Input and Output Impedance
An amplifier can be considered, in many ways, a kind of "black box", which, as shown, has two entry terminals and two output terminals. This idea provides a simple h-parameter model of the transistor, which we can use to find the DC setting point and operating parameters. In reality, one of the terminals is common between the inlet and output, which represents soil or zero volts. From the outside, these terminals have an input impedance, ZIN and an output impedance, ZOUT. The input and output impedance of an amplifier is the ratio of voltage in or out of these terminals to the current. The input impedance may depend on the source source that feeds the amplifier, while the output impedance may also vary by load impedance, RL, along the output terminals. Processed, i.e. elevated input signals are usually alternating currents (AC) that represent a load whose amplifier circuit is Z to the source. The input impedance of an amplifier can be from tens of ohm (Ohm Ω) to several thousand ohms, (kilo-ohm kΩ) for bipolar-based transistor circuits up to millions of ohms, for FET-based transistor circuits (Mega-ohm MΩ). . When an amplifier is connected to a signal source and load, the corresponding electrical properties of the amplifier circuit can be modeled as shown.
Output and Input Impedance Model
Here, VS is the signal voltage, RS is the internal resistance of the signal source, and RL is the load resistance due to the output. We can expand this idea further by looking at how the amplifier is connected to the source and load. When a load is connected to a signal source, the source "sees" the amplifier's input impedance, Zin, as a load. Similarly, the input voltage, Vin, is what the amplifier sees along the input impedance Zin. It can then be modeled as a simple voltage dividing circuit, as shown in the amplifier input.
Amplifier Input Circuit Model
The same applies to the riser's output impedance. When a load resistance is connected to the amplifier's output, the amplifier becomes the source that feeds the load. Therefore, the output voltage and impedance automatically become weld voltage and weld impedance for the load, as shown.
Amplifier Output Circuit Model
Then we can see that the input and output properties of an amplifier can both be modeled as a simple voltage dividing network. The amplifier itself can be connected in Common Emitter, Common Collector, or Common Base configurations. In this lesson, we will look at the bipolar transistor connected in a common emitter configuration seen earlier.
Common Emiterli Amplifier
The so-called classic common emitter configuration uses a potential divisive network to guide the base of my transistors. The power supply Vcc and front polaration resistors adjust the transistor working point to transmit it in advanced active mode. In the absence of signal current flow to the base, the collector current does not flow (in transistor cutting) and the voltage on the collector is the same as the supply voltage Vcc. A signal current to the base causes a current to flow in the collector resistance, creating a voltage drop on the Rc, causing the collector voltage to decrease. Then the direction of change of the jollctor voltage is the opposite of the direction of change on the base, that is, the polarity is reversed. Thus, the common emitter configuration produces a large voltage amplification and a well-defined DC voltage level by taking the output voltage along the collector, as shown by the resistance RL representing the load along the output.
Single Stage Common Emitter Amplifier
We can calculate the values of the resistors needed for the transistor to work in the middle of its linear active zone, called the sedentary point or Q point, but we may need to make a quick recall. First, let's start by making a few simple assumptions about the single-stage common emitter amplifier circuit above to define the working point of the transistor. Voltage drop during transmitter resistance, VRE = 1.5V, still current, IQ = 1mA, current gain of the NPN transistor (Beta) is 100 (β = 100), and the amplifier's corner or breakpoint frequency is given as follows: ε-3dB = 40Hz. Since the immobile current without input signal passes through the collector and transmitter of the transistor, we can say that: IC = IE = IQ = 1mA. Using the Ohm Law: When the transistor is fully positioned (saturation), voltage drop along the collector resistance will be half of the Rc Vcc – VRE and will allow maximum top-to-top output signal oscillation around the center point without trimming the output signal. Note that the amplifier's DC voltage gain can be found from –RC/RE. Also note that the voltage gain is negative in value because the output signal is inverted according to the original input signal. Since the NPN transistor is forward-sided, the Base Transmitter connection acts like an advanced diode, so the Base will be 0.7 volts more positive than the Transmitter voltage ( And + 0.7V ), so the voltage on base resistance R2 will be as follows: If two polarizing resistances have already been given, we can also use the following standard voltage divider formula to find base voltage etc. along R2. The information indicated that the stagnant current was 1mA. Thus, the transistor is polarized with a collector current of 1mA throughout the 12-volt Vcc feed. This collector current is proportional to the base current in Ic = β*Ib. The DC current gain of the transistor was given as Beta (β) 100, after which the base current flowing into the transistor will be as follows:
The DC pre-receding circuit created by the voltage dividing network of R1 and R2 determines the DC operating point. The base voltage was previously calculated at 2.2 volts, then in order to produce this voltage value at 12 volt vcc we need to create the appropriate ratio of R1 to R2.
