解决方案搜寻

LNK304 dropout voltage

Posted by: Damian Musialik on 11/16/2021

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:

https://www.powerint.cn/community/forum/ac-dc-conversion-forum/2018/lnk304-output-voltage-drops?language=zh-hans

 

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 [%]

but...

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

Damian

 

评论

Submitted by PI-BERJ on 11/19/2021

Hi Damian,

LNK304 output voltage does drop with the corresponding increase in load. The highest output voltage is when it is unloaded. This is where Rpl is needed.

However when it is loaded, let's say approximately 10% to 100% load, output voltage variation should be within 5%. In your case, that would be something in the range of 5V to 5.25V.

If your load needs 5V exactly e.g. microcontrollers, you may need to increase the output of LNK304 to let's say 8V and use a post regulator (linear regulator) instead. Otherwise, you will have to make do with the 5% output voltage variation and tweak the feedback resistors to obtain 5V at the expected loading condition.

Regards,

Submitted by Damian Musialik on 11/19/2021

Thanks a lot for answer

So, this converter has to a lot of deviation of the output voltage.

hm, your stabilization / feedback is of low efficiency - why ?

I thought that it has more accuracy. 

 

...first of all, for the LNK304GN, I used for the Uout:5V resistor Rpl = 1.65k (5V/1.65k = 3mA) 

5.01V without load ... and 4.51V with load 44.7mA, load as resistor, input voltage: 233V AC

{(4.51V/5.01V) -1} * 100 [%] = 9.98 [%] deviation

 

5.01V without load ... and 4.54V with load 88.6mA, load as resistor, input voltage: 100V AC

{(4.54V/5.01V) -1} * 100 [%] = 9.38 [%] deviation

 

5.04V without load ... and 4.51V with load 130.1mA, load as resistor, input voltage: 151V AC

{(4.51V/5.04V) -1} * 100 [%] = 10.51 [%] deviation

 

Why your product have not more stability. Now we have difference approximately 500mV. Thats is a big value.

 

I checked also with other value output voltage and with other model converter - LNK305PN:

I used for the Uout:11.5V resistor Rpl = 3.83k (11.5V/3.83k = 3mA) 

11.55V without load ... and 11.05V with load 109.4mA, load as resistor, input voltage: 183V AC

{(11.05/11.55) -1} * 100 [%] = 4.32 [%] deviation

 

....as I see, the LNK304GN has 10% deviation and LNK305PN has about 5% deviation....

What does it mean? Could you respond to me?

 

Summary

I spent a lot of time in Labo and I came to the conclusion that the only solution is when I use LDO stabilizer

I thought that I would not have to do this in my project

 

Submitted by PI-BERJ on 11/19/2021

Hi Damian,

Link-TN products do not work like conventional PWM regulators with analog control. Its feedback functions via an ON/OFF regulation scheme with cycle-by-cycle current limiting, i.e. control is 'digital'. This scheme has several advantages e.g. cost, simplicity, and transient response to name a few. They are however inferior to conventional PWM regulators when it comes to load regulation. These advantages however should be more than enough if load regulation is not that critical to the design.

If your load can tolerate a bit of voltage variation, I suggest you tune the output voltage when it is already loaded, not when it is unloaded. In your case, you should try to increase the regulation voltage from ~4.51V to 5V by changing the feedback resistors with the load on. The output voltage might increase to maybe 5.5V when it is unloaded but I believe these unloaded conditions rarely happen in the field.

Regards,

Submitted by Damian Musialik on 11/20/2021

ok, I understood all 
Thanks for support

Damian