Driver to drive?

[Boris Mohar]

On Sun, 26 Feb 2017 14:59:54 -0600, oldschool@tubes.com wrote:

On Sat, 25 Feb 2017 07:43:21 -0500, Boris Mohar
borism_void_@sympatico.ca> wrote:

On Fri, 24 Feb 2017 21:07:50 -0600, oldschool@tubes.com wrote:

I have a small amplifier (about 15W per channel), that I use for my
computer to listen to music. I have two speakers connected to it, which
are bookshelf sized speakers from an old stereo. It works well, but
lacks bass. I just got a sub-woofer which came from some sort of home
theater system. I want to connect this to what I already have.

However, there is only ONE sub-woofer. How can I get both channels to
work in this sub-woofer without combineing the channels to mono? Maybe I
need a separate amplifier for this sub-woofer, which I could probably
do, but then I have to feed the input of that amp with the signal from
both channels.

I see commercial systems which are stereo, which only have one
sub-woofer, so there has to be some means to do this.



http://www.brainfartz.com/images/Stereo%20Stuff/VTA%20ST-120/st120-build/dynaco-center-channel.jpg



Thats a very bizarre way of connecting speakers. I question whether
there will still be stereo sound, or just some sort of mono????

I had my speakers hooked up this way. There will be full stereo. Just try
it with what you have. It is an elegant solution.



Regards,

Boris Mohar

Got Knock? - see:
Viatrack Printed Circuit Designs (among other things) http://www.viatrack.ca

void _-void-_ in the obvious place



---
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

 

[dcaster@krl.org]

On Sunday, February 26, 2017 at 4:00:44 PM UTC-5, olds...@tubes.com wrote:

I actually think I am going to go about this in a different manner. The
plan is to use a separate amplifier for the sub-woofer.

If you are going to use a separate amp for the woofer then it is an easy solution. Or at least I think so. Have not tried this.

I would connect a 1 kohm resister to each of the stereo amp outputs. Connect the other end of the resistors together. and another 1 k ohm to that junction to signal ground. So the signal from stereo amp a would go thru one 1 k resistor and thru another 1 k resistor to ground. So half the signal would be fed into the woofer amp. THis signal would back feed to the stereo amp B , but be seriously attenuated by the divider of the 1 k ohm and the output impedance of the stereo amp B. So the signal from amp A would hardly appear at the output of amp B. The woofer amp would just need a gain of about 2.

Clear in my mind, but probably not clear to anyone else.

Dan

 

[Spehro Pefhany]

On Sun, 26 Feb 2017 13:34:01 -0800 (PST), the renowned
georgethomasmarriott@gmail.com wrote:

On Thursday, 21 February 2002 18:57:32 UTC, Frank Bemelman wrote:
Patrik Johansson <patrik@patadata.net> schreef in berichtnieuws
6b5955c6.0202210951.55bce44b@posting.google.com...
I would like to calculate the force that a electro magnet is able to
lift. Or rather to know the formula for designing a magnet to lift a
certain weight. Is the "push-force" equal the "lift-force"??

For example:
If I would like the magnet to be able to lift 10kg using 12V. How
would that formula look like??

10 Kg, that magnet will have the size of a bucket, if you want one
with a plunger and - let's say 5 cm of pulling distance. Current at
12V about 65 amps.

Well, these are just my gut feelings about it. Any formula that
would give you smaller solenoid, I would not trust until proven

Perhaps you can search the web for manufacturers of electromagnets
and solenoids. Browsing through their catalogs may give you an
idea of what is possible and what not.

--
Thanks,
Frank Bemelman
(remove 'x' & .invalid when sending email)

Hi frank, could you explain how you would work this out? I have been tasked with finding a power rating and material size for lifting 1.8 tonnes using an iron electromagnet.
Thanks

Just a note- you are responding to a message that is more than 15
years old, and unfortunately Frank has not been seen in these here
parts for quite some time.

--sp


--
Best regards,
Spehro Pefhany

 

On Thursday, 2 March 2017 20:42:31 UTC, olds...@tubes.com wrote:

On 27 Feb 2017 01:53:20 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

Thats a very bizarre way of connecting speakers. I question whether
there will still be stereo sound, or just some sort of mono????