In general, for a standard voltage divider DC pre-redefinition network of a common emitter amplifier circuit, the current flowing from the lower resistance is ten times greater than the DC current flowing to the R2, Base. Then the resistance value R2 can be calculated as follows:
The voltage falling to resistance R1 will be the feed voltage minus the base polarity voltage. In addition, if the R2 resistance carries 10 times the base current, the upper resistance of the series chain R1 must exceed the R2 current plus the actual base current of the transistors, Ib. So, as you can see, the base is 11 times the current. For a common emitter riser, the rectant Xc of the transmitter bypass capacitor is usually one-tenth of the RE value of the Transmitter resistance at the cutting frequency point (1/10). Amplifier properties gave a corner frequency of -3dB of 40Hz, then the value of the CE capacitor is calculated as follows: Now we have the values set for our common emitter amplifier circuit above, now we can look at calculating the values of the amplifier's input and output impedance, as well as the C1 and C2 coupling capacitors.
Basic Emitter Amplifier Model
The generalized formula for input impedance of any circuit is ZIN = VIN/IIN. The DC pre-redefinition circuit adjusts the DC operating "Q" point of the transistor and, as the input capacitor, acts as the C1 open circuit and prevents any DC voltage, in DC (0Hz) the input impedance (ZIN) of the circuit will be extremely high. high. However, when an AC signal is applied to the input, the characteristics of the circuit change due to the fact that capacitors act as short circuits at high frequencies and pass AC signals. Generalized formula for ac input impedance of a base-facing amplifier ZIN = REQ|| Β(RE+ re). REQ is the soil equivalent resistance (0v) of the polarl network along the base and the internal signal resistance of the re, forward-sided Emitter layer. Then, if we short-circuit the 12 volt power supply to Vcc, as it appears as a short circuit to Vcc, we can redraw the common emitter circuit above as follows: Then, when the supply voltage is short circuit, we can see that there is a series of resistances connected parallel to the transistor. By simply taking the input side of the transistor amplifier and treating the C1 capacitor as a short circuit to AC signals, we can redraw the above circuit to define the amplifier's input impedance as follows:
Inlet Impedance of the Riser
In previous Common Emitter Amplifier training, we said that the internal signal resistance of the Transmitter layer is equal to the product Ie ÷ 25mV, and that this 25mV value is the internal voltage drop and IE = IQ. Then the equivalent AC resistance value of emitter diode for our amplifier circuit is given as follows:
Transmitter Leg Signal Resistance
Here re represents a small internal resistance in series with the Transmitter. Because Ic/Ib = β, the value of the Basic impedance of transistors will be equal to β*re. If the bypass capacitor CE is not included in the amplifier design, the value is: β(RE+ re) and significantly increases the amplifier's input impedance.
Our example includes bypass capacitor CE, so the input impedance of the common Transmitter amplifier is the input impedance "seen" by the AC resource running the ZIN amplifier and is calculated as follows:
This 2.2kΩ is the input impedance facing the amplifier's input terminal. If the impedance value of the source signal is known and given as 1kΩ in our simple example above, this value can be added or collected with ZIN if desired.
However, let's assume for a minute that there is no CE-connected bypass capacitor in our circuit. What would be the amplifier's input impedance without it? Since resistance will no longer be short-circuited at high frequencies, the equation will still be the same except for the addition of RE to the β (RE+ re) part of the equation. Then the unpasteurized input impedance of our CE-free amplifier circuit will be as follows:
Bypass Capacitor-Free Input Impedance
Then, since the transmitter leg bypass capacitor decreases from 15.8kΩ to 2.2kΩ without impedance in our sample circuit, we can see that the inclusion of the transmitter leg bypass capacitor makes a big difference in the input impedance of the circuit. Then we will see that the addition of this bypass capacitor (CE) also increases amplifier gain. In our calculations to find the input impedance of the amplifier, we assumed that the capacitors in the circuit had zero impedance (Xc = 0) for AC signal currents and infinite impedance (Xc = ∞) for DC pre-redefinition currents. Since we know the bypassed input impedance of the amplifier circuit, we can use this 2.2kΩ to find the C1 value of the input coupling capacitor, which is required at the specified cutting frequency point previously given as 40Hz. Therefore:
Input Coupling Capacitor Equation
Now that we have a value for the input impedance of our single-stage common Emiter amplifier circuit above, we can obtain a similar expression for the output impedance of the amplifier.
