Transistor Characteristics

Note: this material is a bit advanced. You may want to skip it and come back later. You may want to come back later even if you don't skip it. Some of this material will also be covered again in Analog Circuits.

Bi-polar Junction Transistors

The following circuit is an example of a test circuit to demonstrate the characteristics of a bi-polar junction transistor. Notice that there are two current paths, making this a parallel circuit. We have current flowing into the base of the transistor (the base current, labeled I_B) and current flowing into the collector (the collector current, labeled I_C). These two currents exit the emitter together (the emitter current, I_E, not labeled in the diagram). According to Kirchhoff's Current Law the current exiting the emitter must be the sum of the currents entering the base and collector (see Kirchhoff's Current Law). These currents then split up again at the bottom of the circuit where the emitter connects to the negative terminals of the two batteries. The base current circulates in the left part of the circuit (the base circuit) and the collector current circulates in the right part of the circuit (the collector circuit).

Example test circuit for the following transistor curve.

This is called a common-emitter amplifier because the emitter of the transistor is common to both circuits (the emitter connection is a common-connection type of ground [see What is Ground in DC Circuits]). We have two current meters to show that we are interested in the base current and the collector current. There is also a voltmeter showing that we are interested in the voltage between the collector and the emitter (the collector-to-emitter voltage labeled V_CE). The collector resistor isolates the collector from the collector power supply (the battery labeled V_CC). Without this resistor the voltage at the collector would always be the battery voltage. We need the collector voltage to be able to drop as the collector current increases.^{^[1]} The battery labeled V_BB is there to supply current to the base of the transistor. The variable resistor in the base circuit controls the base current.

Let's start by setting the base current to 20 μA. Holding the base current at 20 μA let's increase the battery voltage until V_CEis 20 volts. The graph below shows what the the collector current (I_C) does as we increase V_CE.

Going from left to right V_CE is increased. Notice that after an initial rise the collector current (I_C) levels off and hardly changes as V_CE increases.

Notice that I_C levels off at about 2 mA and stays there regardless of V_CE. With this information we can calculate h_FE,which is I_C divided by I_B (see Bi-polar Junction Transistors). Here I_C is 2 mA and I_B is 20 μA. Regardless of V_CE, I_C is about 100 times I_B so we have an h_FE of 100.

Now let's repeat the process with the base current set at 40 μA

The same transistor with a base current of 40 μA. Notice that the flat line isn't quite as flat. There is a slight increase in collector current as V_CE is increased.

Everything looks about the same except the horizontal line is a bit higher. This shows that doubling the base current just about doubled the collector current. The collector current is now about 4 mA regardless of V_CE. Well, not really. Notice that the horizontal line isn't quite as horizontal. The slope was negligent with a base current of 20 μA, but now it is quite noticeable. At a V_CE of 20 volts the collector current is pretty close to 4 mA, but at a V_CE of 5 volts I_C is closer to 3.5 mA.

The collector current (I_C) is lower at lower levels of V_CE.

Does that mean that with a V_CE of 20 volts h_FE is 100 and with a V_CE of 5 volts h_FE is about 90? Yes, it does. It would be really nice if h_FE were completely independent of all the other parameters, but it isn't. It's just something we have to deal with.

Now let's start over and draw graphs using base currents of 10 μA through 120 μA in 20 μA steps. When we are finished we have something like this:

A typical transistor curve (this curve is a close approximation of that for a 2N2222 transistor).

In this graph each horizontal line represents the collector current (I_C) where the base (I_B) current is held at a different steady value (indicated to the right of each line). As we increase the base current the horizontal line gets higher, indicating a higher collector current. Going from left to right the collector-to-emitter voltage (V_CE) is increased. The fact that each horizontal line is fairly flat shows that the collector current is much more dependent on the current flowing into the base than the voltage across the transistor.

When designing a transistor circuit we usually want to make sure the transistor operates under conditions where the collector current lines are fairly flat. In the above graph the slopes are reasonably flat throughout all the parameters. However, at higher base currents the slopes will get increasingly steep. You would want to avoid designing an amplifier that operates under those conditions. The output will be significantly distorted compared to the input.

