Transistor basics

Introduction

A whole range of TTL and ECL chips are made solely of bipolar junction transistors (BJT). Also a lot of OPAMPs are designed using either BJTs transistors or a combination of JFETs (junction gate field effect transistor) especially in the input stage and BJTs for current mirrors and output stage.
Modern chips are based on CMOS technology using MOSFETs.
So it's worthwhile to shine a light on widely used device and its practical use.

The bipolar junction transistor has 3 terminals: collector, base and emitter.
In the past, BJTs were made of Germanium (having a PN junction voltage drop of 0.3V). Nowadays transistors are made of Silicon (having a PN junction voltage drop of about 0.7V, so more than double of the Germanium version). Germanium transistors slowly disappeared from the stage because it was more expensive than silicon to make transistors with.

NPN and PNP biasing

There are 2 main types of bipolar transistors : NPN and PNP, indicating the type of layer (either N or P) for the collector, base and emitter respectively.
So a transistor has 2 PN junctions that are combined to form a NPN or a PNP : a collector/base junction and a base/emitter junction.
We'll not dive too deep into all the details, because i want to focus more on the practical side of things.

To make a transistor work, the 2 PN junctions of the transistor have to be biased as shown in the picture below.
At the right in the picture the symbols for an NPN and PNP transistor are shown.

Regardless of the type of transistor (NPN or PNP) the collector-base junction has to be reverse biased and the base-emitter has to be forward biased to do anything useful with a transistor.
When applying this kind of biasing, the current flow in the transistor will be as follows :

Transistor current gain

In the picture above we see the currents Ic (collector current), Ie (emitter curent) and Ib (base current).
Using Kirchhoff's first law, we can state that :

Ie = Ic + Ib

Ic is in the order of magnitude of tens to hundreds of mA for a low power transistor and Ib is in the order of magnitude of tens to hundreds of uA.
That is why Ib is shown with a small arrow, while Ic and Ie are shown with a big arrow.
Because of this, we can also say that :

Ie about the same as Ic
The relation between Ic and Ie is given by alpha (DC) = Ic/Ie, which is almost 1 (about 0.99)

The base-emitter junction acts as a diode with a voltage drop of 0.7V. So when the base-emitter voltage (Vbe) comes close to 0.7V, the current will rise very fast just like with a diode. Let's investigate what happens with Ic when we increase Vbe from 0 to 1V, while we apply a constant voltage of 10V over the collector and emitter (Vce = 10V).
The picture below shows how we do this. We use a common emitter configuration (emitter connected to ground) because this is the most used configuration.

The graph below shows the result.
We see that the collector current raises very fast, from the moment that Vbe is getting close to ca. 0.7V, just like with a diode.
The current suddenly stops raising at 10mA because of Rc (collector resistor that we added in the circuit to protect the transistor from too high currents).

What would happen when we would change Vce and do the previous test for Vce = 1V, then for Vce = 2V etc... up to f.e. : Vce = 10V ?
In the graph below the result is shown.
In the graph above, we see that Ic, even though we change Vce, still follows the same curve of a rising diode current. Ic stops rising when the current reached the maximum determined by Rc, being Vce/Rc. For Vce = 1V, the max. current will be 1V/1k = 1mA. For Vce = 2V, the max. current will be 2V/1k = 2mA, etc...

The fact that Ic keeps following the same rising curve even when changing Vce, means that Ic does not depend heavily on Vce but almost only on Vbe. Otherwise the Ic curve would have shifted to the the left with every increment of Vce.
This is an important property of a transistor that tells us that Ic is defined directly by Ib and not so much by Vce.

So lets investigate the relation between Ic and Ib. In the picture below is shown how we do this. Again we use the common emitter configuration (emitter connected to ground), because this is the configuration that is used the most.

The graph below shows the result of the measurements.
We see that the relation between Ic and Ib starts off rather linear until Ib = 100uA and Ic = 40mA. After that the curve starts rounding off.
This tells us that there seems to be a rather constant ratio between Ic and Ib for a limited range of currents.

The ratio between Ic and Ib is called Beta for DC currents = Ic/Ib.
This ratio is also called hFE (hybrid forward gain in common emitter configuration).
The steeper the slope of the curve below, the higher Beta is. So we see that Beta is decreasing for higher Ic currents.
Beta (DC) for a common low power transistor is in the range of 200 to 400. The transistor in the example has a Beta of ca. 400 for an Ic < 40mA. High power transistors typically have a low Beta that can be as low as 50.

Important to note is that Beta (DC) decreases with higher collector currents (Ic).

Since we know that Alpha (DC) = Ic/Ie and Beta (DC) = Ic / Ib and Ic + Ib = Ie, we can state that:
Ic + Ic/Beta = Ic/Alpha
Ic * (1 + 1/Beta) = Ic/Alpha
(Beta + 1) / Beta = 1 / Alpha
Alpha = Beta / (1 + Beta)
The bigger Beta is, the closer Alpha will be to 1. Since Alpha = Ic/Ic, this means that Ie is almost the same as Ic.

