F
Fart Dicko
Guest
On Sun, 16 Apr 2006 23:13:05 +0100, Spin Dryer <me2@privacy.net>
wrote:
wrote:
Allah thinks otherwise, spamdrier.
M
Michael A. Terrell
Guest
John Fields wrote:
That's because you keep replying to Rich. ;-)
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.
Michael A. Terrell
Central Florida
That's because you keep replying to Rich. ;-)
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.
Michael A. Terrell
Central Florida
R
Roger Dewhurst
Guest
"John Fields" <jfields@austininstruments.com> wrote in message
news:cnv9425kb1iomefi8etf12v340nplvru7r@4ax.com...
R
news:cnv9425kb1iomefi8etf12v340nplvru7r@4ax.com...
Old fashioned meggers do that very well!
R
S
ScottM
Guest
Got this breadboarded; it's working beautifully. Thanks.
J
John Popelish
Guest
ScottM wrote:
"Op" amp refers to the devices ability to be programmed to perform all
sorts of mathematical operations.
Excellent! You have programmed your first analog circuit.
"Op" amp refers to the devices ability to be programmed to perform all
sorts of mathematical operations.
S
ScottM
Guest
*Chuckle* Well, no. I used to hack around with audio circuits, using
discrete, scrounged transistors from AM radios, in my distant youth. I
didn't know what I was doing then either.
Actually, I spoke too soon. The op amp, hooked to a regulated 12v
supply and fed 5v on the (+) input, puts out about 9.8v. No matter how
I fiddle the resistor bridge on the (-) input (eg, swapping one 100k
resistor for a 10k), I can't get this above 9.8v. I can easily get it
lower, by using a partially charged cap instead of the full 5v on the
(+) leg, so I know the basic, voltage doubling idea is sound. But over
2v of drop on the output just doesn't seem right (and means I can't use
the circuit as is.)
Should I be looking at some other op amp? Having just blown $55 on a
12v, switching power supply, I'd rather not have to spring for a 15v.
Guest
ScottM wrote:
1.5 volts of the supply. Of course, if it has to also supply load
current, it needs more head room. Do you have any idea what the load
current or resistance is? The opamp also has an output current limit,
so that if the load resistance is low enough, then no amount of extra
supply voltage will allow more output voltage than that which produces
the limit value of current. There are other, higher output current
opamp types and some that will pull up closer to the positive rail, but
they are a lot less common than the LM324.
By the way, any opamp should have a small capacitor connected directly
between their supply pins, say, .1 uF ceramic or low inductance type
film, to keep the opamp from forming an oscillator with the supply
trace inductance. Such oscillations can also reduce the average output
current.
I was also going to suggest that you add a 100 ohm resistor between
thew opamp output and the load, to keep long wires between them from
causing oscillations, but this will require that the opamp be able to
produuce even more output voltage to compensate for what is dropped
across this resistor.
a milliamp meter in series, if possible), and I'll see if I can suggest
something better.
At zero output current, the opamp should be able to pull up to withingExcellent! You have programmed your first analog circuit.
*Chuckle* Well, no. I used to hack around with audio circuits, using
discrete, scrounged transistors from AM radios, in my distant youth. I
didn't know what I was doing then either.
Actually, I spoke too soon. The op amp, hooked to a regulated 12v
supply and fed 5v on the (+) input, puts out about 9.8v. No matter how
I fiddle the resistor bridge on the (-) input (eg, swapping one 100k
resistor for a 10k), I can't get this above 9.8v. I can easily get it
lower, by using a partially charged cap instead of the full 5v on the
(+) leg, so I know the basic, voltage doubling idea is sound. But over
2v of drop on the output just doesn't seem right (and means I can't use
the circuit as is.)
1.5 volts of the supply. Of course, if it has to also supply load
current, it needs more head room. Do you have any idea what the load
current or resistance is? The opamp also has an output current limit,
so that if the load resistance is low enough, then no amount of extra
supply voltage will allow more output voltage than that which produces
the limit value of current. There are other, higher output current
opamp types and some that will pull up closer to the positive rail, but
they are a lot less common than the LM324.
By the way, any opamp should have a small capacitor connected directly
between their supply pins, say, .1 uF ceramic or low inductance type
film, to keep the opamp from forming an oscillator with the supply
trace inductance. Such oscillations can also reduce the average output
current.
