Hello.
I'm studying transistor for the first time:) and have seen some wired statement. It appears that base-emitter voltage (Vbe) in transistor is always assumed as 0.6 V in circuit analysis. I've seen one figure (see attached image). It seems over the wide range of base current (Ib), Vbe doesn't change much. Thus we can interpret this as it is safe to assume Vbe of 0.6 V.
But I interpret this differently that there is strong dependency of base current to small change of the base-emitter voltage thus precise value of Vbe is important.
What is wrong in my interpretation? Maybe in KCL and KVL analysis of the circuit embedding transistor, small change of Vbe is not important but..I always prefer exact value. Even I'm not sure 0.6 V assumption is valid even in other mode rather than active mode!
In circuit analysis for example, 2nd attached image, is there a way to precisely know the Vbe?
Your interpretation is correct "there is strong dependency of base current to small change of the base-emitter voltage". But as already mentioned every single transistor (even from the same production lot) will have slightly different value of Vbe for a given current. Software dedicated for electronic circuits analysis, e.g. Spice, does not use a fixed value but exponential dependence of the i(Vbe) during calculations. But if you want to evaluate operation of a particular circuit "on paper" (meaning without use of such a software), you need to use some simplifications. Assumption that Vbe is about 0.6-0.7 V is one of such simplifications. Take into consideration that in a manual analysis of operation of a particular circuit the analysis is performed in such a way that such assumptions will not lead to substantial errors (e.g. you never calculate current from the Vbe using the mentioned exponential dependence, but use other circuit constrains and properties that let you evaluate the current with a reasonable accuracy)
As the first figure shows, for any decently large base current the voltage drop between the base and the emitter will be around 0.6 to 0.7 V. Mathematically, this means that the BE voltage is the log of the current. Since the current can vary by orders of magnitude with only a few 100 mV of BE voltage change, it's usually safe to estimate the voltage drop as a constant. This makes manually analyzing the circuit easier.
VBE changes as a function of IBE are negligible in general because you normally use an emitter resistor for feedback that will 'equalize' things (via feedback).
This is even more important because of the temperature dependency of VBE with the "omnois" -25 mV/°C - which really counts for circuits operating at temperatures different from the 25 °C the 0.6 .. 0.7 VBE value is specified at.
BTW: 0.6 .. 0.7 V is true for Si transistors @ 25 °C - slightly depending on manufacturing technology. Other technologies (Ge, Schottky, ...) have different values but same basic characteristics.
Base emitter junction is simply forward biased PN junction of Si semiconductor. If you consider the junction structure based on bandgap of Si, it is nearly stabilized at room temperature around this voltage level as forward biased junction. That`s why it always referred as such level.
It's called a "Rule of Thumb" and is a fine starting point.
Each type of bipolar transistor will have a variation on this up or down a bit, the same as diodes have a bit of variation acording to type.
When testing using the diode range on a DVM, this enables one to quickly see if the part looks good.
Also note there is usually a small but discernable diference in the voltage between the base and the collector or the emitter (with the emitter being the higher).
Thus a simple meter test can also identify npn or pnp and the pinout.
Your interpretation is correct "there is strong dependency of base current to small change of the base-emitter voltage". But as already mentioned every single transistor (even from the same production lot) will have slightly different value of Vbe for a given current. Software dedicated for electronic circuits analysis, e.g. Spice, does not use a fixed value but exponential dependence of the i(Vbe) during calculations. But if you want to evaluate operation of a particular circuit "on paper" (meaning without use of such a software), you need to use some simplifications. Assumption that Vbe is about 0.6-0.7 V is one of such simplifications. Take into consideration that in a manual analysis of operation of a particular circuit the analysis is performed in such a way that such assumptions will not lead to substantial errors (e.g. you never calculate current from the Vbe using the mentioned exponential dependence, but use other circuit constrains and properties that let you evaluate the current with a reasonable accuracy)
its very simple
consider a NPN transistor
1. In the NPN trasistor N-(P-N) (P-N)---> is a diode part.
2. In this diode you are fusing P-type and N-type material to form a junction.
3. After forming a junction, a depletion layer is formed due to the diffusion current.
4.this depletion region act like emf in a coil, but the depletion region is constant at room temperature and without any bias.
5. in-order to make the diode conduct, you need to break the depletion region which is opposing the current flow (forward current)
6. so you in-order to attain a current flow you bias the diode with external voltage.
7.when you increase the external voltage above 0.6V , the depletion region breaks down and make a conductive path for the current flow that is drifted by the current.
this is why it takes 0.6 V to make conductive path.
we are not assuming it. 0.6 V is measured practical value used while designing circuits
Hello Vasile,
the reverse biased BE junction does not act like a zener diode ! - - - It acts like an ordinary diode (meaning: breaking down).
If current limited to some extend (series-R), it may survive this 'attack'.
BTW: the current capability without destroying depends on the current (! - not "power') rating of the transistor.
This is from own/colleagues' practical experience and not specified in the data sheet (at least not to this extend). AND: breakdown may start as low as 5 V for small-signal NPN tranistors.
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