18 Aug
2015
18 Aug
'15
8:40 p.m.
As part of my C-64 RS-232 to M15 60mA CL project, I've put together a
simple MAX232 based interface to connect the TTL levels on the C-64 to the
standard RS-232 +/- levels.
Is there any reason that I can or cannot install LEDs - on either side of
the MAX232 converter - to give some indication of line status?
Specifically, a pair of LEDs, one each for Tx and Rx lines, to blink / etc.
as line status changes and data moves.
I don't see any obvious issue - but I can conceive of a situation where it
might screw up the line in some way I haven't considered.
Comments?
18 Aug
18 Aug
8:50 p.m.
On 8/18/2015 10:40 PM, drlegendre . wrote:
As part of my C-64 RS-232 to M15 60mA CL project,
I've put together a
simple MAX232 based interface to connect the TTL levels on the C-64 to the
standard RS-232 +/- levels.
Is there any reason that I can or cannot install LEDs - on either side of
the MAX232 converter - to give some indication of line status?
Specifically, a pair of LEDs, one each for Tx and Rx lines, to blink / etc.
as line status changes and data moves.
I don't see any obvious issue - but I can conceive of a situation where it
might screw up the line in some way I haven't considered.
Comments?
Probably "best" to feed the TTL RX and TX lines to a 74'04
and then hook
the output to a 330 ohm and then the cathode of a LED, with anode going
to 5V.
But, probably good enough is to just put the diode cathode on the TX and
RX lines, 330 ohm to 5V, and have fun.
Jim
--
Jim Brain
brain at jbrain.com
www.jbrain.com
9:27 p.m.
Those little RS232 testers with LEDs built into a double DB25 connector box are usually
just made of LEDs and resistors connected to each signal line. They can load signals
enough to cause problems at high speeds, with weak drivers, or with long cables, but
usually they don't cause problems. If you're concerned, you could always include
jumpers or switches to disconnect the LEDs when they're not needed.
Of course, buffering the TTL signals eliminate any such problems. But on the other hand,
using two LEDs connected with opposite polarities on each RS232 level signal lets your
discriminate between driven positive, driven negative, and open. That can come in handy
when debugging things where the other end may or may not be driving properly, or may be
mis-wired.
For an example, feel free to take a look at this little modular jack RS232 tester that I
made:
https://github.com/NF6X/YostTester/blob/master/YostTester.pdf
The red/green LED pairs show whether each line is high, low or open. Resistor values may
vary depending on the LEDs that you choose.
--
Mark J. Blair, NF6X <nf6x at nf6x.net>
http://www.nf6x.net/
19 Aug
19 Aug
8:40 p.m.
I think the main issue is that TTL is usually a weak pullup and a hard
pulldown.
RS 232 levels are equal up and down.
Putting it on the RS232 lines could even improve signal quality because
the signal is usually poorly terminated, causing reflections.
The recommended lengths and Baud rates are intended to account for this.
Usually the line is lower impedance than the input termination,
requiring the source to absorb the reflected signal.
This is why 422/485 came to be. The impedance of the loads better
match that of the termination.
Dwight
Subject: Re: RS-232 Tx / Rx monitoring LEDs?
From: nf6x at nf6x.net
Date: Tue, 18 Aug 2015 21:27:19 -0700
To: cctalk at classiccmp.org
Those little RS232 testers with LEDs built into a double DB25 connector box are usually
just made of LEDs and resistors connected to each signal line. They can load signals
enough to cause problems at high speeds, with weak drivers, or with long cables, but
usually they don't cause problems. If you're concerned, you could always include
jumpers or switches to disconnect the LEDs when they're not needed.
Of course, buffering the TTL signals eliminate any such problems. But on the other hand,
using two LEDs connected with opposite polarities on each RS232 level signal lets your
discriminate between driven positive, driven negative, and open. That can come in handy
when debugging things where the other end may or may not be driving properly, or may be
mis-wired.
For an example, feel free to take a look at this little modular jack RS232 tester that I
made:
https://github.com/NF6X/YostTester/blob/master/YostTester.pdf
The red/green LED pairs show whether each line is high, low or open. Resistor values may
vary depending on the LEDs that you choose.
--
Mark J. Blair, NF6X <nf6x at nf6x.net>
http://www.nf6x.net/
9:04 p.m.
On 08/19/2015 08:40 PM, dwight wrote:
I think the main issue is that TTL is usually a weak
pullup and a
hard pulldown. RS 232 levels are equal up and down. Putting it on the
RS232 lines could even improve signal quality because the signal is
usually poorly terminated, causing reflections. The recommended
lengths and Baud rates are intended to account for this. Usually the
line is lower impedance than the input termination, requiring the
source to absorb the reflected signal. This is why 422/485 came to
be. The impedance of the loads better match that of the termination.
My setup with a DM232A doing the level translation lights up the LEDs in
one of those inline LED line monitors just fine without any apparent
signal degradation.
--Chuck
20 Aug
20 Aug
6:27 a.m.
On Aug 19, 2015, at 11:40 PM, dwight <dkelvey at
hotmail.com> wrote:
I think the main issue is that TTL is usually a weak pullup and a hard
pulldown.
RS 232 levels are equal up and down.
