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Am I right?

E

ehsjr

Jan 1, 1970
0
Tom said:
Um, the 1k resistor I suggested goes between the top of the LED string
and the LM317 adj pin. You yank down directly on the adj pin. The
LM317 output will follow 1.25V above that. So if you manage to fully
ground the adj pin, the output voltage will be 1.25V. 1.25V across a
series string of 3 LEDs will cause a current of perhaps nanoamps to
flow in the LEDs, depending on what kind they are. But it will be at
most a tiny current, even it they are pretty hot.

When the adj pin is allowed to "float", it will be held by the 1k
resistor to very nearly the voltage at the top of the LED string. The
LM317 output pin will be 1.25V above that. So to get 300mA current in
the LEDs, you put a resistor between the output pin and the top of the
LED string: 1.25V/300mA is about 4 ohms.

When you pull the adj pin low, the thing which pulls it low must
handle the current out of the adj pin of the LM317, at most 100uA
(from the data sheet), plus the current in the 1k resistor. Since the
LED current is very low now, with the adj pin held low, there will be
essentially no drop in the 4-ohm resistor, so 1.25V drops across the
1k resistor, causing 1.25mA current. So the net current in the "low-
puller" (whatever it may be) is 1.35mA at most. You do want to make
sure you only pull DOWN on the adj pin to turn things off. Do NOT
connect it directly to the output (pin 3) of a 555, that can pull it
UP. That would try to turn on the LEDs harder than you intend.
Instead, put a diode in (anode to the adj pin) so you can only pull
down if you have any doubts about your circuit. If the LEDs you're
driving have even a moderately high voltage drop, you can safely use
just about any silicon signal diode. You don't need a diode if you're
driving the adj pin low using an open-collector or open-drain
transistor.

You shouldn't need any capacitors on the output end of the LM317 for
all this to work.

Cheers,
Tom

It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.

Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |
+------------+ <-- total LED vF when the NPN
| |a is off, Vce when the NPN
| [LED] is conducting.
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"
 
T

Tom Bruhns

Jan 1, 1970
0
Um, the 1k resistor I suggested goes between the top of the LED string
and the LM317 adj pin. You yank down directly on the adj pin. The
LM317 output will follow 1.25V above that. So if you manage to fully
ground the adj pin, the output voltage will be 1.25V. 1.25V across a
series string of 3 LEDs will cause a current of perhaps nanoamps to
flow in the LEDs, depending on what kind they are. But it will be at
most a tiny current, even it they are pretty hot.
When the adj pin is allowed to "float", it will be held by the 1k
resistor to very nearly the voltage at the top of the LED string. The
LM317 output pin will be 1.25V above that. So to get 300mA current in
the LEDs, you put a resistor between the output pin and the top of the
LED string: 1.25V/300mA is about 4 ohms.
When you pull the adj pin low, the thing which pulls it low must
handle the current out of the adj pin of the LM317, at most 100uA
(from the data sheet), plus the current in the 1k resistor. Since the
LED current is very low now, with the adj pin held low, there will be
essentially no drop in the 4-ohm resistor, so 1.25V drops across the
1k resistor, causing 1.25mA current. So the net current in the "low-
puller" (whatever it may be) is 1.35mA at most. You do want to make
sure you only pull DOWN on the adj pin to turn things off. Do NOT
connect it directly to the output (pin 3) of a 555, that can pull it
UP. That would try to turn on the LEDs harder than you intend.
Instead, put a diode in (anode to the adj pin) so you can only pull
down if you have any doubts about your circuit. If the LEDs you're
driving have even a moderately high voltage drop, you can safely use
just about any silicon signal diode. You don't need a diode if you're
driving the adj pin low using an open-collector or open-drain
transistor.
You shouldn't need any capacitors on the output end of the LM317 for
all this to work.
Cheers,
Tom

It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.

Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |
+------------+ <-- total LED vF when the NPN
| |a is off, Vce when the NPN
| [LED] is conducting.
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"

Hi Ed,

OK, yours will work, BUT you have to sink 300mA to do it! If you put
the resistor in, then you only need to sink 1.25V/1k + the adj pin
current, max 100uA. With the 1k in there, the LED string sees 1.25V
plus Vce(sat), say 1.35V, but so what? Even red LEDs have negligible
current when hot at 0.5 volts each. Much less power hungry to add the
1k in there. The 1k is nominal, of course; it could be 220 ohms, or
2.2k ohms, though with 2.2k, I'd start worrying about the voltage drop
across the 2.2k when the LEDs light; that increases the drop across
the 4 ohm resistor by about 0.1 volts. 1k seemed a reasonable
compromise to get to low current in the "yank down control
transistor." Revised version, as I had posted, below. Thanks for
providing the framework to edit!

Cheers,
Tom
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R] (when they are lit)
| |
+--[1000R}---+ <-- total LED vF when the NPN
| |a is off, 1.25V+Vce(sat)
| [LED] when the NPN
/c | is conducting.
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"
 
T

Tom Bruhns

Jan 1, 1970
0
Um, the 1k resistor I suggested goes between the top of the LED string
and the LM317 adj pin. You yank down directly on the adj pin. The
LM317 output will follow 1.25V above that. So if you manage to fully
ground the adj pin, the output voltage will be 1.25V. 1.25V across a
series string of 3 LEDs will cause a current of perhaps nanoamps to
flow in the LEDs, depending on what kind they are. But it will be at
most a tiny current, even it they are pretty hot.
When the adj pin is allowed to "float", it will be held by the 1k
resistor to very nearly the voltage at the top of the LED string. The
LM317 output pin will be 1.25V above that. So to get 300mA current in
the LEDs, you put a resistor between the output pin and the top of the
LED string: 1.25V/300mA is about 4 ohms.
When you pull the adj pin low, the thing which pulls it low must
handle the current out of the adj pin of the LM317, at most 100uA
(from the data sheet), plus the current in the 1k resistor. Since the
LED current is very low now, with the adj pin held low, there will be
essentially no drop in the 4-ohm resistor, so 1.25V drops across the
1k resistor, causing 1.25mA current. So the net current in the "low-
puller" (whatever it may be) is 1.35mA at most. You do want to make
sure you only pull DOWN on the adj pin to turn things off. Do NOT
connect it directly to the output (pin 3) of a 555, that can pull it
UP. That would try to turn on the LEDs harder than you intend.
Instead, put a diode in (anode to the adj pin) so you can only pull
down if you have any doubts about your circuit. If the LEDs you're
driving have even a moderately high voltage drop, you can safely use
just about any silicon signal diode. You don't need a diode if you're
driving the adj pin low using an open-collector or open-drain
transistor.
You shouldn't need any capacitors on the output end of the LM317 for
all this to work.
Cheers,
Tom

It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.

Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |
+------------+ <-- total LED vF when the NPN
| |a is off, Vce when the NPN
| [LED] is conducting.
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"

Hi Ed,

OK, yours will work, BUT you have to sink 300mA to do it! If you put
the resistor in, then you only need to sink 1.25V/1k + the adj pin
current, max 100uA. With the 1k in there, the LED string sees 1.25V
plus Vce(sat), say 1.35V, but so what? Even red LEDs have negligible
current when hot at 0.5 volts each. Much less power hungry to add the
1k in there. The 1k is nominal, of course; it could be 220 ohms, or
2.2k ohms, though with 2.2k, I'd start worrying about the voltage drop
across the 2.2k when the LEDs light; that increases the drop across
the 4 ohm resistor by about 0.1 volts. 1k seemed a reasonable
compromise to get to low current in the "yank down control
transistor." Revised version, as I had posted, below. Thanks for
providing the framework to edit!

Cheers,
Tom
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R] (when they are lit)
| |
+--[1000R}---+ <-- total LED vF when the NPN
| |a is off, 1.25V+Vce(sat)
| [LED] when the NPN
/c | is conducting.
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"
 
T

Tom Bruhns

Jan 1, 1970
0
I -> ~300 mA when LEDs are on, about 1.35mA when the LEDs are off.
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R] (when they are lit)
| |
+--[1000R}---+ <-- total LED vF when the NPN
| |a is off, 1.25V+Vce(sat)
| [LED] when the NPN
/c | is conducting.
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+
YDC = "Yank Down Control"

Ooops. I failed to edit the top line in that diagram I posted
earlier. The current is about 300mA when the LEDs light, but only
about 1.35mA when they are off. The LED current is essentially zero,
the current in the 1k resistor is 1.25V/1k, and the adj pin current is
100uA max.