If anything this will ephasise the difference between left and right.
maybe stick 100uF parallel with the sub-woofer

I'll give it a try wiring it in that manner...

What is the 100uf cap for? Is that a simple form of crossover?

Makes L&R speakers see the whole frequency, woofer only sees LF

I'd assume that if I use a separate amp, I can filter out the mids and
highs at the INPUT of the subwoofer amp, but how???

Thanks

RC filter.


NT

 

On Thursday, 2 March 2017 20:53:05 UTC, olds...@tubes.com wrote:

On Mon, 27 Feb 2017 17:25:10 -0800 (PST), "dcaster@krl.org"
dcaster@krl.org> wrote:

I would connect a 1 kohm resister to each of the stereo amp outputs. Connect the other end of the resi
stors together. and another 1 k ohm to that junction to signal ground. So the signal from stereo amp a
would go thru one 1 k resistor and thru another 1 k resistor to ground. So half the signal would be fed
into the woofer amp. THis signal would back feed to the stereo amp B , but be seriously attenuated by
the divider of the 1 k ohm and the output impedance of the stereo amp B. So the signal from amp A w
ould hardly appear at the output of amp B. The woofer amp would just need a gain of about 2.

Clear in my mind, but probably not clear to anyone else.

Dan

So then the separate woofer amp would connect to that junction between
the resistors, right?

Yes, but 1k is rather low. Some things will feed that fine, some won't. 10k would be better. If using historic equipment, 47k.


NT

 

On Thursday, 2 March 2017 22:36:38 UTC, bitrex wrote:

On 02/24/2017 10:07 PM, oldschool@tubes.com wrote:
I have a small amplifier (about 15W per channel), that I use for my
computer to listen to music. I have two speakers connected to it, which
are bookshelf sized speakers from an old stereo. It works well, but
lacks bass. I just got a sub-woofer which came from some sort of home
theater system. I want to connect this to what I already have.

However, there is only ONE sub-woofer. How can I get both channels to
work in this sub-woofer without combineing the channels to mono? Maybe I
need a separate amplifier for this sub-woofer, which I could probably
do, but then I have to feed the input of that amp with the signal from
both channels.

I see commercial systems which are stereo, which only have one
sub-woofer, so there has to be some means to do this.

Your speaker + sub system will sound like dogshit if you don't have any
way to adjust the crossover of the combined 2.1 system.

Get one of these and feed one main output to your small amp for the
bookshelf speakers, and then feed the "mono" output to your sub amp.

https://www.bhphotovideo.com/bnh/controller/home?O=&sku=473021&gclid=Cj0KEQiAot_FBRCqt8jVsoDKoZABEiQAqFL76PtY1ESn1LaCh9NHt4vEZNuz2e1_laMfhPZo2T1taX4aAq_S8P8HAQ&is=REG&ap=y&m=Y&c3api=1876%2C92051677682%2C&Q=&A=details

madness.


NT

 

[dcaster@krl.org]

On Thursday, March 2, 2017 at 3:53:05 PM UTC-5, olds...@tubes.com wrote:

On Mon, 27 Feb 2017 17:25:10 -0800 (PST), "dcaster@krl.org"
dcaster@krl.org> wrote:

On Sunday, February 26, 2017 at 4:00:44 PM UTC-5, olds...@tubes.com wrote:





So then the separate woofer amp would connect to that junction between
the resistors, right?

Right

Dan

 

On 27 Feb 2017 01:53:20 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

Thats a very bizarre way of connecting speakers. I question whether
there will still be stereo sound, or just some sort of mono????

If anything this will ephasise the difference between left and right.
maybe stick 100uF parallel with the sub-woofer

I'll give it a try wiring it in that manner...

What is the 100uf cap for? Is that a simple form of crossover?
I did want to obtain a crossover of sorts, so only the bass goes to the
subwoofer. But I dont really need to buy a complete crossover unit,
since I am not needing the mids and high freqs. The MAIN speakers will
cover the FULL freq range. So, I just need one or two passive components
to separate the low freqs.

I actually think I am going to go about this in a different manner. The
plan is to use a separate amplifier for the sub-woofer. Those small amp
modules can be gotten on ebay for $5 or less. That way I will get full
wattage on the main speakers and can power the sub-woofer separately.

yeah, and put the "sound card" into a mode that provides a sub-bass
output.

