Analog Dimming by Driving DC Voltage on the Feedback Pin

Hi,

Any more numbers to work from or are they a secret? Saves people having to use their time to guess how to arrive at the result for you:

VDC
Vin LEDs
If LEDs

Isn't this what Thevenin, KVL, etc., are for?

Hi,

I would like to have 100mA on the LEDs, so the I led = 100mA and Vfb is by the datasheet 200mV.The Rsense (R1) in the schematic should be 2 ohms and Vdc will be 5V. I need to calculate R3 that I would like to put a potentiometer and R2 to achieve a range from 100mA to 0A.

Hi,

I was puzzled. Couldn't see method clearly, anyway... Creative maths that I'm making up so am pretty sure is wrong:

5002/95002 = 0.0526
3.8V * 0.0526 = 0.20007(V) {3.8V by first 5VDC * 0.0526 = not 0.2Vfb so iterative calculations to ascertain/guess 3.8VDC.}

95002/5002 = 18.992
18.992 * 0.2V = 3.79V

For 5VDC, 5/3.8 = 1.315
So, 90k * 1.315 = 118421 ohms.
Nearest easy values are 122k = 100k + 22k. So, maybe trimpot of 220k + small series resistor to limit current through pot at '0' A. Or if very precise current needed, two pots for coarse and fine tuning, like 220k and 10k or 1k.

Ignoring multiplication by 1.31 of 5k and 2R:

127002/5002 = 25.390
25.390 * 0.2 = 5.078V

The method is doubtful, maybe you can understand how to work it out better, hopefully. Sorry if it's nonsense maths and unhelpful, I tried at least...

Hi,

Some years ago I provided this information:

While the headline says "SMPS" the design note is true for many regulation loops.

Basically you can calculate with the two extremes: I_LED = 0mA when V_DC = 5V and ... I_LED = 100mA when V_DC = 0V.
In both cases the FB pin voltage needs to be 200mV.
Now there is one problem: V_DC is 0..5V, thus it's higher and lower than 200mV, means the R3 current is positive as well as negative.
This makes it difficult to work with R_shunt = 2Ohms.
Thus I recommend to first calculate R_shunt. Best when R2 = 0V and R3 = 0V, means V_DC = 200mV.
You want 0.. 5V to represent 100mA ... 0mA. I_LED = 100mA - (V_DC × 100mA / 5V). Thus 200mV represents 96mA
R_shunt = 200mV/96mA = 2.0833 Ohms

Now at 100mA you get 208.3mV V_shunt at 100mA
V_DC = 5V (range)
V_shunt =208.3mV (range)
R3 / R2 = 5V / 208.3 mV
If R2 = 100 Ohms, (me chosen)
R3 needs to be 100 Ohms * 5V / 0.2083V = 2k4
This simplified calculation ignores total shunt resistance thus causes about 0.102% error

*******

Pedantic:
You say "Driving DC Voltage on the Feedback Pin".
But indeed you don't drive, nor change the voltage of the FB pin. The FB voltage does not change (significantly)
You rather introduce current into the FB node. ...
You drive the voltage to one side of a resistor, while the other side of the resistor is fix.

Klaus

Hi,

If it helps at all, to add to above comment about a current more than a voltage, I think the circuit is more or less like so (excuse errors as I'm not great at this kind of analysis):



(Can a moderator make the jpg smaller than ~4 MB, please? Stupid 'phone only allows image modification of clipping or infantile, gimmicky effects...)
[Moderator action: reduced to 37kBytes. I guess every phone can setup initial resolutioon and usually has simple photo editing tools. Please next time do it on your own. Also there are online tools like https://www.resizepixel.com]

I thought 5k was chosen as: 0.2V/5,000R = 0.0000400A
So:
5V/122k = 0.0000409A
and, based on my previous dodgy maths:
3.8V/90k = 0.0000422A (almost 40uA).

So that device regulates I_LED based on ideal I_in_FBpin = 80uA? Then, e.g. 81uA lowers LED drive current and 79uA increases LED drive current, to keep I_LED steady/constant, does it?

Hi,

thanks d123 for pointing out FB pin current.
I ignored it in my simple calculation.
It adds about 1.7% error (typ). Still less than the initial V_FB error of 6%.

Current direction is not clear. I guess it pulls up only. Thus the resulting LED current will be reduced by 1.7% FS.

Klaus