What is a common emitter amplifier
The common emitter amplifier
The common emitter amplifier
Chapter 4 - Bipolar Junction Transistors
At the beginning of this chapter we saw how transistors can be used as switches that operate in either their "saturation" or "shutdown" modes. In the last section we saw how transistors behave in their "active" modes, between the limits of saturation and cutoff. Since transistors are able to control current in an analog (infinitely divisible) manner, they find use as amplifiers for analog signals.
Transistor as a simple switch
One of the simpler transistor amplifier circuits to be examined previously demonstrated the switching capability of the transistor. (Picture below)NPN transistor as a simple switch.
It is known as the emitter configuration because (both the power source and the load) the emitter line is shared as a common connection point, as shown in the figure below. This isn't the only way a transistor can be used as an amplifier, as we'll see in later sections of this chapter.Common emitter amplifier: The input and output signals both have a connection to the emitter.
Before that, a small amount of solar cell current saturated a transistor and lit a lamp. Since transistors are now able to "throttle" their collector currents according to the level of the base current supplied by an input signal source, we should see that the brightness of the lamp in this circuit can be controlled by the exposure of the solar cell. If only a little light shines on the solar cell, the lamp will glow dimly. The brightness of the lamp increases the more light falls on the solar cell.
Let's say we would be interested in using the solar cell as a light intensity instrument. We want to measure the intensity of incident light with the solar cell by using its output current to control a meter movement. For this purpose, a counter movement can be connected directly to a solar cell (picture below). In fact, the simplest photometers for photographic work are designed like this.High intensity light drives the light meter directly.
While this approach is suitable for measurements with moderate light intensity, it would not work as well for measurements with low light intensity. Since the solar cell has to cover the power requirements of the meter movement, the system is necessarily limited in its sensitivity. If we assume that we have to measure very low light intensities here, we are forced to find another solution.
Transistor as an amplifier
Perhaps the most straightforward solution to this measurement problem is to use a transistor (picture below) to boost the solar cell's current so that more meters of deflection can be obtained for less incident light.The cell current must be increased for low intensity light.
The current through the counter movement in this circuit is & bgr; times the solar cell current. With a transistor β of 100, this means a substantial increase in the measurement sensitivity. It is advisable to point out that the extra energy to move the probe comes from the battery on the right side of the circuit, not from the solar cell itself. All of the current from the solar cell controls the battery current to the meter to provide a reading greater than that could provide the solar cell without assistance.
Since the transistor is a current regulating device, and because the pointer movement indications are based on the current through the movement coil, the display of the counter in this circuit should only depend on the current from the solar cell and not on the voltage provided by the battery. This means that the accuracy of the circuit is independent of the battery condition, an essential feature! All that is required from the battery is a certain minimum voltage and current output capability to fully operate the meter.
Voltage output due to current through a load resistor
Another way in which the emitter configuration can be used is to generate an output voltage that is derived from the input signal, rather than a specific output current. Let's replace the counter movement with a simple resistor and measure the voltage between the collector and emitter in figure belowThe common emitter amplifier develops a voltage output due to the current through the load resistor.
When the solar cell is dark (no current), the transistor is in shutdown mode and behaves like an open switch between the collector and emitter. This creates a maximum voltage drop between the collector and emitter for a maximum V- Exitwhich is equal to the full voltage of the battery.
At full power (maximum light irradiation) the solar cell drives the transistor into saturation mode and behaves like a closed switch between collector and emitter. The result is a minimal voltage drop between collector and emitter or almost zero output voltage. In fact, a saturated transistor can never achieve a voltage drop between collector and emitter due to the two PN junctions through which the collector current must travel. However, this "collector-emitter saturation voltage" will be quite low, around a few tenths of a volt, depending on the specific transistor used.
At light exposure values between zero and maximum solar cell output, the transistor is in its active mode and the output voltage is somewhere between zero and full battery voltage. An important property to note here regarding the emitter configuration is that the output voltage is inverted with respect to the input signal. That is, the output voltage decreases as the input signal increases. For this reason, the emitter configuration of the common emitter is called an inverting amplifier.
A quick SPICE simulation (figure below) of the circuit in figure below will verify our qualitative statements about this amplifier circuit.
|* common emitter amplifier i1 0 1 dc q1 2 1 0 mod1 r 3 2 5000 v1 3 0 dc 15 .model mod1 npn .dc i1 0 50u 2u .plot dc v (2, 0) .end|
Common emitter scheme with node numbers and corresponding SPICE netlist.Common emitter: collector voltage output opposite base current input.