I was also going to suggest that you add a 100 ohm resistor between
thew opamp output and the load, to keep long wires between them from
causing oscillations, but this will require that the opamp be able to
produuce even more output voltage to compensate for what is dropped
across this resistor.
Tell me more about the load current at 9.8 volts output (measured with
a milliamp meter in series, if possible), and I'll see if I can suggest
something better.
S
ScottM
Guest
The load in this case was a 15k resistor in series with an LED, and my
test meter in parallel with all that. Hard to believe the total
resistance was much under 10k. In the final application, it will drive
(over short wires) the analog input of a dimmer, and the input
impedience is at least 4.7k, possibly much higher. The op amp is used
as a follower (output tied back to - input) but there's at least 100k
between the output and anything else. Hard to imagine this op amp was
struggling.
A 1.5v drop would be fine - I'm starting at 12v and need to get up to
10v.
I'll look into a small cap across the power pins. I probably should
look into another test meter, as well...
test meter in parallel with all that. Hard to believe the total
resistance was much under 10k. In the final application, it will drive
(over short wires) the analog input of a dimmer, and the input
impedience is at least 4.7k, possibly much higher. The op amp is used
as a follower (output tied back to - input) but there's at least 100k
between the output and anything else. Hard to imagine this op amp was
struggling.
A 1.5v drop would be fine - I'm starting at 12v and need to get up to
10v.
I'll look into a small cap across the power pins. I probably should
look into another test meter, as well...
J
John Popelish
Guest
ScottM wrote:
bit more than the LM324, however.
http://www.national.com/ds/LM/LMC6484.pdf
an output that can swing all the ways to the rails. It costs quite a
bit more than the LM324, however.
http://www.national.com/ds/LM/LMC6484.pdf
S
ScottM
Guest
Looks like I panicked without need. Another test meter (and one that
can show 5 ,10 and 12v all in the same range, so I can check it against
my 5 & 12v power supplies) reports that with 110k between output and
pot, and 100k between pot and ground, and (-) on the wiper, I can use a
20k pot to adjust between about 9v and 11v. Looks like I'm good to go...
can show 5 ,10 and 12v all in the same range, so I can check it against
my 5 & 12v power supplies) reports that with 110k between output and
pot, and 100k between pot and ground, and (-) on the wiper, I can use a
20k pot to adjust between about 9v and 11v. Looks like I'm good to go...
J
John Popelish
Guest
ScottM wrote:
If you want to build a better low pass filter for the faster changing
signals (you did say that some of the PWM outputs were for sound
generation, didn't you?) that change is very simple. Just a couple
more resistors and a couple more capacitors per opamp, and you can get
about 3 times the ripple reduction for the same ability to change the
output, or the same ripple with 3 times the rate of change in the output.
Here is the general form of the filter:
http://jpbedinger.tripod.com/projects/an/ds1-3polelp.htm
Very good.
If you want to build a better low pass filter for the faster changing
signals (you did say that some of the PWM outputs were for sound
generation, didn't you?) that change is very simple. Just a couple
more resistors and a couple more capacitors per opamp, and you can get
about 3 times the ripple reduction for the same ability to change the
output, or the same ripple with 3 times the rate of change in the output.
Here is the general form of the filter:
http://jpbedinger.tripod.com/projects/an/ds1-3polelp.htm
D
Dave Plowman (News)
Guest
In article <Xns97B054C7FADA674C1H4@127.0.0.1>,
Andy <nomail@nomail.com> wrote:
drive the push pull output stage. There'd likely be a matching output one
for a low impedance speaker - although some used approx 40 ohm ones driven
direct.
may not matter in practice. If from a small transistor portable may not
have a very good spec.
likely once were in valve days. Most balanced audio these days is low
impedance out high in - same as domestic.
Ground isolating transformers these days will be usually 10k in and out.
An easy solution is to buy a line level transformer from a good car audio
shop used to isolate the ground to an add on power amp. It will be stereo
too. In the UK they cost about 5 gbp so if things are normal $5 or less in
the US. Make sure it is a 1:1 though as some are designed to match off the
speaker outputs in the head unit.
--
*The severity of the itch is proportional to the reach *
Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.
Andy <nomail@nomail.com> wrote:
Correct way to do it.
Not that I've heard of, but I'm from the UK.
DC measurements can give a guide to the turns ratio but not the impedance.