That depends. Open collector outputs, yes, but those are the less common case. Totem
pole outputs have comparable drive strength in both directions, that's precisely their
purpose (to provide symmetric rise/fall times when driving capacitive loads).
Putting it on the RS232 lines could even improve
signal quality because
the signal is usually poorly terminated, causing reflections.
A consideration with RS232 is that the signals swing to either side of 0, so if you use an
LED referenced to 0, it either needs a series diode or a sufficiently high reverse voltage
rating. And the driver voltage range spec for RS232 is quite loose; it might be just a
couple of volts but it can be as high as 15 volts or so.
paul
8:58 a.m.
On Aug 20, 2015, at 06:27, Paul Koning <paulkoning
at comcast.net> wrote:
Totem pole outputs have comparable drive strength in both directions, that's
precisely their purpose (to provide symmetric rise/fall times when driving capacitive
loads).
That's true for CMOS outputs. TTL outputs pull down much more strongly than they pull
up, which is why older designers are still in the habit of driving LEDs with active low
outputs even though active high outputs work just fine with modern CMOS logic. Look at
nearly any TTL datasheet, and note that VOL is much closer to ground than VOH is to VCC.
--
Mark J. Blair, NF6X <nf6x at nf6x.net>
http://www.nf6x.net/
9:54 a.m.
On Aug 20, 2015, at 11:58 AM, Mark J. Blair <nf6x
at nf6x.net> wrote:
Ok, but when you refer to "drive strength" I assumed you were talking about
current, not voltage. By that measure totem pole outputs are pretty much symmetrical.
palu
On Aug 20, 2015, at 06:27, Paul Koning
<paulkoning at comcast.net> wrote:
Totem pole outputs have comparable drive strength in both directions, that's
precisely their purpose (to provide symmetric rise/fall times when driving capacitive
loads).
That's true for CMOS outputs. TTL outputs pull down much more strongly than they pull
up, which is why older designers are still in the habit of driving LEDs with active low
outputs even though active high outputs work just fine with modern CMOS logic. Look at
nearly any TTL datasheet, and note that VOL is much closer to ground than VOH is to VCC.
10:11 a.m.
Totem pole outputs have comparable drive strength in
both directions, that$
That's true for CMOS outputs. TTL outputs pull down much
more strongly than$ 
10:35 a.m.
On Thu, 8/20/15, Mouse <mouse at Rodents-Montreal.ORG> wrote:
Maybe _rated_ current, but, even there, I don't
think so (my TTL
doc hasn't been unpacked yet, or I'd go check, but I'm fairly sure
they are generally specced to sink more current to GND
than source from Vcc).
It so happens I have a TTL handbook to hand at the moment. From
the 1985 TI TTL data book:
Ioh Iol
7400 -0.4 16
74H00 -0.5 20
74LS00 -0.4 8
74S00 -1 20
all in mA. So as rated, TTL devices can sink a solid order of
magnitude more current than they can source.
BLS
10:53 a.m.
On Aug 20, 2015, at 09:54 , Paul Koning <paulkoning
at comcast.net> wrote:
Ok, but when you refer to "drive strength" I assumed you were talking about
current, not voltage. By that measure totem pole outputs are pretty much symmetrical.
Again, CMOS totem pole outputs are pretty much symmetrical, but TTL totem pole outputs
aren't even close. Looking at a Fairchild 74LS04 datasheet, I see a 20x difference in
recommended drive-high current (Ioh = 0.4mA) vs. recommended drive-low current (Iol =
8mA). And the voltages are correspondingly different, too: at rated drive current of
0.4mA, Voh = 2.7V (min) to 3.4V (typ) with a 5V supply, while Vol is 0.35V (typ) to 0.5V
(max) at 8mA drive current. So that's > 1.6V drop from Vcc while sourcing a mere
0.4mA, vs. < 0.5V rise from GND while sinking 8mA.
Totem pole outputs just mean that the output driver actively drives both up and down, with
two stacked drive transistors. It does not imply that the drive strengths are even close
to being equal, particularly when we're talking about TTL logic in a vintage
computer.
Incidentally, TTL inputs also present asymmetric loads for high vs. low inputs at about
the same ratio (18x input current ratio in the 74LS04 example when driven at the input
thresholds), and have asymmetrical input threshold voltages. So unloaded TTL output
voltages aren't relevant if we assume that the output is driving a TTL input of the
same logic family. Even if a TTL output appears to drive all the way up to Vcc with no
load, it won't once it's driving a typical load.
So you might think of those TTL totem pole drivers as being symmetrical when they're
strictly driving TTL inputs of the same family, since the TTL inputs and TTL outputs are
designed to work together. But they're very strongly asymmetrical when driving things
other than TTL inputs, such as the LEDs in question here.
--
Mark J. Blair, NF6X <nf6x at nf6x.net>
http://www.nf6x.net/
12:40 p.m.
On Thu, Aug 20, 2015 at 10:54 AM, Paul Koning <paulkoning at comcast.net> wrote:
Ok, but when you refer to "drive strength" I
assumed you were talking about current, not voltage. By that measure totem pole outputs
are pretty much symmetrical.
