From: Winfield Hill
Subject: Re: Best way to power array of LEDs?
Date: 8 Dec 2002 19:29:38 -0800
Organization: Rowland Institute
References: <0001HW.B9C26E7400A7161B165FEAC0@news.covad.net> <2Qan9.3129$cS4.firstname.lastname@example.org> <3DC843FE.4A25FFF3@fanwap.com> <3DC99AD4.2AB0FE52@fanwap.com>
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Watson 'Atto Parsec' Name wrote...
> email@example.com mentioned...
>> Watson wrote...
>>> firstname.lastname@example.org wrote,
>>>> 2.2k white white
>>>> | LED LED
>>>> ,----+ x | |
>>>> | | | |
>>>> | R4 C C
>>>> R2 +--- B ----- B
>>>> 1.0k | E E
>>>> | C | Q2 | Q2'
>>>> +--B | +---
>>>> | E Q1 | |
>>>> R1 | R5 R5'
>>>> 2.2k | 6.8 6.8
>>>> | | | |
>> More accurately, R4 = R2/R1 * r_e for Q1, or 36 ohms.
>> The circuit can be easily scaled to handle many LEDs. One possibility
>> is to operate Q1 at higher currents, scaling R1 to R4 appropriately.
>> Another is to add a buffer transistor between Q1 and the bank of current-
>> sink transistors. We can do this because the Vbe reference is adjustable
>> and can easily accommodate another Vbe drop.
>>= 1.69V power supply
>>= ,----+-- R3 --+---- 3.4V to 6V or more
>>= | | 1.8k | multiple 30mA
>>= | R4 R6 white LED driver
>>= R2b 62 27 using 2n4401
>>= 68 | | white white
>>= | | 1.63V | LED a LED b ...
>>= | | C | |
>>= R2a +----- B | 30mA | 30mA
>>= 3.3k | Q2 E | |
>>= | C | 0.96V C Q3a C Q3b..
>>= +--B +------ B ------ B ------- many many more...
>>= | E Q1 | E E
>>= R1 | R5 | | <- 0.20V
>>= 2.2k | 1.0k R7a R7b
>>= | | | | |
>>= '----+--------+---------+--------+--- etc
>>= all 6.8 ohms
>> As before R4 cancels changes in Vbe-Q1 with supply voltage. If you
>> desire, R2 can be trimmed for exactly 30mA with your transistors,
>> this worked out to 3.368k for Spice's 2n4401 parts. R6 protects Q2
>> by limiting the transistor-array base drive to 100mA, which should
>> be enough for perhaps 500 LEDs. :>)
> Well, thank you for the improvement. I found one design deficiency(?)
> with the original circuit. I found that if one LED fails open, the
> other LEDs dim substantially. The 30 mA is no longer flowing thru the
> open LED's 6.8 ohm emitter resistor, so there is no IR drop across it.
> This allows much more current to flow thru the B-E junction of that
> transistor, hogging the base drive current away from the rest of the
> transistors. Hopefully, keeping the LED current at 30 mA will prevent
> that open LED from happening.
Nope, that's a problem or potential problem at one level or another
with all of the shared-reference circuits, because they rely on only
driving Ic/beta for each LED-transistor combination, rather than Ic.
The circuit above, capable of driving 100mA into the base-bus line, is
less vulnerable because in applications with say up to 100 LEDs it has
left-over capability to handle the situation of a few missing LEDs.
You do have the option of using more Q2 transistors, so that each one
is responsible for fewer LED drivers.
One "solution" is to add a separate base resistor with each LED-driver
transistor, thereby limiting the current it can draw if the collector
is open. But be careful in choosing the resistor's value, because it
introduces an error related to the transistor's beta.
For example, assume a beta of 150 creating a base current of 200uA.
Adding a 470-ohm resistor would create a 94mV drop; we would correct
for this by raising the reference by 94mV. However for a transistor
beta of say 250, drawing 120uA for a 30mA output, and dropping 56mV,
we'd see an emitter voltage of 238mV instead of 200mV and thus get a
current of 36mA (actually 34.5mA, but you get the idea). So adding
this resistor can create a likely non-trivial error. In the absence
of an LED, the 470-ohm resistor limits the extra base current draw
to about 0.4mA, which is nice. Perhaps one could be satisfied with
a smaller base resistor, say 220 or even 100 ohms as a compromise.
Alternately a different solution could be used, such as adding a
second transistor to provide a current path in place of an open LED.