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EE101: BJT basics M. B. Patil [email protected] www.ee.iitb.ac.in/~sequel Department of Electrical Engineering Indian Institute of Technology Bombay M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Base pnp transistor Collector Emitter n p n Collector Base npn transistor M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Base pnp transistor Collector Emitter n p n Collector Base npn transistor * Bipolar: both electrons and holes contribute to conduction M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Base pnp transistor Collector Emitter n p n Collector Base npn transistor * Bipolar: both electrons and holes contribute to conduction * Junction: device includes two p-n junctions (as opposed to a “point-contact” transistor, the first transistor) M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Collector Base pnp transistor Emitter n p n Collector Base npn transistor * Bipolar: both electrons and holes contribute to conduction * Junction: device includes two p-n junctions (as opposed to a “point-contact” transistor, the first transistor) * Transistor: “transfer resistor” When Bell Labs had an informal contest to name their new invention, one engineer pointed out that it acts like a resistor, but a resistor where the voltage is transferred across the device to control the resulting current. (http://amasci.com/amateur/trshort.html) M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Collector Base pnp transistor Emitter n p n Collector Base npn transistor * Bipolar: both electrons and holes contribute to conduction * Junction: device includes two p-n junctions (as opposed to a “point-contact” transistor, the first transistor) * Transistor: “transfer resistor” When Bell Labs had an informal contest to name their new invention, one engineer pointed out that it acts like a resistor, but a resistor where the voltage is transferred across the device to control the resulting current. (http://amasci.com/amateur/trshort.html) * invented in 1947 by Shockley, Bardeen, and Brattain at Bell Laboratories. M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Collector Emitter Base pnp transistor n p n Collector Base npn transistor * Bipolar: both electrons and holes contribute to conduction * Junction: device includes two p-n junctions (as opposed to a “point-contact” transistor, the first transistor) * Transistor: “transfer resistor” When Bell Labs had an informal contest to name their new invention, one engineer pointed out that it acts like a resistor, but a resistor where the voltage is transferred across the device to control the resulting current. (http://amasci.com/amateur/trshort.html) * invented in 1947 by Shockley, Bardeen, and Brattain at Bell Laboratories. * “A BJT is two diodes connected back-to-back.” M. B. Patil, IIT Bombay Bipolar Junction Transistors Emitter p n p Collector Emitter Base pnp transistor n p n Collector Base npn transistor * Bipolar: both electrons and holes contribute to conduction * Junction: device includes two p-n junctions (as opposed to a “point-contact” transistor, the first transistor) * Transistor: “transfer resistor” When Bell Labs had an informal contest to name their new invention, one engineer pointed out that it acts like a resistor, but a resistor where the voltage is transferred across the device to control the resulting current. (http://amasci.com/amateur/trshort.html) * invented in 1947 by Shockley, Bardeen, and Brattain at Bell Laboratories. * “A BJT is two diodes connected back-to-back.” WRONG! Let us see why. M. B. Patil, IIT Bombay Bipolar Junction Transistors Consider a pnp BJT in the following circuit: E R1 5V 1 k I1 p n p B I3 C I 2 R2 1 k 10 V M. B. Patil, IIT Bombay Bipolar Junction Transistors Consider a pnp BJT in the following circuit: E R1 1 k I1 C p n p I 2 R2 1 k B I3 5V 10 V If the transistor is replaced with two diodes connected back-to-back, we get, E R1 5V 1 k I1 C D1 B D2 I3 I 2 R2 1 k 10 V M. B. Patil, IIT Bombay Bipolar Junction