Try it. Assume the LED voltages at 20 mA are approximated by 3.5, 3.5, 4 and 2.5v (the unspecified low end of LED voltages). With three LEDs this is 4x3 values.
Then add voltage distribution, assume 12.8, 13.5, 14.4, 14.4, 15.
Then add resistor tolerance, assume a normal distribution with only three values - the center value, + 5% and - 5%.
Plug these values into the circuit and look at your 12x5x3 values of current on a spreadsheet, then rank them by order of likelihood.
A designer doesn’t really need to look in detail at distributions to realise that (with the extremely wide range of LED characteristics you are postulating and the range of voltages you want to accommodate), a design involving three of the LEDs in series is a non-starter – one only needs to look at the ‘worst cases’ ...
For the purpose of calculation, I have assumed that the intention was to get an LED current of 20mA when the LED Vf was the mean of the figures you postulate (3.375V) and the supply voltage is the mean of the figures you quote (14.02V). That equates to a resistor value of 194.75Ω. I will therefore assume that you would select the neared E24 value (200Ω
and, for the purpose of illustration, further assume that all such resistors have values within 5% of that nominal value (i.e. 190Ω-200Ω
.
The worst cases are then:
(a)...Vf = 4V*3 = 12V supply = 12.8V resistor = 210Ω, which gives an LED current of 3.8 mA, and
(b)...Vf =2.5V*3 = 7.5V supply = 15V resistor = 190Ω, which gives an LED current of 39.5 mA.
Both of those extremes are outside of an acceptable range for a ’20 mA LED’. Examining the distribution will indicate what proportion would be outside of the acceptable range, but the designer obviously should design such that the probability of
any being outside of that range is very low.
In contrast, if one adopts a more sensible design using just two LEDs in series, the resistor value would then become 363.5Ω – so use a 360Ω E24 component with an assumed range of possible values of 342Ω-378Ω.
The worst cases are then:
(a)...Vf = 4V*2 = 8V supply = 12.8V resistor = 378Ω, which gives an LED current of 12.7 mA, and
(b)...Vf =2.5V*2 = 5.0V supply = 15V resistor = 342Ω, which gives an LED current of 29.2 mA.
Those extreme values are probably just about acceptable for a ’20 mA LED’. If one worked with a more realistic estimate of variability of LED Vf, the range of currents would certainly be acceptable. Examining the actual distribution of LED currents would add relatively little to this situation.
Having said that, I’ll do some simulations of the distributions and show you shortly what they look like in the two cases.
Kind Regards, John