I have redesigned the rear tail lights of a motor vehicle to work of a micro controller (provide PWM, High, low) and numerous driver circuits to drive 5W dc light bulbs. Each driver (IR2104) is connected to its own power transistor which controls the power sent to a single bulb. During initial testing a single driver cicuit and load was connected and performed flawlessly with minimal heating after a few hour running at 12v - 1A. the circuit was expanded to 6 parallel drivers and loads with and increased supply current of 2.2A at 12V. the increased load resulted in a large ripple of close to 2V p-p in the 12V supply, after 30mins of running 3 drivers failed and I am unable to determine whether it was the increased supply current, the large ripple or some other reason which caused the massive failure of the system.

 

Schematic?

 

 

A friend has suggested a current sink to prevent the transistor from drawing to much current from the chip

 

Easy to fix I think.

IR2104 is a half-bridge driver, and you are using the high side drive for your part (pin 7). You should use the low side (pin 5) instead. So, simply attach the base of your transistor to pin 5, and realize that the logic will be inverted. This will probably solve all of your issues, and you won't need the diode or

Another option would be to try tying pin 6 to ground. This would keep you from having to invert the logic.

 

Where is your current limiting resistor between the IR2104 and transistor's base? If that's how you wired it, as in your schematic, I'm surprised that they didn't all blow from day one. What transistors are you using?

Ken

 

I forgot to include that pin 6 goes to ground. @ john*michael i will try to use the low side drive but it is stated in the date date sheet that a diode should be connected from pin 1 to 8. I am using a power transistor n type, it has a 3 A typical rating with a 5 A max.

 

Kmoffet is also correct! I have gotten used to MOSFETs. If you have pin 6 already tied to ground, you will need a resistor between pin 7 and the base.

 

If your "TIP" is a Darlington, Hfe should be more than 500 and you need 2.2 amps divided by 500 or a minimum of 4.4 milliamps. 1K would provide 10+ milliamps so that seems about right.

 

Also keep in mind that your "on" voltage is about 1.5 volts, so at 2.2 amps your transistor is dissipating 3.3 watts when you are not in PWM mode. You might need a heat sink.

When you say "parallel" you cannot run multiple bipolar transistors on the same load. I think you are saying that you have a separate load for each transistor. Is this correct?

 

View attachment 92004

The files size that you attached were too too big, I was compressed them to the normal size, they are from about 3.15MB compressed to about 54KB, next time before you upload the files, please compress the photos to 800x600 or 640x480, if necessary then you can compress to 1024x768, almost the photos using 800x600 image is enough.

Too much file size will affecting the connecting speed of members to the forums, thanks for your cooperation and understanding.

 

A friend has suggested a current sink to prevent the transistor from drawing to much current from the chip

Do you know what a current sink is? Could you explain?

 

@john*michael i am using one transistor per load as i need independent contol of each. I have added a basic heat sink, also the supply i used was just simulation. A more powerful one is required to stimulate the effects of a car in my next stage (higher V up to 20 V and current in excess of 3A ).@atferrari the current sink he recommended was a basic bjt current sink where i just calc my desired current and buid my Iref accordingly.