Not for true (bipolar) TTL. See the specs for the 7400, 74LS00, 74S00:
http://www.ti.com/lit/ds/symlink/sn74ls00.pdf
Ioh(max) Iol(max)
7400 -0.4 16
74LS00 -0.4 8
74S00 -1 20
CMOS for comparison:
http://www.ti.com/lit/ds/symlink/sn74hct00.pdf
http://www.ti.com/lit/ds/symlink/sn74act00.pdf
Ioh(max) Iol(max)
74HCT00 -4 4
74ACT00 -24 24
12:49 p.m.
On Aug 20, 2015, at 3:40 PM, Eric Smith <spacewar
at gmail.com> wrote:
On Thu, Aug 20, 2015 at 10:54 AM, Paul Koning <paulkoning at comcast.net> wrote:
Ok, clearly my memory was faulty. Thanks for the corrections.
paul
Ok, but when you refer to "drive
strength" I assumed you were talking about current, not voltage. By that measure
totem pole outputs are pretty much symmetrical.
Not for true (bipolar) TTL. See the specs for the 7400, 74LS00, 74S00:
http://www.ti.com/lit/ds/symlink/sn74ls00.pdf
Ioh(max) Iol(max)
7400 -0.4 16
74LS00 -0.4 8
74S00 -1 20
CMOS for comparison:
http://www.ti.com/lit/ds/symlink/sn74hct00.pdf
http://www.ti.com/lit/ds/symlink/sn74act00.pdf
Ioh(max) Iol(max)
74HCT00 -4 4
74ACT00 -24 24
21 Aug
21 Aug
8:36 a.m.
I was going to add something but it has already been
said several times.
I will add that if using a LED on an AC like signal of high voltage,
one should use a diode.
I recommend using a shunt diode rather than a series diode when
high voltages are being dropped by the resistor. It reduces the need
for a high voltage diode but makes the resistor hotter.
Some red LEDs glow orange when not protected from 12VAC.
You can ask how I know.
Dwight
8:36 a.m.
for a high voltage diode but makes the resistor
hotter.
Some red LEDs glow orange when not protected from 12VAC.
I discovered (over 35 years ago) that green LEDs glow orange if massively
overcurrented (you know what I mean). No they don't last long like that. It
doesn't appear to be a thermal effect though.
-tony
8:40 a.m.
On Fri, Aug 21, 2015 at 8:36 AM, dwight <dkelvey at hotmail.com> wrote:
I was going to add something but it has already been
said several times.
I will add that if using a LED on an AC like signal of high voltage,
one should use a diode.
I recommend using a shunt diode rather than a series diode when
high voltages are being dropped by the resistor. It reduces the need
for a high voltage diode but makes the resistor hotter.
Some red LEDs glow orange when not protected from 12VAC.
You can ask how I know.
Dwight
You can also turn on all of the pretty blues lights in a UV-erasable PROM
by putting it in the PROM burner the wrong way around.
-- Charles
8:57 a.m.
On 08/21/2015 08:36 AM, dwight wrote:
I was going to add something but it has already been
said several
times. I will add that if using a LED on an AC like signal of high
voltage, one should use a diode. I recommend using a shunt diode
rather than a series diode when high voltages are being dropped by
the resistor. It reduces the need for a high voltage diode but makes
the resistor hotter. Some red LEDs glow orange when not protected
from 12VAC. You can ask how I know. Dwight
Ever take a close look at a string of Christmas-tree LEDs? Most are
composed of a string of LEDs hooked directly across the AC line--no
rectifier diode to be found.
Some seek to reduce the 60Hz flicker by employing two strings to
illuminate on both half-cycles, reducing the flicker somewhat. Since
my eyes react to the flicker (it's like ants crawling over the string),
I found that simply employing a full-wave bridge rectifier can reduce
the appearance of flicker tremendously.
--Chuck
22 Aug
22 Aug
10:23 p.m.
From: cclist at sydex.com
On 08/21/2015 08:36 AM, dwight wrote:
I would think the reverse voltage sum of the diodes is enough.
Different diodes also can handle different voltages. Since the sum
of the forward voltages is enough to handle AC, I'd suspect the
reverse voltages each would handle is quite small as well.
The problem is when the current limiting is done with a resistor
that in the forward direction drops a lot of voltage.
The diode has to handle the voltage until breakdown when reversed.
If the resistor was handling 1 Watts, with the right break down,
the LED could be taking .5 Watts. This is more than most are designed
for.
Dwight
I was going to add something but it has already
been said several
times. I will add that if using a LED on an AC like signal of high
voltage, one should use a diode. I recommend using a shunt diode
rather than a series diode when high voltages are being dropped by
the resistor. It reduces the need for a high voltage diode but makes
the resistor hotter. Some red LEDs glow orange when not protected
from 12VAC. You can ask how I know. Dwight
Ever take a close look at a string of Christmas-tree LEDs? Most are
composed of a string of LEDs hooked directly across the AC line--no
rectifier diode to be found.
Some seek to reduce the 60Hz flicker by employing two strings to
illuminate on both half-cycles, reducing the flicker somewhat. Since
my eyes react to the flicker (it's like ants crawling over the string),
I found that simply employing a full-wave bridge rectifier can reduce
the appearance of flicker tremendously.
--Chuck
11:17 p.m.
