Current source, hot transistors - Page 1
Forum section: electronics / beginners

author: hermitengineer
author_rank: Regular Contributor
timestamp: April 27, 2020, 07:34:07 pm
content: I have a circuit board where two of the transistors get hot enough to discolor the PCB around them (old fiberglass board). The transistors seem to be part of a 2-output current source circuit that looks something like the attachment.Would you expect the transistors to heat up under this load? If not, what are some likely failure scenarios?For reference, the outputs are alternately pulled to ground by a 7407 controlled by a clock phasing circuit. The consumer of the outputs is a CPU that requires a 2-phase clock. When the clocking circuit is working correctly, the two outputs should both be square waves at exactly 180 degrees apart with 12Vp-p.

author: jmelson
author_rank: Super Contributor
timestamp: April 27, 2020, 08:55:35 pm
content: These transistors should be dissipating somewhere arond 50 mW each, so not enough to heat up the board. I'm guessing some componentfailed. Maybe the transistors, or maybe one of the Zeners, and that has caused a large current increase. Or, maybe the 7407 has failed, or is not getting the right inputs to only momentarily short the outputs.Jon

author: hermitengineer
author_rank: Regular Contributor
timestamp: April 29, 2020, 07:23:00 am
content: I'm also trying to educate myself on this circuit since it's not one that I encounter a lot.If there were no load at all on the circuit, not even the diode/zener series pair, what would its current at various nodes be? I'll try to analyze this myself, and others can feel free to grade the work.1) With no collector current, this is R2 in series with "diode" Q2. That node is in parallel with CR2. And that node is in series with R17.2) (hFE+1)*RE doesn't apply here because no current is drawn through the collector. The reason for the resistance difference is that a small current flow through the base causes hFE* that current flow through the collector, which causes voltage in RE to be (1+hFE)*iB * RE. This causes VB to reach VE with proportionately less current flow, and effectively creates resistance. But there is no current through collector, so VE is entirely decided by iB.3) 16V through R2, Q2, and R17 works out to (16-0.6)/1068 = 14.4mA. With this much current flow, VR2Q2 is 1.7V, so CR2 is off.This indicates that the base must be passing 14.4mA if no load is attached. The transistor claims to handle 100mA peak through the base, but I wonder how much continuous?As a follow-on question, does this mean that any current "surplus" up to 14.4mA will be sunk by the base? In other words, if I calculate this right, the R2/Q2 circuit should provide up to 40mA to the load. If the load takes 25mA or less, does that mean that the base still sinks 14.4mA? On second thought, with 2 current sources sinking to the same R17, the sum of the current must be around 14.8 between them, and the two will split responsibility for summing around 14.8mA (no load).Reanalyzing with at least the diode/zener load:1) CR5+CR3 will sink as much current as they have to, to make the voltage 12V across.2) This means R2+Q2 is 4V. Furthermore, R2 is 2.7V (3.3-.6). So Q2 is 1.3V.3) 2.7V/68 = 39.7mA. 39.7mA * 1.3V = 51mW. Well at least I arrived at the same answer as jmelson.And almost all of that will be going through CR3+CR5, right?Now, the CPU (GI CP-1610), requires 15mA for each clock input. So CR3+CR5 only need to sink about 25mA with the CPU connected. Meanwhile, the 7407 does not list maximum sink current but recommends a max of 40mA. What a coincidence!Just as a side note, this is apparently a level shifter circuit. Much of the system runs on a 5V clock, but the CPU clock input is 12V logic. Well, VDD up to 18V logic that is. Strangely enough, the rest of the CPU's logic is VCC (5V) logic.