Basically when we talk about transistor amplifiers, they are designed to work with AC input signals that goes up and down between a positive value and a negative values.
Now to make sure that the transistor can work properly between these two extreme points, we need to set things up in a specific way.
This setup is often called “presetting” and it usually involves using a common emitter amplifier circuit configuration. To make this all work smoothly, we use something known as the “biasing”.
The Importance of Biasing
Biasing is super crucial when it comes to designing amplifiers.
Why? Because it helps us establish the right operating point for the transistor amplifier so that it is ready to receive signals without messing them up too much.
This means that biasing helps reduce any distortion that might happen in the output signal which is definitely something we want to avoid.
Visualizing Operating Points with Load Lines
Another neat thing we can do is draw something called a static or DC load line on the output characteristics curves of the amplifier.
This line lets us visualize all the possible operating points of the transistor, ranging from being fully “ON” to fully “OFF.” By doing this we can pinpoint where the quiescent operating point or Q-point of the amplifier is located.
The Goal of Small Signal Amplifiers
Now when it comes to small signal amplifiers, the main goal is pretty straightforward: we want to amplify the input signal as much as possible while also keeping the distortion to a minimum.
In simpler terms this means that the output signal should be a larger version of the input signal without changing its original shape.
Selecting the Right Quiescent Point
To achieve low distortion while using an amplifier, it is essential to choose the right quiescent point.
This quiescent point represents the DC operating point of the amplifier and can be placed anywhere along the load line depending on how we set up our biasing arrangement.
The Ideal Q-Point Position
Ideally we want this Q-point to be positioned as close to the center of the load line as we can manage.
This setup allows for what is known as Class A type amplifier operation. In technical terms this means that Vce should equal half of Vcc (Vce = 1/2Vcc).
So if you take a look at the Common Emitter Amplifier circuit shown below, you will see how all these concepts come together!
Understanding The Common Emitter Amplifier Circuit
The single stage common emitter amplifier circuit that you see up there makes use of something that is often referred to as “Voltage Divider Biasing.”
Now what this means is that this particular biasing setup involves two resistors working together as a potential divider network connected across the power supply.
The cool part is that, the point right in the middle of these two resistors provides the necessary Base bias voltage for the transistor to operate.
This kind of voltage divider biasing is pretty popular when it comes to designing bipolar transistor amplifier circuits.
One of the big advantages of this method is that it significantly minimizes the impact of any variations in Beta which is represented by the symbol β.
It does this by keeping the Base bias at a nice steady voltage level which leads to better stability overall.
How Quiescent Base Voltage Works
Now lets talk about how we can figure out the quiescent Base voltage which we call Vb.
You can calculate this voltage by the potential divider network created by those two resistors R1 and R2 along with the power supply voltage which we is the Vcc.
In this setup we have current flowing through both these resistors.
To understand how this works mathematically we can calculate the total resistance RT. This total resistance is simply R1 + R2.
From there we can find the current flowing through the circuit using the formula i = Vcc/RT.
The voltage that gets generated at the junction where resistors R1 and R2 meet helps keep the Base voltage (Vb) constant at a level that is below the supply voltage.
The Function of the Potential Divider Network
The potential divider network used in our common emitter amplifier circuit does a neat job of splitting the supply voltage in proportion to the resistance values of R1 and R2.
If you want to calculate this bias reference voltage it can be done quite easily using a straightforward voltage divider formula that you can find below:
Formula for Calculating Transistor Bias Voltage
VB = (VCC * R2) / (R1 + R2)
So you know how the supply voltage which we call Vcc plays a big role in figuring out the highest Collector current which is represented as Ic when the transistor is completely switched on, or in other words when it is in saturation mode and at that point Vce equals zero.
Now to find out the Base current which is the Ib for the transistor, you actually take the Collector current Ic and divide it by the DC current gain which is denoted by Beta and we write that as β for the transistor, as shown below:
What is Beta Value of a BJT
β = ΔIC /ΔIB
Okay so lets talk about the Beta value of a transistor which you might see referred to as hFE on datasheets.
