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I'm trying to figure out what will be the total current consumption of a TIP120.

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Looking at the internal schematic I see that the second BJT's base is following the emitter of the first one so it will consume current proportional to the base current of the first one but I'm having a hard time determining the exact values. The ones given in the datasheet are for the entire pair, right? How can I determine what current will be consumed by the base-emmiter paths with a 500mA load on the collector of the pair?

Additionally what is the role of the voltage divider? If current follows the path of least resistance then this must mean there will not be any current going through the resistors? (which is certainly not the case)

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  • \$\begingroup\$ The statement of current following the path of least resistance, it isn't technically true, otherwise parallel circuits just wouldn't work! Are you wanting to solve for each individual transistor in the Darlington pair? Or are you just interested in the part as a whole? You are correct that the datasheet values are for the pair, as the TIP120 has a Darlington pair built inside. The transistors inside should be well matched, so you can solve the circuit as if it is a single transistor with a B, C and E. Unfortunately, the datasheet doesn't give an exact beta value \$\endgroup\$ – MCG Aug 8 '19 at 11:45
  • \$\begingroup\$ @MCG basically I'm curious to find out how much on top of my load's 500mA will be drawn, because the second transistor of the pair gets its base current from the same current source as my load. \$\endgroup\$ – php_nub_qq Aug 8 '19 at 11:48
  • \$\begingroup\$ In that case, yes, you should be able to solve that as if you were using a single transistor. As Bimpelrekkie mentions in his answer, the current gain for each part can vary, as you can see in the datasheet, only a minimum value is provided. \$\endgroup\$ – MCG Aug 8 '19 at 11:52
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    \$\begingroup\$ If it's a relatively benign 500mA maximum load (no short circuit tolerance or other such requirement) I would just use an SOT-23 N-channel MOSFET like the AO3400A with 100K gate pull-down. It's less than 48m\$\Omega\$ with >2.5V drive so at 0.5A it will dissipate less than 12mW, just about nothing. Just make sure that the brown-out condition is locked out (eg. enable BOR on your MCU and put the aforementioned pull-downs in place). \$\endgroup\$ – Spehro Pefhany Aug 8 '19 at 14:27
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    \$\begingroup\$ @php_nub_qq If you're switching no more than 250uA it's fine with 2V, yes. Between 2V and 4V nothing is guaranteed. Worse at temperature extremes of course. That's what "roll the dice" means. \$\endgroup\$ – Spehro Pefhany Aug 8 '19 at 14:44
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Start with the information that is given for the current gain \$h_{FE}\$, you should be able to extract a "ballpark figure" from the datasheet:

enter image description here

Note how that information is quite "basic", all that the manufacturer guarantees is that \$h_{FE}\$ will be larger than 1000.

So you will not be able to find an exact value because \$h_{FE}\$ will never have an exact value, unless you take a transistor and measure it, then you have that transistor's \$h_{FE}\$ at the conditions at which you measured it. The next transistor in the box might have a completey different \$h_{FE}\$!

So stop trying to find an exact value for beta. What engineers do is that they design the circuit around this transistor such that it will work reliably for \$h_{FE}\$ = 1000 and higher values. (And most engineers would also add some margin to that so would design for \$h_{FE}\$ > 500 just to be sure).

The resistors don't really work as a a voltage divider as there are also the base-emitters in parallel with these resistors. The resistors are there to make sure the transistors switch off when no input current (at the base) is applied. Since the TIP120 is designed for switching applications, this makes sense. Without these resistors it would take longer to discharge the base of the "big" transistor when we want to switch it off.

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