Right. That's an inter stage driver much used on early portable radios to
drive the push pull output stage. There'd likely be a matching output one
for a low impedance speaker - although some used approx 40 ohm ones driven
direct.
Have a read of a book on basic AC theory or do a Google on it.
No - it is an audio transformer, but will drop the level somewhat. This
may not matter in practice. If from a small transistor portable may not
have a very good spec.
600- 600 ohm transformers are used on balanced pro equipment. Or more
likely once were in valve days. Most balanced audio these days is low
impedance out high in - same as domestic.
Ground isolating transformers these days will be usually 10k in and out.
An easy solution is to buy a line level transformer from a good car audio
shop used to isolate the ground to an add on power amp. It will be stereo
too. In the UK they cost about 5 gbp so if things are normal $5 or less in
the US. Make sure it is a 1:1 though as some are designed to match off the
speaker outputs in the head unit.
--
*The severity of the itch is proportional to the reach *
Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.
J
JLU_EE
Guest
I agree with Luhan.
But it is interesting.
But it is interesting.
J
JLU_EE
Guest
I agree with Luhan.
But it is interesting.
But it is interesting.
N
no_one
Guest
it is AC? I don't believe that you can build a diode-capacitor voltage
multiplier with DC.
"temp@temp.com" <not@not.com> wrote in message
news:Xns97B2B98C4773Etenotempcom@140.99.99.130...
multiplier with DC.
"temp@temp.com" <not@not.com> wrote in message
news:Xns97B2B98C4773Etenotempcom@140.99.99.130...
C
Chris
Guest
temp@temp.com wrote:
needs to source *and* sink current -- the ULN2003 only sinks. Not only
that, but the darlington output means the output voltage never goes
below 1V, no matter how little current it's sinking. Once you subtract
another diode drop, you'll need a lot of stages to get to 16V.
Before anything else, though, it's odd you say you've got 3V. The spec
sheet says you get 5V at 200mA max at the output I/O connector. This
comes direct from the USB, I think.
Now if you were on a desert island and a ULN2003 was all you had, or if
an evil genie or professor were holding your feet to the fire, you
might be able to get away with switching an inductor on and off without
using the ULN2003 internal diode, and then use a diode, cap and zener
to give you a very low current 16V supply like this (view in fixed font
or M$ Notepad):
|
| VCC
| +
| |
| C|
| C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
You need a good value inductor to give you the inductive kick, and it
should have a DC resistance of at least 20 ohms (4V / .2A). If that's
a problem, you might want to play with the duty cycle, and have a local
supply for the inductor separated by a 20 ohm resistor and, say, a
220uF cap like this:
|
| ___
| VCC+o-|___|-o------.
| 33 ohm +| |
| 220uF--- |
| 16V --- |
| | C|
| GND C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
This will allow you to choose a DC coil resistance that keeps current
as less than the max for the ULN2003. Again, you'll have to leave pin
16 open (which might mess up your other plans with the ULN2003, if
you're switching anything else inductive).
If this is a real-world project, this is a little cheesy -- you
definitely should use a DC-to-DC converter instead of this unless you
only need a couple of mA and good regulation isn't a problem.
Don't worry about having spare stuff left over when you're done with a
project. It's not like the broccoli your mom served you. You should
be crafty, but you don't have to use 'em all.
And make sure to read the data sheets, sir. They're not printed
because the manufacturers hate trees. And you can always get the .pdf
file if you're concerned about the clutter of paper, or have
environmental reservations.
Go Aztecs!
Chris
Hi, Tia Sal. Your idea won't work. The input to a voltage multiplier
needs to source *and* sink current -- the ULN2003 only sinks. Not only
that, but the darlington output means the output voltage never goes
below 1V, no matter how little current it's sinking. Once you subtract
another diode drop, you'll need a lot of stages to get to 16V.
Before anything else, though, it's odd you say you've got 3V. The spec
sheet says you get 5V at 200mA max at the output I/O connector. This
comes direct from the USB, I think.