This Beta value is all about how much current the transistor can gain when it is set up in what we call the common emitter configuration.
Basically Beta is this electrical parameter that gets built into the transistor when it is made. It does not have any units because it is just a fixed ratio comparing two currents which are Ic and Ib.
This means that even a tiny change in the Base current can lead to a big change in the Collector current.
Now here’s something interesting about Beta. If you take transistors that are the same type and have the same part number you will notice that there can be pretty big differences in their Beta values.
For instance the BC547 NPN Bipolar transistor can have a DC current gain Beta value that ranges anywhere from 110 to 450 according to the datasheet.
So what this means is that one BC547 transistor might have a Beta value of 110 while another one could have a Beta value of 450 but they are both sold as BC547 NPN transistors.
The reason behind this variation is that the Beta value which we also write as β is actually a characteristic that comes from how the transistor is built rather than how it works.
When the Base/Emitter junction is forward-biased this means that the voltage at the Emitter which we call Ve will be one junction voltage drop different from the Base voltage.
If you know what the voltage is across the Emitter resistor then you can easily figure out the Emitter current which we write as Ie using Ohm’s Law.
Plus you can estimate the Collector current Ic because it is pretty much going to be almost equal to the Emitter current.
Solving an Common Emitter Amplifier Problem #1
Let’s consider the following scenario. The supply voltage of a common emitter amplifier circuit is 9 volts, while the load resistance or RL is 1 kΩ.
Let us assume that Vce = 0.
So we can now calculate the highest collector current (Ic) that travels via the load resistor whenever the transistor is completely “ON” (saturated).
Also, if there is a 0.9 volt voltage drop across the emitter resistor RE, let us determine its value as well.
Considering a typical NPN silicon transistor, we will also determine the values of each additional circuit resistor.
Ic_max = (Vcc – Vre) / RL = (9 – 0.9) / 1000 = 0.0081 A = 8.1 mA
Vce = 0 (Saturation)
So this whole process sets up what we call point “A” on the vertical axis that represents the Collector current in the characteristic curves and this happens when Vce is equal to zero.
Now when the transistor is completely switched off it means that there is no current flowing through either of the resistors which are RE and RL.
Because of this lack of current there is also no voltage drop across those resistors. As a result the voltage drop across the transistor which we refer to as Vce ends up being equal to the supply voltage which is the Vcc.
This situation establishes point “B” on the horizontal axis of those characteristic curves.
Now typically when we talk about the quiescent Q-point of the amplifier we are looking at a scenario where there is no input signal being applied to the Base.
In this case the Collector is positioned roughly halfway along the load line that stretches between zero volts and the supply voltage so we can say it is at about Vcc divided by 2.
Therefore if we want to figure out what the Collector current is at the Q-point of the amplifier we can express it as:
Ic_Q = (9 – 0.9) / 2 x 1000 = 8.1 / 1000 = 0.00405 A = 4.05 mA
The equation for a straight line produced by this static DC load line has a slope of -1/(RL + RE) and intersects the vertical Ic axis at a position equal to Vcc/(RL + RE). The average value of Ib determines the Q-point’s true position within the DC load line.
Since the transistor’s collector current Ic is additionally identical to its DC gain (Beta) times its base current (β * Ib), when we consider that the transistor’s Beta (β) value is for example 100 (one hundred is a logical average value for low power signal transistors), the base current Ib entering the transistor is going to be as follows:
β = Ic / Ib
Ib = Ic / β = 4.05 mA / 100 = 40.5 µA
It is customary to feed the Base Bias Voltage via the primary supply line (Vcc) by means of a dropping resistor R1, as opposed to utilizing an extra Base bias supply.
Currently R1 and R2 resistors may be selected to provide an appropriate quiescent base current of 40.5μA or 40μA, adjusted to the closest whole number.
In order to prevent the base current flow from overloading the voltage divider network, the current passing across the potential divider circuit must be greater than the real base current Ib.
