Maker Pro
Maker Pro

led dimmer circuit

P

Pooh Bear

Jan 1, 1970
0
mc said:
Will someone explain *why* resistors "won't do the job"?

This is not a basic question. I am not the original poster. I understand
about PWM and its advantages (saving power). But if an op-amp adjustable
current source (which is not PWM) will do the job, surely so will a humble
rheostat.

Is there something strange about the current-vs-brightness relation of these
LEDs?

It's partly to overcome non-linearity caused by the led's forward voltage.

Graham
 
D

Don Bowey

Jan 1, 1970
0
Will someone explain *why* resistors "won't do the job"?

This is not a basic question. I am not the original poster. I understand
about PWM and its advantages (saving power). But if an op-amp adjustable
current source (which is not PWM) will do the job, surely so will a humble
rheostat.

Is there something strange about the current-vs-brightness relation of these
LEDs?

With just resistors, the LED will abruptly extinguish when the voltage goes
down to a low enough level, but just before that point it may not be very
well dimmed.

With a PWM the delivered voltage amplitude is constant, but the time it is
on becomes shorter and shorter causing it to appear dimming.

Don
 
M

Michael A. Terrell

Jan 1, 1970
0
mc said:
Will someone explain *why* resistors "won't do the job"?

This is not a basic question. I am not the original poster. I understand
about PWM and its advantages (saving power). But if an op-amp adjustable
current source (which is not PWM) will do the job, surely so will a humble
rheostat.

Is there something strange about the current-vs-brightness relation of these
LEDs?


Actually, it is a very basic question. LEDs are current operated
devices, and quite non-linear. PWM dimming is the way to go.


--
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
 
D

Don Lancaster

Jan 1, 1970
0
Michael said:
Actually, it is a very basic question. LEDs are current operated
devices, and quite non-linear. PWM dimming is the way to go.
So, use a SERIES RHEOSTAT (not a potentiometer) to regulate the CURRENT.
LINEARLY with resistance.

Be sure to leave a small fixed resistance in series.



--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email: [email protected]

Please visit my GURU's LAIR web site at http://www.tinaja.com
 
J

Jonathan Kirwan

Jan 1, 1970
0
i would like a simple circuit that could be used to dim a high
brightness white led for installation inside a bird box for a colour
camera. i don't want to scare the birds by a simple on / off . A
gradule increase in brightness would be good. i have tried a resistor
in series followed by another variable resistor which works to a degree
but there comes a point when the brightness suddenly increases with a
small turn of the pot. could anyone post a simple cicuit please ?

This one might be a little easier, less battery drain and more readily
available parts- use any small signal transistors- the variable resistor
is totally linearized:
View in a fixed-width font such as Courier.

.
.
. .---------+-----------------+---------.
. | | | |
. | | [10K] |
. | | | |
. | | | |
. | | | |<
. | | +-------|
. | [10K] | |\
. | | | |
. | + | | .----+-----.
. ----- | | | |
. | | | | | ---
. | | +-[2.2K]-. | [2.2k] \ / ~~~
. | 9V | | | | | --- LED
. | | >| | |/ | |
. | | |------+------| '----+-----'
. ----- /| | |> |
. | - | | | |
. | | |_ +---------'
. | | |/| |
. | | [10K] |
. | | /| [100]
. | | | |
. | | | |
. | | | |
. '---------+--------+--------'
.
.
.

For those interested in one hobbyist design walkthrough, I've redrawn
this with an additional resistor, some labels added, no resistor
values to start, and some commentary.
: +V
: |
: |
: Current \
: source / R2
: going \
: into R3 /
: + R4 |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \
: | /
: | |
: | | +V
: e>| | |
: Q1 |------+ |
: c/| | \ +V
: | | / R6 |
: | | \ |
: | | / |
: | | : current | |
: | | : set by | |
: gnd | : R1 | |<e
: | : +------| Q3
: | v | |\c
: | | |
: | | +-------,
: | | | |
: | | | ---
: | |/c \ \ / LED
: +-----------| Q2 / R7 --- D1
: | |>e \ |
: ,---+ | / |
: |pot| emitter| | |
: | \ follower| | |
: | / R3 +--------+-------'
: '-> \ |
: / \
: Uses above | / R5
: current | \
: source to \ /
: set Q2 base / R4 |
: voltage \ |
: / |
: | |
: gnd gnd
:

Q1, R1, and R2 form a current source, with the current set by Q1's Vbe
across R1. (Q1's Ic through R2 allows voltage compliance for the node
at Q2's base.) This current source drives a current through R3 and R4
and thereby sets the linearly adjustable voltage to the base of Q2,
which is operating as an emitter follower.

Q2's emitter voltage sets the current through R5 and therefore also
through the LED. Q2's collector current also supplies the drive
current needed by Q3 to supply the LED's (and R5's) current. But as
this collector current will only be a tiny fraction of Q3's collector
current, it doesn't materially affect the LED current, which is
primary current path for R5's current.