On 08/22/2015 10:23 PM, dwight wrote:
I would think the reverse voltage sum of the diodes is
enough.
Different diodes also can handle different voltages. Since the sum
of the forward voltages is enough to handle AC, I'd suspect the
reverse voltages each would handle is quite small as well.
The problem is when the current limiting is done with a resistor
that in the forward direction drops a lot of voltage.
The diode has to handle the voltage until breakdown when reversed.
If the resistor was handling 1 Watts, with the right break down,
the LED could be taking .5 Watts. This is more than most are designed
for.
...and that's just the nub of it. The success of this depends largely
on the consistent characteristics of every LED in the string. Since
LEDs tend to fail short if submitted to overvoltage, I've often wondered
if a spike in the AC supply would precipitate a cascade failure in the
string. I've looked hard and there are no rectifier diodes in the
string--just the LEDs themselves. Probably saves about 5 cents or so of
manufacturing cost.
I've also seen LED "night lights" from China that employ nothing more
than a safety capacitor (usually about 104) in series with a resistor
connected to two back-to-back LEDs, all across the AC line.
I've wondered what the lifetime of such a setup is.
--Chuck.
11:55 p.m.
I've also seen C-R series voltage dropping circuits, here & there.
Correct me if I'm wrong, but doesn't the series cap dissipate power just as
it would, were it a series resistor? I mean, if the LED is passing 20mA,
the cap is also doing 20mA - and at whatever the Vdrop is.
Right? If not, why?
On Sun, Aug 23, 2015 at 1:17 AM, Chuck Guzis <cclist at sydex.com> wrote:
On 08/22/2015 10:23 PM, dwight wrote:
I would think the reverse voltage sum of the diodes is enough.
Different diodes also can handle different
voltages. Since the sum
of the forward voltages is enough to handle AC, I'd suspect the
reverse voltages each would handle is quite small as well.
The problem is when the current limiting is done with a resistor
that in the forward direction drops a lot of voltage.
The diode has to handle the voltage until breakdown when reversed.
If the resistor was handling 1 Watts, with the right break down,
the LED could be taking .5 Watts. This is more than most are designed
for.
...and that's just the nub of it. The success of this depends largely on
the consistent characteristics of every LED in the string. Since LEDs tend
to fail short if submitted to overvoltage, I've often wondered if a spike
in the AC supply would precipitate a cascade failure in the string. I've
looked hard and there are no rectifier diodes in the string--just the LEDs
themselves. Probably saves about 5 cents or so of manufacturing cost.
I've also seen LED "night lights" from China that employ nothing more than
a safety capacitor (usually about 104) in series with a resistor connected
to two back-to-back LEDs, all across the AC line.
I've wondered what the lifetime of such a setup is.
--Chuck.
23 Aug
23 Aug
2:06 a.m.
On 2015-Aug-22, at 11:55 PM, drlegendre . wrote:
On Sun, Aug 23, 2015 at 1:17 AM, Chuck Guzis
<cclist at sydex.com> wrote:
I doubt if any brief explanation here is going to the topic justice. Look up power factor
or reactive power.
FWIW:
The impedance (capacitive reactance, Z=Xc=1/(2*pi*f*C) of the C does produce the desired
voltage drop but the C also shifts the phase of the current relative to that of the V. To
apply the power equation P=VI properly, you can't just multiply the RMS values
together, you multiply the instantaneous values of the V & I sine waves together
through a cycle. You get a third sine wave, that for power. If V & I are in phase, the
power sine wave will all be in the positive region and is real power consumption. When
they are out of phase, some portion of the power sine wave will be negative: a portion of
the energy the C sucked down the line is being returned during each cycle.
Yes, it does reduce energy consumption relative to a purely R solution.
On a large scale, the power company doesn't like it because it unnecessarily adds to
the currents circulating in the system, but then, this is from C which shifts the current
in one direction, so it's doing some compensation for the inductive wall warts you
have plugged in around the house, which as L shift the current in the other direction.
On 08/22/2015 10:23 PM, dwight wrote:
I would think the reverse voltage sum of the diodes is enough.
I've also seen C-R series voltage dropping circuits, here & there.
Correct me if I'm wrong, but doesn't the series cap dissipate power just as
it would, were it a series resistor? I mean, if the LED is passing 20mA,
the cap is also doing 20mA - and at whatever the Vdrop is.
Right? If not, why? Different diodes also can handle different
voltages. Since the sum
of the forward voltages is enough to handle AC, I'd suspect the
reverse voltages each would handle is quite small as well.
The problem is when the current limiting is done with a resistor
that in the forward direction drops a lot of voltage.
The diode has to handle the voltage until breakdown when reversed.
If the resistor was handling 1 Watts, with the right break down,
the LED could be taking .5 Watts. This is more than most are designed
for.
...and that's just the nub of it. The success of this depends largely on
the consistent characteristics of every LED in the string. Since LEDs tend
to fail short if submitted to overvoltage, I've often wondered if a spike
in the AC supply would precipitate a cascade failure in the string. I've
looked hard and there are no rectifier diodes in the string--just the LEDs
themselves. Probably saves about 5 cents or so of manufacturing cost.
