Typical Transistor Circuits

• Differential Amplifier

  Differential amplifier amplifies the difference $\triangle V=V_1-V_2$ between two voltages $V_1$ and $V_2$. Differential amplification has many applications, such as the first stage of operational amplifiers (Op-amps).

  The two transistors $T_1$ and $T_2$ in the circuit are identical with the same properties, and their emitters are connected to a current source with constant current so that $I_E=I_1+I_2$. If $I_1$ increases, $I_2$ will decrease, and vice versa. Consider these three cases:

  • When the two input voltages are the same $V_1=V_2$, i.e., $\Delta V=V_1-V_2=0$, then $I_1=I_2=I_E/2$ and the output voltage is $V_{out}=V_{CC}-I_2 R_C$, which is treated as a reference level corresponding to $\Delta V=0$.
  • When both $V_1$ and $V_2$ become higher/lower, $I_1$ and $I_2$ are driven to go higher/lower, but as their sum $I_1+I_2=I_E$ is a constant, they remain the same and so does the output $V_{out}$.
  • If $V_1$ becomes higher but $V_2$ remains the same, the following changes take place:

    $$V_1 \uparrow \Longrightarrow I_1 \uparrow \Longrightarrow I_2 \downarrow \Longrightarrow V_{out}=V_{CC}-I_2 R_c \uparrow$$ (127)

  • If $V_2$ becomes higher but $V_1$ remains the same, the following changes take place:

    $$V_1 \downarrow \Longrightarrow I_1 \downarrow \Longrightarrow I_2 \uparrow \Longrightarrow V_{out}=V_{CC}-I_2 R_c \downarrow$$ (128)

  In summary, the output only reflects the difference between the two inputs $V_1$ and $V_2$, but it remains unchanged if both inputs become higher or lower, i.e., this is a differential amplifier. The output voltage $V_{out}$ can be further amplified by the subsequent circuit.

  A simple current source is also shown in the figure. The base voltage $V_B$ of the transistor is fixed at approximately $0.7\times 3=2.1V$, so that the load current $I_L=I_{ce}$ is also approximately constant, independent of the load, i.e., the circuit can be used as a current source providing a current determined by $V_B$ but independent of the load. A better way to hold $V_B$ constant is to replace the diodes by a reverse biased Zener diode. When a zener diode is reversely biased by a voltage exceeding its breakdown voltage, the voltage drop across it, $V_B$ in the circuit, is held at the breakdown voltage, a constant value independent of any other variables in the circuit. Consequently $I_L$ is also constant.

  

• Current Mirror Circuits

  The current mirror circuit shown below is a simple current source that provides a constant current $I_L$ independent of the load $R_L$.

  

  This circuit is composed of two matching transistors $Q_1$ and $Q_2$ with identical behaviors such as the input and output characteristics and $\beta_1=\beta_2=\beta$. They are the input and output stages of the circuit, respectively. As the input, the reference current $I_{ref}$ can be determined as

  $$I_{ref}=\frac{V_{CC}-V_B}{R_C}$$ (129)

  Applying KCL to the collector of $Q_1$, and realizing $\beta\gg 1$, we also get

  $$I_{ref}=I_C+2I_B=I_C(1+2/\beta)\approx I_C$$ (130)

  • In the input stage, as $Q_1$'s collector and base are short-circuited, it behaves like a diode, in terms of the relationship between $V_B=V_C$ and $I_B=I_C/\beta$, the voltage across and current through the base-emitter PN junction:

    $$V_B=V_T\ln \left(\frac{I_B}{I_0}+1\right) \approx V_T\;\ln \left(\frac{I_B}{I_0}\right)$$ (131)

    where $I_0$ is the reverse saturation current of the BE PN-junction, $V_T$ is the thermal voltage. Transistor $Q_1$ can be therefore be considered as a current-voltage converter by which the collector current $I_C=\beta I_B$ is converted to an output voltage $V_B$, which is held constant due to negative feedback loop:

    $$V_B\uparrow \Longrightarrow I_B\uparrow \Longrightarrow (I_C=\beta I_B) \uparrow \Longrightarrow (V_C=V_B=V_{CC}-I_C R_C)\downarrow$$ (132)

    As $V_B=V_{C1}$ is solely determined by $I_{C1}\approx I_{ref}=(V_{CC}-V_B)/R_C$, $I_{B2}$ and therefore $I_{C2}=\beta I_{B2}=I_L$ will be constant independent of the load $R_L$.

  • In the output stage, as $Q_2$ and $Q_1$ are identical and $V_{B2}=V_{B1}=V_B$, we have $I_{B2}=I_{B1}=I_B$ and $I_{C2}=\beta I_B=I_{C1}=I_C$. The load current is determined by $R_C$ but independent of the load $R_L$:

    $$I_L=I_C=I_{ref}/(1+2/\beta) \approx I_{ref} =\frac{V_{CC}-V_B}{R_C}$$ (133)

    Note that the discussion above is valid only if $I_C=\beta I_B$ holds, i.e., both $Q_1$ and $Q_2$ must be working in the linear (active) region away from either the cutoff or saturation region.

