common-emittertransistors
For sillicon the value of Vbe is approximately 0.7 V. So why is the input characteristics of such a transistor like this
Why not a straight vertical line. Or is this curve plotted for I Vs Vbb. I have seen nowhere plotting this curve against Vbb.
So what is wrong here?
The diode is in the best position, and is of an appropriate type.
It conducts when the input is negative, the same as the transistor base conducting when the input is positive. The 47K resistor is about 1/10 of a normal RS-232 load. One could also block the voltage, but then a -100V spike (ESD say) could break down the 1N4148 and break down the E-B junction, causing irreversible damage.
Also, a 1N4148 is an appropriate diode for this application. It's a "switching diode", low capacitance and fast reverse recovery. A 1N4001 would also likely work okay, at least at slow baud rates. The 200mA rating means that even if a very high voltage were to appear at the input the transistor is fully protected, at least up until the resistor arcs over,.
This is a more suitable diagram of a BJT's voltage and current characteristic: -
In the picture there are several curves; each for a different base current injected. So here's the first difference - base is normally depicted as being fed with a current not a voltage because this is more relevant - the BJT is likened to a device which amplifies current and has a current gain (Ic/Ib).
The next difference - imagine Ib was held at 150uA and you varied Vce and looked at Ic. The curve is right there in the picture above. If it were a resistive characteristic you'd see Ic rise linearly with applied Vce. As you can hopefully see, this is far from the case.
Typically once Vce has reached about 2 or 3 volts, Ic is very flat for quite significant changes in Vce - between 5 and 15 volts applied, Ic varies from about 16mA to about 17mA.
This is because a BJT acts like a current amplifier - a fixed base current of 150uA produces a largely fixed collector current of about 17mA i.e it has a current gain of about 113.
For a base current of 200uA, the collector current is about 23mA i.e. there is a current gain of about 115 - pretty much the same for the base current of 150uA.
Hopefully you can understand it's not a simple as the article seems to say.
Best Answer
What's wrong here is that you have read that 'VBE is approximately 0.7v', and now you're wondering why transistors don't follow that rule exactly.
The thing is, transistors do what they do. Then we make up simplified descriptions of their behaviour, called models.
A model gets used by a lot of people if it's found to be useful. George Box, all models are wrong, but some are useful
For 95% of my transistor use, the approximation 'VBE=0.7v' is enough for me to make a good design. In the remaining 5%, it's a valuable first pass, and a good cross-check on whether a more detailed model gives a plausible answer.
To explain the shape of, and the difference between, those transistor curves, you can elaborate the transistor model, with some dynamic resistance, with some current density in the junction, with temperature, with some feedback from the collector voltage.
Because all of those complications means VBE varies by as much as 100mV in either direction, a good transistor design will have bias arrangements that allow for that variation in VBE without too large a variation in the collector current.