Electronic – Circuit analysis for bicolor LED

As described, you could place a resistor from CHG to either battery+ or to Vin and an LED from CHG to ground. Resistor value depends on desired LED current.

When CHG is low LED is off.
When CHG goes O/C CHG is pulled high by R and LED is driven.

If you use say Vin = 16 V and want say 2 mA I_LED the R ~~= ( Vin - V_LED)/I_LED
~= (16-2) / .002 =~~ 6k8.

The LTM8062 comes in an LGA package which is liable not to be overly home-constructor friendly.

It costs about $17 in 1's which is high if only basic functionality is going to be used.

However, it will work with a range of battery chemistries and it has MPPT capability. If you need the fancier features it's a bargain.

LTM8062 data sheet here.

15 x 9mm LGA package.

You could return the LED to a fixed negative voltage of (say) -1V to control where it turns on. Or (better) put the LED in a feedback loop so that you get mA out for volts in.

The latter will eliminate the LED 'on' threshold entirely- if that's acceptable (it might glow very slightly with zero input).

Or add an additional resistor to get a threshold, as below:-

^{simulate this circuit – Schematic created using CircuitLab}

The maximum current is set (mostly) by R1 so if the maximum input voltage is 13V, you'd have a maximum current of about 25mA minus about 0.8mA or about 24mA

Threshold current is set (roughly) by R2 so if the LED starts to come on at around 1.5V, and you want it to light up at about 650mV in you could set R2 to about 2.2K

Edit: D2 is just to prevent the op-amp from imposing a high negative voltage across the LED if it's presented with a negative voltage at the input. If R2 is not present, even a tiny voltage (or the offset of the op-amp) would be enough to rail the op-amp which would exceed the reverse voltage rating on the LED.

The op-amp drives a constant current proportional to the input voltage through the parallel combination of D1||D2||R2. I = Vin/R1. For positive input voltage we can ignore D2 since it will be reverse biased.

D1 needs about 1.5V before it starts conduct much current (and thus to light). Without R2, the voltage across D1 will rise to whatever is required to get even a tiny current because of the feedback mechanism. R2 provides an alternative path when the current is low so, say at 0.1mA, the voltage across R2 will be only 220mV, so the LED will not light at all. So a minimum current of Vf(min)/R2 is required to get the LED to begin to light significantly.