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Basic BJT Differential Amplifier Construction and Analysis

A differential amplifier amplifies the difference between two input signals using mirrored amplifier stages. It is a key building block in operational amplifiers (op-amps), commonly used in integrated circuits (ICs). In addition to ICs, a differential amplifier can also be constructed with discrete components, such as Bipolar Junction Transistors (BJTs) or Field Effect Transistors (FETs). Learn more about the types of differential amplifiers and their construction methods for various applications.

dual input balanced output differential amplifier circuit diagram

A BJT differential amplifier is utilized in a range of applications, including audio amplifiers, instrumentation amplifiers, voltage-controlled oscillators (VCOs), and differential modulator circuits. Its main function is to amplify the difference between two input signals while rejecting common-mode signals, ensuring high precision and efficient signal processing. To better understand the differential and common-mode operation, explore their significance in amplifier design.

Principles of BJT Differential Amplifier

The BJT differential amplifier operates as a two-input circuit designed to amplify the difference between two signals (V1 and V2), applied to the base terminals of a pair of BJTs configured in a differential pair. Key configurations include:

For insights into specialized configurations like single-input unbalanced-output amplifiers, check out their practical uses and benefits. Additionally, improving CMRR (Common-Mode Rejection Ratio) is essential for enhancing differential amplifier performance, especially in high-precision applications.

To dive deeper into how differential amplifiers work, their role in integrated circuits, and ways to optimize their functionality, refer to trusted resources in electronics and amplifier design.

Dual Input Balanced Output differential amplifier circuit diagram

dual input balanced output differential amplifier circuit diagram

There are four main types of differential amplifiers:

  1. Dual Input, Balanced Output
  2. Dual Input, Unbalanced Output
  3. Single Input, Balanced Output
  4. Single Input, Unbalanced Output

The circuit shown is a dual input, balanced output configuration, where two input signals are applied to the bases of the BJTs (Q1 and Q2), and the outputs are taken from the collectors of these transistors. This setup is also known as an emitter-coupled differential amplifier.

The following diagrams illustrate the input waveforms V1 and V2, along with the resulting output waveforms Vout1 and Vout2.

dual input balanced output differential amplifier circuit diagram with waveform

The BJT differential amplifier operates on the principle that the current through the two BJTs is proportional to the difference between the input voltages, V1 and V2. When the input signals are equal, the current through both transistors is balanced, and the output voltage is zero. However, when the input signals differ, the current through each BJT becomes unequal, creating a voltage drop across the load resistor and producing an output voltage.

The amplifier’s performance can be analyzed through both DC and AC analysis. This analysis helps determine the operating point (Vceq and Icq), input and output resistances, and both differential and common mode gains.

DC Analysis

To find the Q-point collector current (Icq), we note that it is equal to the emitter current (Ie). Thus, deriving an expression for the emitter current (Ie) is equivalent to finding the collector current (Icq).

Under DC conditions, the input voltage V1 is grounded, and we analyze the base-emitter loop to derive an expression for the emitter current Ie.

base-emitter loop differential amplifier

Using KVL around the base-emitter loop, we have:

VEE=IBR1+VBE+(2IE)REV_{EE}

Since IB=IEβI_B = \frac{I_E}{\beta}(since IC=IEI_C = I_E), we can express the equation as:

VEE=(IEβ)R1+VBE+(2IE)REV_{EE} = \left(\frac{I_E}{\beta}\right) R_1 + V_{BE} + (2I_E) R_E

Rearranging:

VEE=IE(R1β+2RE)+VBEV_{EE}

Or:

IE(R1β+2RE)=VEEVBE

Thus:

IE=VEEVBER1β+2RE

Because 2RER1β, we can approximate:

IEVEEVBE2RE(1)

Thus, the equation for the quiescent point collector current (ICQ) is:

ICQ=VEEVBE2RE(2)

The emitter current (IE) or the Q-point collector current (ICQ) in a differential amplifier depends on the emitter resistor (RE) and is independent of the collector resistor (RC).