  Wilson Current Mirror:

  

  Again, here transistor $Q_2$ can be considered as a current-voltage converter by which the current $I_{C2}$ through $Q_2$ is converted to the base voltage $V_B$ shared by both $Q_1$ and $Q_2$. The following negative feedback hold the load current $I_L=I_{C3}$ constant:

  $$(I_L=I_{C3})\uparrow \Longrightarrow I_{C1}=I_{C2}\uparrow \Longr... ...wnarrow \Longrightarrow I_{B3}\downarrow \Longrightarrow (I_L=I_{C3})\downarrow$$ (134)

• Darlington Transistor

  

  Darlington transistor (Darlington pair) is a compound structure composed of two transistors, of which the emitter current of the first transistor becomes the base current of of the second transistor. The main advantage of the Darlington transistor is its high current gain $\beta_{Darlington}$, which can be found by the following steps:
  

  $$I_{C1} = \beta_1 I_{B1}=\beta_1 I_{in}$$

  $$I_{E1} = I_{B2}=I_{B1}+I_{C1}=(\beta_1+1)I_{in}$$

  $$I_{C2} = \beta_2 I_{B2}=\beta_2(\beta_1+1)I_{in}$$

  $$I_{out} = I_{C1}+I_{C2}=\beta_1 I_{in}+\beta_2(\beta_1+1)I_{in} =(\beta_1\beta_2+\beta_1+\beta_2+1)I_{in}$$

  $$\beta_{Darlington} = \frac{I_{out}}{I_{in}} =\beta_1\beta_2+\beta_1+\beta_2+1\approx\beta_1\beta_2+\beta_1+\beta_2 \approx\beta_1\beta_2$$

  
  The base-emitter voltage is $V_{BE}=V_{BE1}+V_{BE1}\approx 1.4$.

• Different Classes of Amplifier

  By properly settng the DC operating point of the transistor circuit, it can be working in any one of the following modes:

  • Class A: The transistor remains conducting in the entire sinusoidal cycle (conduction angle $\theta=360^\circ$). The DC operating point is in the middle of the linear range of the transistor to minimize distortion (clipping). However, the DC power consumption is maximized even the AC sinusoidal signal is zero.

  • Class B: The transistor is conducting and amplifies the AC signal only in half of the sinusoidal cycle (conduction angle $\theta=180^\circ$), while it is turned off and consumes no energy for the other half.

  • Class AB: this is intermediate between class A and B, the two transistors are active and conducting current more than half of the time.

  • Class C: Less than half of the signal cycle is used (conduction angle $\theta<180^\circ$)

• The Push-Pull Circuits

  This circuit can be considered as a class AB amplifier that is typically used as the last stage of an amplification system, such as in an op-amp circuit, for power amplification with large current and low output resistance to drive a heavy load (small $R_L$). A push-pull circuit is composed of a pair of two transistors that work in alternation during the two half cycles of the sinusoidal signal. The circuit can be implemented in either of the following two ways:

  • The push-pull pair (one NPN, the other PNP) receives the same input signal from their bases. During the positive half cycle, the NPN transistor is conductive and drives current through the load $R_L$, while PNP transistor is cutoff; during the negative half cycle, the PNP NPN transistor is conductive and draws current from the load $R_L$, while NPN transistor is cutoff. In either polarity, the output resistance, the conducting transistor, is small.

  • The push-pull pair (both NPN) receives the input signal $180^\circ$ out of phase (e.g., from a transformer, or from the collector and emitter of the transistor in previous stage). The transistor receiving positive peak of the input is active and drives current through $R_L$, with small output resistance, while the other transistor receiving negative peak is cutoff (open-circuit). During the next half cycle, the two transistor switch roles with the conducting transistor drawing current from the load.

  • Another advantage of the push-pull circuit is its low power consumption. When the input AC signal is zero, both transistors are close to cut-off, conducting little current and therefore consuming little energy. This can be compared with the class A transistor amplifier where the DC operating point is in the middle of the linear region, i.e., $I_C=I_{sc}/2=V_{CC}/2R_C$, $V_{CE}=V_{oc}/2=V_{CC}/2$, and the power consumption is $P=I_CV_{CE}=V^2_{CC}/2R_C$, even when the AC signal is zero.

  

• Oscillators

  An oscillator is a circuit that receives no input but generates a sinusoidal output at a desired frequency. A typical oscillator circuit is based on an active component (a transistor or an op-amp) with positive feedback and an LC circuit (tank circuit). Initially trigged by switching on the circuit, the LC circuit starts to resonate at frequency $\omega_0=1/\sqrt{LC}$, and the active component with positive feedback compensates for the attenuation due to the inevitable resistance in the circuit and keeps the oscillation going.

  Specifically, the Hartley and Colpitts oscillators are two typical oscillation circuits. In either cases, a transistor amplifier is used to receive positive feedback taken from the LC circuit as a collector impedance $Z_c$, which is maximized at the resonant frequency, thereby the voltage gain of this circuit is also maximized. A fraction of the sinusoidal at the collector is positively fed back to the emitter to prevent attenuation.