R6 acts as a passive pull-up for Q2's collector.

R7, I believe, is present as a simple passive path partly because of
the non-linear active response of any diode and its capacitance, which
may cause oscillation in such a circuit at lower current settings. But
I may be wrong about that, so it would be nice to hear about the exact
thinking. In any case, I believe it should be set so as to account
for only a tenth or twentieth of the LED current.

I added R4 to provide a small voltage at Q2's base even when R3 is
completely at the 0-ohm end of its sweep. I added it to make the
point that, without it, Q2 will pinch off its emitter voltage and shut
down the LED current _before_ R3 reaches that end of its range, --
leaving a small part of its sweep "dead". It's not in any way
important, though. Mostly, I'm just using it to make that behavior
manifest.

Before calculating values, one might have considered just doing it
this way:
:
: +V
: |
: |
: \
: / R2
: \
: /
: |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \
: | /
: | |
: | |
: e>| |
: Q1 |------+
: c/| |
: | | +V
: | | |
: | | : current |
: | | : set by |
: | | : R1 ---
: gnd | : \ / LED
: | v ---
: | |
: | |
: | |
: | |/c
: +-----------| Q2
: | |>e
: ,---+ |
: |pot| emitter|
: | \ follower|
: | / R3 |
: '-> \ |
: / \
: | / R5
: | \
: \ /
: / R4 |
: \ |
: / gnd
: |
: gnd

But then the base current drive required for Q2 would have siphoned
off substantially more current and this would then have required that
the Q1 current source be much "stiffer" to keep the Q2 base drive
impact at the same relatively small proportion. It also would then
affect the range of values allowed for the R3 potentiometer (and
almost certainly excluded the possibility of a 10k pot, which is
common.) The slightly more complex drive stage suggested requires
only a very modest current drive by allowing a third transistor handle
sourcing the current into the LED and requiring Q2 to only have to
supply its base current drive.

Returning to the very slightly modified topology I presented, the
design approach might be like this, the inputs being:

(A) LED maximum current = 20mA
(B) LED voltage at this current = 4V (typical for white)
(C) Expected betas of Q2 and Q3 of, say, >= 50. (This is likely
true, if they will be operating outside their saturated regions.)

If Q3 is to operate outside of saturation, then Q3's Vce >= 1V (or
so.) With VLED at 4V, this means there is 5V from the supply to Q2's
emitter. Allowing at least a few volts of control across R5, this
suggests a supply of at least 8V or so. 9V is just fine. This
confirms Fred's choice of a 9V rail, I think. A higher voltage would
be used, though, if several LEDs are to be used in series.

So,

(D) Supply voltage is 9V.

The Q1 current source needs to supply substantially more current (very
much greater than) than Q2's base requires. Assuming this should be
about 50 times more (let's say) and assuming that Q2 and Q3 can be
relied upon for beta's of 50 each, this means that the current source
should supply (LED current / 50 / 50) * 50 or about 400uA. Different
assumptions could be used, but this represents a reasonable choice, I
think. So,

(E) Q1 current source should roughly target 400uA.

Q1 could operate with its minimum Ic designed to be close to zero, but
doing so would mean that it's Vbe would vary substantially over its
compliance range as more Ic becomes required and then represents, say,
orders of magnitude change. (Vbe would vary 60mV/decade of Ic, 18mV
per doubling.) To keep this variability within say 18mV or so, the
minimum Ic should be about one half of the targeted 400uA, or about
200uA.

With 5V used up by Q3's minimum Vce of 1V plus the LED's required 4V,
this leaves about 4V or so at Q2's emitter. To allow a little room
for error margin, assume about 3.6V and let Q3's Vce have a little
more. This means that Q2's base will be about 4.25V, or so, when the
LED is at 20mA. Since Q1's Vbe will be close to 0.65V itself, this
means that R2 will have (9V-4.25V-0.65V) or about 4.1V across it.
Given that it will also have to carry the 400uA for the current source
plus another 200uA of Q1's Ic current, just established, the value of
R2 should be 4.1V/(400uA+200uA) or about 6800 ohms. Luckily, this is
a standard value so use it.

(F) R2 = 6800 ohms.

Also, R1 will be 0.65V/400uA or 1625 ohms. A lower standard value is
okay:

(G) R1 = 1500 ohms.

Now, it is appropriate to consider R3. Assuming R4 will only have a
few tenths of a volt across it, R3 will have the rest or about 4V.
Given a current source of 400uA or so, this suggests a potentiometer
value of 4V/400uA or 10k. Not bad. Looks like a 10k pot is a good
choice under the circumstances and luckily widely available. Use it.

(H) R3 = 10k potentiometer

R3 can be a linear potentiometer for linear control of the current or
else a logarithmic potentiometer if apparent brightness due to our
eye's log response is appropriate.