"Sound card"???? You lost me.....

I'd assume that if I use a separate amp, I can filter out the mids and
highs at the INPUT of the subwoofer amp, but how???

Thanks

 

On Mon, 27 Feb 2017 17:25:10 -0800 (PST), "dcaster@krl.org"
<dcaster@krl.org> wrote:

On Sunday, February 26, 2017 at 4:00:44 PM UTC-5, olds...@tubes.com wrote:


I actually think I am going to go about this in a different manner. The
plan is to use a separate amplifier for the sub-woofer.


If you are going to use a separate amp for the woofer then it is an easy solution. Or at least I think so.
Have not tried this.

I would connect a 1 kohm resister to each of the stereo amp outputs. Connect the other end of the resi
stors together. and another 1 k ohm to that junction to signal ground. So the signal from stereo amp a
would go thru one 1 k resistor and thru another 1 k resistor to ground. So half the signal would be fed
into the woofer amp. THis signal would back feed to the stereo amp B , but be seriously attenuated by
the divider of the 1 k ohm and the output impedance of the stereo amp B. So the signal from amp A w
ould hardly appear at the output of amp B. The woofer amp would just need a gain of about 2.

Clear in my mind, but probably not clear to anyone else.

Dan

So then the separate woofer amp would connect to that junction between
the resistors, right?

 

[bitrex]

On 02/24/2017 10:07 PM, oldschool@tubes.com wrote:

I have a small amplifier (about 15W per channel), that I use for my
computer to listen to music. I have two speakers connected to it, which
are bookshelf sized speakers from an old stereo. It works well, but
lacks bass. I just got a sub-woofer which came from some sort of home
theater system. I want to connect this to what I already have.

However, there is only ONE sub-woofer. How can I get both channels to
work in this sub-woofer without combineing the channels to mono? Maybe I
need a separate amplifier for this sub-woofer, which I could probably
do, but then I have to feed the input of that amp with the signal from
both channels.

I see commercial systems which are stereo, which only have one
sub-woofer, so there has to be some means to do this.

Your speaker + sub system will sound like dogshit if you don't have any
way to adjust the crossover of the combined 2.1 system.

Get one of these and feed one main output to your small amp for the
bookshelf speakers, and then feed the "mono" output to your sub amp.

<https://www.bhphotovideo.com/bnh/controller/home?O=&sku=473021&gclid=Cj0KEQiAot_FBRCqt8jVsoDKoZABEiQAqFL76PtY1ESn1LaCh9NHt4vEZNuz2e1_laMfhPZo2T1taX4aAq_S8P8HAQ&is=REG&ap=y&m=Y&c3api=1876%2C92051677682%2C&Q=&A=details>

 

[rickman]

On 3/2/2017 3:41 PM, oldschool@tubes.com wrote:

On 27 Feb 2017 01:53:20 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:


Thats a very bizarre way of connecting speakers. I question whether
there will still be stereo sound, or just some sort of mono????

If anything this will ephasise the difference between left and right.
maybe stick 100uF parallel with the sub-woofer

I'll give it a try wiring it in that manner...

What is the 100uf cap for? Is that a simple form of crossover?
I did want to obtain a crossover of sorts, so only the bass goes to the
subwoofer. But I dont really need to buy a complete crossover unit,
since I am not needing the mids and high freqs. The MAIN speakers will
cover the FULL freq range. So, I just need one or two passive components
to separate the low freqs.

I actually think I am going to go about this in a different manner. The
plan is to use a separate amplifier for the sub-woofer. Those small amp
modules can be gotten on ebay for $5 or less. That way I will get full
wattage on the main speakers and can power the sub-woofer separately.

yeah, and put the "sound card" into a mode that provides a sub-bass
output.

"Sound card"???? You lost me.....

I'd assume that if I use a separate amp, I can filter out the mids and
highs at the INPUT of the subwoofer amp, but how???

Maybe you should get a new amp with provision for a subwoofer. Then it
will have the filter and the amps for all channels. I see some one eBay
for under $30.

--

Rick C

 

[George Herold]

On Sunday, March 5, 2017 at 10:44:25 AM UTC-5, John Larkin wrote:

On Sun, 5 Mar 2017 14:48:51 -0000 (UTC), Cursitor Doom
curd@notformail.com> wrote:

https://www.sparkfun.com/products/11005

Replace the existing connector with a BNC, add a bit of series resistance
and you have a *very* cheap current probe for your scope.

