Sinking
Figure 1. Section 9.2 shows TL5940 sinking LED current.
Sinking means that the chip will connect the negative leg or cathode of the LED to ground. This is the opposite of your schematic. Figure 1 shows the LED anodes connected to a positive supply, \$ V_{(LED)} \$.
Minimising current
Provided \$ V_{(LED)} \$ is high enough, LEDs can be series connected to have the same current power multiple LEDs. This will reduce power dissipation in the chip. The chip can work with \$ V_{(LED)} \$ up to 17 V (Section 1, Features).
Maximum voltage
Pay attention to Section 10 Power Supply Recommendations for the \$V_{LED} \$ voltage.
[OP's comment] If I try to drive 10 LEDs (each with a voltage drop of 3.2) in series, with 38 Volts, will the voltage be dropped to 4 volts before the current enters the chip?
No. While it seems it might be true, consider what happens when the output transistor turns off. The full 38 V will be fed through the LEDs to the transitor collector but it is rated at 17 V max. The transistor will break down due to electric stress.
Current control
The chip also has current control which allows you to set the maximum channel current (Section 8.3.7). By adding \$R_{(IREF)} \$ the you get
$$ I_{max} = \frac {1.24}{R_{(IREF)}} \times 31.5 $$
You don't need the current limiting resistors. If you insist on paralleling LEDs or strings of LEDs you might want to add a low value resistor to drop maybe 0.2 V or so at the individual LED current to balance out the currents a bit.
Power calculation
Run the power calculation given in section 11.3. The max power dissipation will be based on ideal heatsink mounting. You'll need to derate this based on your ability to match this. Anyway, remember that those are best case.
How many LED's can a TLC5940 16-Channel LED Driver support (without additional driver chips or transistors)?
That depends on LED current, supply voltage and series / parallel connection.
Can this chip actually supply 120ma to all 16 channels at the same time?
The datasheet says so. See Power calculation above.
Additional: My current is intended to enforce uniformity across all LEDs. I am aware that the chip can regulate current, but I doubt that each channel will have the same number of LEDs.
You will have a problem here. The chip will try to drive the same current through each channel as determined by Current Control above.
Does an LED's voltage drop vanish when current stops flowing?
Figure 2. LED forward and reverse voltage curve. Source: Wikipedia.
Yes. The relationship between voltage drop and current is non-linear but when you get to very low currents the voltage drop falls off rapidly. At zero current there will be zero volts. Similarly when the output transistor switches off the voltage on the collector will be \$ V_{LED} \$. If it was any lower then current would flow through the LEDs.
Constant current and PWM
It appears that this chip uses both constant current and PWM to control the LED brightness. The constant current control (discussed above) sets the maximum current the LEDs will ever see. The PWM controls the brightness. Let's look at the current control and the thermal problem it causes.
simulate this circuit – Schematic created using CircuitLab
Figure 3. \$ V_{LED} \$ set at 6 V. \$ I_{OUT} \$ set at 50 mA. (a) Driving two green LEDs. (b) Driving one green LED.
In Figure 3a we can calculate that at 100% PWM the current sink \$I_0\$ will adjust it's "resistance" to sink 50 mA from the \$ V_{LED} \$ supply. Let's say the green LEDs drop 2.2 V at 50 mA then we can see that OUT_0 will be at 1.6 V. We can calculate the power dissipated in the chip as \$ P = VI = 1.6 \times 50m = 80~mW\$.
If we examine the case in 3b we find \$ P = 3.8 \times 50m = 190~mW \$. Driving one LED makes the IC heat more than twice as much as it would for two LEDs! (for the voltage I chose).
It should now be clear that the selection of the \$ V_{LED} \$ value is critical to maximising the number of LEDs which can be run. Looking at the power dissipation calculation formula in section 11.3
$$ P_C = (V_{CC} \times I_{CC}) + ( V_{OUT} \times I_{MAX} \times \frac {DC_n}{63} \times d_{PWM} \times N) $$
The first part is the power dissipated by the logic and you can't do much about that. (It's small anyway.) The second part tells us that power will be proportional to \$ V_{OUT} \$. This confirms our need to keep this value low.
Choosing \$ V_{LED} \$
Neatly tucked away in Table 4 we find the answer.
Figure 4. The calculation for \$ V_{LED} \$.
- Decide on an LED current.
- Look up the IC knee voltage for that. (See below.)
- Figure out your worst case LED forward voltage at the desired current.
- Figure out how many of these you can get in series with your chosen power supply and still have a little voltage for the IC knee voltage (step 2).
- Re-check your calculations and then set the current limit for the chip using the resistor calculated in section 8.3.7.
The knee voltage
Figure 5. Section 6.7, Figure 3 shows the knee voltage at various output currents. If we were setting for 120 mA output current we would need to allow 1 V for the voltage drop across the chip as indicated at (1). If \$ V{LED} is a volt higher we would end up at (2), etc. For minimum power keep just to the right of the knee at your chosen current._
Power dissipation ability
Figure 6. Section 6.7, Figure 2, shows the maximum power dissipation. I haven't studied this much but if you select a maximum temperature rise of 60°C relative to ambient then you can work out the maximum heat dissipation before thermal shutdown.
Again heatsinking to the PCB will be critical here.
RGB LEDs
simulate this circuit
Figure 7. Converting an RGB LED to a white LED.
If you still want to use your RGB LEDs you can use a couple of resistors on each to compensate for the different forward voltage drops.
$$ R_{RED} = \frac {(V_{blue} - V_{red})}{I_{red}} $$
and similarly for the green.
Best Answer
The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.