As R3's resistance nears zero, Q2's base will go to zero volts if R4
weren't present. Before that, Q2's emitter will be driven to ground
and the LED current will halt. So if some of the wiper range of R3 is
to be recovered for light control (say 10% of it), then R4 might be
set to take up perhaps 0.2V or so at 400uA. This means 0.2V/400uA or
500 ohms. A 470 would do.

(I) R4 = 470 ohms.

So with 400uA driven through (10k + 470), Q2's base should be able to
reach almost 4.2V. The emitter voltage will likely be able to get
close to the 3.6V mentioned before -- perhaps 3.55V. This means that
R5 can be set, assuming the LED current and a small portion more for
R7. Let's say R7 will have 1mA when the LED is operating at 20mA:

(J) R7 = 4V / 1mA = 4k... use 3.9k ohms.

Then, R5 will have 20mA from the LED, 1mA from R7, and about 1/50th of
21mA (sum of LED + R7) for Q3's base drive arriving at R5 via Q2's
collector/emitter path. That's a total of 21mA+21mA/50 or about
21.4mA. So figure R5:

(K) R5 = 3.55V/21.4mA = 166 ohms... use 150 to reach over 20mA.

Selecting a slightly lower value for R5, than calculated, means that
the actual control range will reach closer to 23mA (the 3.55V estimate
divided by the 150 ohm value.) But that's just fine, probably.

What's left is R6. In this case, pulling up at about one tenth of the
required Q3 base drive is probably okay. Since Q3's base drive is
about 1/50th of the 21mA, this means about 1/500th of 21mA. The Vbe
of Q3 will be about 0.7V, so this suggests 0.7V/(21mA/500) or 16.6k. I
think anything from 10k to 22k would be a reasonable choice.

(L) R6 = 10k ohms.

That's it. Substituting, we have:
: +9
: |
: |
: \
: / R2
: \ 6800
: /
: |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \ 1500
: | /
: | |
: | | +9
: e>| | |
: Q1 |------+ |
: c/| | \ +9
: | | / R6 |
: | | \ 10k |
: | | / |
: | | | |
: | | | |
: gnd | | |<e
: | +------| Q3
: | | |\c
: | | |
: | | +-------,
: | | | |
: | | | ---
: | |/c \ \ / LED
: +-----------| Q2 / R7 --- D1
: | |>e \ 3900 |
: ,---+ | / |
: |pot| | | |
: | \ | | |
: | / R3 +--------+-------'
: '-> \ 10k |
: / \
: | / R5
: | \ 150
: \ /
: / R4 |
: \ 470 |
: / |
: | |
: gnd gnd
:

Not much different.

Just to re-iterate, the reason I think that R7 is used in Fred's
circuit is because I've tried to wire up simple constant current
circuits and drive a diode at tiny currents with oscillation
resulting. Worked just fine with various resistors, but oscillated
with a diode and the current control set low. My guess is that this
is the reason for including it. But I'd like to see the reasoning and
calculations exposed, if anyone wants to add that. I'd learn from it.

Jon
 
F

Fred Bloggs

Jan 1, 1970
0
Jonathan said:
i would like a simple circuit that could be used to dim a high
brightness white led for installation inside a bird box for a colour
camera. i don't want to scare the birds by a simple on / off . A
gradule increase in brightness would be good. i have tried a resistor
in series followed by another variable resistor which works to a degree
but there comes a point when the brightness suddenly increases with a
small turn of the pot. could anyone post a simple cicuit please ?

This one might be a little easier, less battery drain and more readily
available parts- use any small signal transistors- the variable resistor
is totally linearized:
View in a fixed-width font such as Courier.

.
.
. .---------+-----------------+---------.
. | | | |
. | | [10K] |
. | | | |
. | | | |
. | | | |<
. | | +-------|
. | [10K] | |\
. | | | |
. | + | | .----+-----.
. ----- | | | |
. | | | | | ---
. | | +-[2.2K]-. | [2.2k] \ / ~~~
. | 9V | | | | | --- LED
. | | >| | |/ | |
. | | |------+------| '----+-----'
. ----- /| | |> |
. | - | | | |
. | | |_ +---------'
. | | |/| |
. | | [10K] |
. | | /| [100]
. | | | |
. | | | |
. | | | |
. '---------+--------+--------'
.
.
.


For those interested in one hobbyist design walkthrough, I've redrawn
this with an additional resistor, some labels added, no resistor
values to start, and some commentary.

