multi-led hi-side constant-current question

[JIW]

In my circuit at http://pat7.com/jp/q1.png (and .jpg)
I put a well-known constant-current circuit (2 transistors,
2 resistors, current ~ Vbe/R3 ~ .7V/24ohms ~ 29mA)
on the high side, with the intention of controlling
current to several LED's with 1 CC circuit, rather
than using several. I only turn on one LED at a time -
red, green, or yellow.

Figure 70 in AT90S2313 pdf shows a drop of .6
to .7V when an output sinks 30mA and Vcc=5V;
the CC circuit will drop about 1.4V; D1 drops .5
to .7V; which from 5.6V B1 leaves over only 2.8V,
too low for my green LED, so I have B2 ~ 1.4V
in series.

I haven't built this version yet, and would like to
know if anyone sees any dumb problems with it.
Also, when micro is in power-down sleep
(drawing < 1 microamp) with PD0,1,2 high, how
much leakage will there be through the LED's?
-jiw (posting w/ google while traveling)

 

[Ostry]

I has 2 batteries - it's rather unusual. It could be a problem when one of
those is low and you don't know which one. I think you should use only 1
battery and voltage stabilizer.
AT90S2313 can work at lower supply voltage, i.e. 3V, you could use just 2
AA batteries if these LEDs aren't white or blue.
D1 and C4 are not needed.
Is current source really needed? Use just 3 resistors (or even single
resistor if only one LED is on at time) connected to VCC.
You could add a simple reset circuit.
Add ISP interface to simplify software develop.

 

[[email protected]]

[Re http://pat7.com/jp/q1.png

[...] 2 batteries [...] could be a problem when one of those
is low and you don't know which one. I think you should use
only 1 battery and voltage stabilizer.

The 5.6 battery pack will wear out first (if both are installed
new at the same time) so I'm not concerned much about
that first problem. I'd rather not put a voltage regulator
in the circuit for a couple of reasons - 1, extra current
draw - I think I'd get only half the battery life. I measure
8 to 12 mA without a regulator (LEDs off) and about twice
that with a low-dropout 5V regulator in circuit. LEDs are
off about 90% of the time in my application. 2 - As shown
in http://pat7.com/jp/2313s2.sch.jpg (which shows more
of the whole circuit) I have a transistor controlling aux
power to 16x2 LCD. When I turn off aux power and go
into power-down sleep for a week between uses, current
drops to less than a microamp. With a regulator in
there, I would have to figure out a way to hibernate
it also, or a way to bootstrap power, etc etc

AT90S2313 can work at lower supply voltage, i.e. 3V, you
could use just 2 AA batteries if these LEDs aren't white or blue.

The AT90S2313-4 is ok at 2.7 to 6.0V but the few
hundred units I have on hand are AT90S2313-10's,
spec'ed for 4.0 to 6.0V. Also, the green LEDs I have
take over 3V to get enough brightness.

D1 and C4 are not needed.

Battery pack B1 starts out around 6.2 volts, a little
above the 6.0V spec for the AT90S2313.
However I see now the spec shows an absolute
max of 6.6V. Maybe I could get by without D1
in there to drop .5 - .7V, but it isn't a big deal.

Is current source really needed? Use just 3 resistors (or
even single resistor if only one LED is on at time) [...]

I built several units with resistors to limit current,
but LED's weren't bright enough over the whole
range of battery voltages, and weren't matched
well enough either.

Add ISP interface to simplify software develop.

I program them as in the AT90S2313s2 URL above,
ie with ponyprog and parallel cable.
-jiw

 

[ehsjr]

In my circuit at http://pat7.com/jp/q1.png (and .jpg)
I put a well-known constant-current circuit (2 transistors,
2 resistors, current ~ Vbe/R3 ~ .7V/24ohms ~ 29mA)
on the high side, with the intention of controlling
current to several LED's with 1 CC circuit, rather
than using several. I only turn on one LED at a time -
red, green, or yellow.

