In bipolar transistor circuit analysis (NPN - 2n3904) is Vbe always 0.7V?

Question

of course I am not talking about the small variation depending on the materials used in the transistor. I built a simple circuit with multisim 10. and it showing me 1.25V for Vbe. I would post the circuit but I am thinking it is a fluke, as everywhere I looked it seems Vbe is always assumed 0.7V whether in saturation or active modes. thanks.

Answer 1

Vbe can be less than .7 Vdc but never greater than .7 Vdc.

The 1,25V is probably from base to ground. There is probably .55 Vdc from emitter to ground that is being dropped across an emitter resistor. This would leave .7 Vdc between base and emitter. This would be considered normal.



However If the base Voltage is really 1.25 Vdc more positive than the emitter Voltage then the base junction is open. Of course if this is true the transistor is kapoot and will have to be replaced.

Answer 2

Vbe Transistor

Answer 3

For silicon transistors, the base emmitter voltage and the base current are exponentially related (Ib=k*e^(a*Vbe) I cannot recall the constants now!). For most 'normal' ranges of current it turns out that Vbe is about 0.7 volts. Because of the exponential relationship, Vbe doesnt change much even with large changes in current, that is why it always 'seems' to be 0.7 volts. But for very large currents (several amps), it can go higher than one volt. It depends on the exact parameters of the transistor. The transistor you mention, 2N3904 is a small signal transistor, so the base current wont be more than a few milliamps in most applications. So probably you are reading the base voltage with respect to ground: the emmitter may not be at ground potential, so the base may be at 1.25V, and the emmiter at 0.55 volts (there must be a resistor in the emitter circuit?) so the base emmitter voltage is 0.7 volts.

Answer 4

The simulator is measuring voltage from base to ground with a normal voltage probe. The base voltage Vbe varies with temperature, and the device and the current. There is an internal series base resistance, sometimes called Rbb. The actual junction voltage is around 630mV @ some temperature like 25C, with a typical base emitter voltage being more like 700mV, and sometimes up to 1V. It does seem 1.25V is higher than normal, but it is still in the possible range. It is not due to temperature, it would be well below normal specified temperature range. It could be excess current flowing, or other external voltage drops (in the emitter circuit for example). Check the base current in the simulator, and compare with the expected base current for that transistor. In normal circuits with small signal types it is unlikely to need more than 10mA, and 100mA to a few amps in a larger power transistor. A simulator could have 1000A base current without any funny smells or noises, especially with generic models, and the transistor just behaves like it was meant to have that.



Like real circuits, it is always a good idea to check the circuit supply current to make sure there are no shorts. A simulator may maintain the voltage with mega-amps flowing. Many simulators try to detect these cases, but that doesn't mean they always do.

Answer 5

Vbe is always assumed 0.7V whether in saturation or active modes

Answer 6

0.7 is the turn-on voltage for a silicon junction. You need more than that for it to work to maximum... 1/2 volt more anyway. Turn-on is not saturation.