: +V
: |
: |
: Current \
: source / R2
: going \
: into R3 /
: + R4 |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \
: | /
: | |
: | | +V
: e>| | |
: Q1 |------+ |
: c/| | \ +V
: | | / R6 |
: | | \ |
: | | / |
: | | : current | |
: | | : set by | |
: gnd | : R1 | |<e
: | : +------| Q3
: | v | |\c
: | | |
: | | +-------,
: | | | |
: | | | ---
: | |/c \ \ / LED
: +-----------| Q2 / R7 --- D1
: | |>e \ |
: ,---+ | / |
: |pot| emitter| | |
: | \ follower| | |
: | / R3 +--------+-------'
: '-> \ |
: / \
: Uses above | / R5
: current | \
: source to \ /
: set Q2 base / R4 |
: voltage \ |
: / |
: | |
: gnd gnd
:


Q1, R1, and R2 form a current source, with the current set by Q1's Vbe
across R1. (Q1's Ic through R2 allows voltage compliance for the node
at Q2's base.) This current source drives a current through R3 and R4
and thereby sets the linearly adjustable voltage to the base of Q2,
which is operating as an emitter follower.

Q2's emitter voltage sets the current through R5 and therefore also
through the LED. Q2's collector current also supplies the drive
current needed by Q3 to supply the LED's (and R5's) current. But as
this collector current will only be a tiny fraction of Q3's collector
current, it doesn't materially affect the LED current, which is
primary current path for R5's current.

R6 acts as a passive pull-up for Q2's collector.

R7, I believe, is present as a simple passive path partly because of
the non-linear active response of any diode and its capacitance, which
may cause oscillation in such a circuit at lower current settings. But
I may be wrong about that, so it would be nice to hear about the exact
thinking. In any case, I believe it should be set so as to account
for only a tenth or twentieth of the LED current.

I added R4 to provide a small voltage at Q2's base even when R3 is
completely at the 0-ohm end of its sweep. I added it to make the
point that, without it, Q2 will pinch off its emitter voltage and shut
down the LED current _before_ R3 reaches that end of its range, --
leaving a small part of its sweep "dead". It's not in any way
important, though. Mostly, I'm just using it to make that behavior
manifest.

Before calculating values, one might have considered just doing it
this way:

:
: +V
: |
: |
: \
: / R2
: \
: /
: |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \
: | /
: | |
: | |
: e>| |
: Q1 |------+
: c/| |
: | | +V
: | | |
: | | : current |
: | | : set by |
: | | : R1 ---
: gnd | : \ / LED
: | v ---
: | |
: | |
: | |
: | |/c
: +-----------| Q2
: | |>e
: ,---+ |
: |pot| emitter|
: | \ follower|
: | / R3 |
: '-> \ |
: / \
: | / R5
: | \
: \ /
: / R4 |
: \ |
: / gnd
: |
: gnd


But then the base current drive required for Q2 would have siphoned
off substantially more current and this would then have required that
the Q1 current source be much "stiffer" to keep the Q2 base drive
impact at the same relatively small proportion. It also would then
affect the range of values allowed for the R3 potentiometer (and
almost certainly excluded the possibility of a 10k pot, which is
common.) The slightly more complex drive stage suggested requires
only a very modest current drive by allowing a third transistor handle
sourcing the current into the LED and requiring Q2 to only have to
supply its base current drive.

Returning to the very slightly modified topology I presented, the
design approach might be like this, the inputs being:

(A) LED maximum current = 20mA
(B) LED voltage at this current = 4V (typical for white)
(C) Expected betas of Q2 and Q3 of, say, >= 50. (This is likely
true, if they will be operating outside their saturated regions.)

If Q3 is to operate outside of saturation, then Q3's Vce >= 1V (or
so.) With VLED at 4V, this means there is 5V from the supply to Q2's
emitter. Allowing at least a few volts of control across R5, this
suggests a supply of at least 8V or so. 9V is just fine. This
confirms Fred's choice of a 9V rail, I think. A higher voltage would
be used, though, if several LEDs are to be used in series.

So,

(D) Supply voltage is 9V.

The Q1 current source needs to supply substantially more current (very
much greater than) than Q2's base requires. Assuming this should be
about 50 times more (let's say) and assuming that Q2 and Q3 can be
relied upon for beta's of 50 each, this means that the current source
should supply (LED current / 50 / 50) * 50 or about 400uA. Different
assumptions could be used, but this represents a reasonable choice, I
think. So,

(E) Q1 current source should roughly target 400uA.

Q1 could operate with its minimum Ic designed to be close to zero, but
doing so would mean that it's Vbe would vary substantially over its
compliance range as more Ic becomes required and then represents, say,
orders of magnitude change. (Vbe would vary 60mV/decade of Ic, 18mV
per doubling.) To keep this variability within say 18mV or so, the
minimum Ic should be about one half of the targeted 400uA, or about
200uA.

With 5V used up by Q3's minimum Vce of 1V plus the LED's required 4V,
this leaves about 4V or so at Q2's emitter. To allow a little room
for error margin, assume about 3.6V and let Q3's Vce have a little
more. This means that Q2's base will be about 4.25V, or so, when the
LED is at 20mA. Since Q1's Vbe will be close to 0.65V itself, this
means that R2 will have (9V-4.25V-0.65V) or about 4.1V across it.
Given that it will also have to carry the 400uA for the current source
plus another 200uA of Q1's Ic current, just established, the value of
R2 should be 4.1V/(400uA+200uA) or about 6800 ohms. Luckily, this is
a standard value so use it.