Now if you were on a desert island and a ULN2003 was all you had, or if
an evil genie or professor were holding your feet to the fire, you
might be able to get away with switching an inductor on and off without
using the ULN2003 internal diode, and then use a diode, cap and zener
to give you a very low current 16V supply like this (view in fixed font
or M$ Notepad):
|
| VCC
| +
| |
| C|
| C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
You need a good value inductor to give you the inductive kick, and it
should have a DC resistance of at least 20 ohms (4V / .2A). If that's
a problem, you might want to play with the duty cycle, and have a local
supply for the inductor separated by a 20 ohm resistor and, say, a
220uF cap like this:
|
| ___
| VCC+o-|___|-o------.
| 33 ohm +| |
| 220uF--- |
| 16V --- |
| | C|
| GND C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
This will allow you to choose a DC coil resistance that keeps current
as less than the max for the ULN2003. Again, you'll have to leave pin
16 open (which might mess up your other plans with the ULN2003, if
you're switching anything else inductive).
If this is a real-world project, this is a little cheesy -- you
definitely should use a DC-to-DC converter instead of this unless you
only need a couple of mA and good regulation isn't a problem.
Don't worry about having spare stuff left over when you're done with a
project. It's not like the broccoli your mom served you. You should
be crafty, but you don't have to use 'em all.
And make sure to read the data sheets, sir. They're not printed
because the manufacturers hate trees. And you can always get the .pdf
file if you're concerned about the clutter of paper, or have
environmental reservations.
Go Aztecs!
Chris
C
Chris
Guest
temp@temp.com wrote:
needs to source *and* sink current -- the ULN2003 only sinks. Not only
that, but the darlington output means the output voltage never goes
below 1V, no matter how little current it's sinking. Once you subtract
another diode drop, you'll need a lot of stages to get to 16V.
Before anything else, though, it's odd you say you've got 3V. The spec
sheet says you get 5V at 200mA max at the output I/O connector. This
comes direct from the USB, I think.
Now if you were on a desert island and a ULN2003 was all you had, or if
an evil genie or professor were holding your feet to the fire, you
might be able to get away with switching an inductor on and off without
using the ULN2003 internal diode, and then use a diode, cap and zener
to give you a very low current 16V supply like this (view in fixed font
or M$ Notepad):
|
| VCC
| +
| |
| C|
| C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
You need a good value inductor to give you the inductive kick, and it
should have a DC resistance of at least 20 ohms (4V / .2A). If that's
a problem, you might want to play with the duty cycle, and have a local
supply for the inductor separated by a 20 ohm resistor and, say, a
220uF cap like this:
|
| ___
| VCC+o-|___|-o------.
| 33 ohm +| |
| 220uF--- |
| 16V --- |
| | C|
| GND C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
This will allow you to choose a DC coil resistance that keeps current
as less than the max for the ULN2003. Again, you'll have to leave pin
16 open (which might mess up your other plans with the ULN2003, if
you're switching anything else inductive).
If this is a real-world project, this is a little cheesy -- you
definitely should use a DC-to-DC converter instead of this unless you
only need a couple of mA and good regulation isn't a problem.
Don't worry about having spare stuff left over when you're done with a
project. It's not like the broccoli your mom served you. You should
be crafty, but you don't have to use 'em all.
And make sure to read the data sheets, sir. They're not printed
because the manufacturers hate trees. And you can always get the .pdf
file if you're concerned about the clutter of paper, or have
environmental reservations.
Go Aztecs!
Chris
Hi, Tia Sal. Your idea won't work. The input to a voltage multiplier
needs to source *and* sink current -- the ULN2003 only sinks. Not only
that, but the darlington output means the output voltage never goes
below 1V, no matter how little current it's sinking. Once you subtract
another diode drop, you'll need a lot of stages to get to 16V.
Before anything else, though, it's odd you say you've got 3V. The spec
sheet says you get 5V at 200mA max at the output I/O connector. This
comes direct from the USB, I think.
Now if you were on a desert island and a ULN2003 was all you had, or if
an evil genie or professor were holding your feet to the fire, you
might be able to get away with switching an inductor on and off without
using the ULN2003 internal diode, and then use a diode, cap and zener
to give you a very low current 16V supply like this (view in fixed font
or M$ Notepad):
|
| VCC
| +
| |
| C|
| C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
You need a good value inductor to give you the inductive kick, and it
should have a DC resistance of at least 20 ohms (4V / .2A). If that's
a problem, you might want to play with the duty cycle, and have a local
supply for the inductor separated by a 20 ohm resistor and, say, a
220uF cap like this:
|
| ___
| VCC+o-|___|-o------.
| 33 ohm +| |
| 220uF--- |
| 16V --- |
| | C|
| GND C| L
| 1/7 C|
| ULN2003 |
| |\ |1N4148
| -| >O---o->|-o----o----o
| |/ +| |
| 47uF --- /-/1N4745A
| 35V --- ^ Vz=16V
| | |
| === ===
| GND GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
This will allow you to choose a DC coil resistance that keeps current
as less than the max for the ULN2003. Again, you'll have to leave pin
16 open (which might mess up your other plans with the ULN2003, if
you're switching anything else inductive).