Generally speaking the resistor R2 should have a value that is no less than ten times Ib. When the silicon transistor’s base/emitter voltage Vbe is set at 0.7V, the value of R2 is as follows:
R2 = (Vre + Vbe) / (10 x Ib)
R2 = (0.9 + 0.7) / (405 x 10^-6) = 3950.61 = 3.95 kΩ
In the divider network, the current passing through resistor R1 must equal eleven times the base current whenever the current passing over resistor R2 is ten times the base current. In other words, IR2 + Ib.
R1 may be calculated as follows since the voltage across it is the same as Vcc – 1.7v (VRE + 0.7 for silicon transistors), or 10.3V.
R1 = (Vcc – (Vre + Vbe)) / (11 x Ib)
R1 = (9 – 1.6) / (446 * 10^-6) = 17 kΩ
Ohm’s Law makes it simple to determine the value of the emitter resistor or the RE. The base current Ib and the collector current Ic combine to form the current that travels via RE which is expressed as follows:
Ie = Ic + Ib = 4.05 mA + 0.0405 mA = 4.09 mA
A voltage drop of 0.9 volt is present over the resistor RE, which has a connection between the transistor’s emitter pin and ground. As a result, the emitter resistor’s (RE) value is determined as follows:
Re = Vre / Ie = 0.9V / 4.09 mA = 220 Ω
Therefore the resistor values that are selected for the situation at hand above and that provide a 5% tolerance (E24) are:
R1 = 17 kΩ, R2 = 3.95 kΩ, RL = 1 kΩ, Re = 220 Ω
The values of all the parts we just determined above can subsequently be included into a revised version of our original Common Emitter Amplifier circuit.
The Finalized Common Emitter Circuit
The Importance of the Amplifier Coupling Capacitors
So when we are talking about Common Emitter Amplifier circuits there are these capacitors called C1 and C2 that play a really important role as Coupling Capacitors.
What these capacitors do is they help to keep the AC signals separate from the DC biasing voltage.
This separation is highly important because it makes sure that the bias conditions we set up for the circuit to work properly are not messed up by any extra amplifier stages that might come after it.
The good thing about these capacitors is that they only allow the AC signals to pass through while blocking any DC components.
This way the output AC signal gets added on top of the biasing for the next stages in line.
Now there is also this other capacitor called CE which is included in the Emitter leg of the circuit.
This particular capacitor acts like an open circuit when it comes to DC biasing conditions.
What this means is that the currents and voltages used for biasing are not influenced by the presence of this capacitor, which helps keep everything stable at what we call the Q-point.
Now, this bypass capacitor that is connected in parallel actually behaves like a short circuit to the Emitter resistor, when we are dealing with high frequency signals because of its reactance.
So at those higher frequencies, only the load made up of RL and a tiny internal resistance really comes into play, and this setup helps boost the voltage gain to its highest possible level.
Typically when choosing the value for this bypass capacitor CE we aim for it to have a reactance that is no more than one-tenth of RE at the lowest frequency where we want the circuit to operate efficiently.
Analyzing Output Characteristics Curves
Alright, everything is looking pretty good so far. Now we are ready to create a bunch of curves that illustrate the Collector current which we refer to as Ic, plotted against the Collector/Emitter voltage, known as Vce.
We will do this for various values of Base current or Ib, in our straightforward common emitter amplifier circuit.
These curves that we are talking about are commonly called the “Output Characteristic Curves.” They serve an important purpose by demonstrating how the transistor behaves across its dynamic range.
To make things clearer, we will draw a static or DC load line on these curves for our load resistor which has a value of 1 kΩ. This line will help us visualize all the potential operating points of the transistor.
Now let us consider what happens when we switch the transistor “OFF.” In that scenario Vce will be equal to the supply voltage which we denote as Vcc.
This specific situation corresponds to point “B” on our load line.
On the other hand when we have the transistor fully “ON” and in saturation mode, the Collector current is determined by that load resistor RL and this condition is represented by point “A” on our line.
