2 projects of BJT common emitter amplifiers

In this post, it is shown how to design a BJT amplifier with voltage divider on common emitter mode. Two examples of projects are presented.

To know BJT transistor operation, click on this link.

BJT operationClick here

Common emitter amplifier with NPN transistor

The first step is to choose the transistor. In this example, was chosen the BC547B.

Calculating collector resistor

The value of supply voltage $Vcc$ is 9 V and the voltage gain $A_{v}$ is -100. These values were chosen arbitrarily. You must look at characteristic curve graphic of BC547, to find the Q point. This point must be where the input AC signal can be amplified without distortion.

This is BC547 characteristic curve, used to find Q or quiescent point on active region. Source: BC547 datasheet.

The following figures show what happens with the output waveform, when quiescent point is too close to cut off and saturation regions.

Q point too close — Source: Tutorials point.

The Q point must be where $V_{CE}$ is half of power supply. In this case, $V_{CE}$ is equal to 4.5 V. Picking up the base current curve $I_{B}=200\mu A$ , collector current ( $I_{C}$ ) is approximately 48 mA. Below, the voltage gain formula for common emitter amplifiers.

$A_{v}=-\frac{R_{C}}{r_{e}}$

Equation to find $r_{e}$ value.

$r_{e}=\frac{26mV}{I_{E}}$

$I_{E}$ is emitter current, that’s the sum of base ( $I_{B}$ ) and collector ( $I_{C}$ ) currents.

$I_{E}=I_{C}+I_{B}=48m+0.2m=48.2mA$

Finding $r_{e}$ and collector resistor ( $R_{C}$ ).

$r_{e}=\frac{26m}{48.2}=0.539\Omega$

$A_{v}=\frac{-R_{C}}{r_{e}}$

$-100=\frac{-R_{C}}{0.539}$

$R_{C}=53.9\Omega$

The closest commercial value that I have for Rc is 100 Ω. But could use a 56 Ω or 75 Ω.

Calculating emitter resistor

Plotting a load line on BC547 characteristic curve, connecting Q point to axis point where $V_{CE}$ equals 9, the saturation collector current ( $I_{C(sat)}$ ) is approximately 85 mA. Calculating emitter resistor $R_{E}$ .

$I_{C(sat)}=\frac{V_{CC}}{R_{C}+R_{E}}$

$85m=\frac{9}{100+R_{E}}$

$R_{E}=5.88\Omega$

The closest available commercial value is 10 Ω.

Calculating base resistors

The transistor’s β or $h_{fe}$ can be measured with a multimeter or a LCR meter.

$R_{b2}$ value must attend the following criterion.

$R_{b2}\leq \frac{\beta R_{E}}{10}$

$R_{b2}\leq 318$

The $R_{b2}$ commercial value is 220 Ω. It’s necessary to use the voltage divider equation to find $R_{b1}$ .

$V_{B}=\frac{R_{b2}V_{CC}}{R_{b1}+R_{b2}}$

Finding $V_{B}$ .

$V_{E}=I_{E}\cdot R_{E}=48.2m\cdot 10=0.48V$

As $V_{BE}=0.7V$ ,

$V_{BE}=V_{B}-V_{E}$

$0.7=V_{B}-0.48$

$V_{B}=1.18V$

Calculating $R_{b1}$ .

$1.18=\frac{220\cdot 9}{220+R_{b1}}$

$R_{b1}=1457.9\Omega$

The commercial value is 1.5 kΩ.

Calculating capacitors

The 3 capacitors determine common emitter amplifier’s frequency response. The Ci and Co capacitors serve to filter direct current (DC) signal and allow passage of alternate current (AC).

The Ci works as high-pass filter, that blocks signals below a determined cutoff frequency, determined by the equation.

$f=\frac{1}{2\pi(R_{s}+R_{i})Ci}$

$R_{s}$ is the signal resistance. If input signal comes from a function generator, you can use a multimeter to connect with the probes of the turned down function generator and measure the resistance, which is 840 Ω.

measuring resistance of function generator

To calculate $R_{i}$ .

$R_{i}=R_{b1}\left\| R_{b2}\right\|\beta r_{e}$

$R_{b1}||R_{b2}=\frac{R_{b1}\cdot R_{b2}}{R_{b1}+R_{b2}}=\frac{1500\cdot 220}{1500+220}=191.8\Omega$

$\beta \cdot r_{e}=171.94$

$R_{i}=\frac{191.8\cdot 171.94}{191.8+171.94}=90.66\Omega$

The value of cutoff frequency f is 1 kHz, it was chosen arbitrarily. Calculating Ci.

$f=\frac{1}{2\pi (R_{s}+R_{i})Ci}$

$1000=\frac{1}{2\pi (840+90.66)Ci}$

$Ci=171nF$

The closest available commercial value is 100 nF. In this case, the cutoff frequency is 1.71 kHz. This is the formula to calculate Co.

$f=\frac{1}{2\pi (R_{o}+R_{L})Co}$

Where $R_{o}$ , is the Thévenin resistance saw by output circuit without Co and $R_{L}$ , which is the load resistance linked to Co. Since it won’t be linked to a load, $R_{L}$ can be omitted and $R_{o}=R_{C}$ . The chosen frequency was 20 kHz. Applying the equation above,

$20000=\frac{1}{2\pi 100\cdot Co}$

$Co=79.5nF$

Using the commercial value 100 nF, the frequency value is 15.91 kHz. The capacitor in parallel with $R_{E}$ serves to filter high frequency noises. The cutoff frequency formula:

$f=\frac{1}{2\pi R_{E}Ce}$

If Ce is 1 μF, the cutoff frequency is 15.9 kHz.

common emitter amplifier with NPN transistor — The complete amplifier.

A video demonstrating the operation of common emitter NPN amplifier. Due to input signal amplitude, the output signal saturates and reminds a square wave. Would have to increase Vcc or low the input signal amplitude.