4047 inverter substituting BD139 and issue with power transistor damage  - Page 1
Forum section: electronics / projects

author: dave_j_fan
author_rank: Regular Contributor
timestamp: May 14, 2017, 04:53:49 pm
content: Originally this was the circuit ,http://www.eleccircuit.com/wp-content/uploads/2007/10/inverter_100w_12v-220v_by_ic-4047_irf540.gifhere as the mosfets cost some money , I have replaced with bc547 and bd139 ( mainly i had many of the bd series in parts bin)But i have skipped R5 and R6I get transistor cut after some time ... is it that these resi are very much needed ?- ihave even added protection diodes 1N4007 reverse connected at the collectors

author: Zero999
author_rank: Super Contributor
timestamp: May 14, 2017, 05:56:17 pm
content: It should work. I've certainly seen worse circuits.The transformer is not big enough for a 5W LED. You should use a 1A transformer.

author: dave_j_fan
author_rank: Regular Contributor
timestamp: May 14, 2017, 06:01:26 pm
content: no the 500mA worked for some time .when i was trying with changeouver the transistors get cuti am using a adapter as 12v source for testsi suspect it might give spikes !

author: Benta
author_rank: Super Contributor
timestamp: May 14, 2017, 06:10:50 pm
content: R3 and R4 are unnecessary.But I don't like the circuit.First, BD139 are not the greatest switching transistors made.Without active turn-off drive (negative base current), the BD139s will take ages to turn off, which means you have a lot of cross-conduction over the transformer.The original circuit is better, as the 0...+12 V drive to the MOSFETs fits much better.

author: SeanB
author_rank: Super Contributor
timestamp: May 14, 2017, 06:42:10 pm
content: At a minimum add a resistor across the base emitter of each BD139, so they have at least faster turn off. You can use 220R resistors there, though you would need at least some dead time so both do not conduct.Better is to use a common TL494 or SG3524 SMPS controller, at least there you get a base driver and dead time for free, plus you can get one out of most dead ATX power supplies for free, they almost never fail when it goes boom. Just set frequency to 120Hz and duty cycle can be maximum , it will add the dead time automatically and the transistors will run cooler with no cross conduction. Of course with a PC ATX power supply you get 2 free power devices on the input side, which will also work at 12V with ease, and they also have a heatsink.

author: Zero999
author_rank: Super Contributor
timestamp: May 14, 2017, 07:08:33 pm
content: Quote from: dave_j_fan on May 14, 2017, 06:01:26 pmIt's not surprising the transformer outlasted the transistors. Still, 9V×0.5 = 4.5W is undersized for a 5W LED. I suggested 1A to give you plenty of headroom because the LED will have a poor power factor and is likely to draw more than 5VA.Most of those resistors are unnecessary and the 220R in series with the CD4047's power supply rails with no decoupling is retarded.The switch off can be improved with a small capacitor could be added to bypass the emitter followers to the main driver transistors. A snubber network may also be required across the transformer, which was an afterthought, hence it's not on the attached schematic.

author: diyaudio
author_rank: Frequent Contributor
timestamp: May 14, 2017, 07:35:29 pm
content: Quote from: dave_j_fan on May 14, 2017, 04:53:49 pmSome pointers.You need a scope.You need to check the if you NOT saturating the BJT.Consider using a totem pole output (see attachment)You need a scope. your circuit NOT allowing the transformer flux to reset, hence your transistor maybe ON all the time, the BJT needs to be forced off.You need a scope.I find it difficult to believe you cannot obtain even a IRF540 jellybean MOSFET, however even if you did, your drive circuit will only yield sub optimal performance and probably destroy the FETs.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 14, 2017, 07:48:07 pm
content: For just a 5 W LED one could use smaller MOSFETs than the IRF540, more like IRF510 or a more modern one rated for about 40-60 V. With a smaller FET is might be OK to directly drive the gate from the IC.Anyway a SMPS chip is preferred, as it would include a dead time.

author: dave_j_fan
author_rank: Regular Contributor
timestamp: May 15, 2017, 01:49:37 am
content: good info,ordered d mosfets

author: dave_j_fan
author_rank: Regular Contributor
timestamp: May 15, 2017, 01:51:36 am
content: yes but wnted to keep simple..my b mosfet wll b the better option

author: dave_j_fan
author_rank: Regular Contributor
timestamp: May 15, 2017, 02:02:26 am
content: hero999,thnk u fr the new changes.i will try this at later time.now quickly hd to give it fr demo so mosfet were ordered