# How should I understand the amount of base current in BJT transistors?

I'm having trouble understanding how much current flows through the base of a BJT. With a MOSFET the answer is easy: 0. How do I calculate the amount of base current in a BJT and am I doing it wrong if my circuit cares?

• What kind of circuit? Do you want to saturate the transistor (logic circuit, on/off), or use it as an amplifier? – tcrosley Sep 29 '10 at 13:57

## 2 Answers

In this circuit, the base current would be approximately (Vs - 0.7) / R2. The BE junction (with the arrow) behaves like a diode, so as long as Vs is greater than 0.7 V, then current will flow through the diode and there will be a drop of about 0.7 V (depending on the part) and then the current is just limited by the resistor according to Ohm's law.

• I find when building circuits using .7 works a lot better. – Kortuk Sep 29 '10 at 15:48
• This is one thing that I can really teach well in person, the internet is hard. – Kortuk Sep 29 '10 at 15:48
• 0.65V is more commonly found on the 'net and is a compromise. – Thomas O Sep 29 '10 at 20:19
• The exact value varies depending on the part, the amount of current, and the temperature. – endolith Sep 29 '10 at 21:33
• 0.7 volts is for silicon. 0.2 volts for germanium – Phil Freedenberg Jan 11 at 21:30

## A functional explanation

The base-emitter junction behaves, like a diode, as a "dynamic resistor" that decreases its resistance when the voltage across it increases and v.v., increases its resistance when the voltage decreases.

So, the input circuit consists of two resistors (constant RB and dynamic RBE that is seen from the input source as differential resistance rbe) in series driven by the input voltage source... and the "amount of base current in a BJT" is determined by the dominant resistor having a higher resistance. There are two typical cases:

Voltage-driven BJT. If RB << rbe, the "amount of base current in a BJT" is determined only by rbe according to the input IV characteristic... and it can be graphically solved by the so-called "load line" technique.

Current-driven BJT. If RB >> rbe, the "amount of base current in a BJT" is determined only by RB... and it can be easily calculated by Ohm's law (look at Glorfindel's answer).

• I am always unhappy when I see that the symbol for a differential resistance is written in capitals (RBE). I think, it is very important always to discriminate between STATIC resistances R (for DC) and dynamic (differential) resistances "r" . Otherwise, misinterpretations are very likely. – LvW Jan 12 at 9:11
• @LvW, Exactly... I perfectly know that... but small letters look bad in the text... this is the reason... Okay, I formatted them in italics. Just to note that for me "dynamic resistance" and "differential resistance" are not the same. The first means "changing ohmic resistance RBE"; the second means "virtual resistance rbe" seen by the input source through RB. – Circuit fantasist Jan 12 at 13:57
• OK - agreed. With other words: "rbe" is for sinusoidal signals only. – LvW Jan 12 at 14:26
• @LvW, ... and for any (both AC and DC) signal change... (it does not depend on time). Very interesting... I continue thinking about this phenomenon. To realize what the differential resistance 'rbe' means, I suppose that there is an ohmic (static) resistance RBE = VBE/IBE that changes. As a result, the input source "sees" a differential resistance 'rbe' that (almost) does not change. So, the paradox is that RBE is changing while 'rbe' is constant... – Circuit fantasist Jan 12 at 15:11
• The quantity "rbe" is certainly the inverse slope of the Ib=f(Vbe) characteristic. Together with all the other BJT parameters we also can write rbe=beta/gm with transconductance gm=d(Ic)/d(Vbe) (slope of the Ic=f(Vbe) curve) – LvW Jan 12 at 16:06