I'm actually working with a darlington pair of BJTs.

My question is if I need to put a 1kohm resistor between my MCU pin and the BASE to use it as a switch. Since the Hfe ratio for my transistor is quite large, and the collector current will be limited to a reasonable level, does that mean that the base current is dependent on the collector current at all times and thus I can eliminate the usage of this current limiting resistor?


2 Answers 2


No, it's the collector current that is dependent on the base current, not the other way around. No matter what the collector current is, the base current is \$\dfrac{V_{MCU} - V_{BE}}{R}\$.
Keep in mind that \$V_{BE}\$ will be twice the value of another transistor, as there are two junctions between base and emitter.

But it's true that the collector current is what you want in the end. So to find the resistor value (don't just pick 1k), you calculate \$I_B=\dfrac{I_C}{H_{FE}}\$. If you want \$I_C\$ = 2A and \$H_{FE}\$ = 400, then your \$I_B\$ will have to be \$\dfrac{2A}{400}=5mA\$. This is a value your microcontroller will be able to deliver, but always check the datasheet.

To put it all together, \$R=\dfrac{H_{FE}}{I_C}\times (V_{MCU} - V_{BE})\$.

Olin is right about the resistor value being the maximum, i.e. the base current being minimum. For many parameters in a datasheet you'll find more than one value, like typical and maximum or minimum. You should always calculate for worst case conditions, and it may require some logical thinking to find out whether worst case is minimum or maximum for a particular parameter.

Take \$H_{FE}\$. In my example I picked a value of 400. As higher is usually better datasheets often mention a minimum value. What if it's higher? The base current won't be different, so the collector current will be higher. If you drive the transistor in saturation \$I_C\$ will no longer be determined by the transistor, but by the load's impedance will be a limiting factor. So, while the transistor would very much like to draw a larger collector current, it won't change. So you think you're safe; the minimum specified \$H_{FE}\$ is fine, higher is still OK. There's something else to consider, however: \$H_{FE}\$ is not constant, it varies with \$I_C\$, and the datasheet should have a graph for this. So check this for the wanted collector current.
\$V_{BE}\$. Two PN junctions, so that's 2 x 0.65V = 1.3V. Olin found that a 300\$\Omega\$ base resistor should be fine, in fact leaves some margin. But when I look at the datasheet for the TIP110 it says \$V_{BE}\$ may be as high as 2.8V! That would result in a base current of \$\dfrac{3.3V - 2.8V}{300 \Omega}=1.7mA\$, and that's too little to get the wanted \$I_C\$ of 2A: \$400 \times 1.7mA\$ is only 670mA.

You're getting the idea. Don't simply use typical values, but make sure that your circuit still works with components with extreme parameter values. This is not so much of a problem with projects where you only build 1 device: you can see what's wrong and adjust. For production you have no choice: always design for worst case.

  • \$\begingroup\$ Vbe would actually be a single transistor voltage drop. regarding the use of the resistor, it could also depend on whether the MCU IO pin is in push/pull mode or not. If it is, a resistor is probably not needed. \$\endgroup\$
    – dhsieh2
    Jun 19, 2011 at 16:09
  • \$\begingroup\$ @dhsieh - The datasheet for the TIP110 mentions a \$V_{BE(ON)}\$ of maximum 2.8V. You could never get such a high voltage with just a single PN junction. \$\endgroup\$
    – stevenvh
    Jun 19, 2011 at 17:08
  • \$\begingroup\$ 2.8V looks extremely high, even for a double junction. Can you explain this high value somehow? I think I would try to measure it and if I can't, call my supplier's FAE for confirmation. \$\endgroup\$ Jun 19, 2011 at 17:20
  • \$\begingroup\$ Wat's the difference with dfrac? \$\endgroup\$
    – clabacchio
    Apr 3, 2012 at 7:39
  • \$\begingroup\$ @clabacchio - dfrac displays larger. frac may be hard to read, especially if numerator or denominator have terms with subscripts. Check edit history to see what frac looked like \$\endgroup\$
    – stevenvh
    Apr 3, 2012 at 7:43

To go a little further, Stevenh's calculation shows you the maximum base resistor. It's a good idea to allow for some margin and provide a little more base current (a little lower base resistor) than the absolute minimum required to get the desired collector current.

Let's expand the Stevenh's example and get some real numbers. Let's say the processor is running from a 3.3V supply. A darlington has two B-E junctions between its base and emitter, so let's say the overall B-E drop is 1.3V. That leaves 2.0V accross the base resistor. 2V / 5mA = 400 Ohms. If you're really sure about Hfe being 400 over your operating range and that you don't need more than 2A collector current, then you can use only a little lower resistor, like the common value of 360 Ohms. For more margin, use less, like 300 Ohms maybe.

Now you need to go back and see what the load on the micro is. 2V / 300 Ohms = 6.7mA. That will be OK for most micros, especially if its a PIC which tend to have particularly good output current capability. However, I've seen some micros that are specified for less than 6.7mA, so you have to check and possibly adjust things.

One thing to consider with darlingtons is that they are slow to turn off. You say this is for a switching application, so turn off time could matter. If you're just driving a relay then this is no problem, but if you're trying to do 10s of kHz PWM then this is probably not what you want to use.

Another problem with darlingtons is the rather high on-state voltage. It is one B-E drop plus one saturated C-E drop, maybe 900mV but could easily be more at high currents. At 1V the transistor would dissipate 2W with 2A collector current. That will require some sort of heat sinking or at least something like a TO-3 case mounted to the chassis or some metal.

There may very well be better ways to switch what you want without using a darlington.


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