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Why the diodes?

J

James Sweet

Jan 1, 1970
0
For entertainment/educational purposes I'm reverse engineering a cheap
fluorescent nightlight that died and I found it uses an arrangement of
diodes and resistors in the input that I'm unfamiliar with. Excuse the
poor ASCII schematic, hopefully the formatting holds up. The combination
of D1 and R1 is in parallel with D2 and R2, of equal value wired such
that current flowing in either direction will pass through one diode and
one resistor. Why have the diodes there in the first place instead of
just one resistor?




---------------------------------------------------------
Rest of circuit
D1 R1
Mains -------|<-------/\/\--------------------
| |
| |
------------------------->|-------/\/\---------
D2 R2


The nightlight uses a very simple capacitor ballast to limit tube
current, and the thing failed very quickly, it didn't even last 3
months. I suspect the poor photocontrol causing it to flicker near dusk
killed it but the whole thing is not very well made.
 
R

Rich.

Jan 1, 1970
0
Passing current through the diodes reduces the noise that the fluorescent
light introduces to the 120v wiring.
 
J

James Sweet

Jan 1, 1970
0
Salmon said:
Are you sure that what you think are resistors are not in fact
capacitors? You might be describing a form of a voltage doubler
rectifier.

Bill


Not sure if my other post went through, I got a news server error.

Yes, I'm positive they are not capacitors. They are 56 ohm 1W flameproof
resistors and are wired exactly as shown in my crude schematic, each in
series with a silicon diode with the whole thing in series with one leg
of the mains. I'm well familiar with various voltage multiplier
topologies and this is certainly not one of them.

I had a sneaking suspicion it may be a sort of interference filter as
someone else suggested, but I'm unclear on just how this works.

I've drawn up most of the rest of the schematic of the nightlight in a
CAD program just for fun, I need to finish with the photocontrol portion
and then I'll post it.
 
A

Archimedes' Lever

Jan 1, 1970
0
Are you sure that what you think are resistors are not in fact
capacitors? You might be describing a form of a voltage doubler
rectifier.

Bill

It is DIRECTLY on the mains, idiot. They are NOT doubling the mains
voltage.
 
A

Archimedes' Lever

Jan 1, 1970
0
Not sure if my other post went through, I got a news server error.

Yes, I'm positive they are not capacitors. They are 56 ohm 1W flameproof
resistors and are wired exactly as shown in my crude schematic, each in
series with a silicon diode with the whole thing in series with one leg
of the mains. I'm well familiar with various voltage multiplier
topologies and this is certainly not one of them.

I had a sneaking suspicion it may be a sort of interference filter as
someone else suggested, but I'm unclear on just how this works.

I've drawn up most of the rest of the schematic of the nightlight in a
CAD program just for fun, I need to finish with the photocontrol portion
and then I'll post it.


The original responder was correct. It is spike suppression keeping
noise from being placed on the local line.
 
R

Ross Herbert

Jan 1, 1970
0
On Thu, 16 Apr 2009 15:27:14 -0700, Archimedes' Lever

:On Thu, 16 Apr 2009 10:07:27 -0700, James Sweet
:
:>Salmon Egg wrote:
:>> In article <[email protected]>,
:>>
:>>> For entertainment/educational purposes I'm reverse engineering a cheap
:>>> fluorescent nightlight that died and I found it uses an arrangement of
:>>> diodes and resistors in the input that I'm unfamiliar with. Excuse the
:>>> poor ASCII schematic, hopefully the formatting holds up. The combination
:>>> of D1 and R1 is in parallel with D2 and R2, of equal value wired such
:>>> that current flowing in either direction will pass through one diode and
:>>> one resistor. Why have the diodes there in the first place instead of
:>>> just one resistor?
:>>>
:>>>
:>>>
:>>>
:>>> ---------------------------------------------------------
:>>> Rest of circuit
:>>> D1 R1
:>>> Mains -------|<-------/\/\--------------------
:>>> | |
:>>> | |
:>>> ------------------------->|-------/\/\---------
:>>> D2 R2
:>>>
:>>>
:>>> The nightlight uses a very simple capacitor ballast to limit tube
:>>> current, and the thing failed very quickly, it didn't even last 3
:>>> months. I suspect the poor photocontrol causing it to flicker near dusk
:>>> killed it but the whole thing is not very well made.
:>>
:>> Are you sure that what you think are resistors are not in fact
:>> capacitors? You might be describing a form of a voltage doubler
:>> rectifier.
:>>
:>> Bill
:>>
:>
:>
:>Not sure if my other post went through, I got a news server error.
:>
:>Yes, I'm positive they are not capacitors. They are 56 ohm 1W flameproof
:>resistors and are wired exactly as shown in my crude schematic, each in
:>series with a silicon diode with the whole thing in series with one leg
:>of the mains. I'm well familiar with various voltage multiplier
:>topologies and this is certainly not one of them.
:>
:>I had a sneaking suspicion it may be a sort of interference filter as
:>someone else suggested, but I'm unclear on just how this works.
:>
:>I've drawn up most of the rest of the schematic of the nightlight in a
:>CAD program just for fun, I need to finish with the photocontrol portion
:>and then I'll post it.
:
:
: The original responder was correct. It is spike suppression keeping
:noise from being placed on the local line.


