The easiest way you can repair it is flipping the chips backwords, where Vout is now in place of the existing Adj. If you mounting the heatsink with epoxy then....depending on the amount of overhead space it might be better to stand the chips up, exposing more surface area for passive air convection to those heatsinks.
Much much kudos illum for the help and the kind words. Aside from my slipup how does the circuit look?
Each regulator will run two LEDs. I opted to use a 1.8ohm resistor so my curent will be around 700 ma. Does this all make sense to yea?
yep
circuit looks great! I would have never considered the use of solid copper as conducting bridges...mine would have jumper wires traversing across the perf board.
I'm not sure about the LM338 but some power supplies have a different reading between load and no load. That is, if there's an open circuit on the output end, it may give you erratic results. For much complicated power supplies, an open circuit on the output end will fry its electronics, not sure how this works. I can only see how this may happen if the circuit is closed then suddenly opened, where the output capacitors could dump its stored power back to the IC.
What multimeter do you have?
Some digital multimeters have more than one voltage selection as well as current selection...in addition to three ports for probes instead of two.
If I know what multimeter you use I can walk you though it easier
For basic understanding of meters, measuring volts you would parallel your multimeter with some type of a load, where you will measure the potential without the majority of the output current rushing through it.. Measuring current is exact opposite, the meter is wires in-line, or in series with your load, where the only path for current to go is through your multimeter.
For other noteworthy things to consider adding on to your generously wide board... Good for reference, or you can just skip it...your choice.
Some of the stuff down here originates directly from my own sketchbooks back when I used LM317 almost anywhere that needed a cheap point of load current regulated supply.
Measuring power loss [or heat output] of the LM338 use
Code:
P = (Vin - Vload) x I
P = Power loss
Vin = supply voltage
Vload = voltage consumed by your output end
I = current
you noted that you will be wiring two LEDs in series and you waqnt to drive them at 800ma.
LEDs are current regulated creatures, depending on the current applied the Vf will vary as a function [close to a parabola] with current. Say if your using white CREE XREs, the Voltage input is 3.3V at 350ma but will shift to 3.5V when driven at 700ma. There will be slight changes between batches but this is just illustrative.
Suppose you are using two XREs, you would go and pull up the datasheet to see what the draft looks like. In this case, while the typical forward voltage [or Vf] to the LED is 3.3V the Vf is in actuality 3.7V at 800ma
Okay, now you can solve your power loss
The "differential = 2V" part means at minimum the regulator will consume 2V, and its primarily used to solve for minimum power input. In this case you are using 12V, which might be a tad too big a value to promote minimum heat output
Two LEDs in series:
3.7V as Vf [learned from datasheet, each LED is different]
3.7V x 2 = 7.4V
2V differential exists on the LM338, now add that to it
7.4V + 2V = 9.4V, at a 9.4V input the regulator will incur minimum power loss
Where you are with 12V exists that
12-9.4 = 2.6V excess
Excess voltage [2.6V] multiplied by 0.8A of current will contribute an additional 2.08W of power loss.
Code:
P = [12V – (3.7V x 2)] x 0.8A
P = [4.6V] x 0.8A
P = 3.68W
3.68 here represent total power loss for each LM338 in this configuration, noting that 2.08W is from excess voltage applied. By reducing your input voltage, you could have effectively eliminated approximately [2.08/3.68=0.5652] 57% of the power loss we see here.
Choosing your heatsink...in this case you would have wanted a heatsink capable of at minimum dissipating 5W of heat, and depending on whether or not the device is enclosed, you might want much bugger heatsinks
you've got current adjustment covered, so I'll omit it here
looks like you've got the resistor power rating for the adj pin covered as well
Code:
P = UI
P = Power across the Resistor
U = 1.25V for the LM338/317
I = desired current output
This is the power that ill travel through the resistor
Code:
Pd = I^2 x R
Pd = Power disspiated
I = current across resistor
R = resistance value
Pd = (0.8)^2 x 1.6
Pd = 1.024W
this value is the power that's dissipated from the resistor
At minimum use an at least 10% higher power value for resistors, I use 25% or more...resistors are cheap, but some values may not be as common as others. A 1/4W resistor jolted with 1/4W of heat will actual light up like a toaster, a 1W carbon composition resistor jolted with 1W will actually burst into flames
Also...be sure to add some capacitors to your LM338s for stability
read though this segment of the datasheet
External Capacitors
An input bypass capacitor is recommended. A 0.1 µF disc or 1 µF solid tantalum on the input is suitable input bypassing for almost all applications. The device is more sensitive to the absence of input bypassing when adjustment or output capacitors are used but the above values will eliminate the possiblity of problems.
The adjustment terminal can be bypassed to ground on the LM138 to improve ripple rejection. This bypass capacitor prevents ripple from being amplified as the output voltage is increased. With a 10 µF bypass capacitor 75 dB ripple rejection is obtainable at any output level. Increases over 20 µF do not appreciably improve the ripple rejection at frequencies above 120 Hz. If the bypass capacitor is used, it is sometimes necessary to include protection diodes to prevent the capacitor from discharging through internal low current paths and damaging the device.
In general, the best types of capacitors to use are solid tantalum. Solid tantalum capacitors have low impedance even at high frequencies. Depending upon capacitor construction, it takes about 25 µF in aluminum electrolytic to equal 1 µF solid tantalum at high frequencies. Ceramic capacitors are also good at high frequencies; but some types have a large decrease in capacitance at frequencies around 0.5 MHz. For this reason, 0.01 µF disc may seem to work better than a 0.1 µF disc as a bypass.
Although the LM138 is stable with no output capacitors, like any feedback circuit, certain values of external capacitance can cause excessive ringing. This occurs with values between 500 pF and 5000 pF. A 1 µF solid tantalum (or 25 µF aluminum electrolytic) on the output swamps this effect and insures stability.
basically...you'll need something like this
Electrolytic caps are used for low frequency filtering of power due to their slow response times, high frequency filtering is facilitated by tantalum capacitors.
Here's some of the caps the datasheet talked about [your only going to need one electrolytic, one "red" tantalum, and one regular "yellow" tantalum]
left to right: Electrolytic, tantalum, poly-film, ceramic, disc
After reading the datasheet you should have something like this
caps make things stable, caps needs diodes to protect what your trying to keep stable
Then it gets interesting, recall what I said about a sudden open circuit on the output end damaging chips? well...you might want to plan for that in the event one of your LED chains burn out and opens the circuit.
capacitors can provide an exceptional amount of current within a very small time...
the datasheet warns this
Most 20 µF capacitors have low enough internal series resistance to deliver 20A spikes when shorted. Although the surge is short, there is enough energy to damage parts of the IC.
and provides a diagram of where to put diodes where that power will flow through instead of forcing its way through the IC
I redrew it below. IN400x diodes are general purpose diodes that are found in many places, like radioshack, etc, but if your interested I can supply you with sources for these parts
5 additional components per IC will populate that perf board in no time =P