I've also seen LED "night lights" from China that employ nothing more than
a safety capacitor (usually about 104) in series with a resistor connected
to two back-to-back LEDs, all across the AC line.
I've wondered what the lifetime of such a setup is.
--Chuck.
7:10 a.m.
I've used the capacitor method to provide most of the drop
in the past. I don't usually max out the LEDs at 20ma. I find
there is little difference between 10 and 20ma.
Yes, the 10 ( or 20ma ) is current flow through the capacitor.
It is necessary to have some resistor in series as well to suppress
line spikes.
You do have to be careful about mounting the LED packages, as they are
not rated for line voltage. That is, if you are expecting a UL rating.
Dwight
9:06 a.m.
On 08/23/2015 07:10 AM, dwight wrote:
I've used the capacitor method to provide most of
the drop in the
past. I don't usually max out the LEDs at 20ma. I find there is
little difference between 10 and 20ma. Yes, the 10 ( or 20ma ) is
current flow through the capacitor. It is necessary to have some
resistor in series as well to suppress line spikes.
Another similar dirty trick back in the day was to run a 6SL7
dual-triode form the line using a 1.0 uF nonpolar capacitor in series
with the line to provide a supply for the 500 ma 6.3 v heater and then
use one of the triode sections as a half-wave rectifier. You thus had
the other triode section for whatever stupid purpose. Of course, this
was horrible abuse of the tube, particularly in the area of
heater-cathode voltage ratings. It probably wouldn't work as well in
220VAC countries, but it worked well enough in the 120VAC ones.
--Chuck
11:32 a.m.
On 2015-Aug-23, at 9:06 AM, Chuck Guzis wrote:
On 08/23/2015 07:10 AM, dwight wrote:
Those ubiquitous motion detectors for outdoor lights use capacitive dropping to supply the
low voltage for the ICs and electronics. Cap and small R in series with the AC input to a
bridge rectifier.
Works out well in that the bridge rectifier permits current flow in both directions, which
is necessary to get current flow through the cap (the cap has to charge and discharge).
I've used the capacitor method to provide
most of the drop in the
past. I don't usually max out the LEDs at 20ma. I find there is
little difference between 10 and 20ma. Yes, the 10 ( or 20ma ) is
current flow through the capacitor. It is necessary to have some
resistor in series as well to suppress line spikes.
Another similar dirty trick back in the day was to run a 6SL7 dual-triode form the line
using a 1.0 uF nonpolar capacitor in series with the line to provide a supply for the 500
ma 6.3 v heater and then use one of the triode sections as a half-wave rectifier. You
thus had the other triode section for whatever stupid purpose. Of course, this was
horrible abuse of the tube, particularly in the area of heater-cathode voltage ratings.
It probably wouldn't work as well in 220VAC countries, but it worked well enough in
the 120VAC ones.
11:46 a.m.
Those ubiquitous motion detectors for outdoor lights use capacitive dropping to supply
the
low voltage for the ICs and electronics. Cap and small R in series with the AC input to a
bridge rectifier.
Works out well in that the bridge rectifier permits current flow in both directions,
which is
necessary to get current flow through the cap (the cap has to charge and discharge).
Over here that sort of circuit was common for the power supply for the control
electronics
in washing machines and the like. It may still be, I've not worked on anything that
recent.
-tony
9:54 a.m.
-----Original Message-----
From: cctalk [mailto:cctalk-bounces at classiccmp.org] On Behalf Of Brent
Hilpert
Sent: Sunday, August 23, 2015 3:07 AM
To: General Discussion: On-Topic and Off-Topic Posts
Subject: Re: RS-232 Tx / Rx monitoring LEDs?
On 2015-Aug-22, at 11:55 PM, drlegendre . wrote:
I doubt if any brief explanation here is going to the topic justice. Look up
power factor or reactive power.
FWIW:
The impedance (capacitive reactance, Z=Xc=1/(2*pi*f*C) of the C does produce
the desired voltage drop but the C also shifts the phase of the current
relative to that of the V. To apply the power equation P=VI properly, you
can't just multiply the RMS values together, you multiply the instantaneous
values of the V & I sine waves together through a cycle. You get a third
sine wave, that for power. If V & I are in phase, the power sine wave will
all be in the positive region and is real power consumption. When they are
out of phase, some portion of the power sine wave will be negative: a
portion of the energy the C sucked down the line is being returned during
each cycle.
Yes, it does reduce energy consumption relative to a purely R solution.
On a large scale, the power company doesn't like it because it unnecessarily
adds to the currents circulating in the system, but then, this is from C
which shifts the current in one direction, so it's doing some compensation
for the inductive wall warts you have plugged in around the house, which as
L shift the current in the other direction.
-----------------------
Brent, that is an excellent explanation in just a few sentences. One
quibble however. The power company does indeed like components that shift
the current in the capacitive direction. Taken as a whole for the power
grid, the power source sees the load as inductive because of all the
industrial motors it powers, including the ones that exist in almost every
home (washer, dryer, heating/air conditioning, mixer, disposal, etc.)
Capacitor banks are frequently installed in large industrial operations to
shift the inductive load more toward the capacitive power factor. This is
because the power company, as you have implied, charges more for power that
is current shifted away from zero %. I have even seen large motors
installed in industrial situations that run continuously without load,
because such motors appear as a capacitive load, and indeed are called
'rotary capacitors'. I recall one time when the CFO ordered such a motor
turned off because "it is wasting power." It took a little plain and fancy
instruction by the engineer to let him know that it was actually saving the
company money.