I have a clamp-on ammeter that pretty much does that, although it just
indicates amps, and doesn't allow waveform snooping. 60 Hz waveforms
aren't terribly interesting.

I've got a plastic box, AC in/out and a double banana jack
across an R in the neutral line. I've got an ohm in there
(solder in 0.1 ohm if needed) and measure Vrms with DMM

My real problem with current measurement is DC, on PC boards. We want
to know how much current, say, an FPGA is using. Sometimes I include
current shunts in a layout, but sometimes I don't.

One can use existing switcher inductors as current shunts. I wish I
had a PCB trace current probe, but that's probably not posssible. You
can measure millivolt and microvolt drops across traces and vias.

For an outer layer you don't really know the thickness...
but I think for an inner layer the thickness is more consistent.
(typically 1/2 oz?) So maybe a better current shunt...
George H.

--

John Larkin Highland Technology, Inc

lunatic fringe electronics

 

[George Herold]

On Monday, March 6, 2017 at 5:28:11 AM UTC-5, piglet wrote:

On 06/03/2017 01:43, George Herold wrote:
On Sunday, March 5, 2017 at 10:44:25 AM UTC-5, John Larkin wrote:
One can use existing switcher inductors as current shunts. I wish I
had a PCB trace current probe, but that's probably not posssible. You
can measure millivolt and microvolt drops across traces and vias.

For an outer layer you don't really know the thickness...
but I think for an inner layer the thickness is more consistent.
(typically 1/2 oz?) So maybe a better current shunt...
George H.


If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

Huh, push current back through the voltage leads of the Kelvin connection.
I drew that but now I've got all the current flowing through the voltage
leads... seems like I need yet another pair of leads.. A six wire "Kelvin"
connection. Or I didn't draw it right.

George H.

piglet

 

[George Herold]

On Monday, March 6, 2017 at 9:34:46 AM UTC-5, Sjouke Burry wrote:

On 06.03.17 15:08, piglet wrote:
On 06/03/2017 13:47, George Herold wrote:
On Monday, March 6, 2017 at 5:28:11 AM UTC-5, piglet wrote:
If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

Huh, push current back through the voltage leads of the Kelvin connection.
I drew that but now I've got all the current flowing through the voltage
leads... seems like I need yet another pair of leads.. A six wire "Kelvin"
connection. Or I didn't draw it right.

George H.


piglet


So imagine a pcb trace carrying the unknown load current. Place four
test probes in a line along that trace. Measure the voltage between the
inner pair and inject a current through the outer pair such that the
voltage measured is reduced to zero. The current you have to inject will
be the same magnitude but opposite direction to the unknown load current.

piglet

AAAAHHHHHH!!!!!! What a perfect way to destroy electronics!
Forcing current into a print, where the current to be measured,
can change without prior notice, and your current is going the
other way......

No with six connections I think it's OK. If the input current changes,
then the voltage across the trace changes, and the external circuit compensates.
It's a lot of work to measure a current. But a new idea for me,
Thanks piglet.

George H.

 

[piglet]

On 06/03/2017 01:43, George Herold wrote:

On Sunday, March 5, 2017 at 10:44:25 AM UTC-5, John Larkin wrote:
One can use existing switcher inductors as current shunts. I wish I
had a PCB trace current probe, but that's probably not posssible. You
can measure millivolt and microvolt drops across traces and vias.

For an outer layer you don't really know the thickness...
but I think for an inner layer the thickness is more consistent.
(typically 1/2 oz?) So maybe a better current shunt...
George H.

If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

piglet

 

[Ralph Barone]

piglet <erichpwagner@hotmail.com> wrote:

So imagine a pcb trace carrying the unknown load current. Place four
test probes in a line along that trace. Measure the voltage between the
inner pair and inject a current through the outer pair such that the
voltage measured is reduced to zero. The current you have to inject will
be the same magnitude but opposite direction to the unknown load current.

piglet

And how do you measure the injected current?

"You can't fool me! It's turtles all the way down."

