The following circuit uses an inexpensive current shunt monitor (AD8210) and a series of comparators to meet your requirements. I am using a separate 9V battery to power the test circuitry, so that even if the main battery gets completely shorted out, the test circuitry will still function. This also allows the test circuitry to work with a main battery ranging from 3V to 12V without requiring a buck-boost converter to generate the 5V needed. Instead I am using inexpensive 7805-type regulator in a TO-92 case, since the current drain is only a few tens of milliamps when one or more of the LEDs are on, and microamps otherwise.
(Right click and select View Image to see a larger version of this schematic.)
The 0.2 Ω shunt resistor R1 puts a minimal load on the circuit, dropping the battery voltage only 0.1V with a 500 mA load. The current shunt chip IC3 measures the voltage across the resistor, amplifies it with a gain of 20, and outputs a voltage proportional to the current on pin 5.
For example, that 100 mV drop across the resistor for the 500 mA load causes the IC to output a voltage of 2V. The proportion is 20 volts per volt across the shunt resistor, or 20V/V as I indicated on the diagram.
Likewise, a load of only 50 mA will result in a output voltage of 200 mV. So I set up the green LED to turn on for this threshold, namely 50 mA, indicating some activity. If you want it lower than that, you can of course adjust the value of the resistors R11/R12.
All of the comparator reference values are set with voltage dividers, using high value resistors to avoid loading the battery.
The comparator for the short condition IC2B is set up so it will trip with a load of 500 mA (2V). Obviously this is not a full short, but represents a lot of current. Again, you can adjust things as needed. With the current shunt resistor, you can only measure up to 1.25 A accurately. If you need to trip on a value higher than that, then you will want to switch the shunt resistor to 0.1 Ω and adjust all the voltage dividers accordingly. I picked 0.2 Ω so there would be enough voltage for the low-current measurement for the green LED.
Rather than lighting the red LED directly, the output of IC2B sets the flip-flop IC5A. This in turn turns off Q2, which in turn turns off Q1, breaking the path to the circuit under load so the battery will not be shorted anymore, avoiding possible damage to the circuitry. The flip-flop, being set, also turns on the red LED, and turns off the green LED. To restore battery power to the circuit, the RESET button must be pressed, resetting the flip-flop.
The bottom comparator is for the yellow LED. Depending on the resistor used for R8, it will light if the battery voltage is either over 3v (threshold is actually 3.5V) with a resistor value of 105K, or over 6v (threshold of 6.5V) with a resistor value of 237K. The circuit allows for battery voltages anywhere from 3V to 12V, since the battery voltage is divided by 4 by the resistor divider R4/R5 before being compared.
Although I didn't show it, you could add a DPDT switch to turn both batteries on and off at the same time.
Obviously this could be done with a microcontroller, but you will still need much of the I/O circuitry: the shunt resistor and IC, two MOSFET's controlling power, three LEDs (and probably three more MOSFETs to drive them), plus the batteries and pushbutton. So not a lot is saved (three comparators, a NAND gate, flip-flop, and some resistors and caps). I believe a circuit like this demonstrates the solution better than showing a bunch of code (if that's even included in the answer). After all this is a EE site.
According to your datasheet, the claimed brightness is the same at 20mA for either LED. So if you calculate a resistor to allow 20mA to flow through the BLUE LED then the current through the RED LED will be excessive and its life will be shortened.
Suggest you try it with a resistor calculated to allow 20mA for the RED LED (say 150 ohms) and see if the brightness is visually okay on the blue. It may be just fine. The current through the BLUE LED will then be about 15mA, which should be plenty bright. I suspect this is all you have to do.
If it isn't you could do something else that wouldn't involve hacking the board- you could parallel the RED LED with a resistor that would steal some of the current. Perhaps a small surface-mount resistor between the LED leads. So suppose the resistor you use is 100 ohms, then the current through the BLUE LED will be about 18mA, but the current through the RED LED would be more like 28mA. You could put a 220 ohm resistor in parallel with the RED LED to shunt away 10mA. But I doubt this is necessary- a 50% increase in brightness is really not that visually noticeable.
Best Answer
Okay. With the inclusion of your own results that closely mirrors my "mental hand-waving" result, I'll give this a shot. First, though, by getting a precise theoretical result for the short-circuit current, based upon the comment I made.