At the beginning of the simulation in the figure above, where the current source (solar cell) is outputting zero current, the transistor is in shutdown mode and the full 15 volts from the battery are displayed at the amplifier output (between nodes 2 and 0). When the current of the solar cell begins to increase, the output voltage decreases proportionally until the transistor reaches saturation at 30 μA base current (3 mA collector current). Note that the output voltage curve in the graph is perfectly linear (1 volt steps from 15 volts to 1 volt) up to the saturation point, where it never reaches all zero. This is the previously mentioned effect where a saturated transistor can never achieve a voltage drop between collector and emitter of exactly zero due to internal knot effects. What we see is a sharp decrease in the output voltage from 1 volt to 0.2261 volts when the input current is 28µm. A to 30 µ A increases, and then the output voltage continues to decrease from then on (albeit in increasingly smaller steps). The lowest output voltage ever achieved in this simulation is 0.1299 volts and asymptotically approaches zero.
Transistor as an AC amplifier
So far we have seen that the transistor is used as an amplifier for DC signals. In the example of the solar cell light meter, we were interested in amplifying the DC output of the solar cell to drive a DC meter movement or to generate a DC output voltage. However, this is not the only way a transistor can be used as an amplifier. An AC amplifier is often desired for amplifying AC and voltage signals. A common application of this is in audio electronics (radios, televisions and public address systems). Before that, we saw an example of the audio output of a tuning fork that activates a transistor switch. (Image below) Let's see if we can modify this circuit to send power to a speaker rather than a lamp in FigureBelow.Transistor switch activated by audio.
In the original circuit, a full-wave bridge rectifier was used to convert the microphone's AC output signal to DC voltage to drive the input of the transistor. All we cared about here was turning on the lamp with a beep from the microphone, and this arrangement was sufficient for the purpose. But now we actually want to reproduce the AC signal and drive a loudspeaker. This means that we can no longer correct the output of the microphone as we need an undistorted AC signal to run the transistor! Let's remove the bridge rectifier and replace the lamp with a speaker:Common emitter amplifier drives speakers with audio frequency signal.
Since the microphone can generate voltages that exceed the forward voltage drop of the base-emitter PN (diode) junction, I put a resistor in series with the microphone. Let's simulate the circuit in FigureBelow using SPICE. The netlist is included in (figure below)SPICE version of the traditional Emitter audio amplifier.
|Common emitter amplifier vinput 1 0 sin (0 1, 5 2000 0 0) r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 8 v1 4 0 dc 15 .model mod1 npn .tran 0, 02 m 0, 74 m .plot tran v (1, 0) i (v1) .end|
Signal cut off at collector due to lack of DC base bias.
The simulation records both the input voltage (an AC signal of 1.5 volts peak amplitude and 2000 Hz frequency) and the current through the 15 volt battery (the same is the current through the speaker). What we see here is a full AC sine wave that changes in both positive and negative directions and a half-wave output current waveform that only pulsates in one direction. If we were to actually run a speaker on this waveform, the sound produced would be terribly distorted.
What's wrong with the circuit "// www.allaboutcircuits.com/vector-lectures/diode-characteristics-circuits/"> served the current source model in FigureBelow.The model shows that the base current flows in one direction.
The collector current is controlled or regulated by the constant current mechanism according to the speed set by the current through the base-emitter diode. Note that both current paths through the transistor are monodirectional: only one direction! Despite our intent to use the transistor to amplify an AC signal, it is essentially a DC device that can handle currents in a single direction. We can apply an AC input signal between the base and emitter, but electrons cannot flow in this circuit during the part of the cycle that reverse biases the base-emitter diode junction. Therefore, the transistor remains in shutdown mode during this part of the cycle. It only turns on in its active mode when the input voltage is of the correct polarity to forward feed the base-emitter diode and only when that voltage is high enough to overcome the diode's forward voltage drop. Remember that bipolar transistors are current controlled devices: they regulate collector current based on the presence of base-emitter current, not base-emitter voltage.