Riser's Output Impedance
The output impedance of an amplifier can be considered impedance (or resistance) that the load sees when it "looks back" at the amplifier when the input is zero. Working on the same principle as what we do for input impedance, the generalized formula for output impedance can be given as follows: ZOUT = VCE/IC. However, the signal current flowing in collector resistance also flows in RL load resistance, as it is connected serially throughout the VCC. Then, by simply taking the output side of the transistor amplifier and treating the C2 output coupling capacitor as a short circuit to the AC signals, we can redraw the above circuit to identify the amplifier's output impedance as follows: Then we can see that the amplifier output impedance is equal to RC in parallel with the RL and gives us an output resistance: this 833Ω value is due to the fact that the load resistance depends on the transistor. If RL is omitted, the amplifier's output impedance will be equal only to RC Collector resistance. Now that we have a value for the output impedance of our amplifier circuit above, we can calculate the value of the C2 output coupling capacitor as before at the 40Hz cutting frequency point.
Output Coupling Capacitor Equation
Again, the value of the coupling capacitor C2 can be calculated by including or including load resistance RL.
Common Transmitter Voltage Gain
The voltage gain of a common emitter circuit is given as Av = ROUT/REMITTER; here ROUT represents the output impedance, as seen in the collector's leg, and REMITTER is equal to the equivalent resistance in the Transmitter leg, whether or not the bypass capacitor is connected. When the bypass capacitor is not connected to ce, (RE+ re). and only when the bypass capacitor CE is connected. Next, we can see that the inclusion of the bypass capacitor in the amplifier design makes a dramatic change in the voltage gain of our common emitter circuit Av from 0.5 to 33. It also indicates that the common emitter gain does not go away forever. external emitter resistance is short-circuited by the bypass capacitor at high frequencies, but instead the gain goes to the finite value of ROUT/re. We also saw the input impedance drop from 15.8kΩ without it to 2.2kΩ with it as earnings increased. The increase in voltage gain can be considered an advantage in most amplifier circuits at the expense of lower input impedance.
In this tutorial, we found that a common emitter amplifier input impedance can be found by short-circuiting the feed voltage and treating the voltage divider pre-redefinition circuit as parallel resistors. When looking at the divisive network (R1|| R2) The "seen" impedance is usually much less than the impedance facing directly to the transistor Base, β (RE+ re), while the AC input signal changes the bias at the Base of the transistor control, the current passes through the transistor. There are many ways to direct the transistor. Therefore, there are many practical single transistor amplifier circuits, each with its own input impedance equations and values. If you need the input impedance and welding impedance of the entire stage, then you will also need to consider Rs in series with base pre-reclaid resistors, (Rs + R1|| R2). The output impedance of a common emitter stage, if connected, can be added to the load resistance (RC|| RL) in parallel equals collector resistance, otherwise it is only RC. The voltage gain of the amplifier depends on Av, RC/RE. Transmitter bypass capacitor CE can provide an AC path to the soil for the Transmitter by short-circuiting the emitter resistance at RE high frequencies and thus leaving only the Transmitter resistance signal in the Transmitter leg circuit. The effect of this causes an increase in the voltage gain of the amplifier (from 0.5 to 33) as the signal frequency increases. However, this also has the effect of reducing the input impedance value of the amplifier from 18.5kΩ to 2.2kΩ, as shown. When this bypass capacitor is removed, the voltage gain of the amplifier decreases the Prey and the ZIN increases. One way to maintain a constant amount of gain and input impedance is to add additional resistance in series with CE to create a "split emitter" amplifier circuit, an exchange between an amplifier circuit that has not been bypassed and is completely bypassed. . Note that the addition or removal of this bypass capacitor has no effect on the amplifier output impedance. Next, we can see that the input and output impedances of an amplifier can play an important role in defining the transfer characteristics of an amplifier in terms of the relationship between output current, Ic and input current, Ib. Knowing the input impedance of an amplifier can help graphically create a series of output characteristic curves for the amplifier.