## Introduction: How to Design Common Emitter Amplifier

Amplifiers are used to increase the voltage and current of a weak signal to desired level.There are two types of amplifiers.They are given below.

- DC to DC Amplifiers.
- AC Amplifiers.

**DC to DC Amplifier**

In DC amplifiers,if you increase the voltage of DC Signal then the current will drop.If you increase the current of DC signal,then the voltage will drop.You can't amplify both the voltage and current of DC signal at same time using DC to DC amplifiers.DC amplifiers involves capacitors for boosting operation.

**AC Amplifiers**

AC amplifiers can increase the voltage and current both at the same time.AC amplifiers is used to amplify only AC signals,It can't amplify DC signals.AC amplifiers involves transistors to increase the voltage and current of weak AC signals.AC amplifiers consists of three configurations.They are given below.

- CB Amplifier(Common Base)
- CE Amplifier(Common Emitter)
- CC Amplifier(Common Collector)

Out of three configurations we are going to see only about designing CE Amplifier. Because CE amplifier have greater efficiency in increasing voltage and current of AC signal.The common emitter amplifier is one of the most common transistor amplifier configurations.

**Transistors**

Transistors are used either as a switching or amplifying purpose.There are many types of transistors.They are given below.

- BJT(Bipolar Junction Transistor)
- MOSFET(Metal Oxide Semiconductor Field Effect Transistor)
- CMOS( Complementary Metal Oxide Semiconductor Transistor)

Out of three transistors we are going to use only BJT transistors.

## Step 1: Fixing the Q-point of Transistor

Transistor can operate in three regions.They are given below

- Cutoff Region
- Active Region
- Saturation Region

Amplification of AC signal can be only done in active region of transistor.To operate transistor in active region Q-point need to be fixed at the center of DC load line.The red line in the graph indicates DC load line.Fixing Q-point at center of DC load line gives maximum amplification.To fix Q-point at the center of DC load line, designing of CE amplifier with correct choosing of resistor and capacitor values must be done.Designing of CE amplifier and fixing Q-point is clearly explained in step 3.

## Step 2: Finding HFE of BC547 Using Multimeter

To calculate hFE value of BC547 NPN transistor,turn the multimeter knob to hFE.Insert the BC547 transistor in the blue color port.The blue color port contains two segments.One port is for NPN type of transistor and other port for PNP type transistor.While inserting the transistor keep the terminals of transistor correctly in that port.Transistor has Base,Collector and Emitter terminals.

## Step 3: Designing CE Amplifier

Let's get started with the session of designing CE Amplifier.Let us take some example values to design CE amplifier for understanding.

- Supply Voltage= 15V,Frequency=95Mhz.

**Choosing transistor**

As before, the transistor type should be chosen according to the anticipated performance requirements.Iam using NPN transistor **BC547**.In order to have greater amplification you need to choose transistor with high **β**(Beta) Value.

The **β **value is also called as **hfe** value.You can see the **hfe **at the knob of multimeter.β is called amplification factor.With high β value,the transistor can be turned ON with low base current.We can easily determine the **β**(Beta) value using multimeter.

**Calculate collector resistor**

It is necessary to determine the current flow required to adequately drive the following stage. Knowing the current flow required in the resistor, choose a collector voltage of around half the supply voltage to enable equal excursions of the signal up and down.You can choose collector current at your own to have high current at the output,but the collector current should be within the supply current.This will define the resistor value using Ohms law.

**For Example**

To find out collector resistor

I had used the supply voltage and current as 15V and 1A. I need to have 0.5A current at output that is at collector.The voltage need to be half the supply voltage.So finally V=7.5V and I=0.5A(500mA).Using Ohms law

V=IR

7.5=0.5xR

R=7.5/0.5

The collector resistor value is R=15Ω.

**Calculate the emitter resistor **

Generally a voltage of around 1 volt or 10% of the supply voltage is chosen for the emitter voltage. This gives a good level of DC stability to the circuit. Calculate the resistance from a knowledge of the collector current (effectively the same as the emitter current) and the emitter voltage that is 10% of supply voltage.

