Class A RC coupled amplifier theory and design.

This article is about a Class A RC coupled amplifier, its theory of operation, working and design. RC coupled Class A amplifier is one of the most simple and commonly used small signal amplifier. RC coupled means the input signal is coupled to the input of the amplifier through an RC network. Capacitor connected at the input forms the “C” and the collector resistance forms the “R”. To be more precise, in a multi stage amplifier design, the capacitor connecting the output of the first stage to the input of the second stage will be the “C” and the collector resistor of the transistor in the second stage will be the “R”.

Class A operation means the amplifying element ie; the transistor conducts for the full cycle of the input signal. That means the complete cycle of the input waveform will be available at the output. No part of the input information is lost in the output. Circuit diagram of a typical Class A RC coupled amplifier is given below.

RC coupled amplifier circuit diagram.

RC coupled amplifier circuit

Function of each component in the above circuit diagram is as follows. Cin is the input decoupling capacitor. It blocks any DC component that may be present in the input signal from entering the amplifier. If DC components in the input are not blocked, it may change the biasing conditions of the amplifier and this will alter the performance.

R1 and R2 are biasing resistors. Their function is to give proper forward bias to the base of the transistor. The forward base-emitter voltage of a Silicon transistor is 0.7V. That means the transistor starts conducting only after the base-emitter voltage has risen above 0.7V. If biasing is not given, all parts of the input waveform where amplitudes are below 0.7V will be lost in the output. The function of the biasing resistors are to keep the base of the transistor at a suitable voltage so that even the least amplitude parts of the input wave form are conducted.

RC is the collector resistor. Its primary job is to set the collector current of the transistor. Usually 40% of the Vcc drops across the RC.

RE is the emitter resistance. Usually 10% of Vcc is dropped across RE. RE in conjunction with RC makes sure that the operating point is located at the center of the load line.

CE is the emitter by-pass capacitor. Its job is to by-pass the ac component of the emitter current. if this capacitor is not there, the ac component of the collector current will get added to the base-emitter voltage (VBE) of the transistor and this will disrupt the biasing conditions.

Cout is the output capacitor. Its job is to de-couple the dc component of the collector current from the output. If this is not done, the dc component of the collector current will flow freely through the low impedance load and this is a waste. Also the dc component may affect the succeeding stage if any.

Q1 is the transistor itself. It is the active element that performs the task of amplification. The transistor performs this job by amplifying the small fluctuation in the collector current by a factor called amplification factor into a large fluctuation in the collector current.

RL is the load resistor. It is used to set the gain of the RC coupled amplifier.

Designing the RC coupled amplifier.

Next part shows the design of an RC coupled amplifier with following parameters. Few parameters of the amplifier has to be set previously set before designing. They are Gain(Av), collector current(IC), supply voltage(Vcc), and input frequency(F)

Here we have IC=5mA, Vcc=12V , Av=10 and F=1KHz. β of the transistor is assumed to be 100. Now lets see the design of each components.

1.RC and RE.

RC and RE must be so selected that 40% of Vcc drops across RC and 10% of Vcc drops across RE. remaining 50% drops across the transistor itself.

That means RC=0.4(Vcc/IC) and RE=0.1(VCC/IC)

Substituting the values of Vcc and IC we get

RC= 960Ω ≅ 1K 

RE= 240Ω ≅ 220Ω

2.R1 and R2

Current through R1 must be 10 times the base current. Also voltage across R2 must be equal to the sum of the VBE of the transistor and the voltage drop across emitter resistor RE. So we have,

IR1=IR2=10IB

IR1=IR2=10 (IC/β)=10(5mA/100)=50uA

VR2=VBE+VRE

VBE=0.7V and VRE=ICxRE=5mAx220Ω=1.2V

VR2=0.7V+1.2V=1.9V

 

Vcc=VR1+VR2

VR1=VCC-VR2=12-1.9=10.1V

 

therefore

R1=VR1/IR1 and R2=VR2/IR2. Substituting the values in hand we get,

R1=202.2K ≅200K

R2=38K

 

3.Cin, CE and Cout.

Cin should be so selected that its impedance should be 1/10th of the input resistance of the amplifier.

XCin = (1/10) Rin

Rin=R1//R2//(1+βre)

re is the resistance looking in to the emitter terminal of the transistor and here it is 25mV/IE

also IC≅IE

so re=25mV/5mA = 5Ω

substituting the values in the equation for Rin we get

Rin=200k//38K//(1+100×5)=200k//38K//501Ω

so Rin=492Ω

XCin =1 / 2π F Cin

From this we get Cin=1/2πFXcin   and F is 1KHz

Substituting the values we got in to the above equation we get

Cin= 3.2uF≅ 3.3uF

 

CE should be so selected  that its impedance is equal to 1/10th of the emitter resistor (RE).

XCE=1 / 10 (RE)

XCE=1 / 2π F CE

CE=1 / 2π F XCE

substituting the values we got into the above equation we get

CE=7.2uF≅ 10uF

 

 

Cout should be so selected that its impedance  is equal to 1/10th of the output resistance of the amplifier.

XCout=1/10th Rout   ; where Rout is the output resistance of the amplifier.

and we know that Rout= RC   the collector resistance.

XCout= 1 / 2π F Cout

therefore Cout= 1 / 2π F XCout

Substituting the values we got in to the above equation, we get the value of Cout as

Cout= 16.5uF ≅ 22uF

4. Load resistor RL.

Load resistor RL can be used to set the gain of the transistor. We know that the voltage gain of a common emitter transistor amplifier, Av= -(rc / re)

Av = – (rc / re)

rc = RC // RL

re = 25mV / IE

Whe know the voltage gain Av required is 10 and re is 5Ω.

Using these values in the above equations we get

RL= 52.74Ω ≅ 56Ω