Frequency Response of Common Emitter Amplifier
To study the frequency response of Common Emitter Amplifier and calculate its Bandwidth.
|1||Transistor BC107||1(One) No.|
|2||Resistors (100K%u2126, 10K%u2126, 1K%u2126)||1(One) No. Each|
|3||Resistors (2.2 K%u2126)||2(Two) No.|
|4||Capacitors (10µF)||2(Two) No.|
|5||Capacitors (100µF)||1(One) No.|
|6||Bread board||1(One) No.|
|1||Dual DC Regulated Power supply (0 – 30 V)||1(One) No.|
|2||Cathode Ray Oscilloscope (CRO) (0-20MHz)||1(One) No.|
|3||Function Generator (0-1MHz)||1(One) No.|
|4||Connecting wires (Single Strand)|
The common emitter configuration is widely used as a basic amplifier as it has both voltage and current amplification.
Resistors R1 and R2 form a voltage divider across the base of the transistor. The function of this network is to provide necessary bias condition and ensure that emitter-base junction is operating in the proper region.
In order to operate transistor as an amplifier, biasing is done in such a way that the operating point is in the active region. For an amplifier the Q-point is placed so that the load line is bisected. Therefore, in practical design VCE is always set to VCC/2. This will confirm that the Q-point always swings within the active region. This limitation can be explained by maximum signal handling capacity. For the maximum input signal, output is produced without any distortion and clipping.
The Bypass Capacitor:
The emitter resistor RE is required to obtain the DC quiescent point stability. However the inclusion of RE in the circuit causes a decrease in amplification at higher frequencies. In order to avoid such a condition, it is bypassed by a capacitor so that it acts as a short circuit for AC and contributes stability for DC quiescent condition. Hence capacitor is connected in parallel with emitter resistance.
The Input/ Output Coupling (or Blocking) Capacitor: An amplifier amplifies the given AC signal. In order to have noiseless transmission of a signal (without DC), it is necessary to block DC i.e. the direct current should not enter the amplifier or load. This is usually accomplished by inserting a coupling capacitor between two stages.
XCC << ( Ri hie )
CC - Output Coupling Capacitor
CB - Input Coupling Capacitor
Frequency response of Common Emitter Amplifier:
Emitter bypass capacitors are used to short circuit the emitter resistor and thus increases the gain at high frequency. The coupling and bypass capacitors cause the fall of the signal in the low frequency response of the amplifier because their impedance becomes large at low frequencies. The stray capacitances are effectively open circuits.
In the mid frequency range large capacitors are effectively short circuits and the stray capacitors are open circuits, so that no capacitance appears in the mid frequency range. Hence the mid band frequency gain is maximum.
At the high frequencies, the bypass and coupling capacitors are replaced by short circuits. The stray capacitors and the transistor determine the response.
Characteristics of CE Amplifier:
- Large current gain.
- Large voltage gain.
- Large power gain.
- Current and voltage phase shift of 1800.
- Moderated output resistance.
- Connect the circuit as shown in the circuit diagram.
- Set source voltage VS = 50mV (say) at 1 KHz frequency using the function generator. Observe the phase difference between input and output by giving these two signals to the dual channels of CRO.
- Keeping input voltage constant, vary the frequency from 50 Hz to 1 MHz in regular steps and note down the corresponding output voltage. Calculate gain in dB as shown in the tabular column.
- Plot the graph: gain (dB) verses Frequency on a semi log graph sheet.
- Calculate the 3-dB bandwidth from the frequency response.
(a) The Input & Output Waveforms at 1 KHz
(b) Frequency Response Curve
In the usual application, mid band frequency range are defined as those frequencies at which the response has fallen to 3dB below the maximum gain (|A| max). These are shown as fL and fH and are called as the 3dB frequencies (Lower and Upper Cut-Off Frequencies respectively). The difference between higher cut-off and lower cut-off frequency is referred to as bandwidth (fH - fL).
Fig: Frequency Response Curve
Calculations from the graph
Bandwidth = fH – fL (in Hz)
VS = 50mV
|Frequency||Vo(Volts)||Gain = Vo/Vs||Gain(dB) = 20 log(Vo/Vs)|
Common Emitter Amplifier is studied and its Bandwidth is calculated.
