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**Common Base Amplifier: **

The common base amplifier circuit is shown in The dc equivalent circuit is obtained by reducing all ac sources to zero and opening all capacitors. The dc collector current is same as I V |

These current and voltage fix the Q point. The ac equivalent circuit is obtained by reducing all dc sources to zero and shorting all coupling capacitors. r'_{e} represents the ac resistance of the diode as shown in **Fig. 2**

** **

**Fig. 3**, shows the diode curve relating I_{E} and V_{BE}. In the absence of ac signal, the transistor operates at Q point (point of intersection of load line and input characteristic). When the ac signal is applied, the emitter current and voltage also change.** **

If the ac signal is small, the points A and B are close to Q, and arc A B can be approximated by a straight line and diode appears to be a resistance given by

If the input signal is small, input voltage and current will be sinusoidal but if the input voltage is large then current will no longer be sinusoidal because of the non linearity of diode curve. The emitter current is elongated on the positive half cycle and compressed on negative half cycle. Therefore the output will also be distorted.

r'_{e} is the ratio of Î”V_{BE} and Î”I_{E} and its value depends upon the location of Q. Higher up the Q point, smaller will be the value of r'_{e} because the same change in V_{BE} produces large change in I_{E}. The slope of the curve at Q determines the value of r'_{e}. From calculation it can be proved that.

r'_{e} = 25mV / I_{E}

_{COMMON BASE AMPLIFIER}

**Proof:**

In general, the current through a diode is given by

Where q is the charge on electron, V is the drop across diode, T is the temperature and K is a constant.

On differentiating w.r.t V, we get,

The value of (q / KT) at 25Â°C is approximately 40

Therefore,

or,

To a close approximation the small changes in collector current equal the small changes in emitter current. In the ac equivalent circuit, the current â€˜i_{C}' is shown upward because if â€˜i_{e}' increases, then â€˜i_{C}' also increases in the same direction.

**Voltage gain: **

Since the ac input voltage source is connected across r'_{e}. Therefore, the ac emitter current is given by

i_{e} = V_{in} / r'_{e}

or, V_{in} = i_{e} r'_{e}

The output voltage is given by V_{out} = i_{C} (R_{C} || R_{L})

**Example-1 **

Find the voltage gain and output of the amplifier shown in **fig. 4**, if input voltage is 1.5mV.

**Fig. 4 **

**Solution: **

or, A_{V}= 56.6

and, V_{out} = 1.5 x 56.6 = 84.9 mV

**Example-2 **

Repeat example-1 if ac source has resistance R_{s} = 100 W .

**Solution: **

The ac equivalent circuit with ac source resistance is shown in fig. 5.

The emitter ac current is given by

or,

Therefore, voltage gain of the amplifier =

and, V_{out} = 1.5 x 8.71 =13.1 mV

**Small Signal CE Amplifiers:**

CE amplifiers are very popular to amplify the small signal ac. After a transistor has been biased with a Q point near the middle of a dc load line, ac source can be coupled to the base. This produces fluctuations in the base current and hence in the collector current of the same shape and frequency. The output will be enlarged sine wave of same frequency.

The amplifier is called linear if it does not change the wave shape of the signal. As long as the input signal is small, the transistor will use only a small part of the load line and the operation will be linear.

On the other hand, if the input signal is too large the fluctuations along the load line will drive the transistor into either saturation or cut off. This clips the peaks of the input and the amplifier is no longer linear.

The CE amplifier configuration is shown in **fig. 1**

The coupling capacitor (C_{C} ) passes an ac signal from one point to another. At the same time it does not allow the dc to pass through it. Hence it is also called blocking capacitor.

**Fig. 2**

For example in **fig. 2**, the ac voltage at point A is transmitted to point B. For this series reactance X_{C} should be very small compared to series resistance R_{S}. The circuit to the left of A may be a source and a series resistor or may be the Thevenin equivalent of a complex circuit. Similarly R_{L} may be the load resistance or equivalent resistance of a complex network. The current in the loop is given by

As frequency increases, decreases, and current increases until it reaches to its maximum value V_{in} / R. Therefore the capacitor couples the signal properly from A to B when X_{C}<< R. The size of the coupling capacitor depends upon the lowest frequency to be coupled. Normally, for lowest frequency X_{C} < 0.1R is taken as design rule.

The coupling capacitor acts like a switch, which is open to dc and shorted for ac.

The bypass capacitor C_{b} is similar to a coupling capacitor, except that it couples an ungrounded point to a grounded point. The C_{b} capacitor looks like a short to an ac signal and therefore emitter is said ac grounded. A bypass capacitor does not disturb the dc voltage at emitter because it looks open to dc current. As a design rule X_{Cb }< 0.1R_{E} ** **Analysis of CE amplifier:

In a transistor amplifier, the dc source sets up quiescent current and voltages. The ac source then produces fluctuations in these current and voltages. The simplest way to analyze this circuit is to split the analysis in two parts: dc analysis and ac analysis. One can use superposition theorem for analysis .

**AC & DC Equivalent Circuits:**

For dc equivalent circuit, reduce all ac voltage sources to zero and open all ac current sources and open all capacitors. With this reduced circuit shown in **fig. 3** dc current and voltages can be calculated.

For ac equivalent circuits, reduce dc voltage sources to zero and open current sources and short all capacitors. This circuit is used to calculate ac currents and voltage as shown in **fig. 4**.

The total current in any branch is the sum of dc and ac currents through that branch. The total voltage across any branch is the sum of the dc voltage and ac voltage across that branch.**Phase Inversion:** Because of the fluctuation in base current; collector current and collector voltage also swings above and below the quiescent voltage. The ac output voltage is inverted with respect to the ac input voltage, meaning it is 180Âº out of phase with input.

During the positive half cycle, base current increases causing the collector current to increase. This produces a large voltage drop across the collector resistor; therefore, the voltage output decreases and negative half cycle of output voltage is obtained. Conversely, on the negative half cycle of input voltage less collector current flows and the voltage drop across the collector resistor decreases, hence collector voltage increases. We get the positive half cycle of output voltage as shown in fig. 5

Lowest frequency.

**AC Load line: **

Consider the dc equivalent circuit fig. 1.

Assuming I_{C} = I_{CE}(approx), the output circuit voltage equation can be written as

The slop of the d.c load line is .

When considering the ac equivalent circuit, the output impedance becomes R_{C} || R_{L} which is less than (R_{C} +R_{E}).

In the absence of ac signal, this load line passes through Q point. Therefore ac load line is a line of slope (-1 / (R_{C} || R_{L}) ) passing through Q point. Therefore, the output voltage fluctuations will now be corresponding to ac load line as shown in fig. 2. Under this

condition, Q-point is not in the middle of load line, therefore Q-point is selected slightly upward, means slightly shifted to saturation side.

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