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Apr 14, 2022 at 19:00 comment added Audioguru The graph shows a device with "typical" specs. Some devices have better specs and some devices have worse specs. I design circuits so all passing devices work, not just the "typical and better" ones.
Apr 13, 2022 at 16:40 comment added Nihat @Audioguru I just found time to see the working principles of the transistor in practice. Due to the limit of the power supply I have, the current through the LEDs could only reach 2.6A, but the Vce value was 1.31V as in the graph. But even though, I was having a 470R in my base leg, I measured the current around 60mA. That only confused me a little bit but other than that, I feel alright :) Do you think that matches the graph in the datasheet?
Apr 13, 2022 at 16:31 vote accept Nihat
Apr 7, 2022 at 0:17 comment added Audioguru hFE is used only when a transistor or darlington IS NOT SATURATED. The hFE spec has a fairly high Vce. Most transistors have a maximum Vce saturation voltage when the base current is 1/10th the collector current even if the non-saturated hFE is very high. A darlington has a base current of 1/250th the collector current for its guaranteed max VCE when saturated.
Apr 6, 2022 at 4:26 comment added Nihat @LvW I understand this. But still, I need to know which \$h_{FE}\$ value I should rely on since it makes a difference on the \$I_{B}\$ which is important for me not to harm my controller.
Apr 5, 2022 at 16:42 comment added LvW @Nihat, remeber the definition for saturation: The voltage Vce is so small (0.2...0.3 volts) that not only the B-E junction but also the B-C junction is forward biased. This RESULTS in a pretty large base current - however, it is not a current that brings a BJT into the saturation state. It is the voltage developed by the current across the collector resistor which is responsible.
Apr 5, 2022 at 14:34 comment added Nihat @GodJihyo thanks for the effort. appreciated it. Your explanations brought another question to my mind. Is it necessary to exceed 12mA to fully saturate the transistor? My concern is that there would be very little space for my Pi's maximum current rating (16mA) if I would put such a resistor value. From various documents, I saw that people take \$h_{FE}\$ value into account which is specified as 750 in my case but you derive it from Figure 2. When I calculate by that value I end up with 4mA (3A from the LED divided by 750) in the base leg. Which is more reliable you think?
Apr 5, 2022 at 13:38 comment added GodJihyo @nihat hFE will vary with collector current, it will be lower when the transistor is saturated than when it is operating in it's linear region. The data sheet is taking Ib = Ic/250 as it's test point for saturation. I've added the graphs to my answer.
Apr 5, 2022 at 13:32 history edited GodJihyo CC BY-SA 4.0
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Apr 5, 2022 at 10:44 comment added LvW @Nihat, why should they specify a minimum of current? Current is the RESULT of voltage - hence, a minimum/maximum/typical value for the voltage makes sense.
Apr 5, 2022 at 9:07 comment added Nihat @LvW I also wonder where this minimum current is stated for the base in the datasheet. There are \$I_{B}\$ values that belong to testing rates but couldn't find the minimum need for my setup.
Apr 5, 2022 at 9:05 comment added Nihat @GodJihyo tbh, the values shared in the datasheet are a bit confusing. They don't truly match. On one side, they share \$h_{FE}\$ as 750, then, in Figure 2, they take it as 250. Besides this, I can't see a graph for the transistor behavior under the 12V/3A circumstance.
Apr 5, 2022 at 8:20 history undeleted GodJihyo
Apr 5, 2022 at 8:03 history deleted GodJihyo via Vote
Apr 5, 2022 at 7:40 comment added LvW Quote: "The base current needs to be enough to drive it into saturation,". The base current cannot "drive" the BJT into saturation. In contrary: A large base current is an INDICATION for the state of saturation. Saturation means, that there is a base voltage which is large enough to open not only the B-E junction but also base-collector pn junction. That is the reason for the large current into the base. And the design must ensure that this current can go through the base resistor into the BJT.
Apr 5, 2022 at 7:37 comment added GodJihyo Ah... I think I mixed up the Vbe and Vce graphs. Answer edited. Looking at it again it looks like Audioguru used the Vbe(sat) value instead of Vce(sat) so you'd only lose 2.0V to the LED instead of 2.5V. Of course those are the max values, both of those are lower in the graphs which should be more typical values, so you might want to figure on Vce(sat) being 1.3V. I would try it with 12V and see how it works, it might be fine, As for the 12mA value, that's the value they tested Vce(sat) at, and the graph says Ic = 250 * Ib.
Apr 5, 2022 at 7:11 history edited GodJihyo CC BY-SA 4.0
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Apr 5, 2022 at 6:44 comment added Nihat @GodJihyo may I ask which graph you looked at for the \$V_{BE(SAT)}\$ value? Additionally, is the 12mA value the threshold value for the transistor needed to be exceeded to be able to saturate?
Apr 5, 2022 at 6:38 comment added Nihat @Audioguru I do not have a datasheet for the LED, unfortunately. It works by 12V and according to that, I measured a 3A load from my power supply. What if I provide 14.5V to my collector's leg so that I compensate for the 2.5V drop while the saturation?
Apr 5, 2022 at 4:02 comment added GodJihyo @Audioguru True, they'd probably be better off with a MOSFET for this application.
Apr 5, 2022 at 0:50 comment added Audioguru The Vce(sat) of the darlington is 2.5V max at 3A. Then the "12V" LED gets only 9.5V and might not light or might be very dim. Please post the datasheet for the LED.
Apr 4, 2022 at 20:58 history answered GodJihyo CC BY-SA 4.0