Saturation

As you increase the base current the collector current increases proportionally. However, you eventually reach a point where further increasing the base current no longer causes an increase in collector current. This condition is called saturation. The saturation point depends largely on the value of the collector resistor. Let's add some parameters to the above circuit and see what happens in both nominal operation and in saturation. We'll make the battery voltage 20 volts and increase the base current until the collector current is 5 mA. (We don't need to know the base current, We'll just turn the variable resistor until we see 5 mA for I_C.)

Here is our test circuit in nominal operation.

With a battery voltage (V_CC) of 20 volts and a collector current of 5 mA we have 10 volts across the 2 k collector resistor. This leaves 10 volts for the collector-to-emitter voltage (V_CE). That's 10 volts across the collector resistor and 10 volts for V_CE. This adds up to the battery voltage and Kirchhoff's Voltage Law is satisfied (see Kirchhoff's Voltage Law).

Now let's increase the base current until the collector current is 7.5 mA.

The same circuit with the collector current increased slightly.

Now the voltage across the collector resistor has increased to 15 volts. There is a corresponding drop in V_CE to 5 volts. Kirchhoff's Voltage Law is still satisfied.

Wait a minute. The current through the transistor went up yet the voltage across it went down. How can this be? In DC Circuits we learned that as you increase the current through a resistor the voltage across it increases. Did the resistance of the transistor go down as the current went up?

It did. That's what transistors do.

Remember that a BJT acts like a current-controlled variable resistor. Let's apply Ohm's law to the transistor and see what's going on. We have 5 volts across the transistor (V_CE) and 7.5 milliamps flowing through it (I_C). 5 volts divided by 7.5 milliamps gives us 666 ohms.

The base current is adjusted to where the collector current is 7.5 mA. The transistor is acting like a 666 ohm resistor and the collector circuit is acting like any series circuit.

If we apply the rules for series circuits to the collector circuit we will see that it is acting like a 2k resistor and a 666 ohm resistor in series with a 20 volt battery. The transistor is acting like any 666 ohm resistor. All the rules we learned in DC circuits still apply.

Now lets increase the base current until the collector current is 10 mA.

Now the collector current has reached its limit with this collector resistor.

Now the voltage across the collector resistor is 20 volts. That's all the voltage the battery has to give. There is nothing left for V_CE. It has to be 0 volts. Let's see what happens if we increase the base current further.

Increasing the collector current further causes a Kirchhoff's Voltage Law violation.

If we increase the base current until the collector current is 12.5 mA the voltage across the collector resistor will increase to 25 volts. But it can't. This violates Kirchhoff's Voltage Law. We cannot get more voltage than than the battery has to supply. With this 2 k resistor it is impossible to increase the collector current beyond 10 mA. We can still increase the base current, but regardless of what h_FE says the collector current will not go above 10 mA. The circuit is saturated.

Actually, you cannot even get 20 volts across the collector resistor. The resistance of the transistor will try to go to 0 ohms when the transistor is saturated, but this is impossible. The collector-to-emitter path always has some resistance, so when there is current flowing into the collector there will be some voltage between the collector and the emitter. This is called the saturation voltage and is labeled V_CE(sat) on transistor datasheets. The saturation voltage is typically around 0.4 volts but may be more for power transistors. For example, the 2N3055 has a saturation voltage of around 2 volts. Here is the above circuit with realistic saturation parameters.

Here are realistic parameters for a saturated transistor circuit.^[2]

Now we have a V_CE of 0.4 volts and the voltage across the collector resistor is 19.6 volts.

Troubleshooting tip: If you have a base-to-emitter voltage that is significantly above 0.7 volts, don't automatically assume the transistor is bad. Check the rest of the circuit for voltages like those above. If it looks like that, the circuit is probably saturated and the transistor is doing exactly what it is expected to do. However, if you see signs of low collector current (like a V_CE equal to the battery voltage) you probably have a bad transistor or a bad connection to the transistor.