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

For transistor Q1Q_1, we have:

VCE=VCVEV_{CE} = V_C - V_E

Since VCC=VC+ICRC, we can write:

VCE=VCCICRCVE

Now, VBE=VBVE. Neglecting the voltage drop across R1, VB=0, so:

VBE=VE

Thus, we can rewrite:VCE=VCCICRC(VBE)

Or:VCE=VCCICRC+VBE

At the QQ-point, we have:

VCEQ=VCC+VBEICQRC(3)V_{CEQ} = V_{CC} + V_{BE} - I_{CQ} R_C \tag{3}
which is the expression for the collector-emitter voltage at q-point

AC Analysis

For the AC analysis of a differential amplifier, we can use one of the following transistor models:

  • (a) h-parameter model
  • (b) re model
  • (c) hybrid-pi model

In this case, we will use the h-parameter model for the AC analysis of the BJT differential amplifier. For AC analysis, the two input signals should be of equal magnitude and 180 degrees out of phase with respect to each other. We assume that V1=V2=Vin/2V_1 = V_2 = V_{in}/2. The AC signal across the emitter resistor (ReR_e) is zero, so it is effectively short-circuited in the AC equivalent circuit and connected to ground as shown below.


h-parameter ac equivalent circuit

Using the h-parameter circuit model, we will perform AC analysis to determine the following:

  • (a) Differential gain (AdA_d)
  • (b) Common mode gain (AcmA_{cm})
  • (c) Input resistance (RiR_i)
  • (d) Output resistance (RoR_o)

Input Loop Analysis

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

Vin2=RSIB+hieIB\frac{V_{\text{in}}}{2} = R_S I_B + h_{ie} I_B

That is:

IB=Vin2(RS+hie)(4)I_B = \frac{V_{\text{in}}}{2 (R_S + h_{ie})} \tag{4}

where RS=R1=RBR_S = R_1 = R_B for Q1Q_1.

Output Loop Analysis

Applying KVL in the output loop, we have:

Vout=hfeIBRCV_{\text{out}} = -h_{fe} I_B R_C

Using IB from equation (4), we get:

Vout=hfeVinRC2(RS+hie)V_{\text{out}} = -\frac{h_{fe} V_{\text{in}} R_C}{2 (R_S + h_{ie})}

Or:

VoutVin=hfeRC2(hie+RS)(5)\frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{h_{fe} R_C}{2 (h_{ie} + R_S)} \tag{5}

The negative sign in (5) shows that Vout and Vin are 180° out of phase.

Differential Mode Analysis

In differential mode, the two input signals are equal in magnitude but opposite in phase. The differential input voltage Vd is:

Vd=V1V2=Vin2(Vin2)=VinV_d = V_1 - V_2 = \frac{V_{\text{in}}}{2} - \left(-\frac{V_{\text{in}}}{2}\right) = V_{\text{in}}

Hence, the differential gain is given by:

Ad=VoutVd=VoutVinA_d = \frac{V_{\text{out}}}{V_d} = \frac{V_{\text{out}}}{V_{\text{in}}}

So:

Ad=hfeRC2(hie+RS)(6)A_d = -\frac{h_{fe} R_C}{2 (h_{ie} + R_S)} \tag{6}

This differential gain given by equation (6) is valid when the output is taken at the collector of BJT transistor Q1Q_1 or Q2Q_2 with respect to ground. However, if the output is taken between the collectors of the two transistors, the differential gain will be twice the differential gain of equation (6).

Therefore, for balanced output, the differential gain is:

Ad=hfeRChie+RS(7)A_d = -\frac{h_{fe} R_C}{h_{ie} + R_S} \tag{7}
 

(b) Common Mode Gain (Acm)

In common-mode operation of a differential amplifier, the two input signals are of the same magnitude and in phase. Thus, we have:

V1=V2=VinV_1 = V_2 = V_{\text{in}}

The common-mode signal VcmV_{cm} is the average of the two input signals:

Vcm=V1+V22=Vin+Vin2=2Vin2=Vin(8)V_{cm} = \frac{V_1 + V_2}{2} = \frac{V_{\text{in}} + V_{\text{in}}}{2} = \frac{2V_{\text{in}}}{2} = V_{\text{in}} \tag{8}

The output voltage in the common-mode is given by:

Vout=AcmVinV_{\text{out}} = A_{cm} V_{\text{in}}

Thus, the common-mode gain AcmA_{cm} is:

Acm=VoutVin(9)A_{cm} = \frac{V_{\text{out}}}{V_{\text{in}}} \tag{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 operation, the emitter resistance is 2Re2R_e. The common-mode gain AcmA_{cm} is defined as:

Acm=VoutVinA_{cm} = \frac{V_{\text{out}}}{V_{\text{in}}}

The output voltage Vout is the voltage across RcR_c due to the collector current Ic:

