Electronic – Is a resistor needed between logical pin and base/gate transistor

You don't need the series resistors to control the current; the TLC5940 is a current source, and doesn't give a damn about the resistance. If you program it to source 120 mA it will source 120 mA, whether the series resistor is 0 Ω or 15 Ω (15 Ω is the equivalent of six 90 Ω resistors in parallel). Your calculation of the 90 Ω is correct: 3.2 V for the LED plus 20 mA \$\times\$ 90 Ω = 5 V, so a larger resistor would need a higher voltage to keep 20 mA flowing. If the 5 V is all you have increasing the resistance will decrease the current. A 100 Ω resistor for instance will allow

\$ \dfrac{5 V - 3.2 V}{100 \Omega} = 18 mA \$

Further increasing the resistance will also further decrease the current.

There are two reasons why you should use series resistors:

• balance the currents between the LEDs. Small differences in LED voltage will be compensated by the voltage drop across the resistors. Since the resistors will all see the same voltage from the TLC5940 a 90 mV higher voltage for LED 1 will cause 90 mV less for its series resistor, and hence a 90 mV / 90 Ω = 1 mA less current, which won't be noticeable
• but more importantly, they reduce the power dissipation in the TLC5940. If it supplies 120 mA per output, and you wouldn't have series resistors, then it would dissipate (5 V - 3.2 V) \$\times\$ 120 mA = 220 mW. Not too much it seems, but the TLC5940 has 16 outputs, then that 220 mW may become 3.5 W!! That's too much for the poor device (see power rating on page 3 of the datasheet, at 70 °C, not at 25 °C). So the resistors literally take some of that power out of the device. At 90 Ω there would not be any dissipation in the IC at all, but the current source wouldn't have any headroom either. So it's better to choose the resistance a bit lower: 68 Ω will leave 5 V - (3.2 V + 20 mA \$\times\$ 68 Ω) = 440 mV for the IC. 16 outputs driving 120 mA each will then cause an acceptable 850 mW dissipation.

You are missing one easy thing: voltage across the resistor won't be 5V.

Transistor side you have a \$V_{BE}=V_\gamma\approx0.7V\$, so the base of the resistor is about 0.7V lifted from ground.

Microcontroller side, when the output is high you don't have full \$V_{CC}\$ but a voltage that is somewhat lower. For a cmos chip that can be quite low, some 100mV, but to make the calculations easier let's say you have 0.3V from \$V_{CC}\$, meaning that the arduino output voltage is only 4.7V.

Let's do the math again:

$$R=\frac{(5-0.3-0.7)\text{V}}{10\text{mA}}=400\Omega$$

Since higher base current is better in this case, just stick with the lowest nearest standard value, i.e. \$390\Omega\$.

Note: that 0.3V that I wildly guessed is actually written down in the microcontroller datasheet and it's called \$\mathbf{V_{OH}}\$, as in Voltage Output High. You also can find the \$\mathbf{V_{OL}}\$ and... Yes, you guessed it, that's Voltage Output Low.
The first one is the minimum output voltage when the pin is set high, if you are within current specs of course, while the second value is the maximum output voltage when the pin is set low.