The only way to get the transistor to reproduce the entire waveform as a current through the speaker is to keep the transistor in its active mode all the time. This means that we need to maintain current through the base throughout the waveform cycle of the input signal. Consequently, the base-emitter-diode junction must always remain forward-biased. Fortunately, this can be achieved with a DC bias added to the input signal. By placing sufficient DC voltage in series with the AC signal source, the forward bias can be maintained at all points during the wave cycle. (Picture below)V. preload keeps transistor in the active region.
|Common emitter amplifier vinput 1 5 sin (0 1, 5 2000 0 0) vbias 5 0 dc 2, 3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 8 v1 4 0 dc 15 .model mod1 npn .tran 0 , 02m 0, 78m .plot tran v (1, 0) i (v1) .end|
Undistorted output current I (v (1) due to Vbias)
With the 2.3 volt bias source in place, the transistor will remain in its active mode throughout the cycle of the wave and faithfully reproduce the waveform on the speaker. (Figure above) Note that the input voltage (measured between nodes 1 and 0) fluctuates between about 0.8 volts and 3.8 volts, a peak-to-peak voltage of 3 volts exactly as expected (source voltage = 1.5 Volts peak). The output current (loudspeaker) varies between zero and almost 300 mA, 180 ° out of phase with the input signal (microphone).
The figure in figure below shows another view of the same circuit, this time with some oscilloscopes ("scope meters") connected at crucial points to display all relevant signals.The entrance is biased upwards at the base. The output is inverted.
The need to bias a transistor amplifier circuit in order to obtain full waveform reproduction is an important consideration. A separate section of this chapter will be devoted solely to the subject's pre-tensioning and pre-tensioning techniques. For now, it is enough to understand that biasing may be necessary for proper voltage and current output from the amplifier.
Now that we have a working amplifier circuit we can study the voltage, current, and power gains. The generic transistor used in these SPICE analyzes has a & bgr; out of 100, as indicated by the short transistor statistic expression in the text output in the table below (these statistics have been cut from the last two analyzes for brevity).BJT SPICE model parameters. type npn is 1.00E-16 bf 100,000 nf 1,000 br 1,000 no 1,000
β is listed under the abbreviation "bf", which actually stands for " b eta, f orward ". If we wanted to insert our own β-ratio for an analysis, we could have done this in the .model line of the SPICE netlist.
Since β is the ratio of collector current to base current and our load is connected in series to the collector terminal of the transistor and our source is connected in series to the base, the ratio of output current to input current is beta. Thus, our current gain for this exemplary amplifier is 100 or 40 dB.
The voltage gain is a little more complicated than the current gain for this circuit. As always, voltage gain is defined as the ratio of the output voltage divided by the input voltage. To determine this experimentally, we modify our final SPICE analysis to plot the output voltage rather than the output current so that we have two voltage curves to compare in the figure below.
|Common emitter amplifier vinput 1 5 sin (0 1, 5 2000 0 0) vbias 5 0 dc 2, 3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 8 v1 4 0 dc 15 .model mod1 npn .tran 0 , 02m 0, 78m .plot tran v (1, 0) v (3) .end|
V (3), the output voltage across R spkr, compared to the entrance.
Plotted on the same scale (from 0 to 4 volts) we see that the output waveform in Figure above has a smaller peak-to-peak amplitude than the input waveform, in addition to a lower bias, not increased by 0 volts as the input. Since the voltage gain for an AC amplifier is defined by the ratio of the AC amplitudes, we can ignore any DC bias that separates the two waveforms. Even so, the input waveform is still larger than the output, which tells us that the voltage gain is less than 1 (a negative dB value).
To be honest, this low voltage gain is not characteristic of all common emitter amplifiers. This is a consequence of the large differences between the input and load resistances. Our input resistance (R 1 ) is 1000 Ω here, while the load (loudspeaker) is only 8 Ω. Since the current gain of this amplifier is determined solely by the β of the transistor, and since this β number is fixed, the current gain for this amplifier does not change with variations in either of these two resistors. However, the voltage gain depends on these resistances.If we change the load resistance to get a larger value, a proportionally larger voltage will drop for its range of load currents, resulting in a larger output waveform. Let's try another simulation, this time with only a 30 Ω load in the figure below instead of an 8 Ω load.
|common emitter amplifier vinput 1 5 sin (0 1, 5 2000 0 0) vbias 5 0 dc 2, 3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .tran 0 , 02m 0, 78m .plot tran v (1, 0) v (3) .end|
Becomes r spkr increased to 30 Ω, the output voltage increases.
This time, the output voltage waveform in the figure is significantly larger than the input waveform. If we look closely we can see that the output waveform is between 0 and about 9 volts: about 3 times the amplitude of the input voltage.
