LNK304 dropout voltage
Hi, my name is Damian
I need create AC to DC converter
I have the same problem with my output voltage stability (5V) when have non Load and when Converter LNK304 is loaded current approximately 50...100mA...even 150mA.
I prepared Converter with two solutions: Inverted polarisation and non-inverted. So, I checked both types of the configurations: "High-Side Buck – Direct Feedback" and "High-Side Buck-Boost – Direct Feedback"
As you see, I need a few power consumption, only 0.5[W] (5V*0.1A)
Here is the similar topic that I found on the web site:
So....first I added the basically load current about 50mA on the converter output and second I adjusted output voltage on the 5.10V by the Rfb resistor (with 3.83kOhm into 4.43kOhm).
Resistor Rbias is exactly 2kOhm.
When I addded additional loading (10...50...100mA), voltage on the outside converter dropout from 5.10V to 5.00V
If we assume that:
5.10V = 100 [%]
5.00V = x
we will get accuracy/tolerance output voltage:
5.10V * x = 100 [%] * 5.00V
x = (5.00V / 5.10V) * 100%
x = value -100 [%]
x = 1.96 [%]
You wrote in datasheet LNK that the Rpl resistor should take the value: Rpl = Vout / Base Load
where "Base Load" should equals 3[mA]
Rpl = 5.1[V] / 3[mA] = 1.7[kOhm]
...and with power dissipation on the resistor on the level 100[mW] - eg. 0603 package
Pdiss = I* I * R = 3mA * 3mA* 1.7kOhm = 15.3mW
I had to apply Rpl = 100[Ohm]
now, my base load current equals: 5.1V/100[Ohm] ~ 50[mA]
my power dissipation on the resistor equals: Pdiss = I* I * R = 50mA * 50mA* 100Ohm = 250mW (I used in this moment 1W resistor - 2512 package)
hm, give me support, please