Linkswitch-PL as Buck converter

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Hi!

I have a project:

Uin=230Vac (185Vac-265Vac) Uled=64V (60V-72V) Irmsled = 150mA, Ipkled = 230mA (maksimum led peak current).

I want to rebuild a devboard DER328. The PIexpert calculate the transformer (inductor) for my input data. The problem is with peak current which is flown by LED. I found parameter:

Peak Drain current at maximum input voltage

And I wonder if this current will be the same what i'm looking for in PIexpert? Is it possible to put this parameter like peak current on the LED inside calculator sheet for Buck converter (Linkswitch-PL). I want to buy trafo with enough high inductance to reduce ripple and peak current. The PIexpert calculator don't give a possibility to change a trafor's inductance and reduce peak current.  Tell me, how can I calculate the inductance of transformer if I want to reduce the peak curren on the LED to 340mA (max).

With best regards.

TS

Increasing transformer inductance have little effect on peak output current. If you want to reduce the peak current on the led(or output current ripple), increase the output capacitance.

You can  override the inductance value computed by the PIXls by entering the value on LTYP ( the input column line 44 ). LTYP is the typical inductance value of the buck inductor.

Thanks for reply,

How can I calculate the minimum capacitance which will be enough for my project's requirements?

I cannot exceed the maximum forward current on the LED (230mA).

With best regards.

TS

You can use the approximation

 

Cout = (Io_avg/Io_pkpk) * (1/(2*pi*fline*Zled))  where

Io_avg - average output current

Io_pkpk - peak to peak ripple current

fline - line ferquency, 60Hz or 50Hz

Zled - LED dynamic resistance