Figure 70 in AT90S2313 pdf shows a drop of .6
to .7V when an output sinks 30mA and Vcc=5V;
the CC circuit will drop about 1.4V; D1 drops .5
to .7V; which from 5.6V B1 leaves over only 2.8V,
too low for my green LED, so I have B2 ~ 1.4V
in series.

I haven't built this version yet, and would like to
know if anyone sees any dumb problems with it.
Also, when micro is in power-down sleep
(drawing < 1 microamp) with PD0,1,2 high, how
much leakage will there be through the LED's?
-jiw (posting w/ google while traveling)

Can you eliminate the CC? Then you can go with just B1,
and use a single current limiting resistor for the LEDs.

Can you use a different green LED with a lower Vf?
Ed

 

[J W]

In my circuit at [ http://pat7.com/jp/2313q1.png and http://pat7.com/jp/2313q2.png ]
I put a well-known constant-current circuit (2 transistors,
2 resistors, current ~ Vbe/R3 ~ .7V/24ohms ~ 29mA)
on the high side, with the intention of controlling
current to several LED's with 1 CC circuit, rather
than using several. I only turn on one LED at a time -
red, green, or yellow.

....
Can you eliminate the CC? Then you can go with just B1,
and use a single current limiting resistor for the LEDs.

I need constant current to get adequate LED drive for battery
voltages from 4 to 6V and LED voltages from 1.6V up to say 3.6V.
I've built and compared several units with and without CC circuitry.
Anyhow, to concentrate the discussion on the CC circuit I drew
a simpler version - see http://pat7.com/jp/2313q2.png - leaving
out the 2nd battery and the diode.

Can you use a different green LED with a lower Vf?

For this application (a compact, battery-powered speech timer with
red, yellow, and green signal lights intended to be obvious up to
10m away)the important issues are LED brightness and good battery
life. If a brighter LED has lower Vf, I can use it, but generally
not vice versa.
-jiw

 

[Bob Monsen]

In my circuit at http://pat7.com/jp/q1.png (and .jpg)

Your constant current source requires at least two diode voltage
drops, plus some. So, add the voltage across the LEDs to this, and you
end up with current decreasing when the voltage gets to about 3.5V.

If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.
| |
68R |
| |
e e
.-b .----b (PNP Transistors
| c | c
| | | |
'--o---' '--- LEDS & PORTS
|
'------------- YOUR CURRENT SINK

This arrangement will allow your vbatt to drop to about 2.5V before
the LEDs start to dim. This means you can probably get rid of that
1.4V battery.

Note that if you make the current sink supply about 1mA, that
will allow the LEDs to have 10mA. The 68 ohm resistor makes that happen.

Also, you should check the datasheet of your microcontroller to ensure
that the ports stay high-impedance when sleeping. I'd guess they do,
but you should check.

--
Regards,
Bob Monsen

A prude is a person who thinks that his own rules of propriety are
natural laws. You are almost entirely free of this prevalent evil.

 

[Jim Thompson]

Your constant current source requires at least two diode voltage
drops, plus some. So, add the voltage across the LEDs to this, and you
end up with current decreasing when the voltage gets to about 3.5V.

If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.
| |
68R |
| |
e e
.-b .----b (PNP Transistors
| c | c
| | | |
'--o---' '--- LEDS & PORTS
|
'------------- YOUR CURRENT SINK

This arrangement will allow your vbatt to drop to about 2.5V before
the LEDs start to dim. This means you can probably get rid of that
1.4V battery.

Note that if you make the current sink supply about 1mA, that
will allow the LEDs to have 10mA. The 68 ohm resistor makes that happen.

[snip]

Until the temperature changes ;-)

...Jim Thompson

 

[J W]

JIW wrote:

[re http://pat7.com/jp/2313q1.png ]

so if you lose D1 you can lose B2?