(F) R2 = 6800 ohms.

Also, R1 will be 0.65V/400uA or 1625 ohms. A lower standard value is
okay:

(G) R1 = 1500 ohms.

Now, it is appropriate to consider R3. Assuming R4 will only have a
few tenths of a volt across it, R3 will have the rest or about 4V.
Given a current source of 400uA or so, this suggests a potentiometer
value of 4V/400uA or 10k. Not bad. Looks like a 10k pot is a good
choice under the circumstances and luckily widely available. Use it.

(H) R3 = 10k potentiometer

R3 can be a linear potentiometer for linear control of the current or
else a logarithmic potentiometer if apparent brightness due to our
eye's log response is appropriate.

As R3's resistance nears zero, Q2's base will go to zero volts if R4
weren't present. Before that, Q2's emitter will be driven to ground
and the LED current will halt. So if some of the wiper range of R3 is
to be recovered for light control (say 10% of it), then R4 might be
set to take up perhaps 0.2V or so at 400uA. This means 0.2V/400uA or
500 ohms. A 470 would do.

(I) R4 = 470 ohms.

So with 400uA driven through (10k + 470), Q2's base should be able to
reach almost 4.2V. The emitter voltage will likely be able to get
close to the 3.6V mentioned before -- perhaps 3.55V. This means that
R5 can be set, assuming the LED current and a small portion more for
R7. Let's say R7 will have 1mA when the LED is operating at 20mA:

(J) R7 = 4V / 1mA = 4k... use 3.9k ohms.

Then, R5 will have 20mA from the LED, 1mA from R7, and about 1/50th of
21mA (sum of LED + R7) for Q3's base drive arriving at R5 via Q2's
collector/emitter path. That's a total of 21mA+21mA/50 or about
21.4mA. So figure R5:

(K) R5 = 3.55V/21.4mA = 166 ohms... use 150 to reach over 20mA.

Selecting a slightly lower value for R5, than calculated, means that
the actual control range will reach closer to 23mA (the 3.55V estimate
divided by the 150 ohm value.) But that's just fine, probably.

What's left is R6. In this case, pulling up at about one tenth of the
required Q3 base drive is probably okay. Since Q3's base drive is
about 1/50th of the 21mA, this means about 1/500th of 21mA. The Vbe
of Q3 will be about 0.7V, so this suggests 0.7V/(21mA/500) or 16.6k. I
think anything from 10k to 22k would be a reasonable choice.

(L) R6 = 10k ohms.

That's it. Substituting, we have:

: +9
: |
: |
: \
: / R2
: \ 6800
: /
: |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \ 1500
: | /
: | |
: | | +9
: e>| | |
: Q1 |------+ |
: c/| | \ +9
: | | / R6 |
: | | \ 10k |
: | | / |
: | | | |
: | | | |
: gnd | | |<e
: | +------| Q3
: | | |\c
: | | |
: | | +-------,
: | | | |
: | | | ---
: | |/c \ \ / LED
: +-----------| Q2 / R7 --- D1
: | |>e \ 3900 |
: ,---+ | / |
: |pot| | | |
: | \ | | |
: | / R3 +--------+-------'
: '-> \ 10k |
: / \
: | / R5
: | \ 150
: \ /
: / R4 |
: \ 470 |
: / |
: | |
: gnd gnd
:


Not much different.

Just to re-iterate, the reason I think that R7 is used in Fred's
circuit is because I've tried to wire up simple constant current
circuits and drive a diode at tiny currents with oscillation
resulting. Worked just fine with various resistors, but oscillated
with a diode and the current control set low. My guess is that this
is the reason for including it. But I'd like to see the reasoning and
calculations exposed, if anyone wants to add that. I'd learn from it.

