A series resistor on a high-voltage string of LEDs can really be problematic 
due to really lousy current regulation.

Assumptions (just a random example, no real world parts involved):

* We have a chain of LEDs with a total Vf of 100V.
* The design current is 100ma.
* Thus, the power dissipated/converted to light is 100V*10ma=10W
* The nominal supply voltage is 110VDC.
* So, our series resistor will have to be 10V/100ma = 100 ohms
* The dissipation in the resistor is 1W.

So far it all sounds pretty reasonable. Now, however, what happens if:
Q? The input voltage drops to 105V?
A: We only have 5V across the resistor, so the current flow drops to 50ma.

Q? The input voltage rises to 115V?
A: We now have 15V across the resistor, so the current flow rises to 150ma.
And by the way, dissipation in the resistor jumps to 2.25W.

[Yes, I know I have made a simplifying assumption here: that the Vf is 
constant with current]

So, in other words a lousy 5% variation in the input voltage results in a 
50% variation in current.

Unless Vf varies a lot more with current than I think it tends to, this is 
not a good thing (tm).

But what if we make the resistor a bigger part of the total voltage drop? I 
created a little spreadsheet to do the computations. Again, we assume  a 
110VDC nominal supply, 10W of dissipation/output in the LEDs. But now we 
vary the voltage across the resistor in steps of 5V. The columns below are:

Vr nom -voltage drop across resistor when Vin = 110
Pr nom - power dissipation in resistor when Vin = 110
Eff % - percent of total power dissipated/converted in LEDs.
Err % - percent change in current when Vin=105 or 115

      Vr nom Pr nom Eff % Err %

      5 0.48 95.45 100.00
      10 1.00 90.91 50.00
      15 1.58 86.36 33.33
      20 2.22 81.82 25.00
      25 2.94 77.27 20.00
      30 3.75 72.73 16.67
      35 4.67 68.18 14.29
      40 5.71 63.64 12.50
      45 6.92 59.09 11.11
      50 8.33 54.55 10.00
      55 10.00 50.00 9.09
      60 12.00 45.45 8.33
      65 14.44 40.91 7.69
      70 17.50 36.36 7.14
      75 21.43 31.82 6.67
      80 26.67 27.27 6.25
      85 34.00 22.73 5.88
      90 45.00 18.18 5.56
      95 63.33 13.64 5.26
      100 100.00 9.09 5.00
      105 210.00 4.55 4.76


So, assuming that a current error of 20% is reasonable for the 5V step, the 
best we can do is an efficiency of about 77% where we are dropping 25V 
across the resistor, which is dissipating about 3 watts at nominal Vin (and 
more than 4 watts at Vin=115).

So, I am guessing that a simple linear current source is a good idea. 
Probably just a couple of resistors and one transistor. Basically, the 
errors will then be reduced by a factor of the transsistors beta, which 
could easily be 50 or more. (of course new errors would be introduced in the 
temperature coefficients of the transistor, etc).

-- Bob Ammerman
RAm Systems


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