 ### Network E

#### ed

Jan 1, 1970
0
Hey Everyone

I'm looking at building a simple bench top power supply with a DC output.
I've found a simple circuit which will regulate the output using LM338's.
Under the circuit diagram though it has the text:

Am I right in thinking that this means that any device connected to my
homemade power supply must draw a minimum of 100mA? How can I ensure that a
minimum of 100mA is always being drawn? I guess I could use a resistor
going from the output to ground. But I want the output voltage to be
variable. So would I have to select a resistor which worked for all
possible voltage outputs?

Thanks you for any help.

J

#### John Fields

Jan 1, 1970
0
Hey Everyone

I'm looking at building a simple bench top power supply with a DC output.
I've found a simple circuit which will regulate the output using LM338's.
Under the circuit diagram though it has the text:

Am I right in thinking that this means that any device connected to my
homemade power supply must draw a minimum of 100mA? How can I ensure that a
minimum of 100mA is always being drawn? I guess I could use a resistor
going from the output to ground. But I want the output voltage to be
variable. So would I have to select a resistor which worked for all
possible voltage outputs?

Thanks you for any help.

J

#### Jonathan Kirwan

Jan 1, 1970
0
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

J

#### Jonathan Kirwan

Jan 1, 1970
0
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.

That worst case calculation for 105mA assumes that both regulators A and B
precisely share the current through R1. In reality, that won't happen. So it
would be further helpful to ask how bad the split up between A and B can get.

The data sheet says that the difference between the adjust pin and the output
pin of the LM338 can be anywhere from 1.19V to 1.29V. The design shown has the
LM307 opamp output driving their adjust pins through a 2k resistor to each of
regulators A and B. The data sheet says the adjustment current can be as much
as 100uA and doesn't say how low it can be -- let's say it can be as little as
0uA, just to make this as bad as possible.

So, let's assume all the errors accumulate in the worst case directions for A
and B. A will have 0uA adjustment current and will also have a 1.19V
difference. B will have 100uA adjustment current and will also have a 1.29V
difference. What will the output pins of A and B have?

V(A,out) = V(opamp,out) + 0uA*2k + 1.19V = V(opamp,out) + 1.19V
V(B,out) = V(opamp,out) + 100uA*2k + 1.29V = V(opamp,out) + 1.49V

This means that there must be (1.49-1.19)/0.1 or 3 Amps difference in the
currents through R5 and R6 to compensate. Since we have to have at least 5mA on
one of them, this means that the other could have to be as bad as 3.005A, to
cover this worst case assumption. (And then add the other one.)

Yuk.

Jon