Calculating Base Resistor for Darlington Pair

Question

So I know lots of people have asked this question but the more answers/threads I read the more confused I get because people explain it so differently.

My confusion is why do I have to even use the hFE Current Gain in any part of my calculation for the base resistor value?

My thought process is since Vbe is 1.4V (0.7*2 transistors) and my arduino output is 5V then lets say "I choose" to have Ib be 2mA then by ohms law R=(5-1.4)/2mA=1.8K ohms.

Does this thought process not work? The purpose of the resistor is to limit current to the transistor, well if I decide to limit it to 2mA then I have all my parameters to choose R without having to use hFE at all.

Answer 1

You are, in one respect, correct in not trusting hfe, but not for the reason you think.

hfe is specified for a certain voltage and current. For the TIP120, this is a Vce of 3 volts. Problem is, you don't want to run your transistor at 3 volts. You want the transistor to be in saturation, with Vce as small as possible.

Looking at saturation numbers, you'll see a specified base current and collector current. As a good rule of thumb, use the closest numbers to your application. Then, you can compute your base resistor as you suggest.

When using single transistors, the general rule is to assume an hfe of 10 to drive the transistor into saturation. This is not a hard and fast rule, but it will consistently give solid performance and low saturation voltages.

Answer 2

The TIP120 incorporates two base-emitter resistors (120 ohms and 8K), which minimize leakage, but reduce the effective hFE (and, more importantly here, give it a kind of threshold that is relatively high compared to an ordinary darlington).

Figure 2 in the datasheet shows typical Vce with Ic/Ib = 250, as does the guaranteed Vce(sat) numbers in the datasheet - (both guaranteed numbers are Ic/Ib = 250) so I would suggest that as a useful guideline.

You will get almost no improvement in Vce(sat) numbers by going for more than 1/250 of the collector current, and you may get significantly worse if you go much less, so it's safest to go with 1/250 in the 1A-3A 'wheelhouse' of this part (here is the only place the hFE curve Fig 1 comes in handy), perhaps a bit more at higher or lower currents. That's probably close enough unless you need to operate at low temperatures where hFE will drop.

So, at 1A out you need about 4mA base current. The microcontroller output won't be quite 5V when supplying 4mA, so if you use (4V-1.5V)/0.004 = 625 ohms (pick the next lowest E24 value 620 ohms) and you should be fine. Note that the 1.5V is the typical value from the graph in Fig 2 above, I didn't calculate it. The worst case value is 2.5V at 3A current (vs. 1.7 typically)- using that would be more conservative again, but it gets you down into the 330 ohm range and is probably unnecessary for an ordinary design.

Answer 3

Transistors are current controlled devices that is Ic=Ib*Hfe so to give the right value to base resistor you should start from current required by pump so you will know Ic. From Ic you could define Ic= 10*Ib for giving minimum Vce drop at saturation, but with darlingtons probably you could use Ic=100*Ib. Anyway mosfet are better switches in these applications because voltage drop is much less.

Answer 4

Well, first off, the LM137 is a negative voltage regulator and it's a 3 terminal device, so I don't know why that ground - or a positive power supply for that matter - is connected to it.

Assuming you made a trypo and it's supposed to be an LM317, it can't source more than about 1.5A, so if we assume your motor draws about an amp once it's running, we can start to think about the TIP120 as a switch.

More often than not, when you design, you have to work backwards, and since we know the motor's the load and we have to - somehow - drive it, this is one of those cases.

I have some other business to take care of so I can't finish this now, but I will tomorrow.

It's tomorrow, and it looks like Spehro pretty much covered what I was going to, so the only other thing I'd advise you to watch out for is whether the LM317 can supply the starting/running current for the pump motor. If it can't supply the starting current, then a big enough BFC from the LM317's output to ground to store the charge to start the pump before the TIP120 kicks in will be required.

If it can't supply the running current, then it's a new ball game.