On Sun, Aug 23, 2015 at 1:17 AM, Chuck Guzis
<cclist at sydex.com> wrote:
> On 08/22/2015 10:23 PM, dwight wrote:
>
> I would think the reverse voltage sum of the diodes is enough.
>> Different diodes also can handle different voltages. Since the sum
>> of the forward voltages is enough to handle AC, I'd suspect the
>> reverse voltages each would handle is quite small as well.
>> The problem is when the current limiting is done with a resistor
>> that in the forward direction drops a lot of voltage.
>> The diode has to handle the voltage until breakdown when reversed.
>> If the resistor was handling 1 Watts, with the right break down, the
>> LED could be taking .5 Watts. This is more than most are designed
>> for.
>
> ...and that's just the nub of it. The success of this depends
> largely on the consistent characteristics of every LED in the string.
> Since LEDs tend to fail short if submitted to overvoltage, I've often
> wondered if a spike in the AC supply would precipitate a cascade
> failure in the string. I've looked hard and there are no rectifier
> diodes in the string--just the LEDs themselves. Probably saves about 5
cents
or so of manufacturing cost.
I've also seen LED "night lights" from China that employ nothing more
than a safety capacitor (usually about 104) in series with a resistor
connected to two back-to-back LEDs, all across the AC line.
I've wondered what the lifetime of such a setup is.
--Chuck.
I've also seen C-R series voltage dropping circuits, here & there.
Correct me if I'm wrong, but doesn't the series cap dissipate power
just as it would, were it a series resistor? I mean, if the LED is
passing 20mA, the cap is also doing 20mA - and at whatever the Vdrop is.
Right? If not, why?
10:31 a.m.
Heh, all you had to say was "power factor" and I've have understood. I
suspected that was the case, but it seemed too easy..
Guess I'm too used to thinking in simple ohmic terms, with Watt & Kirchhoff
always looming large. Something told me that, in the end, there was no way
around dealing with the E^2/R heat - anything else seemed like a
thermodynamic "cheat".
On Sun, Aug 23, 2015 at 11:54 AM, Steve Moulding <fti1983 at xmission.com>
wrote:
-----Original Message-----
From: cctalk [mailto:cctalk-bounces at classiccmp.org] On Behalf Of Brent
Hilpert
Sent: Sunday, August 23, 2015 3:07 AM
To: General Discussion: On-Topic and Off-Topic Posts
Subject: Re: RS-232 Tx / Rx monitoring LEDs?
On 2015-Aug-22, at 11:55 PM, drlegendre . wrote:
I doubt if any brief explanation here is going to the topic justice. Look
up
power factor or reactive power.
FWIW:
The impedance (capacitive reactance, Z=Xc=1/(2*pi*f*C) of the C does
produce
the desired voltage drop but the C also shifts the phase of the current
relative to that of the V. To apply the power equation P=VI properly, you
can't just multiply the RMS values together, you multiply the instantaneous
values of the V & I sine waves together through a cycle. You get a third
sine wave, that for power. If V & I are in phase, the power sine wave will
all be in the positive region and is real power consumption. When they are
out of phase, some portion of the power sine wave will be negative: a
portion of the energy the C sucked down the line is being returned during
each cycle.
Yes, it does reduce energy consumption relative to a purely R solution.
On a large scale, the power company doesn't like it because it
unnecessarily
adds to the currents circulating in the system, but then, this is from C
which shifts the current in one direction, so it's doing some compensation
for the inductive wall warts you have plugged in around the house, which as
L shift the current in the other direction.
-----------------------
Brent, that is an excellent explanation in just a few sentences. One
quibble however. The power company does indeed like components that shift
the current in the capacitive direction. Taken as a whole for the power
grid, the power source sees the load as inductive because of all the
industrial motors it powers, including the ones that exist in almost every
home (washer, dryer, heating/air conditioning, mixer, disposal, etc.)
Capacitor banks are frequently installed in large industrial operations to
shift the inductive load more toward the capacitive power factor. This is
because the power company, as you have implied, charges more for power that
is current shifted away from zero %. I have even seen large motors
installed in industrial situations that run continuously without load,
because such motors appear as a capacitive load, and indeed are called
'rotary capacitors'. I recall one time when the CFO ordered such a motor
turned off because "it is wasting power." It took a little plain and fancy
instruction by the engineer to let him know that it was actually saving the
company money.
On Sun, Aug 23, 2015 at 1:17 AM, Chuck Guzis
<cclist at sydex.com> wrote:
> On 08/22/2015 10:23 PM, dwight wrote:
>
> I would think the reverse voltage sum of the diodes is enough.
>> Different diodes also can handle different voltages. Since the sum
>> of the forward voltages is enough to handle AC, I'd suspect the
>> reverse voltages each would handle is quite small as well.
>> The problem is when the current limiting is done with a resistor
>> that in the forward direction drops a lot of voltage.
>> The diode has to handle the voltage until breakdown when reversed.