I was wrong to include D1 drop in the calculation, since it isn't
in the LED current path. I'm more concerned about whether the CC
circuit will work ok. See http://pat7.com/jp/2313q2.png that
removes the distractions of D1 and B2.

You could replace Q2 with a diode, and could then look at what diode
will give you low drop if you want.

Q2 is an active part of the constant current circuit. If current
thru R3 is below setpoint, Q2 partly shuts off, putting more current
into base of Q3, causing it to put more current thru R3. If R3
current is too high, Q2 shunts more current, causing Q3 to wane, AIUI.

If your batteries are removable, dont forget a parallel diode across
them, and preferably a fusible link as well.

Do you mean as protection in case batteries are reversed?

See micro's spec sheet for total consumption figure..That will include
any LED i.

Most currents are obvious -- eg 8-12mA with LEDs off, and 30-40mA with one
LED on, and less than 1 uA to the micro in powerdown sleep. What I don't
know is how much leakage there would be due to the about 2V difference
between top of R2 and bottom of LEDs in http://pat7.com/jp/2313q1.png .
My application is a small, battery-powered speech timer with red, yellow,
and green signal lights, intended to be bright and obvious up to 10m
away. LED brightness, good battery life, portability, and no-mains
operation are important. A set of alkaline AA's can power it for a few
years (with an hour or two of on time per week). 100 uA of leakage during
off time would cut battery life in half. (Power switch is electronic,
to allow automatic shutoff if no buttons are pushed for an hour.)

Returning to the B1-B2-D1 question, I plan to try using 3 AA's rather
than 4 for B1, 1 AA for B2, and leave out D1, because of no risk of
exceeding 6V operating voltage. End of useful life battery voltage
would be a little higher but that cost would be compensated by needing
only 4 AA's rather than 5; and leakage (if any) lower due to 1.5V
rather than 2V difference. In layout http://pat7.com/jp/2313q3.png
I put a header JP3 next to JP1 for easy battery/measurement hookups.
-jiw

 

[J W]

Bob Monsen wrote in sci.electronics.design:[ http://pat7.com/jp/2323q1.png or http://pat7.com/jp/2323q2.png ]

Your constant current source requires at least two diode voltage
drops, plus some. So, add the voltage across the LEDs to this, and you
end up with current decreasing when the voltage gets to about 3.5V. ....
If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.

[snipped circuit]

Thanks for the PNP suggestion - I'll try it out. In looking around
I've also seen some simple FET CC circuits I might try.

Also, you should check the datasheet of your microcontroller to ensure
that the ports stay high-impedance when sleeping. I'd guess they do,
but you should check.

To the best of my knowledge, the data sheet (eg page 27 of
http://www.atmel.com/dyn/resources/prod_documents/doc0839.pdf )
doesn't explicitly say that, except for "IO memory is retained". I've
verified that pins set to be inputs retain their characteristics
during powerdown sleep - either high-impedance or pulled high.
[Before sleep I set most pins to be high-impedance ("tri-stated")
inputs, and set PD2 (INT0) pulled high by its internal pullup. Wakeup
by INT0 requires a low level on PD2, which I get by a switch to
ground.]

For my circuit with a second battery, B2, where I was concerned about
leakage, for some reason I forgot that I set most pins to inputs for
sleep, i.e., was thinking of leakage from 7V on the CC high side to 5V
on output pins rather than from 7V to high-impedance. Anyway, either
case might be invalid because the Absolute Maximum Ratings (p. 72 in
doc0839.pdf) require IO pin voltages between -1V and Vcc+0.5V.

-jiw

 

[ehsjr]

[ http://pat7.com/jp/2313q1.png and http://pat7.com/jp/2313q2.png ]

...

Can you eliminate the CC? Then you can go with just B1,
and use a single current limiting resistor for the LEDs.

I need constant current to get adequate LED drive for battery
voltages from 4 to 6V and LED voltages from 1.6V up to say 3.6V.
I've built and compared several units with and without CC circuitry.
Anyhow, to concentrate the discussion on the CC circuit I drew
a simpler version - see http://pat7.com/jp/2313q2.png - leaving
out the 2nd battery and the diode.