Transistors Consider a pnp BJT in the following circuit: E R1 1 k I1 C p n p I 2 R2 1 k B I3 5V 10 V If the transistor is replaced with two diodes connected back-to-back, we get, E R1 5V 1 k I1 C D1 B D2 I 2 R2 1 k I3 10 V Assuming Von = 0.7 V for D1, we get 5 V − 0.7 V I1 = = 4.3 mA, R1 I2 = 0 (since D2 is reverse biased), and I3 ≈ I1 = 4.3 mA. M. B. Patil, IIT Bombay Bipolar Junction Transistors Using a more accurate equivalent circuit for the BJT, we obtain, E R1 5V 1 k I1 p n p B C E I 2 R2 1 k R1 I3 5V 10 V 1 k I1 α I1 B C I2 R2 1 k I3 10 V M. B. Patil, IIT Bombay Bipolar Junction Transistors Using a more accurate equivalent circuit for the BJT, we obtain, E R1 5V 1 k I1 p n p B C E I 2 R2 1 k R1 I3 5V 10 V 1 k I1 α I1 B C I2 R2 1 k I3 10 V We now get, 5 V − 0.7 V I1 = = 4.3 mA (as before), R1 M. B. Patil, IIT Bombay Bipolar Junction Transistors Using a more accurate equivalent circuit for the BJT, we obtain, E R1 5V 1 k I1 p n p B C E I 2 R2 1 k R1 I3 1 k I1 5V 10 V α I1 B C I2 R2 1 k I3 10 V We now get, 5 V − 0.7 V I1 = = 4.3 mA (as before), R1 I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and M. B. Patil, IIT Bombay Bipolar Junction Transistors Using a more accurate equivalent circuit for the BJT, we obtain, E R1 5V 1 k I1 p n p B C E I 2 R2 1 k R1 I3 1 k I1 5V 10 V α I1 B C I2 R2 1 k I3 10 V We now get, 5 V − 0.7 V I1 = = 4.3 mA (as before), R1 I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and I3 = I1 − I2 = (1 − α) I1 ≈ 0 A. M. B. Patil, IIT Bombay Bipolar Junction Transistors Using a more accurate equivalent circuit for the BJT, we obtain, E R1 5V 1 k I1 p n p B C E I 2 R2 1 k R1 I3 1 k I1 5V α I1 B C I2 R2 1 k I3 10 V 10 V We now get, 5 V − 0.7 V I1 = = 4.3 mA (as before), R1 I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and I3 = I1 − I2 = (1 − α) I1 ≈ 0 A. The values of I2 and I3 are dramatically different than the ones obtained earlier. M. B. Patil, IIT Bombay Bipolar Junction Transistors Using a more accurate equivalent circuit for the BJT, we obtain, E R1 5V 1 k I1 p n p B C E I 2 R2 1 k R1 I3 1 k I1 5V α I1 B C I2 R2 1 k I3 10 V 10 V We now get, 5 V − 0.7 V I1 = = 4.3 mA (as before), R1 I2 = αI1 ≈ 4.3 mA (since α ≈ 1 for a typical BJT), and I3 = I1 − I2 = (1 − α) I1 ≈ 0 A. The values of I2 and I3 are dramatically different than the ones obtained earlier. Conclusion: A BJT is NOT the same as two diodes connected back-to-back (although it does have two p-n junctions). M. B. Patil, IIT Bombay Bipolar Junction Transistors What is wrong with the two-diode model of a BJT? M. B. Patil, IIT Bombay Bipolar Junction Transistors What is wrong with the two-diode model of a BJT? * When we replace a BJT with two diodes, we assume that there is no interaction between the two diodes, which may be expected if they are “far apart.” Emitter p p n Collector Base Emitter Collector D1 Base D2 M. B. Patil, IIT Bombay Bipolar Junction Transistors What is wrong with the two-diode model of a BJT? * When we replace a BJT with two diodes, we assume that there is no interaction between the two diodes, which may be expected if they are “far apart.” Emitter p p n Collector Base Emitter Collector D1 Base D2 * However, in a BJT, exactly the opposite is true. For a higher performance, the base region is made as short as possible (subject to certain constraints), and the two diodes therefore cannot be treated as independent devices. Emitter p n p Collector Base M. B. Patil, IIT Bombay Bipolar Junction Transistors What is wrong with the two-diode model of a BJT? * When we replace a BJT with two diodes, we assume that there is no interaction between the two diodes, which may be expected if they are “far apart.” Emitter p p n Collector Base Emitter Collector D1 Base D2 * However, in a BJT, exactly the opposite is true. For a higher performance, the base region is made as short as possible (subject to certain constraints), and the two diodes therefore cannot be treated as independent devices. Emitter p n p Collector Base * Later, we will look at the “Ebers-Moll