Re: how to control LED array? (follow-up)

I forgot to finally address myself to the value of RA in:

: |
: \
: / RA
: \
: /
: |
: |
: RbA |/c QA
: ,--/\/\-+-| NPN
: | | |>e
: | | |
: A \ |
: RaA / gnd
: 47k \
: /
: |
: gnd

What it should be depends on the Vcesat of QA and the Vcesat of Q1 (or
Q1p, from the earlier discussion) and on the expected voltage across
the diode at the desired current. Already, we know that for QA, this
Vcesat is about 0.1V. Also, if you are going with 30mA per LED, the
2N4403 gives you about 0.35V for Vcesat at full-bore 240mA. So, you
lose about .45V (call it .5V) at those two places. Also, the LED
itself at 30mA is about 3.3V from what you've said. So the total loss
of voltage is about 3.8V from 5V. That leaves 1.2V for the resistor,
RA, at 30mA. So, Ra=1.2V/30mA, or 40 ohms. Make it 39, as a standard
value.

In the case of the higher drive, 100mA per diode, more thoughts are
needed. Here, with the TIP30A I mentioned, the Vcesat is still about
0.6V at 800mA. If you use a TIP42A, instead, that drops well below
0.2V with 40mA base current. Whether you use one or the other only
chooses where the heat gets dissipated (RA or Q1p), so long as there
is enough voltage headroom for the LED itself. So let's see what the
LED requires.

I don't have a data *** on your LEDs, so a lot of guessing is in
order. But the diode equation that says that the slope of the Vd vs
Id line (the linearized resistance) of a diode is about (nkT/q) /
(Id+Isat). At Id's of 20-100mA, Isat can be largely ignored. So this
is about n*26mV/Id or something like 2.6 ohms at 30mA with n=3
(nominally 1, but 3 isn't an unreasonable guess) and .78 ohms at
100mA. For pure rough-shod guessing work, I'd take the average of
these two (2.6+.8)/2 or 1.7 ohms as the mean slope and figure that if
there is 3.3V at 30mA, then there is 3.3V+(100mA-30mA)*1.7 or 3.42V at
100mA. Call it 3.5V. So, with 5V supply, we are left with (5V - 3.5V
- .6V - .1V) or 0.8V. That's slim, but it suggests an RA of .8/100mA
or 8 ohms. Use a 7.5, I suppose.

RA would need to dissipate 75mW. In the 100mA per LED design, Q1p
would be dissipating some 550mW, if you recall, with the TIP30. If
you wanted to shift more dissipation to RA, you could use the TIP42
which would cut its Vcesat to say 0.2V and its dissipation to under
200mW. But then the difference in Vcesat (0.6 - 0.2) would have to be
picked up in the resistor, RA. So it's new value would be 12 ohms and
its dissipation would be 120mW. That might not seem quite right,
because you might be wondering how it is that you can go from 550mW to
200mW on Q1p and only go up from 75mW to 120mW on RA. But if you
remember that there are eight of these RA resistors, each of which are
carrying their own part of this difference, then you can see how it
all comes out even again. In other words, by using a slightly more
expensive PNP on the high-side source, you can distribute the power
dissipation over 8 resistors, resulting in lower peak temperatures
overall.

Jon
.