Can you use a different green LED with a lower Vf?

For this application (a compact, battery-powered speech timer with
red, yellow, and green signal lights intended to be obvious up to
10m away)the important issues are LED brightness and good battery
life. If a brighter LED has lower Vf, I can use it, but generally
not vice versa.
-jiw

Here's a possible solution: use DC-DC converter chip to provide
a regulated Vout of your choosing to drive the LEDs. Set the
current limit with a resistor individually for each LED. Use
diodes from pins 2,3 and 6 of your micro to bring the negative
to the chip so that it runs only when you want a LED to light.
You won't need B2, and the chip & components will fit easily in
the footprint formerly dedicated to it.

I don't know if this applies to everyone - but I see a red led
a lot more easily than a green or a yellow LED. My guess is that
equal drive current through those colors will result in different
perceived brightness. Is that acceptable for your device?

Ed

 

[J W]

ehsjr wrote in sci.electronics.design:[Re http://pat7.com/jp/2313q1.png and http://pat7.com/jp/2313q2.png ]
....

Here's a possible solution: use DC-DC converter chip to provide
a regulated Vout of your choosing to drive the LEDs. Set the
current limit with a resistor individually for each LED. Use
diodes from pins 2,3 and 6 of your micro to bring the negative
to the chip so that it runs only when you want a LED to light.
You won't need B2, and the chip & components will fit easily in
the footprint formerly dedicated to it.

My long range plan is to change to a 3V micro with a few more IO pins;
use 2 C cells rather than 4 AA's; and use a micro-controlled switching
power supply for LED power. The AT90S2313 has 15 IO pins, not quite
enough to run a power supply along with keyboard, display, and lights.
Specialized LED driver chips, eg LT1618 from a few years ago, look
like they would work really well but cost a lot and be hard to get.

I don't know if this applies to everyone - but I see a red led
a lot more easily than a green or a yellow LED. My guess is that
equal drive current through those colors will result in different
perceived brightness. Is that acceptable for your device?

Among affordable ($.10-$1) 20-30mA LEDs I've tried, the brightest reds
give around 20000 mCd; yellow, 8000 mCd; green, 12000 mCd. Brightness
variation is undesirable but difficult to avoid at low cost. So, what
I want to do is run high-brightness LEDs at about 25mA over a wide
range of battery voltage. I've also tried a variety of optical
arrangements for least light loss and best visibility (still
researching that) and have used 2 yellows in series, white LEDs with
color filters, & red-green combination LEDs. For example, see last
picture at http://pat7.com/timer . With C or D cells I probably could
move up to 350mA LEDs or parallel LED strings.
-jiw

 

[[email protected]]

[snip]

Until the temperature changes ;-)

...Jim Thompson

Its the temp diff between the 2 trs that would be the issue, and the
LED's tr is only dissipating 75 mW or so max. @50C/watt thats 3.75C. If
we subtract the 10% temp rise of the other tr, thats apx 3.4C temp
diff. How much variation will that cause?


NT

 

[[email protected]]

[re http://pat7.com/jp/2313q1.png ]

so if you lose D1 you can lose B2?

I was wrong to include D1 drop in the calculation, since it isn't
in the LED current path. I'm more concerned about whether the CC
circuit will work ok. See http://pat7.com/jp/2313q2.png that
removes the distractions of D1 and B2.


Q2 is an active part of the constant current circuit. If current
thru R3 is below setpoint, Q2 partly shuts off, putting more current
into base of Q3, causing it to put more current thru R3. If R3
current is too high, Q2 shunts more current, causing Q3 to wane, AIUI.

I had a brain fart re the voltage drops. Since youre using a silicon tr
to supply the LEDs (I was thinking back to the old germanium
implementations) you'd need either 2 diodes in series to replace your
Q2, or one si diode with a ge tr, neither of which is as good as I was
imagining.

