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Chapter 5 – Transistor Bias Circuits Page 1 out of 14
V. Transistor Bias Circuits
5.0 Introduction
- Transistors must be properly biased to work as amplifiers.
- DC biasing is used to establish a steady level of transistor current and voltage.
- That point of operation is called the dc point or quiescent point (Q-point).
- We have three possible states in a transistor: i) Cutoff: Emitter and collector diodes are OFF. ii) Active: Emitter diode is ON, collector diode is OFF. iii) Saturation: Emitter and collector diodes are OFF.
- When we want to amplify a signal, we should operate in the active region.
- In general we want to amplify voltage rather than current.
- For example, the heads on a tape player (or VCR) pick up the magnetic traces on a tape and provide a small voltage output.
- Speakers used to convert the electrical signal to an ac acoustic (sound) signal need large voltages!
- The stereo amplifier, with tape head input and speaker output, provides the voltage amplification necessary
- That is, the output voltage from amplifier is greater than the input voltage to amplifier.
- Amplification is an example of linear applications.
- Amplifiers are the most common linear devices.
- In general, we need to amplify AC signals (time varying signals).
- However, proper operation depends on its DC bias voltages and currents.
- We will not deal with cutoff and saturation.
5.1 DC Operating Point
- A dc operating point must be set so that signal variations at the input terminal are amplified and accurately reproduced at the output terminal.
- The operating point is given by IC and VCE.
- It is referred to as Q-point ( quiescent point).
A) DC Bias
- If an amplifier is not properly biased, it will go either into cutoff or saturation.
- For example, the inverting amplifier:
5.2 Voltage Divider Bias
5.3 Other Bias Methods
Homework 4 (All from chapter 4 problems) 4: 13, 15, 17, 19, 23, 25.
Chapter 5 – Transistor Bias Circuits Page 2 out of 14
- Consider the following circuit:
- Adjusting VBB to get a IB = 200 μA we get: I (^) C = βDC I (^) B = (100)(200 μA) = 20 mA VCE = VCC - I (^) C RC = 10 – (20 mA)(220 Ω) = 5.6 V
- If we adjust VBB so that I (^) B = 300 μA we get: I (^) C = βDC I (^) B = (100)(300 μA) = 30 mA VCE = VCC - I (^) C RC = 10 – (30 mA)(220 Ω) = 3.4 V
- If we adjust VBB so that I (^) B = 400 μA we get: I (^) C = βDC I (^) B = (100)(400 μA) = 40 mA VCE = VCC - I (^) C RC = 10 – (40 mA)(220 Ω) = 1.2 V
- We get three different Q points, one for each case:
- Note that when I (^) B increases, I (^) C increases and VCE decreases.
Chapter 5 – Transistor Bias Circuits Page 4 out of 14
Waveform distortion
- If we operate to close to cutoff or too close to saturation, waveform distortion may occur.
- The top or the bottom of the output wave will appear clipped.
- When the positive peak is clipped, transistor is being driven into cutoff.
- If the negative peak is clipped, transistor is going into saturation.
Example: Determine the Q-point in the following circuit. Assume βDC = 200. Find the maximum peak value of the base current for linear operation.
Chapter 5 – Transistor Bias Circuits Page 5 out of 14 Solution: The Q-point is defined by I (^) C and VCE. We find these values as follows: I (^) B = (VBB – VBE)/RB = (10 V – 0.7 V)/47 kΩ = 198 μA I (^) C = βDC I (^) B = (200)(198 μA)= 39.6 mA VCE = VCC – I (^) C RC = 20 V – 13.07V = 6.93 V
Thus the Q-point is at IC = 39.6 mA and VCE = 6.93 V. Since I (^) C(cutoff) = 0, we need to know IC(sat) to determine how much variation in collector current can occur and still maintain linear operation. I (^) C(sat)=VCC /RC = 20 V/330 Ω = 60.6 mA
The dc load line looks as follows:
Thus, before saturation is reached, I (^) C can increase, ideally, by I (^) C(sat) – I (^) CQ = 60.6 mA – 39.6 mA = 21 mA
And it can decrease by 39.6 mA before cutoff (IC = 0) is reached. The limiting value is, then, 21 mA. Thus we can see that we are closer to saturation than to cutoff.
The maximum peak variation of the base current is: I (^) b(peak) = I (^) c(peak)/βDC = 21 mA/200 = 105 μA
5.1 Voltage-divider Bias
- This is the most widely used method for biasing a transistor for linear operation.