If this is a real-world project, this is a little cheesy -- you
definitely should use a DC-to-DC converter instead of this unless you
only need a couple of mA and good regulation isn't a problem.
Don't worry about having spare stuff left over when you're done with a
project. It's not like the broccoli your mom served you. You should
be crafty, but you don't have to use 'em all.
And make sure to read the data sheets, sir. They're not printed
because the manufacturers hate trees. And you can always get the .pdf
file if you're concerned about the clutter of paper, or have
environmental reservations.
Go Aztecs!
Chris
J
jack
Guest
You set up the transistor array as a multivibrator -- use this to drive a
coil and capacitor in "boost" configuration -- or as a classic push-pull
switching supply.
fwiw, I have an ancient Genrad sound level meter which uses discrete
transistors and a torroidal transformer (about the size of your pinkie
fingernail) to drive all the electronics from a couple "C" batteries.
It's much less expensive, much less noise generating to use an off-the-shelf
switching regulator chip -- and these are almost as inexpensive as a
ULN2003.
coil and capacitor in "boost" configuration -- or as a classic push-pull
switching supply.
fwiw, I have an ancient Genrad sound level meter which uses discrete
transistors and a torroidal transformer (about the size of your pinkie
fingernail) to drive all the electronics from a couple "C" batteries.
It's much less expensive, much less noise generating to use an off-the-shelf
switching regulator chip -- and these are almost as inexpensive as a
ULN2003.
P
Phil Allison
Guest
"Ian Bell"
** Utter nonsense.
No such balancing act exists.
** Bollocks.
** Even worse bollocks !!!
A 2:1 step up tranny provides no noise advantage * AT ALL * since it
increases the source impedance as seen by the pre-amp by 4 times and so
*doubles* the noise voltage along with the signal voltage.
Where the hell do folk get these WACKY ideas ???
** The input impedance is around 1200 ohms direct with most mic pres.
Inserting a 2:1 step up REDUCES the load seen by the mic to 300 ohms, likely
cutting its output voltage in half.
The net effect is a poorer signal to noise ratio.
........ Phil
** Utter nonsense.
No such balancing act exists.
** Bollocks.
** Even worse bollocks !!!
A 2:1 step up tranny provides no noise advantage * AT ALL * since it
increases the source impedance as seen by the pre-amp by 4 times and so
*doubles* the noise voltage along with the signal voltage.
Where the hell do folk get these WACKY ideas ???
** The input impedance is around 1200 ohms direct with most mic pres.
Inserting a 2:1 step up REDUCES the load seen by the mic to 300 ohms, likely
cutting its output voltage in half.
The net effect is a poorer signal to noise ratio.
........ Phil
P
Peter J Ross
Guest
uSXJUrlqVCrJaCCOvDzDOVrLYiHiNYRiOSxJWDkJhIVtdUmCtm
ELDAEvupjiuuhInWvzJgSludhJggRbOWYvqTKZOIUaMqrAURBa
KkRHpqAtZFXLkRmjwxHFZUTPieexlGRRMWVgvJpuikQeFsrhCJ
ySQwiQNGIenDLiNCVjqmIOQolvAhDgxaoLcyAvsoTXumSUXlMM
yODMSNAWOjCAiToEuWDgwXhwCypzIjtgqTPHWcIMQIjuVPMjAD
OGlfxRdstgyKVamVYYqdiQveOygmbJaQxKFHrrMoOBhZyBGZDr
ELDAEvupjiuuhInWvzJgSludhJggRbOWYvqTKZOIUaMqrAURBa
KkRHpqAtZFXLkRmjwxHFZUTPieexlGRRMWVgvJpuikQeFsrhCJ
ySQwiQNGIenDLiNCVjqmIOQolvAhDgxaoLcyAvsoTXumSUXlMM
yODMSNAWOjCAiToEuWDgwXhwCypzIjtgqTPHWcIMQIjuVPMjAD
OGlfxRdstgyKVamVYYqdiQveOygmbJaQxKFHrrMoOBhZyBGZDr