Do you mean as protection in case batteries are reversed?

yes, though of course I dont know what your battery arrangement is.

Most currents are obvious -- eg 8-12mA with LEDs off, and 30-40mA with one
LED on, and less than 1 uA to the micro in powerdown sleep. What I don't
know is how much leakage there would be due to the about 2V difference

This is one of the reasons I suggested a single battery option. From
the max specs quoted it appears you could happily run both micro and
LEDs from the same 6v pack. In which case youre then in known
guaranteed specs territory, and arent risking a failure rate, or
premature battery depletion, due to unspecified operating conditions.
Also it gives the advantage of a single battery. Again I dont know who
your users are, but users can be very dense, and if you have 2
batteries runing out at different times, you will get unnecessary unit
returns.

Re premature batt depletion, if you cant guarantee your current draw
while off, and you need the system to run for so long on your cells,
you have a problem. At any time you may get ICs that greatly exceed
your max leakage specs under these conditions, then you have a whole
batch of problem units. I dont like it.

between top of R2 and bottom of LEDs in http://pat7.com/jp/2313q1.png .
My application is a small, battery-powered speech timer with red, yellow,
and green signal lights, intended to be bright and obvious up to 10m
away. LED brightness, good battery life, portability, and no-mains
operation are important. A set of alkaline AA's can power it for a few
years (with an hour or two of on time per week). 100 uA of leakage during
off time would cut battery life in half. (Power switch is electronic,
to allow automatic shutoff if no buttons are pushed for an hour.)

Returning to the B1-B2-D1 question, I plan to try using 3 AA's rather
than 4 for B1, 1 AA for B2, and leave out D1, because of no risk of
exceeding 6V operating voltage.

From what I half remember, your max permited voltage is a bit below

what you could see from a 6v pack, so this issueful configuration
shouldnt be necessary. IIRC the micro will cope with a bit more than
6v.

End of useful life battery voltage
would be a little higher but that cost would be compensated by needing
only 4 AA's rather than 5; and leakage (if any) lower due to 1.5V
rather than 2V difference.

So a single batt pack looks better on all those points and more.

In layout http://pat7.com/jp/2313q3.png
I put a header JP3 next to JP1 for easy battery/measurement hookups.
-jiw

NT

 

[[email protected]]

JIW wrote:

[re http://pat7.com/jp/2313q1.png ]

Now I'm going to suggest a different approach, see if its got legs. How
about if you first use a fixed R to set i_LED to give you the i you
need with your lowest V_batt and highest LED V_drop, then use the micro
to modulate (chop) the LED drive when i_LED runs higher than this? I'm
not familiar with the chip at all, so dont know if theres an A/D
available, but if there is it should be fairly easy to do. An R senses
i_LED, the micro chops the LED drive to give the desired average i_LED.

If your A/D needs a V to sense referenced to 0v, you can sense the
whole circuit i instead of i_LED, as your task is only to approximately
constantify the i_LED.

If theres no A/D, there are possibly ways round that, maybe.

It begs some questions, but does this go anywhere? If it does, you'd
lose your V drop issues, and cut component count.


NT

 

[J W]

[Re http://pat7.com/jp/2313q1.png ]
....

[snip]

Until the temperature changes ;-)

...Jim Thompson

Its the temp diff between the 2 trs that would be the issue, and the
LED's tr is only dissipating 75 mW or so max. @50C/watt thats 3.75C. If
we subtract the 10% temp rise of the other tr, thats apx 3.4C temp
diff. How much variation will that cause?

Each Vbe would shift about 2 mV/K (for silicon near room temperature
obeying (kT/q)*ln(I/Is) equation) so a 3.4K difference gives about 7
mV or 1% deviation. I'm not concerned about variation that small, but
if I were would use a 2-transistor die.
-jiw

 

[Jim Thompson]

[email protected] wrote in sci.electronics.design:[Re http://pat7.com/jp/2313q1.png ]
...