**For Example**

To determine emitter resistor

I had used the supply voltage as 15V.So the emitter voltage should be 10% of supply voltage.The emitter current should be same as collector current.So finally V= 1.5V and I=0.5A.

Using Ohms Law V=IR

1.5=0.5xR

R=1.5/0.5

The emitter resistor is R=3Ω.

**Determine base current: **

It is possible to determine the base current by dividing the collector current by β (or hfe which is essentially the same).

**For Example**

The collector current is 0.5A.The β value is 300.

Base current =Collector Current/β

Base Current=0.5/300

Base current=0.0016A

The Base current is 1.6mA.

**Determine the base voltage**

This is easy to calculate because the base voltage is simply the emitter voltage plus the base emitter junction voltage. This is taken to be 0.6 volts for silicon and 0.2 volts for germanium transistors.

**For Example**

The emitter voltage is 1.5V.The transistor taken is silicon transistor that has 0.6V base emitter junction voltage.

Base voltage= emitter voltage + 0.6

Base voltage= 1.5 + 0.6

Base voltage= 2.1V

**Determine base resistor**

The voltage required at the base is 2.1V. It can be taken approximately as 2V. Choose the ratio of R1 and R2 resistors to provide the voltage required at the base.For choosing R1 and R2 resistor use voltage divider formula.The concept of voltage divider is clearly explained in **STEP 4**.

Vout=(VsxR2/R1+R2)

**For Example**

The Resistor R1 and R2 is connected between 15V and GND.So we need to use voltage divider formula to find out actual resistor value,so that we can get 2V at the base terminal of transistor.Supply voltage is Vs.Substitute the following value in voltage divider formula. R1=1KΩ,R2=160Ω,Vs=15V.

Vout=(VsxR2/R1+R2)

Vout=(15x160/1000+160)

Vout=(2400/1160)

Vout=2V

Thus we got exactly 2V at base using voltage divider formula.

**Emitter bypass capacitor**

The gain of the circuit without a capacitor across the emitter resistor is approximately R3/R4. To increase the gain of AC signals,the emitter resistor bypass capacitor C3 is added. This should be calculated to have a reactance equal to R4 at the lowest frequency of operation.The formula to calculate bypass capacitor C3 is given below.

C=1/(2πf)Xc

**For example**

Xc is emitter resistor(RE) value,that is 3Ω.

f is the frequency of AC signal to be amplified.Iam taking the frequency value as 95Mhz.That means iam going to amplify the AC signal that has 95Mhz frequency.

Substitute the following values in C=1/(2πf)Xc formula.

π=3.14,Xc=3Ω,f=95Mhz,Mhz=10^6.

C=1/(2πf)Xc

C=1/(2x3.14x95x10^6x3)

C=5.587216x10^-10

C=558.72x10^-12

C=559x10^-12

Thus we can take approximately as 600picofarad.

C=600pF.

**Determine value of input capacitor value**

The value of the input capacitor should equal the resistance of the input circuit at the lowest frequency to give a -3dB fall at this frequency. The total impedance of the circuit will be β times R3 plus any resistance external to the circuit, i.e. the source impedance. The external resistance is often ignored as this is likely to not to affect the circuit unduly.

The formula for calculating the input capacitor value is

C=(1/2πfR)

Where

R is the resistance of input circuit.The input circuit can be a oscillator or signal generator.

f is the frequency of AC signal to be amplified.

**For Example**

Let us take the value of resistance of input circuit **R=500****Ω **for example.

Substitute R=500Ω.and f=95Mhz in the formula C=(1/2πfR).

C=(1/2x3.14x95x10^6x500)

C=3.350x10^-12

C=3.3pF.

The practical method of finding resistance of input circuit for calculating input capacitor is explained in **STEP 5.**

**Determine output capacitor value **

Again, the output capacitor is generally chosen to equal the circuit resistance at the lowest frequency of operation. The circuit resistance is the emitter follower output resistance plus the resistance of the load, i.e. the circuit following.

**For Example**

To calculate the emitter follower resistance,turn the multimeter to resistance mode.Connect the positive probe to collector terminal of BC547 transistor and connect the negative probe to ground where the emitter resistor is grounded.Note down the resistance value using multimeter.This is the method to find emitter follower resistance.

