EE 461G Introduction to Electronics

Homework # 9 Due November 9, 1999

In Sedra and Smith read sections 4.7 - 4.10 and sections 5.5 and 5.7

In Sedra and Smith do problem 4.68 at the end of chapter 4 and 5.60 at the end of chapter 5.

The simplest common emitter bipolar junction transistor (BJT) amplifier biasing scheme is shown in Fig. 1. The time varying part of the input signal is omitted because at this point, we are only considering the bias point. The signal, in actual circuit operation, would be an ac signal added to the V_BB. The transfer characteristic (output as a function of the input) for this circuit can be derived to be:

At the operating point, V_BB and V_out are the DC or quiescent values of the input and output. Ideally, for a given V_BB, V_out should not vary if the temperature varies or if different transistors of the same type are used. Unfortunately the BJT's current gain b_f cannot be controlled well during manufacturing and it also varies with temperature. For the 2N2222 BJT transistor, manufacturers specify that b_f may be anywhere from 50 to 150. Thus a circuit biased correctly for one 2N2222 transistor may not be biased correctly for another 2N2222 transistor. This means there is a need for a biasing scheme that is more robust than the one shown in Fig. 1 so the BJT's quiescent operating point resilient to changes in b_f.

Insensitivity of the BJT's quiescent operating point to changes in b_f can be achieved by adding a resistor R_E into the emitter branch of the circuit as shown in the Fig. 2. Analyzing this new circuit gives

for the collector current. Note that if (b_f + 1)R_E >> R_B that the collector current can be approximated as

which is independent of b_f.

This circuit is still not entirely satisfactory because it requires two power supplies, one for V_BB and another for V_CC. The circuit in Fig. 3 remedies this. The Thevenin equivalent for the circuit consisting of V_CC, R₁, and R₂ in Fig. 3 gives the biasing circuit in Fig. 2 where V_BB = V_th and R_B = R_th. With these Thevenin equivalents substituted into the circuit, the circuit is identical to the circuit in Fig. 2 with the exception that the bias voltage V_BB is now dependent on V_CC. The V_BB voltage is now controlled by the proper choice of R₁ and R₂. This eliminates the need for a separate power supply to control V_BB.

Fig. 1 Basic common emitter Fig. 2 Basic common emitter amplifier Fig. 3 Basic common emitter amplifier biasing

amplifier biasing. with reduced b sensitivity. with reduced b sensitivity and employing

a single DC voltage.

Choosing the resistors R₁ and R₂ such that R_B << (b_f +1) R_E is equivalent to making the current through R₁ and R₂ large enough that the BJT's base current can be neglected in comparison. The base voltage is thus determined only by V_CC and the R₁, R₂ voltage divider.

Problem 1

Derive the equation for the DC load line for the BJT circuit in Fig. 3 (with R₁, R₂ and the emitter resistor R_E used for added circuit stability) in terms of the variables V_CC, R_C, and R_E . Plot the DC load line superposed with the characteristic curves (I_Cversus V_CE) for a NPN BJT (2N2222 with b_f = 150). Choose values of R_C and R_E of 1000 and 470 ohms respectively. Choose V_CC= 10 V.

Determine and indicate the quiescent operating point I_CQ and V_CEQ such that it is at the midpoint of the DC load line (to ensure maximum symmetrical output voltage swing).

Determine the value of R₁ and R₂, which will maintain the transistor at that operating point and provide bias stability in conjunction with the emitter resistor. Make I_R1 = 100 I_B with b_fequal to 150 and assume V_BEf = 0.6V.

Decrease b_f to 100 and then 10. Determine the change in the operating point (I_CQ and V_CEQ) that occurs while keeping the same resistor values.

Create a SPICE model for the circuit in Fig. 3 with the values you calculated above. The BJT model is bf = 100 and Is = 1nA.

Do a transient analysis and plot the voltage from the collector to the ground in this problem with b_f = 100 and 10. Also plot the base current. Let your program run for 1ms with tstep = 20ms.

The analysis of semiconductor circuits operating in their linear range can be accomplished using a two step analysis approach. The first step is the nonlinear DC or quiescent analysis (problem 1). The second step, is an AC incremental analysis where each circuit element is replaced by its small signal linearized model. The small signal models of common circuit elements are summarized in table 1. A simplified small signal model of the BJT is shown in Fig. 5. When doing a small signal analysis each circuit element is replaced with its small signal equivalent

Circuit element	Schematic	Small signal circuit element	Small signal schematic
Wire		Wire
Resistor		Resistor
Capacitor		Capacitor
Inductor		Inductor
DC voltage source		Short
DC current source		Open
BJT

Table 1 Summary of circuit elements and their small signal equivalents.

producing a new small signal equivalent schematic of the original circuit. Because the incremental circuit is linear, all of the linear circuit theory can be brought to bear in analyzing the small signal equivalent circuit including phasors, Fourier analysis, impulse response and superposition. Also, once the linear circuit has been obtained, approximations can be used to simplify it further. For example, capacitors can often be treated as short circuits at the frequencies of interest. To obtain the complete circuit solution, the quiescent and small signal solutions are added together. Thus, any circuit voltage of current is equal to

In the small signal model of the BJT shown in Fig. 4, r_p is the incremental resistance of the base-emitter junction, b is the ratio Di_C/Di_B at constant collector to emitter voltage, and r_o = Dv_CE / Di_C at constant base current. The resistor r_o accounts for the small slope of the I-V characteristics in the forward-active region.

Fig. 4 Basic common emitter amplifier biasing with Fig. 5 Incremental or small signal model

reduced b sensitivity and employing a single DC of a BJT.

voltage with an AC input voltage.

Problem 2

Use V_CC= 10 V, R_C = 1000W, R_E = 510W, R₁ = 3.44KW, R₂ = 975W. Neglect r_o ().

Assume the base current is related to the base emitter voltage by the ideal diode model.

. For typical base to emitter voltages the one in the ideal diode model can be neglected. Here q = the charge of an electron, K = Boltsman's constant, and T is the thermodynamic temperature. At room temperature (

) KT/q = 25mV. At what value of v_BE is the exponential 10 times greater than one. For voltages above this value (forward bias) the base current can be approximated as

. Linearize the ideal diode model for a forward biased base to emitter junction using Taylor’s theorem. Show that the incremental base current is related to the incremental base to emitter voltage by

where

Compute r_p for a b = 100 and the quiescent operating point of the circuit with the component values given above.

What must the values of C_in, C_out, and C_E be greater than, for a 10kHz signal, in order that they may be treated as short circuits?

If there is no C_E (prove to yourself that this is C_E = 0), show that the gain of the amplifier with R_s = 0 and R_L =

is approximately G = - R_C / R_E. When is this approximation valid? What is the gain for the component values above?

With R_s = 0 and R_L =

, what is the gain of the circuit when C_E shorts out R_E (CE =

Repeat the gain calculations in parts f) and g) for R_s = 1kW and R_L = 1kW. What are the maximum input voltages for these two cases to insure the BJT stays in its linear range (forward active region)?

Build a SPICE model part e) simulate it for b_f = 100 and Is = 1nA with a 10kHz sine wave Vs input. Plot Vs and Vout. Choose Vs so the amplifier's output is not distorted. What is your tend and tstep?