Is VB and VBE, of a transistor the same? If not, how is Vbe generated? ....and must the base voltage of a transistor be equal to 0.7 or can it exceed this value.


5 Answers 5


You might be interested to know some more interesting details about this topic:

  • Since "voltage is a difference of two potentials", the voltage Vbe represents a "difference of two differences" (Vb - Vground and Ve - Vground).

  • Such a voltage is "floating" (not connected to ground) and it requires a differential input.

  • It is an output voltage with respect to the other two which are produced by grounded input sources.

  • The two input voltages can be changed in a variety of ways:

    • both simultaneously (differential amplifier)

      • in the same direction (common mode)
      • in opposite directions (differential mode)
    • only one of them keeping the other of constant voltage (single-ended amplifier)

  • If the input voltage Vb changes with the emitter voltage Ve following it (emitter follower), the voltage Vbe and current Ib will not change. As a result, the input source Vb will "think" there is no transistor. This trick is known as bootstrapping and is used to artificially increase the input impedance of amplifiers.

  • If the input voltage Ve changes (decreases) while keeping the base voltage Vb constant (common-base amplifier), the collector current Ic is added to the base current Ib thus "increasing" it. As a result, the input source will "think" that the base-emitter resistance has lowered. This phenomenon is known as re.

  • This is the simplest way to subtract two voltages according to KVL. It is widely used in series negative feedback amplifiers (eg, an op-amp non-inverting amplifier).

  • From this viewpoint, a transistor can be thought as a comparator with two single-ended input voltages (Vb and Ve) and a current output (Ic).


\$V_{XY}\$ is the voltage at node \$X\$ with respect to \$Y\$. Or in other words, the voltage across \$X\$ and \$Y\$ (X positive lead, Y negative lead).

\$V_X\$ is the voltage at node \$X\$ with respect to reference point (GND/0V or whatever it is).

Now \$V_{B}\$ becomes the base node's voltage w.r.t. reference (e.g. GND/0V), and \$V_{BE}\$ becomes the voltage across the base and the emitter.

One other thing: \$V_{XX}\$ is the supply voltage applied towards node \$X\$. For example, \$V_{CC}\$ is the supply voltage applied towards the collector (can come through a collector resistance but doesn't have to, depending on the configuration). Likewise, \$V_{BB}\$ and \$V_{EE}\$ are the supply voltages applied towards the base and the emitter, respectively. BJT-based ICs have their supply terminals named with this convention. CMOS ICs use \$V_{DD}\$ and \$V_{SS}\$ as the voltages applied towards drains and sources of the internal FETs.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ K. I am somewhat surprised about the direction of the arrows you have used; From the lower to the larger potential. I think, this is not the standard convention? \$\endgroup\$
    – LvW
    Sep 23 at 11:23
  • \$\begingroup\$ @LvW I don't know if it's a standard convention or not. Our lecturers at the university in the late '90s and early 2000s tended to use arrows, specifically for DC voltages, pointing towards the higher potential. I also remember that I came across this convention in some books/resources. I somewhat find it practical to use arrows instead of +/- markers. \$\endgroup\$ Sep 23 at 11:42
  • \$\begingroup\$ @LvW Arrows pointing to the lower potential is the European convention; in the US, when arrows are used, they point to the higher potential. \$\endgroup\$
    – Hearth
    Sep 23 at 11:51
  • 2
    \$\begingroup\$ @Circuitfantasist Unless the arrow is pointing along a voltage source. \$\endgroup\$
    – Hearth
    Sep 23 at 12:33
  • 1
    \$\begingroup\$ @Circuitfantasist More logical? For who? \$\endgroup\$
    – G36
    Sep 23 at 12:48

Vb is the convention to state the voltage at base, compared to your 0V reference, which may or may not be emitter voltage Ve.

Vbe is the convention to state the voltage difference between base and emitter, so Vbe = Vb - Ve.

If your emitter has 10V, your base voltage Vb should be at 10.7V, which still means Vbe is 0.7V.


\$V_B\$ and \$V_{BE}\$ are only the same if \$V_{E}\$ is 0V; in other words, they are equal if the emitter is grounded. The \$V_B\$ voltage can exceed 0.7V (with respect to ground) if the emitter is also at higher voltage potential than 0V.


Quote: "how is Vbe generated........and must the base voltage of a transistor be equal to 0.7 or can it exceed this value."

Yes - it must be externally generated with a DC value (bias voltage) of app. 0.7 volts. In most cases, it is advantageous to create the corresponding base voltage with the help of resistive voltage divider at the base node. Because the relationship Ic=f(Vbe) is very temperature-sensitive, the "exact" Vbe value for a certain design value of Ic is not known.

For this reason, current-controlled negative voltage feedback should be provided (emitter resistor RE). It is known that negative feedback can drastically reduce the influence of active parameters (and its tolerances/uncertainties) on such circuits.


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