Jon

Whew- that's way more thinking than I put into it. My approach was to
first enable the circuit to run off a 9V battery with end-of-life
defined as ~7V at 20mA maximum LED current, and use a 10K pot to make a
VCCS. The most common VCCS samples the current as the voltage developed
across an emitter resistor and the main error sources arise from the
base current and series Vbe. The 10K pot requirement pretty much
dictates a high impedance input which in turn dictates current
multiplication either using a Darlington or complementary emitter
follower. The complementary follower puts at least one Vbe error source
inside the feedback loop, in addition to mitigating base current error,
so it is the better choice of the two. The next step is to settle on the
order of the emitter current sampling voltage which should be large
enough to adequately swamp the Vbe variation due to a 20:1 CE current
variation and reasonable estimates of temperature variation, and small
enough for circuit operation down to 7V. So allowing for total Vbe
variation of 100mV and 5% as more than adequate linearity, an emitter
voltage of ~2V is a good choice, and this allows for a minimum of
7V-(2V+3.5V)=1.5V for Q3 Vce. So right off, you can pick R5=100R for ~2V
at 20mA. This then defines the maximum Q2 base voltage of 2V+0.65=2.65V
for something like 2.7V across the 10K rheostat, or 270uA, throw in an
extra 10uA for Q2 Ib. So allowing for Q1 Vbe of 0.65V,
R1=0.65V/(280uA)=2.2K. Then R2 is roughly evaluated as worst case
keep-alive current for Q1, which occurs at Q1 Vb of ~2.8V and 7V power
supply. At R2=10K, you have (7-2.8-0.65)=350uA through it, making for
350uA-280uA=70uA for Q1 Ic, this is adequate. Maximum Q1 current drain
becomes (9V-0.7V)/10K=830uA when supply is 9V and R3 is adjusted to
zero- this is fine. The ~60mV variation in Q1 Vbe makes more
60mV/2.2K=30uA x 10K=300mV and /100R for 3mA nonlinearity in output
current throughout the adjustment range which is again adequate for the
application. R7 is included to linearize the LED characteristic and buy
some headroom for Q2 startup by allowing both Q2/Q3 to be well into
conduction by the time the LED cutin voltage is reached.
 
J

Jonathan Kirwan

Jan 1, 1970
0
Jonathan said:
[email protected] wrote:

i would like a simple circuit that could be used to dim a high
brightness white led for installation inside a bird box for a colour
camera. i don't want to scare the birds by a simple on / off . A
gradule increase in brightness would be good. i have tried a resistor
in series followed by another variable resistor which works to a degree
but there comes a point when the brightness suddenly increases with a
small turn of the pot. could anyone post a simple cicuit please ?

This one might be a little easier, less battery drain and more readily
available parts- use any small signal transistors- the variable resistor
is totally linearized:
View in a fixed-width font such as Courier.

.
.
. .---------+-----------------+---------.
. | | | |
. | | [10K] |
. | | | |
. | | | |
. | | | |<
. | | +-------|
. | [10K] | |\
. | | | |
. | + | | .----+-----.
. ----- | | | |
. | | | | | ---
. | | +-[2.2K]-. | [2.2k] \ / ~~~
. | 9V | | | | | --- LED
. | | >| | |/ | |
. | | |------+------| '----+-----'
. ----- /| | |> |
. | - | | | |
. | | |_ +---------'
. | | |/| |
. | | [10K] |
. | | /| [100]
. | | | |
. | | | |
. | | | |
. '---------+--------+--------'
.
.
.


For those interested in one hobbyist design walkthrough, I've redrawn
this with an additional resistor, some labels added, no resistor
values to start, and some commentary.

: +V
: |
: |
: Current \
: source / R2
: going \
: into R3 /
: + R4 |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \
: | /
: | |
: | | +V
: e>| | |
: Q1 |------+ |
: c/| | \ +V
: | | / R6 |
: | | \ |
: | | / |
: | | : current | |
: | | : set by | |
: gnd | : R1 | |<e
: | : +------| Q3
: | v | |\c
: | | |
: | | +-------,
: | | | |
: | | | ---
: | |/c \ \ / LED
: +-----------| Q2 / R7 --- D1
: | |>e \ |
: ,---+ | / |
: |pot| emitter| | |
: | \ follower| | |
: | / R3 +--------+-------'
: '-> \ |
: / \
: Uses above | / R5
: current | \
: source to \ /
: set Q2 base / R4 |
: voltage \ |
: / |
: | |
: gnd gnd
:


Q1, R1, and R2 form a current source, with the current set by Q1's Vbe
across R1. (Q1's Ic through R2 allows voltage compliance for the node
at Q2's base.) This current source drives a current through R3 and R4
and thereby sets the linearly adjustable voltage to the base of Q2,
which is operating as an emitter follower.

Q2's emitter voltage sets the current through R5 and therefore also
through the LED. Q2's collector current also supplies the drive
current needed by Q3 to supply the LED's (and R5's) current. But as
this collector current will only be a tiny fraction of Q3's collector
current, it doesn't materially affect the LED current, which is
primary current path for R5's current.

R6 acts as a passive pull-up for Q2's collector.

R7, I believe, is present as a simple passive path partly because of
the non-linear active response of any diode and its capacitance, which
may cause oscillation in such a circuit at lower current settings. But
I may be wrong about that, so it would be nice to hear about the exact
thinking. In any case, I believe it should be set so as to account
for only a tenth or twentieth of the LED current.

I added R4 to provide a small voltage at Q2's base even when R3 is
completely at the 0-ohm end of its sweep. I added it to make the
point that, without it, Q2 will pinch off its emitter voltage and shut
down the LED current _before_ R3 reaches that end of its range, --
leaving a small part of its sweep "dead". It's not in any way
important, though. Mostly, I'm just using it to make that behavior
manifest.

