Our customer is studying with your ADD5211 LED driver.
So I got a question.
Our customer will design with 2FB gang of sink(open drain) pins.
Such as Fig-23 following.
And he will sink the current by FB1+FB2 and FB3+FB4.
Max sink current is 200mA per 1FB.
So in this case, FB1+FB2=200mA+200mA=400mA.
Is this correct?
Otherwise should we think the derating as eg;360mA(90%) ?
Can he drive this sink 400mA continuously no problem?
Of course I will comment to put enough heat sink pad on PWB for heat dissipation.
OK, I got it.
Thank you for your additional comment also!
This is allowed. And it will produce 400mA through each LED string. However, at 200mA per sink, the typical voltage on the FBx pins is (see p. 11 of datasheet):
FB_REF=0.23+0.0041*200mA = 1.05V
Therefore, the ADD5211 will be dissipating 1.05V * 800mA = 840mW
That is a fair amount of heat for the ADD5211 to dissipate. A 4 layer board, and thermal vias will help. Alternatively, the ADP8140 requires just 150mV for 200mA of LED current. Giving 120mW of total power dissipation. However, an external boost regulator would be required (see Figure 35).
- Jon Kraft
Also, I forgot to add that the power dissipation of the current sinks will increase for any voltage mismatch between the strings. For example, if LED string 1 is 32.5V and LED string 2 is 33.7V, then the IC (either ADD5211 or ADP8140) power dissipation will increase by 1.2V*400mA = 480mW. So please keep that in mind with your design.
Ah,,,thats a very important!
I will tell this also to our customer.
Sorry I have a lot of info. So wrap up my question here.
How should we calculate for the case of 1 or 2 LED?
On right side lower, page-15 on your datasheet
And page-14 formura
Vout_max = N × V F MAX + 1 V
In this case, Rramp become -(minus) ohm!
Customer is studying 2 type for ADD5211 using.
type-A: 4 ganged , 1 string, 2LEDS (Vf=1.8V LEDs) 800mA
type-B: 2 ganged , 1LED (Vf=1.8V LED) 400mAx2 strings
I need your good comment to fix this.