How is the relation between Ie and Ib : Ic = (Beta * Ib) = (Alpha * Ie), so:
Ie = (Beta * Ib) / Alpha
Ie = (Beta * Ib) / (Beta / (Beta + 1) )
Ie= (Beta + 1) * Ib

So a transistor is a device that provides DC current gain between the base and collector in a common emitter arrangement, given by Beta (DC) = Ic/Ib.

The AC current gain for a common emitter arrangement is given by : Beta (AC) = change of Ic / change of Ib while keeping Vce constant. The AC current gain is also called hfe (note the lowercase fe here).

Transistor output characteristics (common emitter configuration)

Next we want to see what the influence of Vce is on the relation between Ic and Ib. So we want to apply a certain Ib and measure Ic while changing Vce.
This way we get a more complete picture of the interactions of Ic, Ib and Vce.
In the picture below is shown how we can do this:

The graph below shows the result of the measurements. The different graphs show the curves of the relation between Ic and Vce while Ib is kept constant at a certain value. For each curve Ib is stepped up with 50uA starting from 0 to 1mA, so we see a total of 20 different curves.
This is also called the output characteristics of a transistor in common emitter configuration (more about the different configurations later).

The graph tells us multiple things :

• The bottom curves are nearly horizontal while the curves with higher Ic values are not. When the curve is horizontal, this means that Ic does not change while changing Vce and keeping Ib constant. So the ratio of Ic/Ib is constant even when changing Vce. This means that Beta (DC) is constant for low Ic currents. But the higher Ib and thus Ic, the more influence the change of Vce has.

• The active region is the part of the graph where the curves form straight lines (running horizontal or a bit sloped) above Ic = 0.

• The space between the curves seems to increase with higher Ib and thus Ic. This indicates that Beta (DC) of the transistor is dropping down for higher Ic values. We already noticed this in the previous measurements (Ic versus Ib).

• When we take the bottom curve (purple), Ib = 50uA, we see that this results in an Ic = 20mA, independent of the value of Vce as long as we stay in the active region. This tells us that Beta (DC) = Ic / Ib = 20mA / 50uA = 400. So the transistor has a hFE = Beta (DC) of 400.

• The cut-off region is where Ib is so low that there is no current gain at all and Ic = 0. The transistor does not conduct and is cut-off. No current is flowing through the collector.

• The saturation region is where Vce is very small (< 1V) and Ic starts to drop down to 0 fast. In this region the transistor is said to be saturating. It is this region that is used when the transistor is used as a switch and that has to avoided when using the transistor as an amplifier. The voltage Vce where Ic starts dropping down to 0 is called Vce-sat (saturation voltage). The value of Vce-sat depends on the value of Ib. The higher Ib and thus Ic, the higher Vce-sat will be. For an Ib of 50u, Vce-sat is ca. 0.3V. For Ib = 500uA, Vce-sat is ca. 0.5V. This means that the lower Ib and Ic, the lower Vce-sat will be, thus the further the transistor will go into saturation and the more it resembles a closed switch, because the voltage drop over the transistor then is at it's minimum.

• Note that the base current Ib determines the collector current Ic. F.e an Ib of 50uA results in an Ic of 20mA. So when you want a certain Ic, the Ib has to be high enough to accommodate the desired Ic. This is important to remember when designing circuits for high currents and high power. Without taking care that the transistor gets enough base current, it will not be able to drive the desired current in the collector (and thus emitter, because the current in collector and emitter is about the same).

So there is quite a bit of information in this graph that is revealing the inner secrets of transistors.

When Ic increases, the voltage drop Ve over the emitter resistor will increase because Ve = Ie * Remitter = about the same as: Ic * Remitter.
When Ve increases, Ib will decrease again, since Ib = (Vsupply - Vbe - Ve) / Rb.
When Ic decreases, the voltage drop Ve will decrease, thereby increasing Ib again.
So this biasing scheme also adds extra stability using negative feedback.
Ic and Ib are being stabilized by the voltage drop over the emitter resistor Remitter, so temperature or difference in Beta have less effect.

By adding the emitter resistor, the voltage swing of the output is decreased, because the collector can not go lower than the emitter voltage, which is 1V in the example. This reduces the maximum gain of the amplifier. So we sacrifice gain to improve stability. But not only stability is improved by applying negative feedback, also distortion will be less and the input impedance of the circuit will increase.