Vout=IcRc=hfeIbRc(10)V_{\text{out}} = -I_c R_c = -h_{fe} I_b R_c \tag{10}

Applying KVL in the input loop:

Vin=IbRs+Ibhie+2Re(Ic+Ib)(11)V_{\text{in}} = I_b R_s + I_b h_{ie} + 2R_e(I_c + I_b) \tag{11}

Since Ic=hfeIbI_c = h_{fe} I_b, we have:

Vin=IbRs+Ibhie+2Re(1+hfe)IbV_{\text{in}} = I_b R_s + I_b h_{ie} + 2R_e (1 + h_{fe}) I_b

Simplifying:

Vin=Ib(Rs+hie+2Re(1+hfe))V_{\text{in}} = I_b \left( R_s + h_{ie} + 2R_e(1 + h_{fe}) \right)

Thus:

Ib=VinRs+hie+2Re(1+hfe)(12)I_b = \frac{V_{\text{in}}}{R_s + h_{ie} + 2R_e(1 + h_{fe})} \tag{12}

Substituting Ib from equation (12) into equation (10), we obtain:

Vout=hfeRcVinRs+hie+2Re(1+hfe)(13)V_{\text{out}} = -\frac{h_{fe} R_c V_{\text{in}}}{R_s + h_{ie} + 2R_e(1 + h_{fe})} \tag{13}

Therefore, the common-mode gain is:

Acm=VoutVinA_{cm} = \frac{V_{\text{out}}}{V_{\text{in}}} Acm=hfeRcRs+hie+2Re(1+hfe)(14)A_{cm} = -\frac{h_{fe} R_c}{R_s + h_{ie} + 2R_e(1 + h_{fe})} \tag{14}

Differential Input Resistance (RiR_i)

The differential input resistance (RiR_i) is the resistance between one of the input terminals and ground while the other input terminal is grounded. The following circuit can be used to determine the differential input resistance (Ri).

Applying KVL around the input loop:

Vin2=IbRs+Ibhie=Ib(Rs+hie)\frac{V_{\text{in}}}{2} = I_b R_s + I_b h_{ie} = I_b (R_s + h_{ie})

Rearranging:

VinIb=2(Rs+hie)\frac{V_{\text{in}}}{I_b} = 2(R_s + h_{ie})

Thus, the differential input resistance Ri is:

Ri=VinIb=2(Rs+hie)(15)R_i = \frac{V_{\text{in}}}{I_b} = 2(R_s + h_{ie}) \tag{15}

Output Resistance (RoR_o)

To determine the output resistance RoR_o, consider the following conditions:

  1. Vin=0V_{\text{in}} = 0
  2. Ib=0I_b = 0

The output resistance is defined as the resistance measured between the output terminal and ground.

From the circuit:

Ro=Rc(16)R_o = R_c \tag{16}

Applications of BJT Differential Amplifier

BJT differential amplifiers are widely used in electronic circuits due to their ability to amplify differential signals and reject common-mode noise. Key applications include:

  1. Audio Amplifiers: Amplify differences between left and right audio signals for stereo output.
  2. Instrumentation Amplifiers: Enhance small differential signals from sensors while rejecting noise.
  3. Operational Amplifiers: Provide high input impedance and low output impedance, suitable for diverse applications.

Summary

BJT differential amplifiers can be built using discrete components or integrated circuits (ICs) in various packages, such as DIP or SMT. Their versatility makes them essential in applications ranging from audio to instrumentation. They amplify input signal differences while suppressing common-mode signals, ensuring reliable performance across many circuits.

Key Analysis Parameters

DC Analysis (Q-point):

  1. Icq=VeeVbe2ReI_{cq} = \frac{V_{ee} - V_{be}}{2R_e}
  2. Vceq=Vcc+VbeIcqRcV_{ceq} = V_{cc} + V_{be} - I_{cq} R_c

AC Analysis:

  1. Differential Mode Gain: Ad=hfeRchie+RsA_d = -\frac{h_{fe} R_c}{h_{ie} + R_s}
  2. Common Mode Gain: Acm=hfeRcRs+hie+2Re(1+hfe)A_{cm} = -\frac{h_{fe} R_c}{R_s + h_{ie} + 2R_e(1 + h_{fe})}
  3. Input Resistance: Ri=2(Rs+hie)R_i = 2(R_s + h_{ie})
  4. Output Resistance: Ro=RcR_o = R_c


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