If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.
| |
68R |
| |
e e
.-b .----b (PNP Transistors
| c | c
| | | |
'--o---' '--- LEDS & PORTS
|
'------------- YOUR CURRENT SINK ...
Note that if you make the current sink supply about 1mA, that
will allow the LEDs to have 10mA. The 68 ohm resistor makes that happen.

[snip]

Until the temperature changes ;-)

...Jim Thompson

Its the temp diff between the 2 trs that would be the issue, and the
LED's tr is only dissipating 75 mW or so max. @50C/watt thats 3.75C. If
we subtract the 10% temp rise of the other tr, thats apx 3.4C temp
diff. How much variation will that cause?

Each Vbe would shift about 2 mV/K (for silicon near room temperature
obeying (kT/q)*ln(I/Is) equation) so a 3.4K difference gives about 7
mV or 1% deviation. I'm not concerned about variation that small, but
if I were would use a 2-transistor die.
-jiw

Your math is flawed.

WITHOUT even a differential temperature, the variation from 20°C to
60°C (both devices at same temperature) is 21.34%

The dominating effect is delta-VBE = (kT/q)*ln(Iright/Ileft)

NOT the delta-VBE hand-waving rule-of-thumb you quote, which is valid
only for devices at the same current.

...Jim Thompson

 

[Spehro Pefhany]

[email protected] wrote in sci.electronics.design:[Re http://pat7.com/jp/2313q1.png ]
...

If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.
| |
68R |
| |
e e
.-b .----b (PNP Transistors
| c | c
| | | |
'--o---' '--- LEDS & PORTS
|
'------------- YOUR CURRENT SINK ...
Note that if you make the current sink supply about 1mA, that
will allow the LEDs to have 10mA. The 68 ohm resistor makes that happen.

[snip]

Until the temperature changes ;-)

...Jim Thompson

Its the temp diff between the 2 trs that would be the issue, and the
LED's tr is only dissipating 75 mW or so max. @50C/watt thats 3.75C. If
we subtract the 10% temp rise of the other tr, thats apx 3.4C temp
diff. How much variation will that cause?

Each Vbe would shift about 2 mV/K (for silicon near room temperature
obeying (kT/q)*ln(I/Is) equation) so a 3.4K difference gives about 7
mV or 1% deviation. I'm not concerned about variation that small, but
if I were would use a 2-transistor die.
-jiw

How much do you think the *load current* will change with a 3.4K
temperature delta? It's also not all that stable wrt ambient
temperature. Maybe good enough for LEDs.


Best regards,
Spehro Pefhany

 

[Bob Monsen]

[email protected] wrote in sci.electronics.design:

Jim Thompson wrote:
57:59 -0800, JIW wrote:

[Re http://pat7.com/jp/2313q1.png ]
...

If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.
| |
68R |
| |
e e
.-b .----b (PNP Transistors
| c | c
| | | |
'--o---' '--- LEDS & PORTS
|
'------------- YOUR CURRENT SINK ...
Note that if you make the current sink supply about 1mA, that
will allow the LEDs to have 10mA. The 68 ohm resistor makes that happen.

[snip]

Until the temperature changes ;-)

...Jim Thompson

Its the temp diff between the 2 trs that would be the issue, and the
LED's tr is only dissipating 75 mW or so max. @50C/watt thats 3.75C. If
we subtract the 10% temp rise of the other tr, thats apx 3.4C temp
diff. How much variation will that cause?

Each Vbe would shift about 2 mV/K (for silicon near room temperature
obeying (kT/q)*ln(I/Is) equation) so a 3.4K difference gives about 7
mV or 1% deviation. I'm not concerned about variation that small, but
if I were would use a 2-transistor die.
-jiw

Your math is flawed.

WITHOUT even a differential temperature, the variation from 20°C to
60°C (both devices at same temperature) is 21.34%

The dominating effect is delta-VBE = (kT/q)*ln(Iright/Ileft)

NOT the delta-VBE hand-waving rule-of-thumb you quote, which is valid
only for devices at the same current.