The biasing scheme that gives the best temperature stability and Beta independence is where the base is connected to a resistive divider additional to the emitter resistor as described above. The resistive divider and emitter resistor will keep the base-emitter voltage, which is temperature dependent, more stable compared to the other biasing schemes.
In the picture below, this biasing scheme is shown with a practical example:

Common emitter configuration

The common emitter (inverting amplifier, buffer) and the common collector (non-inverting buffer) configurations are the ones that you will encounter most. We'll concentrate on the common emitter configuration, because this is the configuration that is used when you want to make an amplifier.
We saw how to bias a transistor, so now we will apply the biasing scheme and make an common emitter amplifier.
Important for an amplifier is that it amplifies the input signal without distortion or at least with negligible distortion.
To make sure that the distortion is minimal, the transistor should stay in the active region and should stay away from the saturation region and from the cut off region (see explanation of the regions above). So the DC operating point of the transistor should be chosen carefully and the input signal amplitude should not force the transistor into the saturation or cut off region.
Our amplifier will be a common emitter amplifier with the signal input at the base and the signal output at the collector of the transistor.
We will use the biasing scheme as seen before with the resistive divider. We leave out the emitter resistor for simplicity. Without the emitter resistor, the collector voltage is equal to Vce. This makes it easier to work with the output characteristics graph since Vce is the value shown in the X-axis of the graph.
In the picture below, the amplifier is shown:

Because we left out the emitter resistor, the biasing resistors R1 and R2 had to be adapted compared to the biasing example with the emitter resistor.
Using the resistive divider R1 and R2, we have defined the DC operating point for the transistor. This operating point defines the collector current Ic = about 10mA. The collector current causes a voltage drop of 5V over the collector resistor (470 Ohm), so Vce will 5V.
With a Beta (DC) of about 430 a base current of about 24uA is required.
The DC operating point (working point) is shown in the output characteristics (Ic versus Vce for a certain Ib) below:

When we apply an input signal at the base of the transistor, the base current Ib will be modulated by the input signal.
When the base current Ib changes, the collector current Ic will also change. The change of Ic is translated to a change of collector voltage Vce, being the output of the amplifier.
So by modulating the base current, Ic and Vce will vary around the DC operating point.
How Ic and Vce will vary can be shown in the output characteristics curves when we draw a line that represents the collector resistor (load resistor) of our amplifier. This is also called the load line, since in a common emitter configuration, the transistor sees the collector resistor as its load resistor.
To draw this load line, we need to check the 2 extremes of Vce that are defined by this collector resistor:
(1) Vce=0 gives the maximum value of Ic for the given collector resistance and
When Vce is 0, the full power supply voltage is found over Rcollector (470E), resulting in an Ic = 10V / 470E = about 21mA.
We can draw this situation as a point on the output characteristics curves at Vce = 0V and Ic =21mA. This is the left point in the picture below.
(2) Vce = power supply voltage gives the minimum of Ic for the given collector resistance.
When Vce is the power supply voltage, Ic = 0 because the voltage drop over Rcollector = 0. No voltage drop, no current.
We can draw this situation as a point on the output characteristics curves at Vce = 10V and Ic = 0mA. This is the right point in the picture below.
When we connect these both extremes with a line, we have the load line that reflects the relation between Ic, Vce and Ib for the given collector resistance . This is the line over which the DC operating point will move when Ic changes due to an input signal on the base of the transistor.

Now we have to check how big the input signal can be without the transistor going into the saturation or cut off region, because we don't want to have a distorted output signal.
In the saturation region, the collector current will suddenly drop down to zero instead of increasing further.
In the cut off region, the collector current will simply stop flowing. Both will give significant distortion that will be visible at the output of the amplifier.
In the picture below, boundaries for Ib and Ic are indicated so the operating point stays far away from the saturation region and the cut off region.
Within these boundaries, the transistor will amplify the signal without significant distortion.

With the set boundaries, we can see what the minimum and maximum for Vce, Ic and Ib should be, in order to keep the transistor within the active region.
Vce will vary from 3V to 7V, so about 2V under and above the Vce = 5V of the DC operating point.
Ic will vary from about 5mA to 15mA, so about 5mA under and above the Ic = 10mA of the DC operating point.
Ib needs to vary from about 12uA to 36uA, so about 12uA under and above the Ib = 24uA of the DC operating point.
Now we still have to find out what the amplitude of the input signal needs to be to modulate the base current so it increases and decreases with 12uA.
Therefor we need to know the input impedance that is seen by the input signal when looking into the circuit (base of transistor + resistive divider).
The input impedance is mainly determined by the resistive divider that runs a current that is at least 10 times Ib. This impedance is expressed by the parallel circuit of the 2 resistances = R1*R2 / (R1 + R2) = (24k6*2k)/(24k6 + 2k) = about 1.8k.
To increase and decrease the base current with about 12uA, we need the input voltage to raise from 0V to (1.8k * 12uA) = 0.021V = about 21mV.
So with an input signal of 21mV (single peak) * 2 = 42mV peak to peak, we get an output voltage with (7V - 3V ) = 4V peak to peak.
This means that our amplifier has a gain of 4Vpp/42mVpp = about 100x = 40 dB.