I doubt very much that this scheme will prevent spikes from being placed on the
"local" (?) line. On each half cycle of mains frequency one diode will be
clamped hard on and any noise spikes generated on the lamp side of the diode can
pass freely through to the "local" line. Unless the amplitude and polarity of a
spike is such as to bias the conducting diode to the "off" condition (this is
just not possible IMO), it may as well not be there at all.
 
J

James Sweet

Jan 1, 1970
0
Ross said:
On Thu, 16 Apr 2009 15:27:14 -0700, Archimedes' Lever

:On Thu, 16 Apr 2009 10:07:27 -0700, James Sweet
:
:>Salmon Egg wrote:
:>> In article <[email protected]>,
:>>
:>>> For entertainment/educational purposes I'm reverse engineering a cheap
:>>> fluorescent nightlight that died and I found it uses an arrangement of
:>>> diodes and resistors in the input that I'm unfamiliar with. Excuse the
:>>> poor ASCII schematic, hopefully the formatting holds up. The combination
:>>> of D1 and R1 is in parallel with D2 and R2, of equal value wired such
:>>> that current flowing in either direction will pass through one diode and
:>>> one resistor. Why have the diodes there in the first place instead of
:>>> just one resistor?
:>>>
:>>>
:>>>
:>>>
:>>> ---------------------------------------------------------
:>>> Rest of circuit
:>>> D1 R1
:>>> Mains -------|<-------/\/\--------------------
:>>> | |
:>>> | |
:>>> ------------------------->|-------/\/\---------
:>>> D2 R2
:>>>
:>>>
:>>> The nightlight uses a very simple capacitor ballast to limit tube
:>>> current, and the thing failed very quickly, it didn't even last 3
:>>> months. I suspect the poor photocontrol causing it to flicker near dusk
:>>> killed it but the whole thing is not very well made.
:>>
:>> Are you sure that what you think are resistors are not in fact
:>> capacitors? You might be describing a form of a voltage doubler
:>> rectifier.
:>>
:>> Bill
:>>
:>
:>
:>Not sure if my other post went through, I got a news server error.
:>
:>Yes, I'm positive they are not capacitors. They are 56 ohm 1W flameproof
:>resistors and are wired exactly as shown in my crude schematic, each in
:>series with a silicon diode with the whole thing in series with one leg
:>of the mains. I'm well familiar with various voltage multiplier
:>topologies and this is certainly not one of them.
:>
:>I had a sneaking suspicion it may be a sort of interference filter as
:>someone else suggested, but I'm unclear on just how this works.
:>
:>I've drawn up most of the rest of the schematic of the nightlight in a
:>CAD program just for fun, I need to finish with the photocontrol portion
:>and then I'll post it.
:
:
: The original responder was correct. It is spike suppression keeping
:noise from being placed on the local line.


I doubt very much that this scheme will prevent spikes from being placed on the
"local" (?) line. On each half cycle of mains frequency one diode will be
clamped hard on and any noise spikes generated on the lamp side of the diode can
pass freely through to the "local" line. Unless the amplitude and polarity of a
spike is such as to bias the conducting diode to the "off" condition (this is
just not possible IMO), it may as well not be there at all.



I wonder if this arrangement creates a slightly longer dead zone between
cycles, when the voltage is too low to forward bias either diode?
Perhaps it helps ensure the SCR turns off between cycles? I was looking
at the circuit again just now and realized I made an error, R11 is 390K,
not 390 ohm as I had drawn, which would have dissipated far too much
power rather than utilizing the reactance of C1 to limit lamp current.

http://picasaweb.google.com/jamesrsweet/VariousPictures?feat=directlink#5325943974808236930

I may have to fix this thing and scope it to get a full understanding of
how it all works. It's bugging me to have elements of such a simple
circuit not make sense.
 