...Jim Thompson

The problem is aggravated by the fact that I proposed him using his
constant current source, which also has a terrible tempco. Thus, the
difference from 0 to 100C is something like 3x.

At room temp, however, it works fine, and will allow him to dump the 1.5V
battery and go with a 3AA solution that'll provide the LED a constant
current until the battery voltage drops down to about 2.5V.

Using an opamp and a sense resistor with a small TC would overcome this
problem, but the OP needs to determine if temp range matters to him.

Jim may also have a better solution using a few discretes. I recall
him posting a 0 tempco current mirror at one point, but I don't remember
the details. It might not have the compliance the OP needs.

 

[Spehro Pefhany]

[email protected] wrote in sci.electronics.design:
Jim Thompson wrote:
57:59 -0800, JIW wrote:
[Re http://pat7.com/jp/2313q1.png ]
...
If you arranged things a bit differently, and added a couple of PNP
transistors + a resistor, things would be much better:

vbatt----------o---------.
| |
68R |
| |
e e
.-b .----b (PNP Transistors
| c | c
| | | |
'--o---' '--- LEDS & PORTS
|
'------------- YOUR CURRENT SINK
...
Note that if you make the current sink supply about 1mA, that
will allow the LEDs to have 10mA. The 68 ohm resistor makes that happen.

[snip]

Until the temperature changes ;-)

...Jim Thompson

Its the temp diff between the 2 trs that would be the issue, and the
LED's tr is only dissipating 75 mW or so max. @50C/watt thats 3.75C. If
we subtract the 10% temp rise of the other tr, thats apx 3.4C temp
diff. How much variation will that cause?

Each Vbe would shift about 2 mV/K (for silicon near room temperature
obeying (kT/q)*ln(I/Is) equation) so a 3.4K difference gives about 7
mV or 1% deviation. I'm not concerned about variation that small, but
if I were would use a 2-transistor die.
-jiw

Your math is flawed.

WITHOUT even a differential temperature, the variation from 20°C to
60°C (both devices at same temperature) is 21.34%

The dominating effect is delta-VBE = (kT/q)*ln(Iright/Ileft)

NOT the delta-VBE hand-waving rule-of-thumb you quote, which is valid
only for devices at the same current.

...Jim Thompson

The problem is aggravated by the fact that I proposed him using his
constant current source, which also has a terrible tempco. Thus, the
difference from 0 to 100C is something like 3x.

At room temp, however, it works fine, and will allow him to dump the 1.5V
battery and go with a 3AA solution that'll provide the LED a constant
current until the battery voltage drops down to about 2.5V.

Using an opamp and a sense resistor with a small TC would overcome this
problem, but the OP needs to determine if temp range matters to him.

Jim may also have a better solution using a few discretes. I recall
him posting a 0 tempco current mirror at one point, but I don't remember
the details. It might not have the compliance the OP needs.

A whisper of resistance in the other emitter will probably make it
more than good enough for visual indicators. Something like 270R/20R.
That only knocks the compliance down by a couple of hundred mV at 10mA
load current.


Best regards,
Spehro Pefhany

 

[Bob Monsen]

A whisper of resistance in the other emitter will probably make it
more than good enough for visual indicators. Something like 270R/20R.
That only knocks the compliance down by a couple of hundred mV at 10mA
load current.

Yes, that helps. However, the best result is probably going to be using a
temperature compensated low drop out voltage regulator and a resistor. If
he can get a 2.5V LDO, for example, he could use a small resistor to the
LED and have practically constant current as the battery voltage decays.

--
Regards,
Bob Monsen

I had thought I had been told that a 'funny' thing is a thing of a
goodness. It isn't. Not ever is it funny to the person it happens to.
Like that sheriff without his pants. The goodness is in the laughing
itself. I grok it is a bravery . . . and a sharing against pain and
sorrow and defeat.