Previously we calculated that the Base current needed for the average position of the transistor was 40.5 μA.
We marked this point as Q on the load line which signifies what we call the Quiescent point or Q-point of our amplifier.
To make things a bit simpler for ourselves, we could easily round this value up to a nice clean 50 μA without causing any issues with the operating point at all.
Diagram showing Output Characteristics Curves
So when we look at Point Q on the load line, we see that it gives us the Base current Q-point which is around 45.8 microamperes, or if we round it off we can say it is about 46 microamperes.
Now what we want to do is figure out what the maximum and minimum peak swings of this Base current can be.
The goal here is to ensure that any changes we make will lead to a proportional change in the Collector current, which we refer to as Ic, all while keeping the output signal nice and clean without any distortion.
As we trace along the load line, it intersects various Base current values on the DC characteristics curves.
This allows us to pinpoint those peak swings of Base current that are evenly spaced out along the load line. We label these specific points as “N” and “M.”
At these points we find that the minimum Base current is 20 microamperes and the maximum Base current is 80 microamperes.
Now here is the interesting part, these points “N” and “M” can be placed anywhere along the load line that we like as long as they maintain equal spacing from our starting point Q.
This arrangement gives us a theoretical maximum input signal to the Base terminal of 60 microamperes peak-to-peak, which translates to a peak of 30 microamperes.
The best part? We can achieve all of this without causing any distortion in the output signal.
However it is important to note that if we apply any input signal that pushes the Base current beyond this maximum value, then things start to get tricky.
The transistor will move past point “N” and enter what is known as its “cut-off” region or it might go beyond point “M” and dive into its Saturation region. When that happens then we end up getting distortion in the output signal which shows up as what we call “clipping.”
Let us take points “N” and “M” as a reference to illustrate what is happening here. By looking at the load line, we can figure out the instantaneous values for the Collector current and the corresponding Collector-emitter voltage.
What is interesting to note is that the Collector-emitter voltage actually moves in the opposite direction, or in other words, it is in anti-phase (which means it is at –180 degrees) when compared to the collector current.
Now when we talk about the Base current denoted as Ib, if we were to increase it from 50 microamperes (μA) to 80 microamperes (μA) then we would observe a decrease in the Collector-emitter voltage, this voltage is also what we refer to as the output voltage.
Specifically this output voltage drops from a steady state of 5.8 volts down to 2.0 volts.
Moreover it is important to highlight that a single stage Common Emitter Amplifier functions as what we call an “Inverting Amplifier.”
This means that when there is an increase in the Base voltage, it results in a decrease in Vout (the output voltage). Conversely if there is a decrease in the Base voltage, Vout will actually increase.
To put it simply this means that the output signal is 180 degrees out-of-phase with the input signal.
Understanding Voltage Gain of a Common Emitter Amplifier
Next when we talk about the voltage gain of a common emitter amplifier, we are actually referring to how much the amplifier can boost an input signal.
This gain is calculated by looking at the change in the input voltage compared to the change in the output voltage of the amplifier. In simpler terms if we denote ΔVL as Vout (which is the output voltage) and ΔVB as Vin (which represents the input voltage), we can express this relationship as:
Voltage Gain = Vout/Vin = ΔVL/ΔVB = – RL/RE
Now the voltage gain can also be understood as the ratio of the signal resistance found in the Collector to that in the Emitter.
So it is not just about how much you crank up the input, but it also depends on those resistances at play.
We previously touched on a crucial point regarding how things behave at higher frequencies.
As the frequency of the alternating current (ac) signal increases, something cool happens with that bypass capacitor labeled CE.
It starts to act like a short circuit for the Emitter resistor because of its reactance characteristics.
When this occurs at those high frequencies, we can effectively say that RE (the Emitter resistor) becomes zero (RE = 0) which theoretically leads to an infinite gain.
However , hold on a second! There is a catch. The bipolar transistor itself has a small internal resistance that is built right into its Emitter region and we call this resistance r’e.
