Basic BJT Differential Amplifier Construction and Analysis

A differential amplifier is kind of amplifier which uses two mirrored amplifiers to amplify the difference between two input signal applied at each amplifier. It is a basic building used in operational amplifier(op-amps) used inside integrated circuit(IC). Besides using it inside integrated circuit, a differential amplifier can also be build with discrete BJT(Bipolar Junction Transistors) or FET(Field Effect Transistor) transistors. In this blog post, we will discuss the basic principles of the BJT differential amplifier, its construction, and its applications.

A BJT differential amplifier is used in a variety of applications, such as audio amplifiers, differential amplifier modulator, instrumentation amplifiers, voltage controlled oscillator(VCO) etc. The differential amplifier is used to amplify the difference between two input signals, while rejecting any common-mode signal that is present in both input signals. This makes the differential amplifier an important component in many circuits, as it allows for accurate measurements and efficient signal processing.

Principles of BJT Differential Amplifier

The BJT differential amplifier is a two-input circuit that amplifies the difference between two input signals, V1 and V2 applied to the base terminals of two BJTs (Bipolar Junction Transistors), which are configured in a differential pair configuration(see figure below). The basic differential pair configuration consists of two BJTs wherein the two inputs V1  and V2 are applied to the bases of the two BJTs with base resistors R1 and R2, their collectors are connected to the power supply Vcc via collector resistors R1 and R2 and their emitter are connected to a common emitter resistor RE.

Following is circuit diagram of basic BJT differential amplifier which is of called dual input balanced output differential amplifier.

Dual Input Balanced Output differential amplifier circuit diagram

dual input balanced output differential amplifier circuit diagram

There are actually 4 typologies or types of differential amplifiers which are:

(1) Dual inputs balanced output

(2) Dual inputs unbalanced output

(3) Single input balanced output

(4) Single input unbalanced output

The above circuit shown is dual input balanced output since two inputs are applied at the two bases of BJT transistors Q1 and Q2 and two outputs are taken from the collectors of the transistors. Furthermore, it is an emitter coupled differential amplifier.

The following shows the V1, V2 input waveform and the resulting waveform at the output Vout1 and Vout2.  

dual input balanced output differential amplifier circuit diagram with waveform

The operation of the BJT differential amplifier is based on the fact that the current flowing through the two BJTs is proportional to the difference between the two input voltages, V1 and V2. When the input signals are equal, the current through both BJTs is equal, and the output voltage is zero. However, when the input signals are different, the current through the BJTs is different, and this causes a voltage drop across the load resistor, resulting in an output voltage.

The operation of the BJT amplifier can be analyzed using DC and AC analysis. This gives us the operating point(Vceq and Icq) and the input, output resistances and the differential and common mode gain.

DC Analysis

Determination of Q-point collector current(Icq)

The q-point collector current(Icq) is equal to the emitter current(Ie). Thus to derivation and expression for emitter current(Ie) is equivalent to finding expression for collector current(Icq). 

Under DC condition we can ground the input V1 and follow the base to emitter loop to derive expression for the emitter current Ie.

base-emitter loop differential amplifier

 Using KVL around the base-emitter loop we have,
 
Vee= IbR1+Vbe+(2Ie)Re
Ib=Ie/beta since Ic=Ie
so,  Vee=(Ie/beta)R1+Vbe+(2Ie)Re
or, Vee=-Ie(R1/beta+2Re)+Vbe
or,Ie(R1/beta+2Re)=Vee-Vbe

that is, Ie = (Vee-Vbe)/(R1/beta+2Re)
because, 2Re>>R1/beta we can write,
Ie=(Vee-Vbe)/2Re     -------------->(1)
 
Thus the equation for the quscient point collector current(Icq) is,
 
Icq = (Vee-Vbe)/2Re   -------------->(2)

Thus the emitter current(Ie) or the q-point collector current(Icq) in differential amplifier is dependent on the emitter resistor(Re) and is independent on the collector resistor(Rc).

Determination of Q-point collector-emitter voltage(Vceq)

For transistor Q1, we have,
Vce = Vc-Ve
Since, Vcc=Vc+IcRc or Vc=Vcc-IcRc
we have,
Vce = Vcc-IcRc-Ve
Now, Vbe = Vb-Ve and neglecting  voltage drop across R1 we have Vb=0 and so,
Vbe=-Ve and so we can write,
Vce=Vcc-IcRc-(-Vbe)
or, Vce=Vcc-IcRc+Vbe
 
At q-point we have,
Vceq = Vcc+Vbe-IcqRc   ------------------>(3)
which is the expression for the collector-emitter voltage at q-point

AC Analysis

For ac analysis of the differential amplifier we can use one of the following ac model of a transistor
(a) h-parameter model
(b) re model
(c) hybrid-pi model

Here we will use h-parameter model for ac analysis of BJT differential amplifier. For ac analysis the two input signal should be equal in magnitude and 180 out of phase with respect to each other. We assume that V1=V2=Vin/2. The ac signal across the emitter resistor(Re) is zero and therefore it is short short circuited in the ac equivalent circuit and connected to the ground as shown below.

h-parameter ac equivalent circuit

Using this h-parameter circuit model we will perform ac analysis to determine the followings:
(a) Differential gain(Ad)
(b) Common mode gain(Acm)
(c) Input resistance(Ri)
(d) Output resistance(Ro)

In the circuit above applying KVL around the input loop we get,

Vin/2 = RsIb+hie*Ib
that is, Ib = Vin/(2(Rs+hie))  ------------>(4)
where Rs=R1=Rb for Q1

Applying KVL in the output loop we have,

Vout = -hfe*Ib*Rc
Using Ib from equation (4)
Vout = -hfe*Vin*Rc/(2(Rs+hie)) 
or, Vout/Vin = -hfe*Rc/(2(hie+Rs))  --------------->(5)
where Rs=Rin=R1

The -negative sign in (5) shows that Vout and Vin are 180 degree out of phase.
 