model” of a BJT, which is a fairly accurate representation of the transistor action. M. B. Patil, IIT Bombay BJT in active mode E IE p n p IC C E IC IE IB B C E IE n p IC C E IC IE IB IB B n B B C IB M. B. Patil, IIT Bombay BJT in active mode E IE p n p IC C E IC IE IB B C E IE n p IC C E IC IE IB IB B n B B C IB * In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias. - For a pnp transistor, VEB > 0 V , and VCB < 0 V . - For an npn transistor, VBE > 0 V , and VBC < 0 V . M. B. Patil, IIT Bombay BJT in active mode E IE p n p IC C E IC IE IB B C E IE n p IC C E IC IE IB IB B n B B C IB * In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias. - For a pnp transistor, VEB > 0 V , and VCB < 0 V . - For an npn transistor, VBE > 0 V , and VBC < 0 V . * Since the B-E junction is under forward bias, the voltage (magnitude) is typically 0.6 to 0.75 V . M. B. Patil, IIT Bombay BJT in active mode E IE p n p IC C E IC IE IB B C E IE n p IC C E IC IE IB IB B n B B C IB * In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias. - For a pnp transistor, VEB > 0 V , and VCB < 0 V . - For an npn transistor, VBE > 0 V , and VBC < 0 V . * Since the B-E junction is under forward bias, the voltage (magnitude) is typically 0.6 to 0.75 V . * The B-C voltage can be several Volts (or even hundreds of Volts), and is limited by the breakdown voltage of the B-C junction. M. B. Patil, IIT Bombay BJT in active mode E IE p n p IC C E IC IE IB B C E IE n p IC C E IC IE IB IB B n B B C IB * In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias. - For a pnp transistor, VEB > 0 V , and VCB < 0 V . - For an npn transistor, VBE > 0 V , and VBC < 0 V . * Since the B-E junction is under forward bias, the voltage (magnitude) is typically 0.6 to 0.75 V . * The B-C voltage can be several Volts (or even hundreds of Volts), and is limited by the breakdown voltage of the B-C junction. * The symbol for a BJT includes an arrow for the emitter terminal, its direction indicating the current direction when the transistor is in active mode. M. B. Patil, IIT Bombay BJT in active mode E IE p n p IC C E IC IE IB B C E IE n p IC C E IC IE IB IB B n B B C IB * In the active mode of a BJT, the B-E junction is under forward bias, and the B-C junction is under reverse bias. - For a pnp transistor, VEB > 0 V , and VCB < 0 V . - For an npn transistor, VBE > 0 V , and VBC < 0 V . * Since the B-E junction is under forward bias, the voltage (magnitude) is typically 0.6 to 0.75 V . * The B-C voltage can be several Volts (or even hundreds of Volts), and is limited by the breakdown voltage of the B-C junction. * The symbol for a BJT includes an arrow for the emitter terminal, its direction indicating the current direction when the transistor is in active mode. * Analog circuits, including amplifiers, are generally designed to ensure that the BJTs are operating in the active mode. M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB C E IE n p B B α IE E IE C IC B IB n C IC E IC IE IB IB B B α IE E IE C IB C IC IB B M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB C E IE n p B B α IE E IE C IC B n C IC E IC IE IB IB B B α IE E IE IB C IB C IC IB B * In the active mode, IC = α IE , α ≈ 1 (slightly less than 1). M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB C E IE n p B B α IE E IE C IC B n C IC E IC IE IB IB B B α IE E IE IB C IB C IC IB B * In the active mode, IC = α IE , α ≈ 1 (slightly less than 1). * IB = IE − IC = IE (1 − α) . M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB C E IE n p B B α IE E IE C IC B n C IC E IC IE IB IB B B α IE E IE IB C IB C IC IB B * In the active mode, IC = α IE , α ≈ 1 (slightly less than 1). * IB = IE − IC = IE (1 − α) . * The ratio IC /IB is defined as the current gain β of the transistor. IC α β= = . IB 1−α M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB C E IE n p B B α IE E IE C IC B n C IC E IC IE IB IB B B α IE E IE IB C IB C IC IB B * In the active mode, IC = α IE , α ≈ 1 (slightly less than 1). * IB = IE − IC = IE (1 − α) . * The