Darlington pair base voltage question - Page 1
Forum section: electronics / beginners

author: fonograph
author_rank: Frequent Contributor
timestamp: October 22, 2017, 06:04:22 pm
content: I am noob,still learning the basics,right now I am trying to understand how Darlington transistor pair works.Its probably simplest thing in electrical engineering after two resistor voltage divider yet I cant understand it.Specificaly I am looking at live graphical animation in the preset of falstad free online circuit builder and simulator.I am looking how the current is flowing through it,how the voltage is at various points.The specific question I have is,why is the voltage and current at the base of the second transistor so low? I mean the transistor that is little bit lower and on right side,considering there is 5V voltage source and single 300 ohm resistor before it,I would expect the base current and voltage be higher.Almost all the energy goes collector > emiter but very little through base > emiter.I am assuming that both collector > emiter and base > emiter resistance is zero,I know this is not the case in real world,but this is extremly primitive idealized simulator,I dont see anywhere shown that there are any resistances in this bipolar transistor,it acts like there is voltage divider thing going on between base collector and emiter.I would think that the first higher left side transistor have zero collector > emiter resistance and voltage drop,I imagined that the voltage at the base of the second lower transistor should be just as high as voltage at its collector,most transistors I read datasheets would blow up becose they cant handle anywhere near as much voltage at their base as they can on collector.EDIT: I made mistake,I mean why is the current higher,not voltage.

author: Zero999
author_rank: Super Contributor
timestamp: October 22, 2017, 06:24:59 pm
content: That looks right to me. What voltages and currents were you expecting?The base voltage is about 1.2V, so with a 2M resistor to 5V, there should be 1.9µA. I = (5-1.2)/2×106= 1.2×10-6A collector voltage of about 800mV is correct. It's the saturation voltage of a transistor, plus a diode drop for the base voltage.A current gain of 11.67k is also perfectly reasonable for a small Darlington pair, indicating the individual transistors have a gain of around 108.

author: fonograph
author_rank: Frequent Contributor
timestamp: October 22, 2017, 06:33:23 pm
content: I described what I expected in my original post.Why is the voltage at the base of second lower rightside transistor lower than voltage at its collector? They are parallel to each other

author: Zero999
author_rank: Super Contributor
timestamp: October 22, 2017, 06:40:06 pm
content: Quote from: fonograph on October 22, 2017, 06:33:23 pmThere's another transistor, between the base and the collector, dropping some voltage.

author: fonograph
author_rank: Frequent Contributor
timestamp: October 22, 2017, 06:51:31 pm
content: I just looked at it again and I noticed the voltage is same at both collector and base! But the current flow is significantly higher at collector > emiter than base > emiter

author: Zero999
author_rank: Super Contributor
timestamp: October 22, 2017, 07:56:42 pm
content: Please post a schematic showing the values of the current(s) and voltage(s) you're questioning.

author: fonograph
author_rank: Frequent Contributor
timestamp: October 22, 2017, 08:02:04 pm
content: go to this website and go circuits > transistors > darlington and it automaticly creates prebuild darlington simulation,you can click and probe all the points and look at voltage and current.http://www.falstad.com/circuit/You just have to click on that switch to close it so current starts runningThe voltage is same at both paths of this parallel section,the voltage is 807.74 mili volt,but the current going to the left side transistor collector is 136.48 micro amps,while the current going into right side transistor collector is 13.84 mili amp... why?I think the answer might be that the simulator secretly simulates that the collector > emitter path have non zero resistance and voltage drop becose the voltage after exiting the left side emitter is only 663.52 mv,while before it was 807.74 mv.

author: Zero999
author_rank: Super Contributor
timestamp: October 22, 2017, 08:59:46 pm
content: Quote from: fonograph on October 22, 2017, 08:02:04 pmBecause the transistor has a current gain of around 100, so the base current is roughly1/100of the collector current.QuoteYes, there is some voltage loss, when the transistor turns on.

author: fonograph
author_rank: Frequent Contributor
timestamp: October 22, 2017, 09:07:25 pm
content: aaaahhhhhh! I see! Its the gain parameter aka hfe thats doing it! When I set the gain on both transistors to 10,then current at the base od lower transistor is 21.7 ua and at emitter 217 ua.Thank you so much Hero999 for taking time to help,its really honorable act that you spend your time to help stranger for free,you are true hero

author: sharaj
author_rank: Newbie
timestamp: January 11, 2018, 12:45:45 pm
content: @Hero999I found that darlington pair is available in TIP122 package, but when i use TIP122 instead of simple BC547 transistors then I get different results regarding the current gain.Is this TIP122 is availabe on thathttp://www.falstad.com/circuit/for testing. I am following TIP122 pinout andTIP122 example from here.

author: rstofer
author_rank: Super Contributor
timestamp: January 11, 2018, 03:17:08 pm
content: Quote from: sharaj on January 11, 2018, 12:45:45 pmReal transistor don't have a singlular value for hFE. Check figure 9 in the datasheet and you can see how that parameter changes as a function of current.https://www.onsemi.com/pub/Collateral/TIP120-D.PDFWhile you're at it, look at VceSat - the transistor drops quite a bit of voltage when the current is increased. Figures 10 and 11.There is a tendency to try using Darlington transistors as motor drivers in small battery powered robotics projects. The problem is, VceSat is so high it wastes a substantial portion of the battery voltage, leaving little for the motor itself.

author: Zero999
author_rank: Super Contributor
timestamp: January 11, 2018, 04:26:51 pm
content: The TIP122 also has some built-in base-emitter resistors to help increase the turn off time, but will decrease the Hfe slightly.