Electronic – Will these LEDs work for throwies

An extremely simple and low cost constantish current source is shown in the circuit below.

Q1 is initially turned on by R2.

I_LED flowing in R1 produces a voltage per V = IR
of V = I_LED x R1
When V_R1 reaches about 0.6V Q2 is turned on and shunts base drive to Q1, thereby limiting any current increase.

ILED max ~= I_R2 x Beta_Q1 where Beta = current gain.
For say 5V and R2=1k IR2~= (5-0.6)/1k = 4.4 mA.
For Q2 Beta = 100 Imax = 440 mA - ample for most purposes.
This circuit drops a minimum of ABOUT 0.6V

Cost, size and complexity are minimal.
Results are far better than using just a series resistor. ie Current is not "constant" but is far closer than with a resistor alone.

^{simulate this circuit – Schematic created using CircuitLab}

Aka:

E&OE.

When you connect the LEDs in series, the forward voltages of the LEDs add up. If you have to connect 20 LEDs in series, the forward voltage of the string of LEDs is 20 * 3.8V = 76V. But then you only need 20mA per series of LEDs! (You got that backwards)

If you instead contact 20 LEDs in parallel, the forward voltage is 3.8V, but now the current is 20 * 20mA. You now can use a 5V supply (since 5V > 3.8V). In that case, you need a resistor in front of every LED to "take care" of the difference of your 5V supply and the 3.8V LED's forward voltage (5V - 3.8V = 1.2V), that's what all the LED-Resistor-Calculators show you.

So, to make LEDs light up in series, you need to supply 20mA, independent of the number of LEDs. An LED driver does that. It supplies 20mA and "adjusts" the voltage to make that happen. You normally would not put 20 LEDs in series, since your LED driver had to supply ~76V (which the power supply has to supply in the first place).

A 5V supply can do 1 LED in series (5V > 3.8V). A 12V supply can do 3 LEDs in series (12V > 3*3.8V). For every string of LEDs you than provide 20mA each.