I want to use a transistor to control a 5v power source but Im confused on how to control the base pin.

How much voltage can I put on the base pin? I've seen tutorials which use 5v with a 1k resistor, but doesnt this exceed the saturation voltage of 0.7v? Also if I were to use a voltage divider circuit to lower the voltage, would this provide enough current to turn on the transistor?

Im using a tip120 transistor. Thanks for any help

  • \$\begingroup\$ Draw your candidate circuit. \$\endgroup\$
    – Andy aka
    Commented May 26, 2020 at 16:39
  • 1
    \$\begingroup\$ "....enough current to turn on the transistor...". It is the voltage Vbe that turns the transistor on/off. However, in order to know the minimum (and allowed) maximum) of this voltage, we are using the output characteristics for defining the region of saturation and to determine suitable values for Ic and for Ib (with a sufficient safety margin). This method ensures that the transistor is not destroyed. The base current is always the result of the voltage Vbe - not vice versa. \$\endgroup\$
    – LvW
    Commented May 26, 2020 at 18:45

1 Answer 1


When you're using a transistor as a switch, the typical way to turn it on is to apply a suitable current into the base pin, usually from a voltage source through a resistor. The current will then develop about 0.7 V across the base emitter diode junction. Most transistors specify a maximum base current.

In the particular case of the TIP120, this is a darlington transistor, consisting internally of two transistors. The 'base' to emitter voltage will typically be 1.4 V.

The absolute maximum base current for the TIP120 is given as 120 mA in the data sheet. However, as the minimum gain is given as 1000, and the maximum collector current as 3 A, the base current needed to turn it fully on shouldn't need to be much more than 3 mA. This 3 mA to 120 mA window gives you a huge range to hit with your voltage source and series resistor.

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    \$\begingroup\$ That's what silicon diodes drop at a middling current. Note I said 'about 0.7 V', not '0.7 V'. Given the difference between the 5 V and the 'about 0.7 V' (or in the case of the TIP120, 'about 1.4 V'), the small variation in VBE is inconsequential. Read up on diodes, wikipedia would be good, for some of the physics behind their conduction and voltage drop. \$\endgroup\$
    – Neil_UK
    Commented May 26, 2020 at 16:51
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    \$\begingroup\$ Thanks once again Neil, I was getting confused with voltage drop vs supply voltage on the base pin. \$\endgroup\$
    – user248702
    Commented May 26, 2020 at 16:55

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