It's on page 10 of this data sheet:
http://www.farnell.com/datasheets/45176.pdf
It's the 15A regulator.
Any ideas?
Well, first thing that crossed my mind is the offset error on the LM307.
Looking that up, it says a max of 10mV over its temperature range. Which
immediately appears to explain the minimum load requirement to me. Here's the
logic...
There are two inputs to the LM307. The LM307 isn't a theoretically perfect
opamp, so it is NOT able to actually control its output in a way to get any
better than (as much as) 10mV difference between them. For example, the
particular LM307 you happen to buy may be quite happy with the V(+) at 1V and
the V(-) at 0.99V and consider that to be balanced out, when it really isn't.
The LM307, in short, is inaccurate.
R1 is set to exactly 1/2 of R2, since the two LM338s tied off of R1 add their
currents in going through R1. If all the LM338s are supplying an equal portion
of the total current to the load, then the voltage drop across R1 will be
exactly equal to the voltage drop across R2:
Let's call the voltage at the + input of the LM307, V(+), and the voltage at the
- input of the LM307, V(-). Let's name the three LM338 regulators as A and B,
for the two tied off of R1, and C for the one tied off of R2. The current
though R1 will be (V(IN) - V(-))/R1 and the current through R2 will be (V(IN) -
V(+))/R2. If the LM307 is doing its job, then V(-) = V(+), and the current
through R1 will be twice as large as it is through R2, since R1 is half the
value of R2. Assuming that A and B are roughly sharing, then the current
through A will be 1/2 of the current through R1, and so will B's, which will be
exactly what is going through C. Nicely balanced, all around.
However, there's a problem. The LM307 may not really work to set the voltage of
V(-) to V(+). It can be as much as 10mV off of where it should be. Since you
want to make sure that there is at least *some* non-zero current going through
both R1 and R2, so that there is at least *some* work going on everywhere in the
circuit (you don't want some of it shut down), you need to be sure that even
with this possible error that both branches are doing a little something > 0.
Since:
V(+) = V(-) +/-10mV
And:
I(R2) = (V(IN) - V(-)) / R2,
I(R1) = (V(IN) - V(+)) / R1
Then you can see that you want both V(-) and V(+) to be at least a little bit
below V(IN). But they can be as much as 10mV apart, so you have to make sure
that at least one of those two is at least 10mV away from V(IN) in order to
guarantee that the other one won't get any closer to V(IN) than exactly at
V(IN).
This works out to ensuring that V(R2) is at least 10mV. And that means that
I(C) is at least 10mV/R2 = 10mV / 0.1Ohm = 100mA.
There must be at least 100mA to be absolutely positive that there is enough of a
drop across R2 that the voltage at V(+) cannot be above V(IN), in theory.
Actually, I think a worst case estimation should also have to include the fact
that none of the LM338s can properly operate without a worst case guarantee of
at least 5mA going through them. To ensure that, you'd need to be sure that
V(-) is at least (5mA + 5mA)*R1 = 500uV below V(IN). So this means that you
really need:
(10mV + 500uV)/R2 or... 10.5/.1 or 105mA
But these are all worst case, resistor values aren't exact, so its very
reasonable to just estimate and call it 100mA.
At least, that's how I see it.
Jon
