Converter

I removed the duplicate thread for you. You usually have 15 minutes after posting to make any edits.

thanks! bc i tried to edit it before, but couldn't.

Yes, that looks about right. More capacitance will smooth it out. What are you trying to do?

yeah. what do you think about paralleling a few capacitorss to get additional capacitance?

i'm making an LED driver. so i'm trying out the buck converter first :)

I think your circuit needs some adjustments. Based on your scope photos and some of the schematic values, the waveforms should look like the simulation below.

Buck Converter SIM
Vin= 19
Vout= 5
freq= 6.17E+004
RL= 150
Co= 1.00E-006
L= 1.00E-003
hyst= 0.1



Close, but not quite right.

Looks like the switch and/or its drive could be improved to get Vswitch closer to the input supply of 20V. Also looks like the inductor current may be dropping to "zero". The operating modes of continuous and discontinuous inductor current may need additional feedback compensation and/or some changes in the converter component values.

Keep trying and learning. Fine tuning switching power converters takes some skill and experience which are best gained by building and debugging many designs over time. Best wishes.

mjb1962853, you are actually right. i don't know the reason why it worked that time. yesterday, i tried the same circuit because i wanted to increase the output capacitance (but nothing else in the circuit changed). but it wouldn't work. i tried with the original capcitor, but it still wouldn't work. the thing is i don't think the IGBT gate is receiving any signal at all. do u have any suggestions?

http://www.ti.com/product/tl494#technicaldocuments

http://www.ti.com/litv/pdf/slva001e

Suggestions:
1) I would not use an opto isolator/driver for a low voltage circuit like this unless absolutely necessary.
2) An inexpensive PNP transistor or P-channel MOSFET will work as the switch. No need to use an IGBT here.
3) Do not attempt to build a switcher with a standard rectifier diode. It will not work. The diode MUST be a fast recovery type.

If you are trying to build a regulated voltage switching converter, please try to follow the literature links above and calculate the proper component values needed for your design.

It looks like you are only using the TL494 as a simple duty cycle oscillator with no feedback from the output circuit. If you are only trying to design a simple (switching) LED driver to provide a (nearly) constant current supply for LED's, please specify the following:

a) How many LED's do you want to drive in series.
b) What is the rated current (mA) and forward voltage drop (V) of one of the LED's you want to use.

thanks for the suggestions. i'll try to see what i can do about it :)

the input is suppose to be 240Vac then to a bridge rectifier then to the converter.

i'm driving 5 of this leds here: http://uk.farnell.com/hlll/t21d1/led-pwr350-to220-daylight-white/dp/1226709

rated current = 350mA and forward voltage drop = 3.5V.

and i'm using an ultrafast diode like this one : http://www.alldatasheet.com/datasheet-pdf/pdf/107508/STMICROELECTRONICS/STTH12R06DIRG.html

this is okay right? :D

Diode looks acceptable.

Since you have changed the design from a 20Vdc on your schematic to a non-isolated 240Vac rectified (340Vdc?) power source, the IGBT and opto-isolator WILL be absolutely necessary. This information should have been part of your original post.

The spec sheet and application note links provided above should be carefully studied. If you cannot derive a satisfactory design from these documents, a local TI rep/applications engineer may be able to help.

The liability and safety issues and the complexity of your design now exceed the basic limits of this discussion forum. Sorry I am not comfortable offering any further assistance. Be safe and best wishes.

i understand. and thanks for the notes, i'm on it :)

hopefully everything will work out.

& thanks again! :)

It looks to me as though on the output of the FET you've got a smoothing cap first then a series inductor. I think it's more normal to have these the other way round.

What are the main component parts?

do you mean to say that i should have the inductor in series with the R load?

the top one is the buck converter and the bottom ones are for the PWM signal. supposedly to be an LED driver. but this couldn't even work :S

Sorry: I hadn't realised that your circuit diagram was a link to a readable drawing:-

I thought that the diode was a capacitor.

The inductor is in series with the load.

In answer to one of your original questions: yes the bigger you make the smoothing capacitor the flatter the output response will be.

F;i your scope (http://img.photobucket.com/albums/v232/juzz_d13d/DSC_1907.jpg) seems to show some strobe B4 the FET starts to form a channel ??? i gess it's the 1µF cap near HCPL-3140(http://www.datasheetcatalog.org/datasheet2/b/0feg3rrhe9fh16wsfqf6fez9wq3y.pdf) although they show such consistent with no inductive load ... (just noticed)

ci139, i don't understand. what do u mean by 'show some strobe before FET starts to form a channel?