ratio IC /IB is defined as the current gain β of the transistor. IC α β= = . IB 1−α * β is a function of IC and temperature. However, we will generally treat it as a constant, a useful approximation to simplify things and still get a good insight. M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB E IE n p B α IE E IE C IC B IB n C IC E IC IE IB IB B β= C B B α IE E IE C IB C IC IB B IC α = IB 1−α α β 0.9 9 0.95 19 0.99 99 0.995 199 M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB E IE n p B α IE E IE α β 0.9 9 0.95 19 0.99 99 0.995 199 C IC E IC IE B B α IE E IC B IC α = IB 1−α C n IB IB B β= C IE C IB C IC IB IB B * β is a sensitive function of α. M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB E IE n p B α IE E IE α β 0.9 9 0.95 19 0.99 99 0.995 199 C IC E IC IE B B α IE E IC B IC α = IB 1−α C n IB IB B β= C IE C IB C IC IB IB B * β is a sensitive function of α. * Transistors are generally designed to get a high value of β (typically 100 to 250, but can be as high as 2000 for “super-β” transistors). M. B. Patil, IIT Bombay BJT in active mode E IE p n p C IC E IC IE IB E IE n p B α IE E IE α β 0.9 9 0.95 19 0.99 99 0.995 199 C IC E IC IE B B α IE E IC B IC α = IB 1−α C n IB IB B β= C IE C IB C IC IB IB B * β is a sensitive function of α. * Transistors are generally designed to get a high value of β (typically 100 to 250, but can be as high as 2000 for “super-β” transistors). * A large β ⇒ IB IC or IE when the transistor is in the active mode. M. B. Patil, IIT Bombay A simple BJT circuit 1k RC C 100 k B RB 2V VBB VCC β = 100 10 V E A simple BJT circuit 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E RC n 2V VBB 100 k p RB β = 100 n A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE M. B. Patil, IIT Bombay A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB . M. B. Patil, IIT Bombay A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB . VBB − VBE 2 V − 0.7 V IB = = = 13 µA. RB 100 k M. B. Patil, IIT Bombay A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB . VBB − VBE 2 V − 0.7 V IB = = = 13 µA. RB 100 k IC = β × IB = 100 × 13 µA = 1.3 mA. M. B. Patil, IIT Bombay A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB . VBB − VBE 2 V − 0.7 V IB = = = 13 µA. RB 100 k IC = β × IB = 100 × 13 µA = 1.3 mA. VC = VCC − IC RC = 10 V − 1.3 mA × 1 k = 8.7 V . M. B. Patil, IIT Bombay A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB . VBB − VBE 2 V − 0.7 V IB = = = 13 µA. RB 100 k IC = β × IB = 100 × 13 µA = 1.3 mA. VC = VCC − IC RC = 10 V − 1.3 mA × 1 k = 8.7 V . Let us check whether our assumption of active mode is correct. We need to check whether the B-C junction is under reverse bias. M. B. Patil, IIT Bombay A simple BJT circuit 10 V VCC 10 V VCC 1k RC C 100 k B RB 2V VBB 1k VCC β = 100 10 V E 1k RC α IE n 2V VBB 100 k p RB β = 100 n RC IC 2V VBB 100 k IB RB IE Assume the BJT to be in the active mode ⇒ VBE = 0.7 V and IC = αIE = β IB . VBB − VBE 2 V − 0.7 V IB = = = 13 µA. RB 100 k IC = β × IB = 100 × 13 µA = 1.3 mA. VC = VCC − IC RC = 10 V − 1.3 mA × 1 k = 8.7 V . Let us check whether our assumption of active mode is correct. We need to check whether the B-C junction is under reverse bias. VBC = VB − VC = 0.7 V − 8.7 V = −8.0 V , i.e., the B-C junction is indeed under reverse bias. M. B. Patil, IIT Bombay A simple BJT circuit (continued) 10 V VCC 1k IC 2V VBB 10 k p RB I B RC n β = 100 n What happens if RB is changed from 100 k to 10 k? M. B. Patil, IIT Bombay A simple BJT circuit (continued) 10 V VCC 1k IC 2V VBB 10 k p RB I B RC n β = 100 n What happens if RB is changed from 100 k to 10 k? Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and IC = β IB . M. B. Patil, IIT Bombay A simple BJT circuit (continued) 10 V VCC 1k IC 2V VBB 10 k p RB I B RC n β = 100 n What happens if RB is changed from 100 k to 10 k? Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and IC = β IB . VBB − VBE 2 V − 0.7 V IB = = = 130 µA. RB 10 k IC = β × IB = 100 × 130 µA = 13 mA. VC = VCC − IC RC = 10 V − 13 mA × 1 k = −3 V . M. B. Patil, IIT Bombay A simple BJT circuit (continued) 10 V VCC 1k IC 2V VBB 10 k p RB I B RC n β = 100 n What happens if RB is changed from 100 k to 10 k? Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and IC = β IB . VBB − VBE 2 V − 0.7 V IB = = = 130 µA. RB 10 k IC = β × IB = 100 × 130 µA = 13 mA. VC = VCC − IC RC = 10 V − 13 mA × 1 k = −3 V . VBC = VB − VC = 0.7 V − (−3) V = 3.7 V , M. B. Patil, IIT Bombay A simple BJT circuit (continued) 10 V VCC 1k IC 2V VBB 10 k p RB I B RC n β = 100 n What happens if RB is changed from 100 k to 10 k? Assuming the BJT to be in the active mode again, we have VBE ≈ 0.7 V , and IC = β IB . VBB − VBE 2 V − 0.7 V IB = = = 130 µA. RB 10 k IC = β × IB = 100 × 130 µA = 13 mA. VC = VCC − IC RC = 10 V − 13 mA × 1 k = −3 V . VBC = VB − VC = 0.7 V − (−3) V = 3.7 V , VBC is not only positive, it is huge! The BJT cannot be in the active mode, and we need to take another look at the circuit. M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n B p IC C E IC IE IB IB B C E C IE B IB αF IE IC Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n p IC C E IC IE IB B C E C IE IB B B IB αF IE IC Reverse active mode: B−E in r. b., B−C in f. b. IE E p n B p IC αR (−IC ) C E IC IE IB IB B C (−IC ) E IE C IC B IB M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n p IC C E IC IE IB B C E C IE IB B B IB αF IE IC Reverse active mode: B−E in r. b., B−C in f. b. IE E p n B p IC αR (−IC ) C E IC IE IB IB B C (−IC ) E IE C IC B IB In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their original names.) M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n p IC C E IC IE IB B C E C IE IB B B IB αF IE IC Reverse active mode: B−E in r. b., B−C in f. b. IE E p n B p IC αR (−IC ) C E IC IE IB IB B C (−IC ) E IE C IC B IB In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their original names.) The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different. M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n p IC C E IC IE IB B C E C IE IB B B IB αF IE IC Reverse active mode: B−E in r. b., B−C in f. b. IE E p n B p IC αR (−IC ) C E IC IE IB C (−IC ) E IE IB B C IC B IB In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their original names.) The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different. Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5. M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n p IC C E IC IE IB B C E C IE IB B B IB αF IE IC Reverse active mode: B−E in r. b., B−C in f. b. IE E p n B p IC αR (−IC ) C E IC IE IB C (−IC ) E IE IB B C IC B IB In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their original names.) The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different. Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5. The corresponding current gains (βF and βR ) differ significantly, since β = α/(1 − α). M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor Active mode ("forward" active mode): B−E in f. b., B−C in r. b. IE E p n p IC C E IC IE IB B C E C IE IB B B IB αF IE IC Reverse active mode: B−E in r. b., B−C in f. b. IE E p n B p IC αR (−IC ) C E IC IE IB C (−IC ) E IE IB B C IC B IB In the reverse active mode, emitter ↔ collector. (However, we continue to refer to the terminals with their original names.) The two α’s, αF (“forward” α) and αR (“reverse” α) are generally quite different. Typically, αF > 0.98, and αR is in the range from 0.02 to 0.5. The corresponding current gains (βF and βR ) differ significantly, since β = α/(1 − α). In amplifiers, the BJT is biased in the forward active mode (simply called the “active mode”) in order to make use of the higher value of β in that mode. M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor The Ebers-Moll model combines the forward and reverse operations of a BJT in a single comprehensive model. IE E p n B IC p IB E p E IC IE IB B αF I′E I′E C C D1 IE D2 C αR I′C p I′C IB B IC n M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor The Ebers-Moll model combines the forward and reverse operations of a BJT in a