Let the resistance of the load be 1KΩ.Then the resistance to find output capacitor value is given below.

Resistance=Emitter follower resistance+Resistance of circuit following.

After obtaining the resistance value apply in the formula

C=1/2πfR

f is 95Mhz.

## Step 4: Voltage Divider Circuit

A voltage divider circuit is a very common circuit that takes a higher voltage and converts it to a lower one by using a pair of resistors. The formula for calculating the output voltage is based on Ohms Law and is shown below.

Vout=(VsxR2/R1+R2)

where:

- VS is the source voltage, measured in volts (V),
- R1 is the resistance of the 1st resistor, measured in Ohms (Ω).
- R2 is the resistance of the 2nd resistor, measured in Ohms (Ω).
- Vout is the output voltage, measured in volts (V),

## Step 5: Practical Method to Find Resistance of Input Circuit

**Finding resistance of input circuit:**

The input circuit which i used is IC555.The resistance of input circuit can be taken only at input of IC555.Multimeter is used to measure the resistance.Turn the multimeter knob in resistance mode.Connect the positive probe to VCC that is 8th pin of IC555 and the negative probe to GND that is 1st pin.Note down the resistance value .This is the method of finding input resistance.

**Note:-**

I had used IC555 as input circuit.It is a square wave oscillator.It can generate frequency upto 200Mhz.The purpose of amplifying the output frequency of this IC555 is because when you generate high frequency such as 95Mhz.The output voltage and current of AC signal decreases with increases in frequency.So this power cannot be enough to drive the following stages.So there is a need for amplification to drive the following stages.So we need AC amplifiers.

## Step 6: Simulated Result

Let us see the working of CE amplifier with the designed values in simulator.See the above image to understand the result of amplification.I had used signal generator as input circuit.Signal generator acts as input source for CE amplifier.The output AC signal from signal generator is 2V.You can notice this from above picture that is simulated result.At the output of CE amplifier,there is a multimeter connected to measure the amplified voltage.The 2V input signal is amplified to 6V at the output of CE amplifier.

## Step 7: Application

**Increasing the radio coverage of FM transmitter using CE amplifier**

I had designed FM transmitter using IC555.When i transmit signal using it,the signal only goes some 10 feet distance.After that distance my radio receiver cannot receive that transmitted signal.This is because,when you generate high frequency such as 95Mhz,the output voltage and current of AC signal that comes out from IC555 is very less that is approximately amplitude=1.2V and current=80mA.This voltage and current is not enough to transmit longer distance.So there is a need to amplify voltage and current of AC signal to extend the transmission distance.So that i used CE amplifier to extend the range.The voltage and current decreases with increase in frequency.Since iam transmitting the signal in air.The signal could fade away quickly after some distance.This is because,earth has the property of absorbing frequency.The absorption will be more to low frequency and less to high frequencies.That's why to have good communication in cellular networks Ghz frequencies are used.The amplitude(voltage) of transmitted AC signal decreases with increase in distance covered.

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## 16 Comments

One of the best instructables out there. Step by step circuit design without overcomplicated formulas.

what about capacitance .... 95MHz is way too much frequency for BJT ?

The amplification will be greater for low frequencies and lesser for high frequencies.It can amplify 95MHz frequency but the amplification rate will be less.High frequencies are very difficult to amplify.Because high frequency such as 95MHz have zero crossing in very few nano seconds so BJT will be working at high efficiency.Due to high working efficiency BJT will be heated and the Q-point will get moved from active region due to heat dissipation of BJT.So at high frequency amplification will be lower.If you maintain Q-point at active region for high frequency by minimizing heat generation you can get more amplification for high frequency such as 95MHz

Good job.

Keep it up.

Nice tutorial, thanks. Which program did you use for simulate?

Its NI Multisim simulator.You can see this tutorial for understanding.

## https://www.instructables.com/id/Intro-to-NI-Multisim-120-Circuit-Simulator/

Thanks for reply and simulator tutorial. I'll use this simulator.

good job, hope you graduate soon.

Nice transistor tutorial. This was one of the first things that we built in my electronics course in college.

Thank you