Now let us break down what this internal resistance really means.
The semiconductor material that makes up the transistor provides some resistance to current flow, and this is generally illustrated by a small resistor symbol located inside the main symbol of the transistor itself.
If you look at transistor data sheets, you will find that for a small signal bipolar transistor, this internal resistance can be calculated using a specific formula: It is essentially 25mV ÷ Ie (where 25mV represents the internal voltage drop across that Emitter junction layer).
So for our common emitter amplifier circuit that we discussed earlier, this internal resistance value will end up being equal to…
r’e = 25mV / IE = 25mV / 4.09mA = 6 Ω
The transistor’s real gain equation will be adjusted to incorporate this internal resistance after the internal emitter leg resistance is connected in series with the external emitter resistor RE. This will look like this:
Voltage Gain = -RL / (RE + r’e)
When we are dealing with low frequency signals, the total resistance that you find in the Emitter leg is simply the sum of the Emitter resistor RE and the internal resistance, referred to as r’e (RE + r’e).
However when we switch to high frequency signals something interesting happens: the bypass capacitor effectively creates a short circuit around the Emitter resistor.
This means that only the internal resistance r’e remains in play within the Emitter leg. As a result of this change we end up with a significantly higher gain.
Now if we take a look at our common emitter amplifier circuit that we discussed earlier, we can express the gain of this circuit for both low and high signal frequencies as follows:
Amplifier Gain at Low Frequencies:
Gain = -RL / (RE + r’e) = -1000 / (220 + 6) = -4.42
Amplifier Gain at High Frequencies:
Gain = -RL / r’e = -1000 / 6 = -167
Now when we are dealing with really low frequencies for the input signal, the reactance of the capacitor XC, ends up being pretty high.
Because of this high reactance, the external emitter resistance known as RE starts to play a significant role in reducing the voltage gain.
In this particular example we are looking at, it brings the voltage gain down to about 4.42.
On the flip side when we crank up the input signal frequency to very high levels, what happens is that the reactance of that same capacitor effectively shorts out RE, making it equal to zero (RE = 0).
This change allows the amplifiers voltage gain to shoot up significantly, reaching around 167 in our example.
Now let us not forget one important detail: the voltage gain we are discussing here is determined solely by the values of the Collector resistor RL and the Emitter resistance that combines both RE and r’e.
It is interesting to note that this voltage gain does not get influenced by the current gain Beta, represented as β (hFE), of the transistor itself.
So to wrap things up for our straightforward example above, let us summarize all the values we have calculated for our common emitter amplifier circuit with the following table. These values are:
| Minimum | Mean | Maximum | |
| Base Current | 20μA | 50μA | 80μA |
| Collector Current | 2.0mA | 4.8mA | 7.7mA |
| Output Voltage Swing | 2.0V | 5.8V | 9.3V |
| Amplifier Gain | -5.32 | -218 |
Conclusions
The Common Emitter Amplifier circuit features a resistor located in its Collector circuit. When current flows through this particular resistor, it generates the output voltage for the amplifier.
Now the value of this resistor is carefully selected so that at what we call the amplifier’s quiescent operating point or Q-point for short, this output voltage sits right in the middle of its load line.
Moving on to the Base of the transistor that is part of the common emitter amplifier, it gets its biasing through a setup involving two resistors arranged as a potential divider network.
This kind of biasing method is pretty standard when designing bipolar transistor amplifier circuits.
It does a fantastic job of minimizing the impact of changes in Beta (which is represented by β) by keeping the Base bias at a nice, steady voltage.
This approach to biasing really helps achieve maximum stability.
Now if we decide to throw in a resistor in the emitter leg, what happens is that the voltage gain changes to -RL/RE.
On the flip side if there is no external Emitter resistance present then the voltage gain of our amplifier does not become infinite.
This is because there exists a very tiny internal resistance known as r’e in the Emitter leg. To give you an idea of how small this internal resistance is its value can be calculated as 25mV divided by IE.
References:
Understanding the basic common emitter amplifier
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