single comprehensive model. IE E p n B IC p IB E p E IC IE IB B αF I′E I′E C C D1 IE D2 C αR I′C p I′C IB B IC n The currents IE0 and IC0 are given by the Shockley diode equation: » „ « – » „ « – VEB VCB − 1 , IC0 = ICS exp −1 . IE0 = IES exp VT VT M. B. Patil, IIT Bombay Ebers-Moll model for a pnp transistor The Ebers-Moll model combines the forward and reverse operations of a BJT in a single comprehensive model. IE p E n B IC p IB E p E IC IE αF I′E I′E C C D1 IE D2 C αR I′C IB p I′C IB B B IC n The currents IE0 and IC0 are given by the Shockley diode equation: » „ « – » „ « – VEB VCB − 1 , IC0 = ICS exp −1 . IE0 = IES exp VT VT Mode B-E B-C Forward active forward reverse IE0 IC0 Reverse active reverse forward IC0 IE0 Saturation forward forward IE0 and IC0 are comparable. Cut-off reverse reverse IE0 and IC0 are negliglbe. M. B. Patil, IIT Bombay Ebers-Moll model pnp transistor IE E p n IC p IB B E p E IC IE αF I′E I′E C C D1 IE D2 C αR I′C IB B p I′C = ICS [exp(VCB /VT ) − 1] I′C IB B IC I′E = IES [exp(VEB /VT ) − 1] n npn transistor IE E n IC p n IB E B E C D1 n IE IC IE B αF I′E I′E C IB D2 C αR I′C n I′C = ICS [exp(VBC /VT ) − 1] I′C IB B IC I′E = IES [exp(VBE /VT ) − 1] p M. B. Patil, IIT Bombay Ebers-Moll model pnp transistor IE E p n IC p IB B E p E IC IE αF I′E I′E C C D1 IE D2 C αR I′C IB B p I′C = ICS [exp(VCB /VT ) − 1] I′C IB B IC I′E = IES [exp(VEB /VT ) − 1] n npn transistor IE E n IC p n IB E B E C D1 n IE IC IE B αF I′E I′E C IB D2 C αR I′C n I′C = ICS [exp(VBC /VT ) − 1] I′C IB B IC I′E = IES [exp(VBE /VT ) − 1] p For an npn transistor, the same model holds with current directions and voltage polarities suitably changed. M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E IB I′E = IES [exp(VBE /VT ) − 1] C D1 n IE IC IE B αF I′E I′E C D2 C αR I′C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A A BJT is a three-terminal device, and its I -V chatacteristics can therefore be represented in several different ways. The IC versus VCE characteristics are very useful in amplifiers. M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E IB I′E = IES [exp(VBE /VT ) − 1] C D1 n IE IC IE B αF I′E I′E C D2 IC n C αR I′C I′C IB B p I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A A BJT is a three-terminal device, and its I -V chatacteristics can therefore be represented in several different ways. The IC versus VCE characteristics are very useful in amplifiers. To start with, we consider a single point, IB = 10 µA, VCE = 5 V . M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E IB I′E = IES [exp(VBE /VT ) − 1] C D1 n IE IC IE B αF I′E I′E C D2 IC n C αR I′C I′C IB B p I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A A BJT is a three-terminal device, and its I -V chatacteristics can therefore be represented in several different ways. The IC versus VCE characteristics are very useful in amplifiers. To start with, we consider a single point, IB = 10 µA, VCE = 5 V . There are several ways to assign VBE and VCB so that they satisfy the constraint: VCB + VBE = (VC − VB ) + (VB − VE ) = VCE = 5 V . M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E IB I′E = IES [exp(VBE /VT ) − 1] C D1 n IE IC IE B αF I′E I′E C D2 IC n C αR I′C I′C IB B p I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A A BJT is a three-terminal device, and its I -V chatacteristics can therefore be represented in several different ways. The IC versus VCE characteristics are very useful in amplifiers. To start with, we consider a single point, IB = 10 µA, VCE = 5 V . There are several ways to assign VBE and VCB so that they satisfy the constraint: VCB + VBE = (VC − VB ) + (VB − VE ) = VCE = 5 V . Let us consider some of these possibilities. M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B IB αF I′E I′E C I′E = IES [exp(VBE /VT ) − 1] C D1 IE D2 C αR I′C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B αF I′E I′E C C D1 IE D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 1V B C n 6V p M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B