(a) Differential Gain(Ad) 

In differential mode, the two input signals are equal in magnitude but opposite in phase, the differential input voltage Vd is,
 
Vd = V1-V2=Vin/2 - (-Vin/2) = Vin
Hence the differential gain is given by,
Ad = Vout/Vd = Vout/Vin
So, 
Ad = -hfe*Rc/(2(hie+Rs))  --------------->(6)
 
This differential gain given by equation(6) is valid when the output is taken at the collector of BJT transistor Q1 or Q2 w.r.t ground. But if the output is taken between the collectors of the two transistors then the differential gain will twice the differential gain of equation(6).
 
Therefore for balanced output the differential gain is,
 
Ad = -hfe*Rc/(hie+Rs)  --------------->(7)
 
(b) Common Mode Gain(Acm)
 
In common mode differential amplifier mode, the two inputs are of same magnitude and in phase, so we have,
V1=V2=Vin
 
The common modesignal Vcm is the average of the two input signals i.e,

Vcm = (V1+V2)/2 = (Vin+Vin)/2 = (2Vin)/2 = Vin    ------------->(8)

The output voltage in common mode is given by,

Vout = Acm*Vin
or, Acm = Vout/Vin   --------------->(9)

Since we are considering matched transistor circuit we can perform ac analysis of just one transistor. AC equivalent circuit for common mode operation is shown below.

ac circuit for common mode differential amplifier

The h-parameter model for the above ac equivalent circuit above is shown below.

h-parameter model for common mode differential amplifier

In common mode, the emitter resistance is 2Re. The common mode gain Acm is given by,

Acm = Vout/Vin

The output voltage Vout is the voltage across Rc due to collector current Ic,
Vout = -Ic*Rc = -hfe*Ib*Rc    ----------------->(10)

Applying KVL in the input loop we have,
Vin = Ib*Rs+Ib*hie+2*Re(Ic+Ib)     ------------->(11)
Since Ic=hfe*Ib
Therefore, Vin=Ib*Rs + Ib*hie +2*Re(1+hfe)Ib
or, Vin = Ib(Rs+hie+2*Re(1+hfe))
So,Ib =Vin/(Rs+hie+2*Re(1+hfe))    ----------------->(12)

Substituting Ib in equation(10) we have the output voltage,
Vout = -(hfe*Rc*Vin)/(Rs+hie+2*Re(1+hfe))    ----------------->(13)
 
Therefore the common mode gain is,
 
Acm = Vout/Vin
or, Acm =  -(hfe*Rc)/(Rs+hie+2*Re(1+hfe))    ----------------->(14)

(c) Differential Input Resistance(Ri)

The differential input resistance(Ri) is the resistance between on of the input to the ground while the other input terminal is grounded. The following circuit can be used to obtain differential input resistance(Ri).
Applying KVL around the input loop we have,
 
Vin/2 = Ib*Rs +Ib*hie = Ib(Rs+hie)
or, Vin/Ib = 2(Rs+hie)
that is, Ri = Vin/Ib = 2(Rs+hie)  ---------------->(15)

(c) Output Resistance(Ro)
 
The determine the output resistance(Ro) consider the following circuit,
In this case Vin=0 and Ib=0. The output resistance is defined as the resistance measured between the output terminal to the ground. 
 
From the above circuit the output resistance is,
 
Ro = Rc   --------------->(16)

Applications of BJT Differential Amplifier

The BJT differential amplifier has many applications in electronic circuits. Some of the most common applications include:

  • Audio Amplifiers: The BJT differential amplifier is used in audio amplifiers to amplify the difference between the left and right audio signals, resulting in a stereo output.

  • Instrumentation Amplifiers: The BJT differential amplifier is used in instrumentation amplifiers to amplify small differential signals, such as those generated by sensors, and reject common-mode signals.

  • Operational Amplifiers: The BJT differential amplifier is used in operational amplifiers to provide a high input impedance and low output impedance, making them suitable for use in a variety of circuits.

Conclusion

In conclusion, a BJT differential amplifier is constructed using discrete and individual BJTs, resistors, and capacitors. Alternatively, it can be built using integrated circuits (ICs), which are available in a variety of packages, including dual-in-line (DIP) and surface-mount (SMT). The BJT differential amplifier is a versatile circuit that is used in a wide range of applications, from audio amplifiers to instrumentation amplifiers. Its ability to amplify the difference between two input signals, while rejecting any common-mode signal, makes it an important component in many circuits. With the widespread availability of discrete components and integrated circuits, it is now easier than ever to incorporate a BJT differential amplifier into your own electronic projects.

The dc analysis Q-point for dual input balanced output are:

(1)  Icq = (Vee-Vbe)/2Re

(2)  Vceq = Vcc+Vbe-IcqRc 

The ac analysis differential mode gain, common mode gain, the input and output resistance are as follows:

(1) Ad = -hfe*Rc/(hie+Rs)

(2) Acm =  -(hfe*Rc)/(Rs+hie+2*Re(1+hfe))  

(3) Ri = Vin/Ib = 2(Rs+hie)

(4) Ro = Rc



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