>> If the resistor was handling 1 Watts, with the right break down, the
>> LED could be taking .5 Watts. This is more than most are designed
>> for.
>
> ...and that's just the nub of it. The success of this depends
> largely on the consistent characteristics of every LED in the string.
> Since LEDs tend to fail short if submitted to overvoltage, I've often
> wondered if a spike in the AC supply would precipitate a cascade
> failure in the string. I've looked hard and there are no rectifier
> diodes in the string--just the LEDs themselves. Probably saves about 5
cents
or so of manufacturing cost.
I've also seen LED "night lights" from China that employ nothing more
than a safety capacitor (usually about 104) in series with a resistor
connected to two back-to-back LEDs, all across the AC line.
I've wondered what the lifetime of such a setup is.
--Chuck.
I've also seen C-R series voltage dropping circuits, here & there.
Correct me if I'm wrong, but doesn't the series cap dissipate power
just as it would, were it a series resistor? I mean, if the LED is
passing 20mA, the cap is also doing 20mA - and at whatever the Vdrop is.
Right? If not, why?
10:46 a.m.
Guess I'm too used to thinking in simple ohmic
terms, with Watt & Kirchhoff
always looming large. Something told me that, in the end, there was no way
Well, Kirchhoff's laws do apply in AC circuits, but you have to be careful
how you apply them, given that voltages and currents need not be in phase.
around dealing with the E^2/R heat - anything else
seemed like a
thermodynamic "cheat".
Not so. There are many ways of reducing a voltage without generating that
much heat. For AC an obvious one is a transformer. Trivial example, if you
had to run a string of valve heaters totalling, say, 25.2V from 240V AC mains you
could use a dropping resistor (which would consume almost 90% of the incoming
power, overall efficiency around 10%)) or you could use a transformer, overall
efficiency around 80%+. Are you suggesting a transformer is a 'thermodynamic
cheat'?
For AC or DC, chopper type devices (switching regualtors, triac-type lamp dimmers, etc
-- depending on the type of supply, etc) are another way which is a lot more efficient
than a resistor. Are those 'thermodynamoic cheats'?
-tony
11:17 a.m.
On 2015-Aug-23, at 9:54 AM, Steve Moulding wrote:
-----Original Message-----
From: cctalk [mailto:cctalk-bounces at classiccmp.org] On Behalf Of Brent
Hilpert
Sent: Sunday, August 23, 2015 3:07 AM
To: General Discussion: On-Topic and Off-Topic Posts
Subject: Re: RS-232 Tx / Rx monitoring LEDs?
On 2015-Aug-22, at 11:55 PM, drlegendre . wrote:
I doubt if any brief explanation here is going to the topic justice. Look up
power factor or reactive power.
FWIW:
The impedance (capacitive reactance, Z=Xc=1/(2*pi*f*C) of the C does produce
the desired voltage drop but the C also shifts the phase of the current
relative to that of the V. To apply the power equation P=VI properly, you
can't just multiply the RMS values together, you multiply the instantaneous
values of the V & I sine waves together through a cycle. You get a third
sine wave, that for power. If V & I are in phase, the power sine wave will
all be in the positive region and is real power consumption. When they are
out of phase, some portion of the power sine wave will be negative: a
portion of the energy the C sucked down the line is being returned during
each cycle.
Yes, it does reduce energy consumption relative to a purely R solution.
On a large scale, the power company doesn't like it because it unnecessarily
adds to the currents circulating in the system, but then, this is from C
which shifts the current in one direction, so it's doing some compensation
for the inductive wall warts you have plugged in around the house, which as
L shift the current in the other direction.
-----------------------
Brent, that is an excellent explanation in just a few sentences. One
quibble however. The power company does indeed like components that shift
the current in the capacitive direction. Taken as a whole for the power
grid, the power source sees the load as inductive because of all the
industrial motors it powers, including the ones that exist in almost every
home (washer, dryer, heating/air conditioning, mixer, disposal, etc.)
Capacitor banks are frequently installed in large industrial operations to
shift the inductive load more toward the capacitive power factor. This is
because the power company, as you have implied, charges more for power that
is current shifted away from zero %. I have even seen large motors
installed in industrial situations that run continuously without load,
because such motors appear as a capacitive load, and indeed are called
'rotary capacitors'. I recall one time when the CFO ordered such a motor
turned off because "it is wasting power." It took a little plain and fancy
instruction by the engineer to let him know that it was actually saving the
company money.
Sure, L shift predominates in the system, I was just speaking in general terms at that
point.
The power companies may be playing it both ways though:
I was once told a story of a small mill or manufacturing co. and an electro-plating
company situated next door to each other. But the mill was inductive (motors) and the
electro-plating was effectively capacitive (not sure how that worked), so they pretty much
cancelled each other out and weren't contributing to line losses for the power
company. The power company was none-the-less charging both of them for poor power factor
(industrial users having special power meters that track power factor).
As you mention rotary capacitors, a few miles away from me is a large gas-fired generating
station that doesn't get used much for generation as we are primarily hydro in BC. But
the hydro generation is distant from here - here being the primary load centre (Vancouver
area). They unhook some of the generators at the gas-fired plant from the turbines and run
them as rotary capacitors.
