BJT transistors 101 question.. - Page 1
Forum section: electronics / beginners

author: Mo3tasm
author_rank: Contributor
timestamp: February 20, 2017, 08:21:11 pm
content: Hello everybody, I'm playing with BJTs trying to learn how they work and all, but I'm having a trouble understanding what's going on..So I've built two circuits, number (1) works as expected, LED lights according to base voltage and draws a maximum of 10 mA when Vb = Vc.But, when I connect components like (2), it appears as a short, and the PSU switches to C.C mode with highest current that I set it to.LED lights for a second and dimm.Could you please explain what's going on...Thanks..

author: Zero999
author_rank: Super Contributor
timestamp: February 20, 2017, 08:37:09 pm
content: The base of the transistor is a diode junction and can't be connected to a constant voltage source like that. You need to connect a resistor in series with the base. As a general rule of thumb, to saturate the transistor, the base current should be around a tenth of the collector current, so go for 3k3 for now. If you've put too much current through the base of the transistor, then it might be damage and will need replacing.

author: John B
author_rank: Frequent Contributor
timestamp: February 20, 2017, 08:38:24 pm
content: Transistor went boom.The power supply is shorted to ground through the base of the transistor.You'll need to limit the base current to less than 10mA

author: Benta
author_rank: Super Contributor
timestamp: February 20, 2017, 08:38:45 pm
content: In the first schematic, your BJT has been reduced to a B-E diode but the circuit works as: battery+ -> diode -> resistor -> LED -> battery-In the second, you have short-circuited the battery through the B-E diode and nothing works (but you have a dead BJT).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 21, 2017, 02:14:10 am
content: BTW, most power supplies have a capacitor on the output, so when you connected the second circuit, BANG, you probably didn't see anything, but that poor transistor discharged a peak current of probably 10A in a teensy fraction of a second. Blown wide open. (Or shorted, as the case may be. Once it fails shorted, it's just a matter of how much excess energy there is, to blow it open.)The proper way to use a current-limited power supply is to reduce the output to 0V (or disable it), connect the circuit, then increase the voltage to the desired setting. This keeps everything in a zero-energy state, then soft-starts it.Tim

author: danadak
author_rank: Super Contributor
timestamp: February 21, 2017, 02:25:23 am
content: Because some circuits, UPs, have min ramp rates on power, to power upproperly, one way is take output of supply, board disconnected, turnvoltage to desired for board you are working on, then turn current downto 0, short supply output, and turn current up to some expected nominal.Then connect board. I power up a lot of simple boards with 100 mA setting(depends on board and design what you should set/expect). This generallywill prevent damage in case there are shorts or excessive out of spec loads,while satisfying for good boards a startup that will meet its min ramp rates.Note if there are large sized C values on board they may affect min youhave to set. eg. boost the peak current that you would set the current limitto. To charge the large caps at a ramp rate. Q = C x V, I = C x dV/dT,from this you can calc the ramp rate. Again where large caps are involved.Regards, Dana.