.1v saturated on darlington?

ian field

Joined Oct 27, 2012
6,536
And Aristotle said it was obvious that a heavy object would fall faster than a light object.

They both sound plausible but there's a hitch in those deductions.
For the Darlington, when you overdrive the input transistor the excess base current goes through the now forward-biased base-collector junction, not the base-emitter, which keeps the output transistor from saturating.

Below is a simulation of this. You can see that the collector current of Q1 is negative (current out of the collector) until Ibq1 equals Ibq2. It's only after Ibq1 is less the Ibq2 that Q1 actually starts to amplify its base current and act like a Darlington with gain higher than Q2 by itself..
Note that Ibq2 is nearly constant as the base current of Q1 varies over a wide range.
And Vc never goes below 768mV.

So no matter what you do to a Darlington's input current, the Vce minimum will never be lower that about one base-emitter drop.

As to Aristotle's deduction, that was disproved several centuries ago, as Aristotle hadn't realized that the inertia of an object goes up by the same factor as its attraction by gravity. ;)

View attachment 85714
I'd say you've pretty much nailed it there.

Without getting around to doing a simulation, I'd assumed that with the output transitor at least reaching for VCEsat, the collector junction would become forward biassed. That in series with a VCE that's at least reaching for sat, is going to shunt at a little less than the 2x Vb/e needed to get the job done.

VCEsat is specified at a value of collector current, so it will rise a bit as you push more current into it via the input collector junction, but the voltage across that junction will also rise slightly with more current - not a lot, but it narrows the deficit on the 2x Vb/e requirement.

Just in case there's any pedants watching - I think those who said you can't saturate a Darlington may have won by a whisker.

In real world designs, Darlington transistors are usually only used if the driving device just can't source enough current to turn on a straight bipolar transistor. IMO: a pretty serious fault would probably have to happen before the output transistor can be turned on by current forced through the input base.
 

AnalogKid

Joined Aug 1, 2013
12,156
I am absolutely so sorry if this is not a darlington I notice it doesn't exactly say. Just thought it was based on current capacity
I don't know where you learned that assumption, but you should re-examine it. While there are many power darlington transistors out there, there are way many more non-darlington power transistors. Also, there are many darlington transistors that fall into the small signal category, generally speaking with collector currents of less than 1 A. So your assumption linking darlington-ness to current capacity is not supported by the marketplace.

ak
 

WBahn

Joined Mar 31, 2012
32,919
And Aristotle said it was obvious that a heavy object would fall faster than a light object.

They both sound plausible but there's a hitch in those deductions.
For the Darlington, when you overdrive the input transistor the excess base current goes through the now forward-biased base-collector junction, not the base-emitter, which keeps the output transistor from saturating.

Below is a simulation of this. You can see that the collector current of Q1 is negative (current out of the collector) until Ibq1 equals Ibq2. It's only after Ibq1 is less the Ibq2 that Q1 actually starts to amplify its base current and act like a Darlington with gain higher than Q2 by itself..
Note that Ibq2 is nearly constant as the base current of Q1 varies over a wide range.
And Vc never goes below 768mV.

So no matter what you do to a Darlington's input current, the Vce minimum will never be lower that about one base-emitter drop.

As to Aristotle's deduction, that was disproved several centuries ago, as Aristotle hadn't realized that the inertia of an object goes up by the same factor as its attraction by gravity. ;)

View attachment 85714
Interesting. I don't know if I'm quite ready to buy that this is universal, but it would be interesting to look at wildly mismatched transistors and to also take some live measurements. I would also like to look at what a detailed analysis using the mathematical models valid for forward biased BC junctions have to say. One more thing on my to-do wish list.
 

crutschow

Joined Mar 14, 2008
38,533
Interesting. I don't know if I'm quite ready to buy that this is universal, but it would be interesting to look at wildly mismatched transistors and to also take some live measurements. I would also like to look at what a detailed analysis using the mathematical models valid for forward biased BC junctions have to say. One more thing on my to-do wish list.
I'm fairly certain the simulations are valid for any but perhaps some pathological cases, but I would be interested in the results of any live measurements. :)
 

Jony130

Joined Feb 17, 2009
5,598
Why do you doubt that when first transistor enters saturation both junction B-E and B-C will be forward biased. And this prevent the second transistor from saturation.?
 

WBahn

Joined Mar 31, 2012
32,919
Why do you doubt that when first transistor enters saturation both junction B-E and B-C will be forward biased. And this prevent the second transistor from saturation.?
Which of the two junctions will be forward biased more? What will the be the ratio of the base current that goes to the collector versus the emitter? How much current needs to go out the collector in order to prevent saturation of the second transistor? How will highly mismatched transistors affect these answers?

Other than the first question, I can't answer any of these off the top of my head. Can you?
 
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