5mm IR diode and transistor questions... - Page 1
Forum section: electronics / beginners

author: cvriv
author_rank: Frequent Contributor
timestamp: February 24, 2020, 07:17:20 am
content: Sorry about the title. I didn't know how to summarize my question(s).Im building a matrix of IR sensors using 5mm RX/ TX diodes. I'm going to control them using an 8bit mcu. I just found out that these 5mm IR diodes operate at 100mA or less. That's a lot more than I thought. I can't simply turn them on directly with the mcu so I need to switch them on using a transistor.I have a handful 2N3904's so I checked the datasheet and it says that the max collector current is 200mA so I thought these would be fine. I setup a test circuit and got the IR diode running at 70mA and that's when I smelled the transistor cooking. I checked my PSU and it said that the total power was 630mW, which just so happens to be a tad over the max the transistor can handle.Am I missing something here? I only had 70mA running the the collector and it was just starting to burn. Do I need to a larger transistor to do this? I think it's a little weird that I need a much larger transistor to switch on one IR diode. I have a 4 x 4 matrix of these too so I would need 4 of them.Thanks.

author: David Hess
author_rank: Super Contributor
timestamp: February 27, 2020, 01:58:20 am
content: How were you limiting the current to each diode?What base current were you providing to each transistor?When using bipolar transistors, they are often selected to operate at 1/4 of their maximum collector current at most where current gain will be higher. Operating at 1/2 of the maximum collector current rating is generally contraindicated unless a ring or perforated emitter transistor is used because they suffer from less hfe droop at higher currents. I would use 600 milliamp 2N4401s instead of 200 milliamp 2N3904s for a 100 milliamp load not because the higher current rating is required but because they will have reliably higher gain, and besides they are practically the same price.

author: mvs
author_rank: Frequent Contributor
timestamp: February 28, 2020, 02:04:49 pm
content: Quote from: cvriv on February 24, 2020, 07:17:20 amIR LEDs have low forward voltage (around 1.1-1.2V). To power them efficiently you will either need low voltage power supply (say 3V) or connect multiple LEDs in series.

author: Zero999
author_rank: Super Contributor
timestamp: February 28, 2020, 02:41:46 pm
content: The total power dissipation of the entire circuit, exceeding the maximum rating of the BJT is not normally problem, as only a small amount of power will be dissipated by the BJT. The rest should be dissipated by the LEDs and current limiting resistors.The BJT needs an appropriate base resistor and the LEDs, the correct current limiting resistor(s).As mentioned above, the LEDs should be connected in series or a lower power supply voltage used, for maximum efficiency.

author: mvs
author_rank: Frequent Contributor
timestamp: February 28, 2020, 03:15:12 pm
content: Quote from: Zero999 on February 28, 2020, 02:41:46 pmYes, if you use transistor as a switch. But here it seems to work in linear mode from 9V supply (630mW/70mA).

author: Zero999
author_rank: Super Contributor
timestamp: February 29, 2020, 09:40:58 pm
content: Quote from: mvs on February 28, 2020, 03:15:12 pmNo schematic has been posted, so we can only guess, but even if it was operating in the linear region, it would have to be dissipating less than the 630mW, indicated on the power supply, because there are other components in the circuit: LEDs, MCU etc. which all consume power.