Cheers,
Tom
 
J

jasen

Jan 1, 1970
0
I have some 1 watt LED's that I want to flash on and off.

I have a 317 current regulator supply set at 300 ma for 3 of them in series.
I initially thought I'd use a 555 timer to have a constant flash rate
Why not this....which makes more sense to me....

Control the AC to the transformer with the same setup....flash a triac
driver opto LED with a 555, control on/off with the opto LED ground, use a
triac to switch on and off the transformer primary?

triacs don't like transformers.
That's what I have breadboarded now...works fine.

Don't you think it would be the proper way to go as far as a design
decision?......it lets the cc power supply be stable and avoids risk?

This seems easier to me.

+V -----[LM317]--[4R2]-+--->|-->|-->|--- 0V
| | | LED X 3
...... | | (300ma)
. 555. | |
.output---|<-+--[120R]---'
...... 1n914
| etc
|
--+-- 0V

if the 555's discharge pin is unused it can be used instead of the output
pin and the diode can be omitted.
 
J

jasen

Jan 1, 1970
0
It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.

Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |
+---[120R]---+ <--- add resistor here
| |a
| [LED]
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

the resistor reduces the current through the transistor to ~11mA when the
transistor is on, and when the transistor is off has essentially no effect.



Bye.
Jasen
 
E

ehsjr

Jan 1, 1970
0
Tom said:
It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.

Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |
+------------+ <-- total LED vF when the NPN
| |a is off, Vce when the NPN
| [LED] is conducting.
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"


Hi Ed,

OK, yours will work, BUT you have to sink 300mA to do it!

Thanks, Tom. Neat. Picture, words, 1000 !

Yes the circuit I drew always pulls 300 mA. Just
couldn't do the words/picture translation to get
what you were saying.
If you put
the resistor in, then you only need to sink 1.25V/1k + the adj pin
current, max 100uA. With the 1k in there, the LED string sees 1.25V
plus Vce(sat), say 1.35V, but so what? Even red LEDs have negligible
current when hot at 0.5 volts each. Much less power hungry to add the
1k in there.

Yes! Much better, and not just from power consumption,
but heat production is lower with your circuit.

Ed
The 1k is nominal, of course; it could be 220 ohms, or
2.2k ohms, though with 2.2k, I'd start worrying about the voltage drop
across the 2.2k when the LEDs light; that increases the drop across
the 4 ohm resistor by about 0.1 volts. 1k seemed a reasonable
compromise to get to low current in the "yank down control
transistor." Revised version, as I had posted, below. Thanks for
providing the framework to edit!

Cheers,
Tom

I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R] (when they are lit)
| |
+--[1000R}---+ <-- total LED vF when the NPN
| |a is off, 1.25V+Vce(sat)
| [LED] when the NPN
/c | is conducting.
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+

YDC = "Yank Down Control"
 
E

ehsjr

Jan 1, 1970
0
jasen said:
It can't work the way I understand your description - I'm
obviously not getting it. A schematic would help. Here's
a possibility for yanking down - feel free to modify
it to show what you have in mind. The 1K resistor from
the top of the LED string to the adj pin escapes me, so
it's not in the diagram, but it should work as drawn.

Ed
I -> ~300 mA
-----
Vcc + ----Vin|LM317|Vout-----+ <-- 1.25 v higher than
----- | the top of the LEDs.
Adj [4R]
| |

+---[120R]---+ <--- add resistor here
| |a
| [LED]
/c |
YDC >---| [LED]
\e |
| [LED]
| |
Gnd ------------+------------+


the resistor reduces the current through the transistor to ~11mA when the
transistor is on, and when the transistor is off has essentially no effect.



Bye.
Jasen

Thanks!

Ed
 
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