  • Hartley Oscillator The feedback signal is taken from the point between $L_1$ and $L_2$ in series. The resonant frequency is $\omega_0=1/\sqrt{C(L_1+L_2)}$.

    

  • Colpitts Oscillator The feedback signal is taken from the point between $C_1$ and $C_2$ in series. The resonant frequency is $\omega_0=1/\sqrt{LC_1C_2/(C_1+C_2)}$.

  In both circuits, the feedback is a fraction of the output $v_c=v_{out}$

  $$v_{feedback}=v_c\frac{Z_{L_1}}{Z_{L_1}+{Z_{L_2}}}=v_c\frac{L_1}{L... ...\;\;\; v_{feedback}=v_c\frac{Z_{C_2}}{Z_{C_2}+{Z_{C_3}}}=v_c\frac{C_3}{C_2+C_3}$$ (135)

  and it is then sent to the emitter of the transistor, which is in phase with the collector connected to the LC circuit, i.e., the feedback is indeed positive:

  $$v_c\uparrow \Longrightarrow v_{feedback} \uparrow \Longrightarrow... ...htarrow i_b\downarrow \Longrightarrow i_c\downarrow \Longrightarrow v_c\uparrow$$ (136)

• Frequency Mixer

  When a transistor is used for amplification, its DC operating point of a type A amplifier is typically set in the middle of the load line to maximize the linear dynamic range and thereby minimize the signal distortion (by avoiding the nonlinear region of the transistor circuit).

  However, in some applications, the nonlinear behavior of the transistor circuit is taken advantage of, such as in a frequency mixer, used for converting all radio frequencies $\omega_{RF}$ of different radio/TV broaccast channels to an intermediate frequency $\omega_{IF}$, so that the amplification circuit of the receiver can be specialized for this intermediate frequency, instead of a wide range of all possible broadcast frequencies. In radio reception, $\omega_{IF}=455$ KHz for AM (535-1605 KHz) and $\omega_{IF}=10.7$ MHz for FM (88-108 MHz). This method is called the super-heterodyne reception which is widely used in all modern radio and TV broadcasting.

  

  The output current $I_C$ of the transistor in a frequency mixer is approximately an exponential function of the input voltage $V_{BE}$

  $$i_c=\beta i_b=\beta\, I_o\, (e^{v_{be}/V_T}-1)$$ (137)

  and in general an exponential function can be approximated by the first few terms of the Taylor series expansion:

  $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!},\;\;\;\;\; \;\;\;\;\; e^x-1=x+\frac{1}{2} x^2+\frac{1}{3!} x^3+\cdots \approx x+\frac{1}{2} x^2$$ (138)

  If the input voltage $v_{be}=\cos(\omega_1t)+\cos(\omega_2t)$ contains two frequency components, then the output current can be approximated as:
  

  $$i_c \propto (\cos\omega_1t+\cos\omega_2t) +\frac{1}{2}(\cos\omega_1t+\cos\omega_2t)^2 +\cdots$$

  $$\approx \cos\omega_1t+\cos\omega_2t+\frac{1}{2}\left(\cos^2\omega_1t+2\cos\omega_1t\;\cos\omega_2t +\cos^2\omega_2t\right)$$

  $$= \cos\omega_1t+\cos\omega_2t +\frac{1}{4}(1+\cos(2\omega_1t)) +\fr... ...omega_2)t +\frac{1}{2}\cos(\omega_1-\omega_2)t +\frac{1}{4}(1+\cos(2\omega_2t))$$ (139)

  
  where we have used the trigonometry identities:

  $$\cos^2\alpha=\frac{1}{2}(1+\cos 2\alpha),\;\;\;\;\; \cos\alpha\;\cos\beta=\frac{1}{2}(\cos(\alpha+\beta)+\cos(\alpha-\beta))$$ (140)

  We see that $i_c$ contains many new frequency components in addition to the two original frequencies $\omega_1$ and $\omega_2$, including $\omega_1+\omega_2$, $\omega_1-\omega_2$, $2\omega_1$, and $2\omega_2$. This transistor circuit is therefore called a frequency mixer. By properly filtering in the circuit following this mixer, one of such frequencies, such as the difference frequency $\omega_1-\omega_2$ (the “beat frequency”) is amplified, while all other frequency components are suppressed.

  Note that the specific nonlinear behavior of the circuit is not important, as the Taylor series expansion of any nonlinear function will contain constant, first and second order terms as the exponential function assumed above, and the same frequency components will result.

  The circuit diagram of a simple super-heterodyne radio receiver is shown below. Note that the first transistor is an oscillator that also receives signal from the LC tuning circuit at the base, i.e., it is also a mixer that mixes two frequencies. The next two transistors amplify frequency component the signal from the mixer. Here the frequency $\omega_{OS}$ of the local oscillator is determined by a variablecapacitor, which is adjusted jointly with the capacitor of the tuning circuit, so that the $\omega_{OS}$ of the local oscillator changes with the carrier frequency $\omega_{RF}$ (radio frequency) of the broadcast signal received by the antenna so that their difference, the intermediate frequency, is always a the same:

  $$\omega_{IF}=\omega_{OS}-\omega_{RF}=\omega_{IF}$$ (141)