Here's the chain of logic: If there is a high base-to-emitter voltage there must be a high base current. If there is a high base current there must be a high collector current. If there is a high collector current there must be a large voltage drop across the collector resistor. This results in a low V_CE. That's what is happening above. However, if you have a high base current and V_CE equals the battery voltage, this indicates that there is no collector current; there must be no current flowing through the collector resistor. The transistor must be bad, acting like an open circuit (see Open Circuits).^[3]

Cutoff

Just as there is a point where increasing the base current no longer causes an increase in collector current (saturation) there is a point where decreasing the base current no longer causes a decrease in collector current. This condition is called cutoff. Let's take another look at that transistor curve.

Take a look at the bottom collector current line. This is the line for a base current of 10 μA. This is the cutoff current. The cutoff current in this graph is a bit high for a real-world transistor, but every transistor has a collector current that is the lowest possible collector current regardless of how low the base current goes. This is called the collector cutoff current and is labeled I_CEO(current from the collector to the emitter with the base open [no connection to the base]).

FETs

The following circuit is a common-soruce amplifier. It is the FET eqivalent of the common-emitter amplifier. We can use it to create a graph of and FET amplifier characteristics. The main difference is the labeling of the components and other parts of the circuit. Instead of a base current we have a gate-to-source voltage (V_GS). Instead of a collector current we have a drain current (I_D). The equivalent of V_CE is V_DS, etc.

FET version of the test circuit.^[4]

After after a procedure equivalent to the one used for a BJT we get a curve something like this.

FET Characteristic Curve

There are several notable differences between the BJT curver and the FET curve. The spacing of the drain current lines is not even; the relation of I_D to V_GS is not as linear as the relationship of I_C to I_B. There is a large area to the left of the diagram where the drain current lines have notable curves. With this example, with a V_DS under 10 volts the drain current lines are far from flat. In this part of the curve drain current is significantly affected by V_DS. These two parts of the graph are divided by a blue line because they provide two different ways to use an FET.

To the left of the blue line is the ohmic region. If operated in this region—with a VDS under 5 volts—the transistor has a resistance proportional to V_GS rather than a drain current proportional to V_GS. It acts like a voltage-controlled variable resistor. We have said all along that a transistor acts like variable resistor. However, in previous examples it acts like a variable resistor with some intelligence; the resistance changes to follow certain rules. In the ohmic region it acts more as if it simply changes its resistance in response to the gate voltage.

To the right of this the blue line is the saturation region^[5]. In this region the FET operates much like a BJT, only it is voltage controlled instead of current controlled.

What is Saturation

¹	Recall that when there is current flowing through a resistor the voltage where conventional current exits the resistor must be lower than where the current enters the resistor. The right side of the collector resistor is connected directly to the positive terminal of the battery, the voltage there cannot change. The voltage at the left side of the collector resistor must therefore drop as the collector current increases.
²	How about a little puzzler here? What can you do to allow more collector current without causing saturation. The solution is in the Transistor Summary.
³	There could also be a break in the circuit where the transistor is connected to the rest of the circuit. When troubleshooting always doublecheck. Connect your voltmeter probes directly to the leads of the component you are testing. Then connect the probes to other parts of the circuit that should be connected to the component. If the voltages aren't the same look for breaks in the circuit.
⁴	How another puzzler? What is the purpose of RG in the diagram. Hints: no current flows into or out of the gate of the transistor; the gate region has conductive metal and conductive silicon separated by an insulating silicon oxide. The solution is in the Transistor Summary.
⁵	This is a poor choice of names. This circuit will be saturated where an increase in VGS (the input) no longer causes an increase in ID (the output). The saturation region as labeled on an FET curve has nothing to do with that. It this saturation region changing VGS will still cause a change in ID (the height of the horizontal drain current line represents the drain current as caused by VGS). If that's not confusing enough, the circuit is in saturation to the left of the ohmic region (not labeled here). Graphs that show this region label it the ''Saturation Region'', giving you a diagram with two areas labeled ''Saturation Region'' that have nothing to do with each other.