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How a BJT Transistor works (base current version)

How a BJT Transistor works (base current version)


Table of Contents
  1. Introduction
  2. Two types
  3. Symbols
  4. Terminals
  5. How it works
  6. Modes of operation
  7. Varying the base current
  8. Generalising a little
1 Introduction

A Bipolar Junction Transistor is the normal type of transistor that we normally just call "a transistor". Saying "BJT Transistor" is like saying "ATM Machine", or "PIN Number", so you would not normally say it. However this thread is for beginners and you may not have known a few seconds ago what a BJT was :)

The operation of a Bipolar Junction Transistor (BJT) can be described at several levels. This description concentrates on a method which is both simple to understand and apply. That is, that we can use the current passing through the base terminal of the BJT as a control of collector current.

There is an alternative explanation which is given by the voltage difference between the base and emitter terminals. You can read more about it HERE (when it exists).

Unless otherwise specified, if I refer to a "transistor" from here on in, I am talking about a BJT.

If you already know how to identify a transistor on a circuit diagram, skip straight to part 5.

2 Two Types

A BJT consists of three layers of silicon (think of it like liquorice allsorts!).

One variant is called NPN.


It is called NPN because the three layers are N, then P, then N.

The other variant is PNP. (I think you can guess the difference)


The three layers go P, then N, then P.

This kind of tells you how they're made, but we don't draw this in a circuit diagram.

3 Symbols

There are some standard symbols for BJTs. Unfortunately, not everyone uses the same symbols, but once you know what to look for, you can spot them with ease.

But before we look at what is used now, lets look at a symbol for a BJT that is no longer used.

These images match the construction shown above. It is unlikely you'll see this symbology, but note that it has a bar representing the middle layer of silicon, and two lines leaving that bar, one of which has an arrowhead. These are the things to look out for... with one exception.

In the modern symbology, The bar is still there, but it has been turned around and those two lines (one with an arrowhead) are now at angles to it. Other variation (some have circles around them, some have thicker or thinner lines, open or filled in features) is irrelevant.

In all of these cases the arrowhead tells us if the BJT is NPN or PNP.

I like to remember it as "The arrow points to N" (like on a map). In the top row the arrow points out -- this is NPN (with the N's on the outside). In the bottom row the arrow points in -- this is PNP (with the N in the middle).

An alternate method others use is NPN is Not Pointing iN (and PNP is the other one (although a less intuitive mnemonic Pointing iN -- often with another P word added, like "Proudly").

4 Terminals

A BJT has three terminals, these are:
  1. Base
  2. Emitter
  3. Collector
With modern symbology, the Base is the lead connecting at right angles to the bar, the Emitter has the arrowhead, and the collector is the other angled connection (without the arrow).

These are summarized below:

5 How it works

A transistor can be imagined as a device where the resistance between the emitter and the collector are varied in order to maintain a relationship between the base current and the collector current.

Transistor man (Horowitz & Hill 2nd ed., p64) keeps an eye on the base current and uses this to try to keep the collector current some multiple of the base current.

This multiple is called hFE. This is a measure of the current gain of the transistor. You will find it listed in the datasheet for a transistor.

The equation which describes this is

Ic = hFE × Ib

It is important to note that hFE is not actually a constant. Consequently, if you look at a graph of base current to collector current, you will notice that it is curved. Because of this, the datasheet will often give either a range of hFE values, or hFE values at specific collector currents.

6 Modes of operation

A transistor operates in one of three modes:
  • Cutoff
  • Active
  • Saturation
Cutoff refers to the mode of operation where there is a voltage across the collector and emitter of the transistor, but no current flows. For practical purposes "no current" means some small leakage current (called ICBO) which may be measured in nA or μA. In practice, the transistor is turned OFF, and no base current is present.

Active refers to the mode of operation where the collector current varies in response to changes in base current. An increase in base current will allow an increase in collector current, and vice versa. The ratio of the collector current (IC) to the base current (IB) can be referred to as Gain, hFE, or Beta. Note that the relationship is not strictly linear, the gain rises as base current rises, reaches a peak and then drops. Datasheets reflect this by either giving a range of values or by nominating values at particular collector currents.

Saturation refers to a mode where the collector current no longer rises even if the base current increases. This occurs at a range of base currents, not at a point. As the transistor is pushed further and further into saturation the voltage across the transistor (VCE reduces. Datasheets often specify the Collector-Emitter saturation voltage at various collector currents and base currents. This is referred to as VCE(sat). Practically this is when the transistor is turned on completely.

When a transistor is used as a switch, it is operated in the cutoff and saturation regions. We want to turn it (and therefore a load) on and off.

When we use a transistor as an amplifier we operate the transistor in its active region.

7 Varying the base current

In this simple introduction we're only going to look at one transistor configuration -- common emitter. For simplicity we will assume that the emitter is connected to ground.


This is as simple as it gets :)

Cutoff - if the input is left disconnected, or the input voltage is less than about 0.6V (more on this later) then the transistor will be in cutoff and the output voltage will be equal to that of the supply voltage.