We can do another computer analysis of this circuit, this time instructing SPICE to analyze it from an AC point of view, giving them peak voltage values for input and output instead of a time-based representation of the waveforms. (Table below)
SPICE netlist for printing AC input and output voltages.common emitter amplifier vinput 1 5 ac 1, 5 vbias 5 0 dc 2, 3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .ac lin 1 2000 2000 .print ac v (1, 0) v (4, 3) .end freq v (1) v (4, 3) 2, 000E + 03 1, 500E + 00 4, 418E + 00
Peak input and output voltage measurements show an input of 1.5 volts and an output of 4.418 volts. This gives a voltage gain ratio of 2. 9453 (4.418 V / 1.5 V) or 9.3827 dB.
Since the push-pull amplifier's current gain is set by β, and since the input and output voltages are equal to the input and output currents multiplied by their respective resistances, we can derive an equation for the approximate voltage gain:
As you can see, the predicted results for voltage gain are pretty close to the simulated results. With perfectly linear transistor behavior, the two groups of numbers would match exactly. SPICE adequately takes into account the many "quirks" of the bipolar transistor function in its analysis, hence the slight mismatch in voltage gain based on the SPICE output.
These voltage gains stay the same regardless of where we measure the output voltage in the circuit: across the collector and emitter, or across the series load resistance as we did in the last analysis. The amount of output voltage change for a given amount of input voltage remains the same. Consider the following two SPICE analyzes as evidence of this. The first simulation in FigureBelow is time based to get a graphical representation of the input and output voltages. You will find that the two signals turn 180 ° are out of phase. The second simulation in TableBelow is an AC analysis to provide simple peak voltage values for input and output.
|common emitter amplifier vinput 1 5 sin (0 1, 5 2000 0 0) vbias 5 0 dc 2, 3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .tran 0 , 02m 0, 74m .plot tran v (1, 0) v (3, 0) .end|
Common emitter amplifier shows a voltage gain with R. spkr = 30Ω
SPICE netlist for AC analysiscommon emitter amplifier vinput 1 5 ac 1, 5 vbias 5 0 dc 2, 3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .ac lin 1 2000 2000 .print ac v (1, 0) v (3, 0) .end freq v (1) v (3) 2, 000E + 03 1, 500E + 00 4, 418E + 00
We still have a peak output voltage of 4.418 volts with a peak input voltage of 1.5 volts. The only difference to the last simulations is the phase of the output voltage.
So far, the example circuits shown in this section have all used NPN transistors. PNP transistors can be used in any amplifier configuration just as well as NPN, as long as the correct polarity and current direction are maintained, and the emitter amplifier is no exception. The output inversion and gain of a PNP transistor amplifier are the same as its NPN counterpart, only the battery polarities are different. (Picture below)PNP version of the common emitter amplifier.
- Emitter-transistor amplifiers are so called because the input and output voltage points share the emitter line of the transistor and do not take power supplies into account.
- Transistors are essentially DC devices: they cannot directly handle voltages or currents that reverse direction. In order to make them amplify AC signals, the input signal must be biased with a DC voltage in order to keep the transistor in its active mode during the entire wave cycle. This is known as bias.
- If the output voltage is measured between the emitter and collector of an amplifier with a common emitter, it is around 180 ° out of phase with the input voltage waveform. Thus, the emitter with a common emitter is called an inverting amplifier circuit.
- The current gain of an emitter transistor transistor with the load connected in series with the collector is β. The voltage gain of an emitter transistor with emitter is given here as an approximation:
- Where "R out "the resistor in series with the collector, and" R in "is the resistor connected in series with the base.
- What are some famous mexican artists
- How Do I Buy Medicare Supplemental Insurance
- What does Tanupriya mean
- What is a CC camera
- What is CTRL E in Excel
- How useful is Oracle as a graduate
- Where was the Galileo telescope invented?
- How great is the love you have for your partner
- What are some picnic spots near Siliguri
- What are the best screenshots of 2020
- When do fat burning pills start to work
- What most shockingly celebrities have become criminals
- Are Tesla cars overrated
- How are trends recognized at an early stage
- Quantum physics is now being applied in practice
- Love to wear revealing clothing
- Sprinkle perfume every day
- How common is identity theft in India
- One should use water softeners
- How is small cell lung cancer diagnosed
- How do you monetize home comedy videos
- Are there current sales of men's clothing
- Why was the Ottoman Empire renamed Turkey
- What is the GRE requirement for NYU