Before calculating values, one might have considered just doing it
this way:

:
: +V
: |
: |
: \
: / R2
: \
: /
: |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \
: | /
: | |
: | |
: e>| |
: Q1 |------+
: c/| |
: | | +V
: | | |
: | | : current |
: | | : set by |
: | | : R1 ---
: gnd | : \ / LED
: | v ---
: | |
: | |
: | |
: | |/c
: +-----------| Q2
: | |>e
: ,---+ |
: |pot| emitter|
: | \ follower|
: | / R3 |
: '-> \ |
: / \
: | / R5
: | \
: \ /
: / R4 |
: \ |
: / gnd
: |
: gnd


But then the base current drive required for Q2 would have siphoned
off substantially more current and this would then have required that
the Q1 current source be much "stiffer" to keep the Q2 base drive
impact at the same relatively small proportion. It also would then
affect the range of values allowed for the R3 potentiometer (and
almost certainly excluded the possibility of a 10k pot, which is
common.) The slightly more complex drive stage suggested requires
only a very modest current drive by allowing a third transistor handle
sourcing the current into the LED and requiring Q2 to only have to
supply its base current drive.

Returning to the very slightly modified topology I presented, the
design approach might be like this, the inputs being:

(A) LED maximum current = 20mA
(B) LED voltage at this current = 4V (typical for white)
(C) Expected betas of Q2 and Q3 of, say, >= 50. (This is likely
true, if they will be operating outside their saturated regions.)

If Q3 is to operate outside of saturation, then Q3's Vce >= 1V (or
so.) With VLED at 4V, this means there is 5V from the supply to Q2's
emitter. Allowing at least a few volts of control across R5, this
suggests a supply of at least 8V or so. 9V is just fine. This
confirms Fred's choice of a 9V rail, I think. A higher voltage would
be used, though, if several LEDs are to be used in series.

So,

(D) Supply voltage is 9V.

The Q1 current source needs to supply substantially more current (very
much greater than) than Q2's base requires. Assuming this should be
about 50 times more (let's say) and assuming that Q2 and Q3 can be
relied upon for beta's of 50 each, this means that the current source
should supply (LED current / 50 / 50) * 50 or about 400uA. Different
assumptions could be used, but this represents a reasonable choice, I
think. So,

(E) Q1 current source should roughly target 400uA.

Q1 could operate with its minimum Ic designed to be close to zero, but
doing so would mean that it's Vbe would vary substantially over its
compliance range as more Ic becomes required and then represents, say,
orders of magnitude change. (Vbe would vary 60mV/decade of Ic, 18mV
per doubling.) To keep this variability within say 18mV or so, the
minimum Ic should be about one half of the targeted 400uA, or about
200uA.

With 5V used up by Q3's minimum Vce of 1V plus the LED's required 4V,
this leaves about 4V or so at Q2's emitter. To allow a little room
for error margin, assume about 3.6V and let Q3's Vce have a little
more. This means that Q2's base will be about 4.25V, or so, when the
LED is at 20mA. Since Q1's Vbe will be close to 0.65V itself, this
means that R2 will have (9V-4.25V-0.65V) or about 4.1V across it.
Given that it will also have to carry the 400uA for the current source
plus another 200uA of Q1's Ic current, just established, the value of
R2 should be 4.1V/(400uA+200uA) or about 6800 ohms. Luckily, this is
a standard value so use it.

(F) R2 = 6800 ohms.

Also, R1 will be 0.65V/400uA or 1625 ohms. A lower standard value is
okay:

(G) R1 = 1500 ohms.

Now, it is appropriate to consider R3. Assuming R4 will only have a
few tenths of a volt across it, R3 will have the rest or about 4V.
Given a current source of 400uA or so, this suggests a potentiometer
value of 4V/400uA or 10k. Not bad. Looks like a 10k pot is a good
choice under the circumstances and luckily widely available. Use it.

(H) R3 = 10k potentiometer

R3 can be a linear potentiometer for linear control of the current or
else a logarithmic potentiometer if apparent brightness due to our
eye's log response is appropriate.

As R3's resistance nears zero, Q2's base will go to zero volts if R4
weren't present. Before that, Q2's emitter will be driven to ground
and the LED current will halt. So if some of the wiper range of R3 is
to be recovered for light control (say 10% of it), then R4 might be
set to take up perhaps 0.2V or so at 400uA. This means 0.2V/400uA or
500 ohms. A 470 would do.

(I) R4 = 470 ohms.

So with 400uA driven through (10k + 470), Q2's base should be able to
reach almost 4.2V. The emitter voltage will likely be able to get
close to the 3.6V mentioned before -- perhaps 3.55V. This means that
R5 can be set, assuming the LED current and a small portion more for
R7. Let's say R7 will have 1mA when the LED is operating at 20mA:

(J) R7 = 4V / 1mA = 4k... use 3.9k ohms.

