitive sources has capacitors at the inputs and a
feedback capacitor. You make the op-amp output look like a capacitor by
simply putting a capacitor in series with it (you may need 100 ohms as
well, to improve op-amp stability). For the same sensitivity, use the
same value capacitance as that of your piezos. You can adjust the
sensitivity by varying the capacitor value.

ooh! that is how i will do it, then. i have never tried the charge
amplifier before but i will try it out soon and report on my success.
since i don't know the piezo's capacitance, just play with the values
until i get a voltage i like? (could i reliably measure it with an
LCR meter?) and the feedback resistor will now determine how quickly
the dc offset mentioned above drops off?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Jon <[email protected]>
(could i reliably measure it with an LCR
meter?)

Yes. Don't make a noise, though, because it's a microphone, of course.
Probably best measured when detached from the instrument.
and the feedback resistor will now determine how quickly the dc
offset mentioned above drops off?

Yes. The ratio of feedback cap to piezo capacitance determines the gain.
Then the low-frequency -3 dB frequency is given by 1/[2 pi{(feedback cap
x gain) + piezo capacitance} x feedback R}].

If you have Spice, it's worth simulating your circuit after you've
designed it, because it does interesting things at high frequencies
(usually well above 20 kHz).
 
N

N. Thornton

Jan 1, 1970
0
John Woodgate said:
I read in sci.electronics.design that Jon <[email protected]>
and the feedback resistor will now determine how quickly the dc
offset mentioned above drops off?

Yes. The ratio of feedback cap to piezo capacitance determines the gain.
Then the low-frequency -3 dB frequency is given by 1/[2 pi{(feedback cap
x gain) + piezo capacitance} x feedback R}].

Suggest setting your low end response to whatever the minimum a guitar
can put out is, so the dc offset cant float around.


Regards, NT
 
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