# Calculating base resistor value using voltage

I'm pretty new to electronics so my question is I think a relatively simple one.

On transistor datasheets, the base-emitter voltage is typically around 6V, which means that voltage across this connection should not exceed 6V, right? When you calculate the value of the resistor connected to the base, you assume a 0.7V drop across the transistor (because this is saturation voltage), and the rest of the voltage is dropped across the resistor. You use this voltage, along with the current you want going through the base, to calculate the resistor value. But why are we assuming only 0.7V dropped across the transistor if it can have up to 6V dropped? And if it can drop more than 0.7V, then how can we calculate the resistor value?

It would help a lot if you'd post a link to the datasheet and then cite the line and page for your question.

But in general, if you see a maximum $V_{BE}$ value on the datasheet, it is really a $V_{EBO}$ value. This just means that if you apply a reverse biasing voltage to the base-emitter junction, that it won't tolerate any more and will instead "break down."

This is about like what a zener diode might do. The BJT isn't supposed to be operated in a reverse-biased base-emitter junction mode. So this is almost always just a warning about preventing this situation. It means you can tolerate a small amount of reverse-biasing, but not a lot of it. That's all.

As far as forward-biasing goes, the general rule for small signal silicon BJTs is that it takes about $700\:\text{mV}$ to produce perhaps $5\:\text{mA}$ of collector current. This forward-biased base-emitter voltage increases by about $60\:\text{mV}$ if you multiply the collector current by a factor of 10. Simularly, it declines by about $60\:\text{mV}$ if you divide the collector current by a factor of 10.

This does NOT mean the base-emitter voltage is fixed, as you can see. It just doesn't vary a lot. So for a lot of reasoned uses, it's close enough for horse-shoes to say $700\:\text{mV}$ for the base-emitter junction when operating in active mode. (Which means that the collector voltage is at least $1\:\text{V}$ different from the emitter voltage, or more.)

In forward-biased mode, you will not see $6\:\text{V}$ across the base-emitter junction (for very long, anyway.) That would pretty much guarantee that the BJT was rapidly destroyed.

Thats a very Noobie question !:)) The 6 v you refer to from the data sheet are "Maximum values". it means the chip will fry if you apply more than that. Look under typical values. BTW Vbe is always 0.65-0.7 volts (varies slightly as per temperature of the base-emitter junction). Because its always a forward biased P-N junction, imagine a diode. A diode can always have only 0.7 volts drop across it. If you apply more voltage across it (in forward direction of diode) without any current limiting resistor in series with it, you will pass huge current and "fry" the diode or P-N junction or Base-emitter junction. In reverse direction it can take any voltage you apply on it, provided it dosent "break-down". This break-down voltage is also to be found in the data sheet (reverse biased voltage). I hope you know that when a P- type material is connected to positive power suppy, and the N- side silicon slab is connected to negative supply rail (or ground as we say it) its said to be "forward biased"? and the opposite way of connecting is called "reverse biased".

• Thanks for your answer. Yup I am a total n00b with this! So are you saying no matter the voltage I apply across the transistor, it will always only drop 0.7V? If so then why state a breakdown voltage of 6V if the voltage will never reach that (or not even close)? – user1390491 Feb 10 '18 at 5:43

The 6V VEB in the data-sheet is the most reverse bias voltage the makers say that you can put on the transistor without damaging it.

The most forwards voltage is probably well under 2V before something breaks due to high current.

You are correct that 0.7 is appropriate VBE for calculating the base resistor and that is the voltage seen in a simple transistor when it's conducting.

see this question for more on reverse biased B-E junctions. Can I reverse bias the base-emitter of a BJT transistor?

The simple blocking oscillator circuit is an example of a circuit that can cause this state. https://en.wikipedia.org/w/index.php?title=Blocking_oscillator&oldid=734813292#See_also

Base and emitter of the transistor will act as a PN diode. As latter, BJT will turn on at Vbe of 0.7V. After that voltage. depletion region of that diode will disappear and electrons start to flow. As given in the picture, after the base current starts to flow, it rises sharply. The current value can be changed to orders of magnitude for a 100mv change. So, Vbe will be varied from 0.6v to 0.8v usually for a silicon transistor. So, it is safe to assume that Vbe is 0.7v for most of the applications.

That so-called 6V is VBEO (i.e) Maximum reverse voltage that can be passed via a base-emitter junction.