On Sun, Aug 23, 2015 at 1:17 AM, Chuck Guzis
<cclist at sydex.com> wrote:
> On 08/22/2015 10:23 PM, dwight wrote:
>
> I would think the reverse voltage sum of the diodes is enough.
>> Different diodes also can handle different voltages. Since the sum
>> of the forward voltages is enough to handle AC, I'd suspect the
>> reverse voltages each would handle is quite small as well.
>> The problem is when the current limiting is done with a resistor
>> that in the forward direction drops a lot of voltage.
>> The diode has to handle the voltage until breakdown when reversed.
>> If the resistor was handling 1 Watts, with the right break down, the
>> LED could be taking .5 Watts. This is more than most are designed
>> for.
>
> ...and that's just the nub of it. The success of this depends
> largely on the consistent characteristics of every LED in the string.
> Since LEDs tend to fail short if submitted to overvoltage, I've often
> wondered if a spike in the AC supply would precipitate a cascade
> failure in the string. I've looked hard and there are no rectifier
> diodes in the string--just the LEDs themselves. Probably saves about 5
cents
or so of manufacturing cost.
I've also seen LED "night lights" from China that employ nothing more
than a safety capacitor (usually about 104) in series with a resistor
connected to two back-to-back LEDs, all across the AC line.
I've wondered what the lifetime of such a setup is.
--Chuck.
I've also seen C-R series voltage dropping circuits, here & there.
Correct me if I'm wrong, but doesn't the series cap dissipate power
just as it would, were it a series resistor? I mean, if the LED is
passing 20mA, the cap is also doing 20mA - and at whatever the Vdrop is.
Right? If not, why?
7:32 a.m.
I've also seen C-R series voltage dropping circuits, here & there.
Correct me if I'm wrong, but doesn't the
series cap dissipate power just as
it would, were it a series resistor? I mean, if the LED is passing 20mA,
the cap is also doing 20mA - and at whatever the Vdrop is.
Right? If not, why?
Yes, the capacitor passes 20mA. But unlike a resistor the voltage drop across a capacitor
is not in phase with the current. For a pure capacitor, it is out of phase by 90 degrees.
For
a pure inductor, BTW, it is 90 degrees out in the oposite direction.
Now, if you consider the power at any instant, it is, indeed the product of the voltage
and current at that instant. For a resistor, this is alwas positive, so if you add up all
the
instantaneous powers over a full cycle (mathematically, this is integration, of cource),
the total power consumed is positive. But for a capacitor, sometimes the instantaneous
power is negative, in effect the capacitor is supplying energy back to the circuit. And
if
you integrate that over a full cycle, you end up with zero.
Intuitively : With a resistor, voltage and current are either both positive or both
negative,
so the product is positive. With a capacitor, due to that phase shift, there are 4
regions
to consider : V +ve., I +ve; V +ve, I -ve; V -ve, I -ve; V -ve, I+ve. The first and third
of those
give a +ve power, the second and fourth a -ve power and they exactly cancel out.
It turns out that the power in an AC circuit can be calcualted as the product of the
RMS voltage, RMS current and the cosine of the phase angle between them. The
RMS values are the ones normally quotes (115V mains had an RMS value of 115V). The
cosine(phase) term is known as the 'power factor', and is probably the thing to
look up
in a book on AC electric circuits.
-tony
8:56 a.m.
On 08/22/2015 11:55 PM, drlegendre . wrote:
I've also seen C-R series voltage dropping
circuits, here & there.
Correct me if I'm wrong, but doesn't the series cap dissipate power
just as it would, were it a series resistor? I mean, if the LED is
passing 20mA, the cap is also doing 20mA - and at whatever the Vdrop
is.
Right? If not, why?
That's correct--but there's the matter of the voltage. Eventually,
you'll see that the brightness of the LEDs will drop fairly rapidly
before the thing is useless. A common dodge used by Chinese
manufacturers. Saves a few components, doesn't it.
--Chuck
20 Aug
20 Aug
9:13 a.m.
A consideration with RS232 is that the signals swing to either side of 0, so if you use
an LED
referenced to 0, it either needs a series diode or a sufficiently high reverse voltage
rating.
A diode in inverse parallel with the LED Is more normal when you want to run an LED off
effectively
an AC supply. In the case of RS232 signals it is common to use a red and a green LED in
inverse
parallel, either separate LEDs or one of the 2-wire bicolour LEDs which are precisely a
pair of LEDs
in inverse parallel.
Most of the little RS232 terters are just that, I think with 3k series resistors for the
LED pairs.
-tony
10:23 a.m.
A consideration with RS232 is that the signals swing
to either side of 0, so$
It's been a while since I read the spec, but I think it's -3 to -20
volts one way and +3 to +20 the other, with the -3 to +3 range
deliberately left ambiguous. I think there are slew rate limits, too,
but I don't recall what they are even approximately.
Of course, serial lines are normally run over unshielded wires with a
single shared signal ground, so, for noise tolerance, transmitters
usually drive the outputs substantially beyond 3V either side of
ground. (There are also short-circuit tolerance specs which mean you
don't want your output impedance to be too low, either. I think it's
something like, any pins or ccombinations of pins may be shorted to one
another and/or any voltage or voltages from -20 to +20 indefinitely
without damage.)
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