I was using 1N5401s on my schematic (1N4001s in prototype) because I might want to draw at least an amp out of the power supply and I thought that perhaps the diodes wouldn't be able to handle this. If they can function at larger currents for a short amount of time (feeding the smoothing cap), I'm happy to use 1N4001s.
To be honest, I really think that you're pushing your luck to expect more than an amp from this. Here's why:
Worst-case power dissipation is when the output is shorted with the current limit turned to max. Perhaps there is ~20 volts across the regulator with 1 amp flowing, giving 20 watts.
To determine the junction temperature rise, we calculate the total thermal resistance from junction to free air. Perhaps we'll have a good heat sink that has a thermal resistance of 1 watt per degree C. So, in theory, the rise is only 20 degrees, right?
But, we have to also consider the thermal resistance of the case to the heat sink, which could be 1C/W again. OK, so the rise could be 40C. Fine so far...
But the killer here is the thermal resistance between the junction and the case. Which is 5 degrees per watt. So now we have 7C/W in total, so the total rise is 140 degrees C. Ambient might be 25C, so the junction temperature is 165C. Well over the maximum allowed temperature of 125C!
Even if the heat sink was "perfect", the thermal resistance
inside the IC, which we can't do anything about, will be responsible for a 100C rise.
OK, so you probably wouldn't run it continuously into a short, but you never know what might happen accidentally. So, there are a few approaches:
1. Re-specify your goals. Make it 500mA instead, and that brings things back into the allowable limit. That'll be my starting point when I write all this up as an article on my website... Don't forget that you'll be able to connect your two units in parallel (or even offer a "parallel" mode, like the Thurlby units do).
2. Add a big power transistor or two around the unit, as the datasheet suggests. These will have a lower thermal resistance, solving the problem. Been there, done that. It works.
3. Use a thermal cut-out to protect the LM317. This is pragmatic, as it saves over-specifying the heat sinking which probably won't be needed 75% of the time, but protects everything against "accidents". I do this a lot, and recommend it, even if you do think everything will be OK (vents get blocked!). The simple bi-metallic units are pretty good for the money.
There are other ideas, but all a bit complex at this stage. Oh, and if you rely on the in-built thermal protection, don't expect it to be reliable. Well, that's my experience, at least
For the current limit LED, I think I'll either put it on a separate resistor (when both LEDs are on, the current going to each will drop - the difference in brightness is the sort of thing that would get on my nerves) or perhaps use a dual colour LED like you suggested. I believe Dave did a video that applies to the latter about the LED on his Hakko 888 soldering station.
Ah well - that's another trick. I was going to explain it previously, but instead left you a clue with the word "changing"...
A red LED has a Vf of 1.6V typically. Whereas it's ~2V for green. So when the red LED comes on, the green LED goes off. Neat
Although if you like the sound of that, do check with the LEDs you have. The ones on my prototype do appear to have a similar Vf, meaning that the green one does indeed dim, and I agree that is a bit naff. If it was a dual red/green LED, it would go yellow, which isn't so bad.
Typing 'conduction angle' into Google Images brings up a labelled waveform from your website as the first entry!
So it does! Lol!
Question: why have you put 10k resistors on the inputs of the current control op amp? Is it because you needed high input impedance for the inverting input (attached via a 100pF cap to the output) and put another resistor on the other input to balance it out?
As you mentioned it first, I'll plug another article on my site - the one about op-amps.
In summary, op-amps draw a tiny DC current into each of their inputs, and each input takes very nearly the same current. This current is turned into a voltage by the net resistance "seen" by this current, and if this total resistance is different for each input, then a voltage difference exists at the op-amp inputs, which of course becomes an offset at the output.
While we're looking at this part of the circuit, I'll point out another "trick". Note the 10k current pot - which was the first one I found - is working into a low impedance (1k1). This will "distort" the law of the pot, so that it's no longer linear. Which in turn gives you more control over the current at lower settings. Yes, you could buy a linear pot, but they have certain drawbacks. And my prototype uses 10k for voltage (with 3k9 in parallel), and it's always best to use identical components where possible.
Oh, and the resistor values shown won't get you to 1 amp - perhaps only a third of that. I haven't optimised them yet - it was just what was conveniently to hand.