.1v saturated on darlington?

Thread Starter

Mrdouble

Joined Aug 13, 2012
107
I have a simple circuit to experiment with a darlington transistor MJE183.
I am reading Vce of .1v and I only have Ib of 2ma. When I remove base current the expected vcc Is present. I trying to use it to drive a small 3v motor. I swapped that transistor out (thinking inductive kick may have killed it) but new one did same thing. Any ideas?

Thanks In advance.
Micheal.
 

wayneh

Joined Sep 9, 2010
17,496
Even I know that's not true. You may be confusing it if it's in an emitter-follower configuration. Been there, done that.

Nevermind, I misunderstood what we were talking about.
 
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crutschow

Joined Mar 14, 2008
34,280
Any saturated transistor, including a darlington, can have Vce(sat) very close to 0.1 to 0.2V with Vbe at 0.7V. Try it in simulation or on a breadboard with a forced beta of 10.
No, Jony130 is correct.
Look at the Darlington configuration. When fully on, the top transistor will saturate which then is the same as connecting the collector of the bottom transistor to its base. Thus the bottom transistor Vce can not be less than its Vbe.
You should follow you own suggestion.
Below is a simulation. As you can see the fully-on Vce is nearly .77V.

Darlington.gif
 

Jony130

Joined Feb 17, 2009
5,487
Any saturated transistor, including a darlington, can have Vce(sat) very close to 0.1 to 0.2V with Vbe at 0.7V. Try it in simulation or on a breadboard with a forced beta of 10.
I do not have to do any simulations to know for sure.
For me the "classic" Darlington connection look like this:


And for this circuit Vce(sat) cannot be larger then Vbe. You do not see why is that ?

We have completely different situation for this circuit:
12.PNG
 

joeyd999

Joined Jun 6, 2011
5,234
Wikipedia:

Another drawback of the Darlington pair is its increased "saturation" voltage. The output transistor is not allowed to saturate (i.e. its base-collector junction must remain reverse-biased) because the first transistor, when saturated, establishes full (100%) parallel negative feedback between the collector and the base of the second transistor.[3] Since collector/emitter voltage is equal to the sum of its own base/emitter voltage and the collector-emitter voltage of the first transistor, both positive quantities in normal operation, it always exceeds the base-emitter voltage. (In symbols,
always.) Thus the "saturation" voltage of a Darlington transistor is one VBE (about 0.65 V in silicon) higher than a single transistor saturation voltage, which is typically 0.1 - 0.2 V in silicon.
 

WBahn

Joined Mar 31, 2012
29,976
For the classic Darlington Vcesat has to be no less than Vbe + Vcesat for the input transistors and the output transistor can't saturate (it's the input transistor that saturates).

You can break the collector connections, as Jony130 shows above to let the output transistor saturate, but that requires a four-terminal device.

I've been trying to find the datasheet for the MJE183 and have had no luck. Anyone else able to find it?
 

blocco a spirale

Joined Jun 18, 2008
1,546
I think the MJE183 must have been discontinued for some considerable time.

Reading through the TSs original post, I still can't figure out what the question is.
 

ian field

Joined Oct 27, 2012
6,536
Are you sure about that?
That was my understanding too.

If the output transistor were to saturate down to 0.4V - that is also the input transistor's collector. Ignore for the minute, the VCEsat of the input transistor, its emitter needs to achieve at least 0.7V to produce said saturation of the output transistor.
 

WBahn

Joined Mar 31, 2012
29,976
My question, is Vce of .1v possible? Not that I've read. Thank you for all your responses.
It's actually not quite that simple.

The floor on Vcesat for a Darlington is based on the premise that the base current of the output transistor is provided predominantly by the collector current of the input transistor. However, it is possible to overdrive the input transistor so heavily that there is little to no collector current in the input transistor and all of the emitter current is due to the base current. At that point you are effectively replacing the input transistor with a diode, in which it is obvious that the output transistor CAN saturate.

How much current is the Darlington supplying to the motor? If it is anything less than, say, 50 mA to 200 mA then you might well be driving it that hard with 2 mA of base current.
 

ian field

Joined Oct 27, 2012
6,536
My question, is Vce of .1v possible? Not that I've read. Thank you for all your responses.
Short the base/emitter - that should switch the transistor off so it doesn't pass any current.

If the collecter is fed by a voltage through a current limiting resistor, the collector voltage should rise - if not, the transistor is shorted.
 

crutschow

Joined Mar 14, 2008
34,280
It's actually not quite that simple.

The floor on Vcesat for a Darlington is based on the premise that the base current of the output transistor is provided predominantly by the collector current of the input transistor. However, it is possible to overdrive the input transistor so heavily that there is little to no collector current in the input transistor and all of the emitter current is due to the base current. At that point you are effectively replacing the input transistor with a diode, in which it is obvious that the output transistor CAN saturate.
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. ;)

Darlington.gif
 
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Thread Starter

Mrdouble

Joined Aug 13, 2012
107
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
 
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