 

[piglet]

On 06/03/2017 13:47, George Herold wrote:

On Monday, March 6, 2017 at 5:28:11 AM UTC-5, piglet wrote:
If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

Huh, push current back through the voltage leads of the Kelvin connection.
I drew that but now I've got all the current flowing through the voltage
leads... seems like I need yet another pair of leads.. A six wire "Kelvin"
connection. Or I didn't draw it right.

George H.


piglet

So imagine a pcb trace carrying the unknown load current. Place four
test probes in a line along that trace. Measure the voltage between the
inner pair and inject a current through the outer pair such that the
voltage measured is reduced to zero. The current you have to inject will
be the same magnitude but opposite direction to the unknown load current.

piglet

 

[Sjouke Burry]

On 06.03.17 15:08, piglet wrote:

On 06/03/2017 13:47, George Herold wrote:
On Monday, March 6, 2017 at 5:28:11 AM UTC-5, piglet wrote:
If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

Huh, push current back through the voltage leads of the Kelvin connection.
I drew that but now I've got all the current flowing through the voltage
leads... seems like I need yet another pair of leads.. A six wire "Kelvin"
connection. Or I didn't draw it right.

George H.


piglet


So imagine a pcb trace carrying the unknown load current. Place four
test probes in a line along that trace. Measure the voltage between the
inner pair and inject a current through the outer pair such that the
voltage measured is reduced to zero. The current you have to inject will
be the same magnitude but opposite direction to the unknown load current.

piglet

AAAAHHHHHH!!!!!! What a perfect way to destroy electronics!
Forcing current into a print, where the current to be measured,
can change without prior notice, and your current is going the
other way......

 

On Monday, March 6, 2017 at 9:08:58 AM UTC-5, piglet wrote:

On 06/03/2017 13:47, George Herold wrote:
On Monday, March 6, 2017 at 5:28:11 AM UTC-5, piglet wrote:
If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

Huh, push current back through the voltage leads of the Kelvin connection.
I drew that but now I've got all the current flowing through the voltage
leads... seems like I need yet another pair of leads.. A six wire "Kelvin"
connection. Or I didn't draw it right.

George H.


piglet


So imagine a pcb trace carrying the unknown load current. Place four
test probes in a line along that trace. Measure the voltage between the
inner pair and inject a current through the outer pair such that the
voltage measured is reduced to zero. The current you have to inject will
be the same magnitude but opposite direction to the unknown load current.

piglet

You don't need to cancel the trace current--just injecting a calibrated
a.c. current between the two probes would be good enough to measure the
trace resistance, then compute the d.c. current from the d.c. drop.

Cheers,
James Arthur

 

On Tuesday, March 7, 2017 at 12:00:32 AM UTC-5, John Larkin wrote:

On Mon, 6 Mar 2017 18:26:39 -0800 (PST), dagmargoodboat@yahoo.com
wrote:

On Monday, March 6, 2017 at 9:08:58 AM UTC-5, piglet wrote:
On 06/03/2017 13:47, George Herold wrote:
On Monday, March 6, 2017 at 5:28:11 AM UTC-5, piglet wrote:
If it's a kelvin connection to the pcb trace then you could neatly
disregard the trace resistance by injecting a servo current to bring the
IR drop to zero.

Huh, push current back through the voltage leads of the Kelvin connection.
I drew that but now I've got all the current flowing through the voltage
leads... seems like I need yet another pair of leads.. A six wire "Kelvin"
connection. Or I didn't draw it right.

George H.


piglet


So imagine a pcb trace carrying the unknown load current. Place four
test probes in a line along that trace. Measure the voltage between the
inner pair and inject a current through the outer pair such that the
voltage measured is reduced to zero. The current you have to inject will
be the same magnitude but opposite direction to the unknown load current.

piglet

You don't need to cancel the trace current--just injecting a calibrated
a.c. current between the two probes would be good enough to measure the
trace resistance, then compute the d.c. current from the d.c. drop.


Just power the board down and measure the trace resistance!

Sure, but we're talking an automated instrument do that all in one go,
without having to make a separate 4-wire resistance measurement.

Another trick is to measure the drop in a trace, add a dummy load
resistor, note the delta-v, and do the math.

Or measure the trace resistance on a bare board from the same batch.

I like the magnetic probe for qualitative info. Maybe a Hall probe
could do it too.

We need a FLIR for magnetic flux.

Cheers,
James Arthur