Then, R5 will have 20mA from the LED, 1mA from R7, and about 1/50th of
21mA (sum of LED + R7) for Q3's base drive arriving at R5 via Q2's
collector/emitter path. That's a total of 21mA+21mA/50 or about
21.4mA. So figure R5:

(K) R5 = 3.55V/21.4mA = 166 ohms... use 150 to reach over 20mA.

Selecting a slightly lower value for R5, than calculated, means that
the actual control range will reach closer to 23mA (the 3.55V estimate
divided by the 150 ohm value.) But that's just fine, probably.

What's left is R6. In this case, pulling up at about one tenth of the
required Q3 base drive is probably okay. Since Q3's base drive is
about 1/50th of the 21mA, this means about 1/500th of 21mA. The Vbe
of Q3 will be about 0.7V, so this suggests 0.7V/(21mA/500) or 16.6k. I
think anything from 10k to 22k would be a reasonable choice.

(L) R6 = 10k ohms.

That's it. Substituting, we have:

: +9
: |
: |
: \
: / R2
: \ 6800
: /
: |
: |
: ,--------+
: | |
: | |
: | \
: | / R1
: | \ 1500
: | /
: | |
: | | +9
: e>| | |
: Q1 |------+ |
: c/| | \ +9
: | | / R6 |
: | | \ 10k |
: | | / |
: | | | |
: | | | |
: gnd | | |<e
: | +------| Q3
: | | |\c
: | | |
: | | +-------,
: | | | |
: | | | ---
: | |/c \ \ / LED
: +-----------| Q2 / R7 --- D1
: | |>e \ 3900 |
: ,---+ | / |
: |pot| | | |
: | \ | | |
: | / R3 +--------+-------'
: '-> \ 10k |
: / \
: | / R5
: | \ 150
: \ /
: / R4 |
: \ 470 |
: / |
: | |
: gnd gnd
:


Not much different.

Just to re-iterate, the reason I think that R7 is used in Fred's
circuit is because I've tried to wire up simple constant current
circuits and drive a diode at tiny currents with oscillation
resulting. Worked just fine with various resistors, but oscillated
with a diode and the current control set low. My guess is that this
is the reason for including it. But I'd like to see the reasoning and
calculations exposed, if anyone wants to add that. I'd learn from it.

Jon

Whew- that's way more thinking than I put into it. My approach was to
first enable the circuit to run off a 9V battery with end-of-life
defined as ~7V at 20mA maximum LED current, and use a 10K pot to make a
VCCS. The most common VCCS samples the current as the voltage developed
across an emitter resistor and the main error sources arise from the
base current and series Vbe. The 10K pot requirement pretty much
dictates a high impedance input which in turn dictates current
multiplication either using a Darlington or complementary emitter
follower. The complementary follower puts at least one Vbe error source
inside the feedback loop, in addition to mitigating base current error,
so it is the better choice of the two. The next step is to settle on the
order of the emitter current sampling voltage which should be large
enough to adequately swamp the Vbe variation due to a 20:1 CE current
variation and reasonable estimates of temperature variation, and small
enough for circuit operation down to 7V. So allowing for total Vbe
variation of 100mV and 5% as more than adequate linearity, an emitter
voltage of ~2V is a good choice, and this allows for a minimum of
7V-(2V+3.5V)=1.5V for Q3 Vce. So right off, you can pick R5=100R for ~2V
at 20mA. This then defines the maximum Q2 base voltage of 2V+0.65=2.65V
for something like 2.7V across the 10K rheostat, or 270uA, throw in an
extra 10uA for Q2 Ib. So allowing for Q1 Vbe of 0.65V,
R1=0.65V/(280uA)=2.2K. Then R2 is roughly evaluated as worst case
keep-alive current for Q1, which occurs at Q1 Vb of ~2.8V and 7V power
supply. At R2=10K, you have (7-2.8-0.65)=350uA through it, making for
350uA-280uA=70uA for Q1 Ic, this is adequate. Maximum Q1 current drain
becomes (9V-0.7V)/10K=830uA when supply is 9V and R3 is adjusted to
zero- this is fine. The ~60mV variation in Q1 Vbe makes more
60mV/2.2K=30uA x 10K=300mV and /100R for 3mA nonlinearity in output
current throughout the adjustment range which is again adequate for the
application. R7 is included to linearize the LED characteristic and buy
some headroom for Q2 startup by allowing both Q2/Q3 to be well into
conduction by the time the LED cutin voltage is reached.

Thanks for exposing some of your own thinking. Very helpful.

I have actually designed and used the part of this exact configuration
from the emitter follower base to the load, beforehand, using a
different driver for the base voltage. Separately, I've used that
particular current source method (inverted) as part of a current
mirror.

Jon
 
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