- Maximum Gain ( Amax ) = ___________ dB
- 3dB Gain = ___________ dB
- 3dB Lower cut-off frequency, fL = ___________ Hz
- 3dB Upper cut-off frequency, fH = ___________ Hz
- 3dB Bandwidth ( fH - fL ) = ___________ Hz
Outcomes: Students are able to
1. Calculate the Bandwidth of BJT Common Emitter amplifier.
1. What is the equation for voltage gain?
2. What is cut off frequency? What is lower 3dB and upper 3dB cut off frequency?
Ans: In electronics, cutoff frequency or corner frequency is the frequency either above or below which the power output of a circuit, such as a line, amplifier, or electronic filter has fallen to a given proportion of the power in the pass band. Most frequently this proportion is one half the pass band power, also referred to as the 3 dB point since a fall of 3 dB corresponds approximately to half power. As a voltage ratio this is a fall to of the pass band voltage
3. What are the applications of CE amplifier?
Ans: Low frequency voltage amplifier, radio frequency circuits and low-noise amplifiers
4. What is active region?
Ans: The active region of a transistor is when the transistor has sufficient base current to turn the transistor on and for a larger current to flow from emitter to collector. This is the region where the transistor is on and fully operating. In this region JE in forward bias and JC in reverse bias and transistor works as an amplifier
5. What is Bandwidth of an amplifier?
Ans: Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. It is typically measured in hertz, and may sometimes refer to passband bandwidth, sometimes to baseband bandwidth, depending on context. Passband bandwidth is the difference between the upper and lower cutoff frequencies of, for example, a bandpass filter, a communication channel, or a signal spectrum. In case of a low-pass filter or baseband signal, the bandwidth is equal to its upper cutoff frequency.
6. What is the importance of gain bandwidth product?
Ans: This quantity is commonly specified for operational amplifiers, and allows circuit designers to determine the maximum gain that can be extracted from the device for a given frequency.
7. Draw h parameter equivalent circuit of CE amplifier.
8. What is the importance of coupling capacitors in CE amplifier?
Ans: Input Coupling capacitor couples the signal to base of the of transistor , It blocks any DC component present the signal and passes only a.c signal for amplification.Output Coupling capacitor couples the output signal to the load or to the next stage of the amplifier , It blocks any DC component present the signal and passes only a.c part of the amplified signal.
9. What is the importance of emitter by pass capacitor?
Ans: Emitter bypass capacitor provide a low reactance path to the amplified ac signal, if it is not connected in parallel with RE , the aplified ac signal passing through RE will cause more voltage drop across it.
10. What type of feedback is used in CE amplifier?
Ans: Negative feedback is used in CE amplifier.
11. What are the various types of biasing a Transistor?
Ans: Biasing methods generally which we use are
- Fixed bias
- Collector feedback bias
- Emitter feedback bias
- Collector - Emitter feedback bias
- Voltage divider bias.
12. What is Q point of operation of the transistor? What is the region of operation of the transistor when it is working as an amplifier?
Ans: When the transistor is biased, we establish a certain current and voltage conditions for the transistor.These conditions are known as operating conditions or d.c operating point or Quiescent point (Q-point) of the transistor.
Wen transistor is used as an amplifier, the Q-point should be selected at the centre of the DC load line to prevent distortion.
13. Why frequency response of the amplifier is drawn on semi-log scale graph?
Ans: The frequency response of a given frequency dependent circuit can be displayed as a graphical sketch of magnitude (gain) against frequency (ƒ). The horizontal frequency axis is usually plotted on a logarithmic scale while the vertical axis representing the voltage output or gain, is usually drawn as a linear scale in decimal divisions. Since a systems gain can be both positive or negative, the y-axis can therefore have both positive and negative values.
14. If Q point is not properly selected, then what will be the effect on the output waveform?
Ans:If the Q-point is not properly selected , we wont get the faithfull amplification at output.The output will be clipped off at the positive peak or at the negative peak.
15. What are the typical values of the input impedance and output impendence of CE amplifier?
Ans:The input impedance is in the order of 1Kohm and the output impedance is in the order of 40Kohms.
UpdatedNov 14, 2019
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