- It uses a single voltage source and a voltage divider circuit.
- Unlike other methods, this method results in a β (almost) independent circuit.
- That is, βDC and VBE do not affect the stability of the Q-point.
- Considering the following circuit, we can see that the there are two paths between point A and ground: through R 2 and through the BE junction.
Chapter 5 – Transistor Bias Circuits Page 7 out of 14
- The resistance looking into the base of the transistor is RIN(base) = VIN/I (^) IN
- Applying KVL and assuming VBE << I (^) ERE we get VIN = VBE + I (^) ERE VIN ≈ I (^) E RE
- And since I (^) E ≈ I (^) C = βDC I (^) B VIN becomes VIN ≈ βDC I (^) B RE
- The input current is the base current, I (^) IN = I (^) B. Substituting we get RIN(base) = VIN/I (^) IN = βDC I (^) B RE/I (^) B = βDC RE
- Thus, the resistance seen by looking into the base of the transistor is the gain, βDC , times the emitter resistance, RE.
Analysis of a Voltage-Divider Bias Circuit
- Consider the following circuit.
- As shown before, RIN(base) = βDC RE. The resistance from base to ground is R 2 ||RIN(base) = R 2 || βDC RE.
- The voltage at the base is, then, VB = (R 2 ||βDC RE)/(R 1 + (R 2 ||βDC RE))VCC
- Assuming βDC RE >> R 2 (at least ten times): VB ≈ R 2 (R 1 + R 2 ) VCC
- Knowing the base voltage, we can find the emitter voltage: VE = VB – VBE.
- Ohm’s law gives us the emitter current: I (^) E = VE/RE and the rest of the values: I (^) C ≈ I (^) E VC = VCC - I (^) C RC VCE = VC - VE
Chapter 5 – Transistor Bias Circuits Page 8 out of 14
- We can also express VCE in terms of I (^) C by using KVL: VCC – I (^) C RC – I (^) ERE – VCE = 0 VCE ≈ VCC – I (^) C RC – I (^) C RE VCE ≈ VCC – I (^) C (RC + RE)
Example: Determine VCE and I (^) C in the voltage-divider biased transistor circuit shown below. Assume βDC = 100.
Solution: Determine the dc input resistance at the base: RIN(base) = βDC RE = (100)(560 Ω) = 56 kΩ
Since RIN(base) = 10R 2 , we may neglect RIN(base). Thus, VB ≈ R 2 (R 1 + R 2 ) VCC = (5.6 kΩ)(15.6 kΩ) 10 V = 3.59 V So, VE = VB – VBE = 3.59 V – 0.7 V = 2.89 V and I (^) E = VE/RE = 2.89 V/ 560 Ω = 5.16 mA
Thus, I (^) C ≈ 5.16 mA and VCE ≈ VCC – I (^) C (RC + RE) = 10V–5.16mA(1.56kΩ) = 1.95 V
Since VCE>0, the transistor is NOT in saturation.
Stability of Voltage-Divider Bias
- It was said before that the voltage-divider bias is NOT dependent on βDC.
- To prove this, let’s find a Thevenin equivalent for the circuit shown below.
Chapter 5 – Transistor Bias Circuits Page 10 out of 14
5-3 Other Bias Methods
- We will briefly study other biasing methods. i) Base Bias.
- Looking at the figure below, we can see that VCC – VR(B) – VBE = 0
- Or VCC – I (^) B RB – VBE = 0
- Which results in: I (^) B = (VCC – VBE)/RB
I (^) C = βDC (VCC – VBE )/RB
- Apply KCL at collector to get: VCC – I (^) C RC - VCE = 0 VCE = VCC – I (^) C RC
- Note that I (^) C is dependent on βDC.
- Any variation in βDC will result in a variation in I (^) C.
- This makes this bias method very unstable!!
Example: Determine how much the Q-point (I (^) C , VCE) in the circuit below will change over a temperature range where βDC increases from 85 to 100 and VBE decreases from 0.7 V to 0.6 V (both changes happen at the same time).