Active - As the input voltage rises, the base current is given by

IB = (Vin - 0.6) / R1​

The collector current is given by:

IC = IB * hFE

And the output voltage is given by:

Vout = Vsupply - IC * R2​

Putting that all together:

Vout = Vsupply - ( (Vin - 0.6) / R1) * hFE * R2​

And employing a little sleight of hand, the overall gain is equal to

= Vout / Vin
= - (hFE * R2) / R1​

But this is only valid within certain limits.

Firstly Vin needs to be greater than about 0.6V
Secondly Vout must be more than about 1V

So the maximum input voltage is

Vin = ((1 - Vsupply) / Gain) + 0.6​

Let's pick some real figures here...

Let's have:
  • hFE = 200
  • Vsupply = 12
  • R1 = 10k
  • R2 = 1k
The gain of this circuit is about -(200 * 1000) / 10,000, or -20. The gain is negative because an increase in Vin produces a decrease in Vout.

The maximum input voltage is about ((1 - 12) / -20) + 0.6 = 1.15V

So the transistor is in its linear region when the input voltage is between about 0.6 and 1.15V, producing an output which varies from 12V down to 1V.

Saturation - As the voltage across the transistor falls it moves from the Active region to the Saturation region.

For an alternative (and more practical) description of saturation, please see this resource.

I am using 1V across the transistor as the end of the active region, but saturation is often defined as the voltage being less than 0.3V. The transition from active to saturation is a bit like the transition from night to day. We can certainly state when its night, and when it's day, but can we pinpoint the exact time that it changes?

As the base current continues to increase, and the voltage across the transistor falls, it gets harder and harder to turn on more. So the relationship between base current and collector current diminishes and finally all but disappears as the voltage gets asymptotically closer to some minimum value.

For a given collector current (IC) the saturation base current (IB) is given as:

IB(sat) = k * IC(sat) / hFE

Where k is between 2 and 10.

The datasheet I am looking at for the BC548 indicates a VCE(sat) of 0.25V for an IC of 10mA and an IB of 0.5mA. Given that the minimum hFE of this transistor is given as 110, they are using a k of approximately 5.

For our circuit above,

IC(sat) = Vsupply / R2

IB(sat) = k * IC(sat) / hFE

Vin(sat) = IB(sat) * R1 + 0.8​

So given the same values:
  • hFE = 200
  • Vsupply = 12
  • R1 = 10k
  • R2 = 1k
  • and k = 5
= Vsupply / R2
= 12/1000
= 12mA​

= k * IC(sat) / hFE
= 5 * 0.012 / 200
= 0.3mA​

= IB(sat) * R1 + 0.8
= 0.0003 * 10,000 + 0.8
= 3.8V​

The interesting thing here is that in the active region, Vin can range from 0.6V to 1.15V, but a voltage of 3.8V is required to fully saturate the transistor.

So what happens between 1.15V and 3.8V? The answer is that it's the twilight region between active and saturated. And depending on our definition of "saturated" this voltage range might be larger or smaller.

8 Generalising a little

Up to now we have omitted any discussion of an emitter resistor.


Adding an emitter resistor introduces feedback, allowing us to produce a circuit which has a gain that is largely independent of the hFE of the transistor.

This moves us part way to being able to produce a practical amplifier (which will be covered elsewhere)

For now, here are a series of equations similar to the ones given above which will reveal the surprising effect of the emitter resistor.


Ic = Ib × hFE


Vin = (Rb × Ib) + VBE + ( (Ib + Ic) × Re)
= (Rb × Ic ÷ hFE) + VBE + ( ( (hFE + 1) ÷ hFE) × Ic) × Re)
≈ (Rb × Ic ÷ hFE) + VBE + (Ic × Re) -- assuming hFE >> (Rc ÷ Re)​

Vout = Vcc - (Ic × Rc)​
So (rearranged)

Ic = (Vcc - Vout) ÷ Rc


Vin ≈ (Rb × ( (Vcc - Vout) ÷ Rc) ÷ hFE) + VBE + ( ((Vcc - Vout) ÷ Rc) × Re)​


Vout ≈ ( ( (Re × hFE + Rb) × Vcc) - (Rc × hFE × (Vin - VBE) ) ) ÷ (Re × hFE + Rb)​

If Re = 0, this becomes

Vout ≈ ( ( Rb × Vcc) - (Rc × hFE × (Vin - VBE) ) ) / Rb
= Vcc - (Rc × hFE × (Vin - VBE) ) ÷ Rb
= Vcc - ( (Vin - VBE) ÷ Rb) * hFE * Rc

And as I gave above:

Vout = Vcc - ( (Vin - VBE) ÷ Rb) × hFE × Rc -- Albeit with slightly different nomenclature​

Now, if Rb = 0, you get

Vout ≈ ( ( Re × hFE × Vcc) - (Rc × hFE × (Vin - VBE) ) ) ÷ (Re × hFE)
= Vcc - (Vin - VBE) × (Rc × hFE) ÷ (Re × hFE)
= Vcc - (Vin - VBE) × (Rc ÷ Re)​

And therefore the gain is

- Rc ÷ Re -- where hFE >> (Rc ÷ Re)​

Which is surprising because it means that variation in the performance of the transistor is no longer a major factor in determining the relationship between the output and input voltages!

As mentioned above, this is not a practical amplifier, for that we need to "bias" the transistor.
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