Suppose I have a voltage regulator acting as a high-power current source (basically a constant-current driver), with a resistor to set the output current. This drives 27 high-brightness white LEDs, arranged as 9 series chains, with 3 LEDs per series chain (9.6 V min at 20 mA).
Do I need to put resistors on every series chain, or is this not necessary for enforcing current sharing?
simulate this circuit – Schematic created using CircuitLab
I'll address a single LED chain first, and get on to multiple chains later.
With a single chain of LEDs, powered by a current source, you do not even need resistor R2 in your circuit. The function of that resistor is to develop ("consume") whatever voltage is not required by the LEDs. For instance, if you employed a fixed voltage 18V supply, and your chain requires 9.6V, then the resistor is required to "absorb" 8.4V of the total 18V, so the remaining 9.6V is delivered to the LEDs.
By contrast, a 20mA current source (such as your LM317) will adjust its voltage to whatever value is required to pass exactly 20mA through whatever load is connected. If that load happens to develop 9.6V (like your 3 × LED chain), then that will be the exact voltage output by the source:
simulate this circuit – Schematic created using CircuitLab
On the right, where I use a fixed 18V supply, R1 is sized to develop \$18V - 9.6V = 8.4V\$ when passing 20mA, according to Ohm's law:
$$ R = \frac{V}{I} = \frac{18V-9.6V}{20mA} = 420\Omega $$
On the left above, where the power supply is a current source, no resistor is required, since the current source automatically produces 9.6V; that's what the diodes' total voltage would be when passing 20mA.
That means your own circuit with the single LED chain is incorrect; R2 would not be required. Do not confuse the purposes of R1 and R2. R1 in your circuit is used to tell the LM317 how much current to pass, and otherwise plays no role in limiting current by "consuming" unwanted voltage. Rather, R1 is calculated according to the following, assuming \$I\$ is the desired current output from the LM317:
$$ R_1 = \frac{1.25V}{I} = \frac{1.25V}{20mA} = 62.5\Omega $$
The LM317 is responsible for "dropping" whatever voltage is left over, after the total LED voltage and the voltage across R2 (1.25V) is removed from the supply. You won't need another resistor.
All that assumes you have one chain of LEDs, but your application requires 9 chains. This requires some extra understanding. In the following two circuits I power two chains of LEDs, from a current source providing 40mA:
On the left, all diodes are identical, well matched in all their parameters. As you can see, current is shared equally between both paths; 20mA each. That's the ideal scenario.
On the right, though, I've changed one of the diode's characteristics by a very small amount, and the result is that current is no longer shared equally. This is the problem with diodes, unless they are are all identical in every respect, they will have slightly different "forward voltages", and this will cause a large difference in how current is distributed through them, and consequently one chain will glow more brightly than the other.
The easy solution is to give some "elasticity" to the system, using small-value resistors, known as "ballast". Their purpose is to take up some of the voltage difference, giving each chain some flexibility in choosing its own total voltage. Here's the same (right hand) circuit reproduced with some ballast in each chain, which causes currents in each path to be closer in value:
All these facts together mean that you should use ballast resistors (one for each chain), even if you use a current source. Using an LM317 as a current source to power multiple chains of LEDs, while mitigating LED mismatches, will look like this:
If you have 9 chains, and each chain requires 20mA, then total current will be \$I=9 \times 20mA = 180mA\$. R1 will be calculated as follows:
$$ R_1 = \frac{1.25V}{180mA} = 6.9\Omega $$
There are caveats. R1 "consumes" 1.25V. Each ballast resistor "consumes" \$20mA \times 47\Omega = 1V\$. The LM317 regulator has a drop-out voltage of 2.5V. The LED chains each have 9.6V across them. The minimum supply voltage at the input to the LM317 will be the sum of all those:
$$ V_{SUPPLY} = 9.6V + 1V + 1.25V + 2.5V = 14.4V $$
This is bad news if you are planning on using a 12V supply. If this is a problem, but you are sure your voltage source will be fairly stable, then it might be simpler to use a single resistor for each chain, and forego the current source altogether, as in the right-hand circuit at the very top of this answer.
You can add resistors to each, which would help balance the current.
However note that the ancient LM317 is not a very good current regulator. It needs about 3V, on top of the 1.25V Vref, to regulate properly, so if you allow 0.5 or 1V across each resistor to equalize the currents you'll need about 15V to get your 9.6V worth of LEDs to run at constant current. Even without resistors you'll need 14V+. If you have a 12V regulated supply you may as well just use resistors (one per chain) and save power and get similar regulation.
Current sharing with parallel loads is not going to happen to any great extent with or without the resistor. OK it improves with a series resistor but, it's not really a viable solution to use a current source.
The trouble with your idea is that if one of the parallel LED strings goes open circuit, the remaining 8 parallel LED strings will now have to consume the current taken by the failed string and, this can cause an avalanche effect where pretty soon all the strings fail. At the very least, the remaining 8 LED strings get brighter.
So, if this is not what you want, use a global/common voltage regulator and put a current limiting resistor in each string. Forget about the LM317 acting as a constant current generator is my advice.
Use one U1, R1, and omit R2 per string. Every series string needs an independent current source. Even better, use a multi-channel LED source/sink or driver/controller.
There are some LED drivers designed to allow current to be shared between LED strings. E.g. the TI LM3466 Multi-String LED current balancer for use with constant current power Supplies has the following description in the datasheet:
The LM3466 integrates a linear LED driver for lighting systems which consist of multiple LED strings powered by a constant current power supply. It balances the current provided by the supply in a pre-set ratio for each active LED string, where an active string is a fully turned on LED string, regardless of the number of strings connected to the supply or the forward voltage of each LED string. If any LED string opens during operation, the LM3466 automatically balances the supply current through all of the remaining active LED strings. As a result, the overall brightness of the lighting system is maintained even if some LED strings open during operation.
The LM3466 lighting system is simple to design owing to a proprietary control scheme. To minimize the component count, the LM3466 integrates a 70-V, 1.5-A, N-channel power MOSFET with a current limit of 2.06 A. To add one more LED string to the system, only a single resistor, a capacitor, and a LM3466 are required. Other supervisory features of the LM3466 include under-voltage lock-out, fault reporting, thermal latch off, and thermal shutdown protection.
I.e. a dedicated LED driver can have features to help manage one or more LED string opening during operation.
This was the first example device found on a search, and a parametric search for LED drivers might find one more suitable for the application in the question.
This drives 27 high-brightness white LEDs, arranged as 9 series chains, with 3 LEDs per series chain (9.6 V min at 20 mA).
Here is a fun solution: digikey lists 200 mA rated white LEDs from reputable manufacturers starting at less than 3 cents each. For about 75 cents you can make sure that all possible current distributions between the strings are safe and simply ignore current balancing.
Bonus: your circuit will continue to work and produce the same total light even if 8 individual LEDs all fail open circuit.
