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I have been reading the datasheet of TIP122 darlington pair transistor. I am confuse over the term "Collector-Emitter Saturation Voltage". It says it could be 2V or 4V (depending upon Ic and Ib). Ideally, saturation voltage shouldn't be zero? (Ok. It should be atleast Vbe as Vce = Vbe + Vcb).

I also want to know how it is related to Ic and Ib as they are causing a significant variation?

One last question, for switching application darlington pair should be in saturation region. By the voltage "Collector-Emitter Saturation Voltage", can i assume that in real world, this could be minimum Vce and if my pair operates in active region, then Vce will be greater than it (Vce in active region depends upon Rc)?

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    \$\begingroup\$ Darlingtons don't, in general, saturate. They just give you lots of current gain. \$\endgroup\$ – jonk May 8 '17 at 21:30
  • \$\begingroup\$ Can you explain why? \$\endgroup\$ – abhiarora May 8 '17 at 21:32
  • \$\begingroup\$ Then why they have mentioned saturation voltage? \$\endgroup\$ – abhiarora May 8 '17 at 21:32
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    \$\begingroup\$ Darlington is a "circuit topology." If you look at it closely, you will see why it cannot saturate (in the meaning of a BJT, as I think you were asking about.) As far as the words used in a document, I can't argue with the authors. They aren't here to debate with me. There's no point in my trying to debate fixed words printed in a datasheet. Complete waste of time. Same thing as talking to a wall. \$\endgroup\$ – jonk May 8 '17 at 21:35
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    \$\begingroup\$ No, it's important to look at. The maximum CE voltage given in the datasheet for a given collector current gives you an idea of your worst-case expected dissipation in the device. And that is very important to figure out -- especially so with Darlington topologies. For example, I see that with \$I_C=5\:\textrm{A}\$ that the worst case \$V_{CE_{SAT}}=4\:\textrm{V}\$! That could mean is much as \$20\:\textrm{W}\$ dissipation if you are unlucky enough to get a worst-case device. \$\endgroup\$ – jonk May 8 '17 at 22:17
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Suppose you have one of those really bright, white LED strips that is 5 meters long and requires \$12\:\textrm{V}\$ and \$3\:\textrm{A}\$ to operate. But you want to be able to turn it on and off using an LDR to detect day and night. You can work out the LDR circuit but now you need to work out the activating switch between your \$12\:\textrm{V}\$ power supply and the LED strip.

Well, you are kind of messed up with a TIP122. The reason can be seen in the first of the following two schematics:

schematic

simulate this circuit – Schematic created using CircuitLab

The datasheet of the TIP122 tells me that at \$I_C=3\:\textrm{A}\$ I cannot count on better than \$V_{CE_{SAT}}=2\:\textrm{V}\$. So, cripes! This means you need to throw away that nice, easy-to-find, \$12\:\textrm{V}\$ power brick and go find something with \$14\:\textrm{V}\$ or else the ability to adjust it to that value.

But let's say you get past that issue and find yourself a very nice \$14\:\textrm{V}\$ power supply with appropriate current compliance. Power dissipation in your TIP122 will be about \$6\:\textrm{W}\$. So, your LED strip now has access to \$36\:\textrm{W}\$ at the price of wasting \$6\:\textrm{W}\$. Your power supply delivers \$42\:\textrm{W}\$.

Switch efficiency is about 85% and base drive requirements are likely under \$10\:\textrm{mA}\$.


Now look at the bottom diagram. I've chosen the 2SD1060, which is rated for about the same collector currents as the TIP122 (apples to apples.) At \$I_C=3\:\textrm{A}\$ it is guaranteed to have less than \$V_{CE_{SAT}}=300\:\textrm{mV}\$. I've added a drive circuit to supply \$180\:\textrm{mA}\$ into its base (which the 2N3904 can achieve.) Dissipation in the 2SD1060, plus the whole drive circuit behind it, will be under \$3\:\textrm{W}\$. (Most of that in the drive circuit itself, where \$R_4\$ is there to soak up most of the dissipation so that the poor 2N3904 might survive the effort.)

You get to use an easy-to-find \$12\:\textrm{V}\$ power supply (you might already have it) and switch efficiency is now over 92%, with base drive requirements likely under \$2\:\textrm{mA}\$.


(Note that these above schematics are just for illustrating a comparison. They are not intended for robust use.)

Now.. someone will bring in a logic-level N-MOSFET at about this point. Or want to replace the 2SD1060 with a P-MOSFET. But that's a different subject.

Mostly, I just want to point out that while Darlingtons have a place, you are also often better off figuring out something different. (I can't recall the last time I bothered with one. And that includes the ULN2003.)

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