Chapter 5 – Transistor Bias Circuits Page 11 out of 14
Solution: Before the temperature rises, we have βDC = 85 and VBE = 0.7 V. Thus I (^) C(1) = βDC (VCC – VBE)/RB = 85(12 V – 0.7 V)/100 kΩ = 9.61 mA VCE(1) = VCC – I (^) C RC = 12 V – (9.61mA)(560 Ω) = 6.62 V
After the temperature rise, we have βDC = 100 and VBE = 0.6 V. Thus I (^) C(2) = βDC (VCC – VBE)/RB = 100(12 V – 0.6 V)/100 kΩ = 11.4 mA VCE(2) = VCC – I (^) C RC = 12 V – (11.4 mA)(560 Ω) = 5.62 V
The percent change in I (^) C and VCE after the temperature change is: %∆ I (^) C = (I (^) C(2) – I (^) C(1))/I (^) C(1) 100% = 18.6% %∆ VCE = (VCE(2) – VCE(1))/VCE(1) 100% = -15.1%
This shows that the Q-point is very dependent on βDC. Thus, the bias arrangement becomes very unstable.
Base bias is very rarely used if linear operation is required (amplification). However, it may be used in switching applications.
ii) Emitter bias.
- Uses both a positive and a negative power supply voltage.
- KVL yields: VEE + VR(B) + VBE + VR(E) = 0 VEE + I (^) B RB + VBE + I (^) ERE = 0 -VEE = I (^) B RB + VBE + I (^) ERE
- But I (^) C ≈ I (^) E and I (^) C = βDC I (^) B , thus I (^) B = I (^) E/βDC. This results in
Chapter 5 – Transistor Bias Circuits Page 13 out of 14
iii) Collector-feedback bias.
- The base resistor is connected to the collector rather than to VCC (as in the base bias arrangement).
- Collector voltage provides the bias for the base-emitter junction.
- The negative feedback creates an offsetting effect that tends to keep the Q-point stable.
- If I (^) C increases, it drops more voltage across R (^) C , thus causing VC to decrease.
- When VC decreases, there is a decrease in voltage across RB , which decreases I (^) B.
- The decrease in I (^) B produces less I (^) C which, in turn, drops less voltage across RC and thus offsets the decrease in VC.
- We can find I (^) B by: I (^) B = (VC – VBE)RB
- Assume I (^) C >>I (^) B. Thus VC ≈ VCC - I (^) C RC
I (^) C /βDC = (VCC - I (^) C RC – VBE)/RB
- Solving for I^ C I (^) C = (VCC – VBE)/(RC – RB /βDC )
VCE = VCC - I (^) C RC
Q-point Stability
- By making VCC >> VBE and RC >> RB /βDC , we are eliminating the VBE and the βDC dependency.
Chapter 5 – Transistor Bias Circuits Page 14 out of 14
Summary of Transistor Bias Circuits Biasing Circuit
Emitter bias Voltage-Divider Bias
Collector-feedback bias
Base Bias
Schematic Diagram
Circuit Recognition
A split(dual polarity) power supply and the base resistor connected to ground.
The voltage divider in the base circuit.
The base resistor is connected between the base and collector terminals of the transistor.
Base resistor connected to power, emitter to ground.
Advantages β-independent output values.
β-independent (like- emitter bias), but does not require a dual-polarity power supply.
A simple circuit that is relatively β independent.
Simple.
Disadvantages Requires the use of a dual-polarity power supply.
None (as compared to other biasing circuits) in terms of dc operation.
Poor ac characteristics.
β dependent.
Applications Used primarily to set the dc biasing for linear amplifiers.
Used primarily to bias linear amplifiers.
Linear amplifiers Switching applications.
Homework 5 (All from chapter 5 problems) 5: 2, 3, 4, 6, 8, 9, 12, 14, 15, 19, 20, 21, 26, 27 Reading: Chapter 6 Chapter Summary
- The purpose of biasing a circuit is to establish a proper, stable dc-operating point (Q- point).
- The Q-point is defined by the values for IC and VCE.
- A dc load line passes through the Q-point on a transistor’s collector curves intersecting the vertical axis at approximately IC(sat) and the horizontal axis at VCE(cutoff).
- The linear (active) region of a transistor lies along the load line, below saturation, and above cutoff.
- The dc input resistance at the base of a BJT is approximately βDC RE.
- Voltage divider provides good Q-point stability with single polarity supply voltage.
- Voltage divider biasing is the most popular bias circuit.
- Base bias circuit arrangement has poor stability because its Q-point varies widely with βDC.
- Emitter bias generally provides good Q-point stability, but requires both positive and negative supply voltages.
- Collector-feedback bias provides good stability using negative feedback from collector to base.
