Voltage Divider Biasing in BJT By Manas Shah ( ) E.C (Level-2)

Name: Manas Shah Enr. No: Branch: Electronics and Comm. (2nd year ) Subject: Electronic Devices and Circuits (Sem 3)

What is a Bipolar Junction Transistor and how does it work?

●Bipolar Junction Transistor (BJT) is a Semiconductor device constructed with three doped Semiconductor Regions (Base, Collector and Emitter) separated by two p-n junctions which is able to amplify or “magnify” a signal.

Below is the equivalent circuit diagram of a NPN and a PNP Bipolar Junction Transistor respectively, with corresponding voltage notations along with current directions.

Bipolar Transistor Configurations As the Bipolar Transistor is a three terminal device, there are basically three possible ways to connect it within an electronic circuit with one terminal being common to both the input and output. Common Base Configuration – has Voltage Gain but no Current Gain. Common Emitter Configuration – has both Current and Voltage Gain. Common Collector Configuration – has Current Gain but no Voltage Gain.

Common Emitter Configuration ●The common emitter configuration has the emitter terminal common to both the input and output signal. The arrangement is the same for a PNP transistor, except that the power supplies will have the opposite polarity. ●Used in this way the transistor has the advantages of a medium input impedance, medium output impedance, high voltage gain and high current gain.

Common Base Configuration ●When the base is used as the common terminal, the transistor will have a low input impedance, high output impedance, unity (or less) current gain and high voltage gain. ●This configuration also realizes the best high frequency performance, and finds dominant use in RF amplifiers and high frequency circuits.

Common Collector Configuration ●This last configuration is also commonly known as the emitter follower. This is because the input signal is applied to the base and passes out at the emitter with little loss. ●Stage properties are high input impedance, a very low output impedance, a unity voltage gain and high current gain.

What is Transistor Biasing? Transistor biasing is the controlled amount of voltage and current that must go to a transistor for it to produce the desired amplification or switching effect. Transistors must be fed the correct or appropriate levels of voltages and/or currents to their various regions in order to function properly and amplify signals to the correct level. Without appropriate transistor biasing, the transistor may not function at all or amplify very poorly, such as produce clipping of the signal or produce too low gain.

Modes of BJT Biasing 1.) Current biasing: Two resistors RC and RB are used to set the base bias. These resistors establish the initial operating region of the transistor with a fixed current bias. The transistor forward biased with a positive base bias voltage through RB. The forward base-Emitter voltage drop is 0.7 volts. Therefore the current through RB is I B = (V cc – V BE ) / I B.

2.) Feedback biasing: Fig.2 shows the transistor biasing by the use of a feedback resistor. The base bias is obtained from the collector voltage. The collector feedback ensures that the transistor is always biased in the active region. When the collector current increases, the voltage at the collector drops. This reduces the base drive which in turn reduces the collector current. This feedback configuration is ideal for transistor amplifier designs.

3.) Double Feedback Biasing: Fig.3 shows how the biasing is achieved using double feedback resistors. By using two resistors RB1 and RB2 increases the stability with respect to the variations in Beta by increasing the current flow through the base bias resistors. In this configuration, the current in RB1 is equal to 10 % of the collector current.

4.) Voltage Dividing Biasing : Fig.4 shows the Voltage divider biasing in which two resistors RB1 and RB2 are connected to the base of the transistor forming a voltage divider network. The transistor gets biases by the voltage drop across RB2. This kind of biasing configuration is used widely in amplifier circuits.

Voltage divider bias is the most popular and used way to bias a transistor. It uses a few resistors to make sure that voltage is divided and distributed into the transistor at correct levels. One resistor, the emitter resistor, RE also helps provide stability against variations in β that may exist from transistor to transistor. For the circuit here, we're going to assume that β=100 for the transistor.

The base supply voltage, VBB, is calculated by: We calculate RB below, which we will use the next calculation for IE.

Then, we calculate for the emitter current using the following formula: The collector current IC is approximately equal to the emitter current. IC ≈ IE

How Emitter Resistor, RE, Fights Against the Instability of β? The RE provides stability in gain of the emitter current of a transistor circuit. of a transistor, its gain or amplification factor, can vary by large amounts from transistor to transistor, even if they're the same exact type from the same batch. There is no way to replicate the same exact βs across transistors. Therefore, when we are designing transistor circuits where we want roughly the same gain in all of them, we must design them in a way that produces the same gain despite fluctuations in the β values. We do this by carefully choosing the emitter resistance, RE, which provides stability against differences in β. RE provides stability in gain of the output current of a transistor circuit.

DC Analysis of VDB Circuit (An Example) Consider a VDB circuit on the right. 1.Treat the capacitor as an open_circuit since its reactance is (1/jωC)=∞ for DC (ω=0) 2.Determine the open-circuit (Thevenin) voltage of the divider

3) Determine the Thevenin resistance of the divider Rth=R1||R2=2.7kΩ||27kΩ=2.45kΩ 4) Check to see if (β+1)RE>>Rth. If so VB≈Vth, 101k>>2.45k?YES, by a factor of 40+VB≈1.36V 5) Determine V E : V E = V B - V BE = =0.66V 6) Determine IE: IE=VE/RE=0.66V/1kΩ=0.66mA

7) Determine IC: 8) Find voltage across RC: V RC = I C.R C =0.65mA*10k=6.5V 9)Find V OUT(DC) : V OUT(DC) = V CC - V RC = (15-6.5)V=8.5V 10)Verify ACTIVE Region, isV SAT <V OUT(DC) <V CC ? 0.2V<8.5V<15V? YES. ∴ V OUT(DC) = 8.5V

●The significant difference is simply the presence of an additional resistance at the input. ●Input impedance: ● Output impedance: Z O =R C AC Analysis of VDB circuit

●Voltage Gain

References pdfhttp:// 09.pdf 10. divider-biasehttps:// divider-biase

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