First, we need a proper schematic. I wish you'd learn to use the schematic editor that is included with this site. We've all had to spend the time learning it, so we know exactly how little excuse there can be for not using it. Your hand-drawn images do not add any value I can think of, that isn't already available in the editor. So no excuses (for you or me.)
simulate this circuit – Schematic created using CircuitLab
Now everything is numbered (as it should be) to save wasted breath and writing. I've used \$V_n\$ to number each node and I've added a ground symbol so that we know what we are talking about when we specify a voltage, without also saying what it is relative to. These things are important and you should get in the practice when asking questions like this. It's your question, after all, and you should add value to it yourself if you want others to take a moment because it saves them time.
Sorry about the lecture. But it has to be said.
Also note that, as already discussed in comments you received in a different question you asked, it is entirely possible to mentally combine \$R_5\$ and \$R_6\$ as a new resistor value because they are both in a series circuit and therefore the current must be the same in both of them (everything to the right of them is mentally a single "circuit element" that is in series with those two resistors.)
But I intend on leaving things exactly as you have them.
With things detailed out, now we can talk again about the short-circuit current, in the case where node \$V_2\$ is shorted to node \$V_7\$. Here, the right-hand side of the circuit becomes meaningless and we only need to focus on \$R_4\mid\mid \left(R_5+R_6\right)\$. The current through the \$R_5+R_6\$ leg will be:
$$I_{R_5}=I_{R_6}=\frac{480 \:\textrm{mA}}{1+\frac{2\cdot 3.42\:\Omega}{350\:\Omega}} = 470.799\:\textrm{mA}$$
This is formed quite simply by mentally treating resistors as conductances and using the ratio of their conductances, plus 1, to determine how the current splits between them. Just as a note, the same idea also computes the current in \$R_4\$ in a very symmetrical way:
$$I_{R_4}=\frac{480 \:\textrm{mA}}{1+\frac{350\:\Omega}{2\cdot 3.42\:\Omega}} = 9.201\:\textrm{mA}$$
Just by way of i-dotting and t-crossing, here's the schematic we've been talking about:
simulate this circuit
So that's done.
Now, disconnecting the circuit leg between \$V_2\$ and \$V_7\$ leaves me unable to isolate the rest of the circuit. However, given your assumption that each LED string (of 10) will have the exact same voltage across them (this is an assumption and in actual practice with real LEDs [or even just realistic models of them] the assumption would be at least slightly wrong), we can simplify the rest of the circuit.
Let's start with the place you and I are discussing now:
simulate this circuit
There are lots of different techniques to use in solving something like this. The following approach is not the one I'd use, since I'm comfortable with nodal analysis and would greatly prefer it here. But I'm just going to follow a more basic approach and hope it communicates the way you want. (You haven't specified a specific approach you want followed, so I'm free to assume anything I want to.)
Here, we've disconnected the circuit leg between \$V_2\$ and \$V_7\$. Now we can see that \$R_5\$ and \$R_6\$ can be combined into one resistor, \$R_7\$ and \$R_8\$ can be combined into one resistor, and also that \$R_9\$, \$R_{10}\$ and \$R_3\$ can also be combined into one (as they could have been earlier, too.) Let's simplify (and also replace your current source and \$R_4\$ with its Thevenin equivalent):
simulate this circuit
(In the above, the new voltage nodes are no longer the exact same ones as in the original schematic. So I will say \$V_2^{'}\$ when referring to this new node. Just to make sure you understand the distinction.)
But since the voltages across the LED strings are assumed to be exactly the same, this circuit can be further simplified:
simulate this circuit
Here you can now easily see that this is nothing more than a series chain of resistors proceeding from one voltage supply to another. The current will be:
$$I_{TOTAL}=\frac{168\:\textrm{V}-27.4\:\textrm{V}}{R_4+R_5+R_6+R_7+R_8+R_2\mid\mid\left(R_3+R_9+R_{10}\right)}\approx 383.906\:\textrm{mA}$$
From this, we can now easily compute:
$$V_2-V_7=V_2^{'}=168\:\textrm{V}-\left(R_4+R_5+R_6\right)\cdot I_{TOTAL}=31.007\:\textrm{V}$$
And so that is your answer to the other part of your question.
I know you say you want a load line. And now you can make one. But I also have no real clue why you care about that particular load line. So I guess it is now up to you where you go from there.
In summary, I've provided one of many different ways to achieve your goal. I've tried to keep this to a simplified approach that uses your assumption that the voltage across the LED strings is fixed.