2
\$\begingroup\$

TIP122 Darlington Transistor IV Characteristics

I am using a TIP122 Darlington transistor. From the graph provided in the datasheet, I can see that the saturation voltage decreases as the collector current decreases.

Since the load I am driving upstream only draws 400mA, can I assume that the aforementioned trend continues? That is, at Ic of 400mA, the saturation Vce voltage will decrease to below 1V?

Or do transistors have a minimum Ic threshold that disrupts this trend? I can't seem to find such a rating on the datasheet.

Datasheet: https://www.onsemi.com/pub/Collateral/TIP120-D.PDF

|improve this question
\$\endgroup\$
  • \$\begingroup\$ And now, as I try to figure out whether I know enough to answer this, I'm realizing I don't know the actual reason saturation voltage is a thing. Something to look up, I suppose. I imagine some (most? all?) of this variation with collector current is due to parasitic resistances, though. \$\endgroup\$ – Hearth Feb 4 '19 at 13:40
1
\$\begingroup\$

No, you can't assume that, at least, not in general.

A Darlington transistor pair operating at full gain has a minimum VCE that is a combination of the VCE(SAT) of the first transistor and the VBE of the second transistor. In order to conduct current at all, the overall VCE must be greater than the sum of these two. Any less, and the second transistor does not receive the base current that it needs.

However, with sufficient base drive, the Darlington pair can "revert" to single-transistor operation, in which the first transistor ceases to provide any gain, and simply provides a path through its B-E junction for the device base current to flow directly to the base of the second transistor. In this mode, the VCE of the device overall can be the VCE of the second transistor alone. I think this is what Figure 11 in the datasheet is showing.

But if you have that much base drive, then there's no reason to use a Darlington in the first place. You'd be better off overall using just a single transistor.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ If one reduces the Ic/Ib from 250 down to 100, one can expect Vce=0.8V @ Ic=400mA with Ib= 1mA whereas a single transistor with Ic= 400mA could be higher near 0.9V but with Ib=20mA can be lower at 0.2V depending on device power rating as Rce which determines Vce rise when saturated is inverse with lower rating \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 '19 at 14:26
  • 1
    \$\begingroup\$ The reason why saturation is nonlinear is that hFE drops to about 10~20% of hFE max when saturated and ratings reflect this like 10 to 20:1 per transistor or 250:1 for Darlingtons then the slope of Rce above this is inverse to its package power rating Rs=0.5/Pmax +/-50% \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 '19 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.