A

Andrew Gabriel

Jan 1, 1970
0
I wonder if this arrangement creates a slightly longer dead zone between
cycles, when the voltage is too low to forward bias either diode?
Perhaps it helps ensure the SCR turns off between cycles? I was looking
at the circuit again just now and realized I made an error, R11 is 390K,
not 390 ohm as I had drawn, which would have dissipated far too much
power rather than utilizing the reactance of C1 to limit lamp current.

http://picasaweb.google.com/jamesrsweet/VariousPictures?feat=directlink#5325943974808236930

That resistor is just there to safely leak away any charge left in
C1 after the unit is unplugged (which would otherwise remain available
across the plug pins, ready to give some unsuspecting person a belt).
I may have to fix this thing and scope it to get a full understanding of
how it all works. It's bugging me to have elements of such a simple
circuit not make sense.

Purely capacitive ballasts at mains frequency tend to result in very
poor tube life, because they generate an extremely poor crest factor
(peak current to average current through the tube), and this is very
detrimental to the life of the emission coating on the tube filaments.
The high current peaks happen when the tube just starts to conduct each
half mains cycle, and the storged charge in the capacitor has to change
suddenly, resulting in a peak current with little impedance to limit it.

A standard 4W T5 tube has an operating voltage of about 30V. I don't
have details of its hot cathode striking voltage, but let's guess it's
40V. So at the tail-end of one half mains cycle, the tube will stop
conducting when the voltage available drops below 30V, and the capacitor
charge (and voltage) at that point will remain fixed as no current flows
(ignoring leakage in R11, which given the R11/C1 time constant, will be
negligable in this time period). Mains voltage now has to change by
30+40V before the tube will start conducting again in the next half-cycle.
However, because the charge and voltage on the capacitor froze when the
tube stopped conducting, it's now 70V out from where it should be, and
a current peak will flow to correct this mismatch. This is limited by
the impedance of the supply, R1+D7 or R2+D8, and the tube (which
effectively has no impedance). So we're looking at something like 1A.
The design tube current is 150mA, giving a crest factor of over 6.
To achieve full tube life, crest factor should not exceed 1.5, so we're
well outside the limits.

I mention this because it may be that the R1+D7 and R2+D8 networks are
there to limit the crest factor by some means unknown to me to less than
the 56ohm resistors alone would imply, which would have a significant
benefit on tube life. OTOH, manufacturers of such items often care little
about the longevity of their products, so this may be a red herring.

Professional quality mains frequency capacitive ballasts include an
inductor to limit the current flow when the tube starts conducting again.
 
J

James Sweet

Jan 1, 1970
0
Andrew said:
That resistor is just there to safely leak away any charge left in
C1 after the unit is unplugged (which would otherwise remain available
across the plug pins, ready to give some unsuspecting person a belt).


Purely capacitive ballasts at mains frequency tend to result in very
poor tube life, because they generate an extremely poor crest factor
(peak current to average current through the tube), and this is very
detrimental to the life of the emission coating on the tube filaments.
The high current peaks happen when the tube just starts to conduct each
half mains cycle, and the storged charge in the capacitor has to change
suddenly, resulting in a peak current with little impedance to limit it.

A standard 4W T5 tube has an operating voltage of about 30V. I don't
have details of its hot cathode striking voltage, but let's guess it's
40V. So at the tail-end of one half mains cycle, the tube will stop
conducting when the voltage available drops below 30V, and the capacitor
charge (and voltage) at that point will remain fixed as no current flows
(ignoring leakage in R11, which given the R11/C1 time constant, will be
negligable in this time period). Mains voltage now has to change by
30+40V before the tube will start conducting again in the next half-cycle.
However, because the charge and voltage on the capacitor froze when the
tube stopped conducting, it's now 70V out from where it should be, and
a current peak will flow to correct this mismatch. This is limited by
the impedance of the supply, R1+D7 or R2+D8, and the tube (which
effectively has no impedance). So we're looking at something like 1A.
The design tube current is 150mA, giving a crest factor of over 6.
To achieve full tube life, crest factor should not exceed 1.5, so we're
well outside the limits.

I mention this because it may be that the R1+D7 and R2+D8 networks are
there to limit the crest factor by some means unknown to me to less than
the 56ohm resistors alone would imply, which would have a significant
benefit on tube life. OTOH, manufacturers of such items often care little
about the longevity of their products, so this may be a red herring.

Professional quality mains frequency capacitive ballasts include an
inductor to limit the current flow when the tube starts conducting again.


Fascinating, thanks! I knew that capacitor ballasts were very hard on
lamps, but I wasn't entirely clear why, this makes perfect sense.
 
J

James Sweet

Jan 1, 1970
0
Salmon said:
You probably know this already but use an isolation transformer if you
are going to use a single scope probe. You do not want a conductive
connection from the line to the protective ground on the scope. Two
probes in a differential input can be used relatively safely, but life
with an isolation transformer will be easier.

Bill


Yeah I always use an isolation transformer for things like this, there's
no other reasonably safe way to do it, I don't want to risk myself or my
equipment.
 
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