αF I′E I′E C C D1 IE D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 1V B C n D1 and D2 are both off, and we cannot satisfy the condition, IB = 10 µA, since all currents are much smaller than 10 µA. 6V p M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B αF I′E I′E C C D1 IE D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 1V B p 6V C n D1 and D2 are both off, and we cannot satisfy the condition, IB = 10 µA, since all currents are much smaller than 10 µA. ⇒ This possibility (and similarly others with both junctions reverse biased) is ruled out. M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E IB I′E = IES [exp(VBE /VT ) − 1] C D1 n IE IC IE B αF I′E I′E C D2 C αR I′C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E C D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] D1 n IE IC IE B αF I′E I′E C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 6V B C n 1V p M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E C D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] D1 n IE IC IE B αF I′E I′E C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 6V B C n D1 and D2 are both conducting; however, the forward bias for the B-E junction is impossibly large. 1V p M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E C D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] D1 n IE IC IE B αF I′E I′E C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 6V B 1V C n D1 and D2 are both conducting; however, the forward bias for the B-E junction is impossibly large. ⇒ This possibility is also ruled out. p M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B IB αF I′E I′E C I′E = IES [exp(VBE /VT ) − 1] C D1 IE D2 C αR I′C I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B αF I′E I′E C C D1 IE D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 0.7 V B C n 4.3 V p M. B. Patil, IIT Bombay IC -VCE characteristics IE n E IC p n IB E B E n IC IE B αF I′E I′E C C D1 IE D2 C αR I′C IB I′E = IES [exp(VBE /VT ) − 1] I′C IB B p IC n I′C = ICS [exp(VBC /VT ) − 1] αF = 0.99, ISE = 1 × 10−14 A αR = 0.50, ISC = 2 × 10−14 A Constraints: IB = 10 µA, VCE = 5 V . 5V E IC n IE IB 0.7 V B p 4.3 V C n D1 is on, D2 is off. This is a realistic possibility. Since the B-C junction is under reverse bias, IC0 and αR IC0 are much smaller than IE0 , and therefore the lower branches in the Ebers-Moll model can be dropped (see next slide). M. B. Patil, IIT Bombay IC -VCE characteristics 5V E IC n IE 0.7 V B IB C n 4.3 V p n IE αF I′E I′E E D1 C IC IB B n p (The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are VBE = 0.656 V and VCB = 4.344 V .) The BJT is in the active mode, and therefore αF IB = 99 × 10 µA = 0.99 mA. IC = β IB = 1 − αF IC -VCE characteristics 5V IC n IE 0.7 V B IB C n 1 4.3 V p E n IE αF I′E I′E D1 C IC IB B p IC (mA) E n 0 0 1 2 3 VCE (V) 4 5 (The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are VBE = 0.656 V and VCB = 4.344 V .) The BJT is in the active mode, and therefore αF IB = 99 × 10 µA = 0.99 mA. IC = β IB = 1 − αF IC -VCE characteristics 5V IC n IE 0.7 V B IB C n 1 4.3 V p E n IE αF I′E I′E D1 C IC IB B p IC (mA) E n 0 0 1 2 3 VCE (V) 4 5 (The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are VBE = 0.656 V and VCB = 4.344 V .) The BJT is in the active mode, and therefore αF IB = 99 × 10 µA = 0.99 mA. IC = β IB = 1 − αF If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and once again, we find that IC = β IB = 0.99 mA. IC -VCE characteristics 5V IC n IE 0.7 V B IB C n 1 4.3 V p E n IE αF I′E I′E D1 C IC IB B p IC (mA) E n 0 0 1 2 3 VCE (V) 4 5 (The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are VBE = 0.656 V and VCB = 4.344 V .) The BJT is in the active mode, and therefore αF IB = 99 × 10 µA = 0.99 mA. IC = β IB = 1 − αF If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and once again, we find that IC = β IB = 0.99 mA. Thus, the plot of IC versus VCE is simply a horizontal line. IC -VCE characteristics 5V IC 0.7 V B IB C n p E n IE αF I′E I′E D1 C IC IB B p 1 1 4.3 V IC (mA) n IE IC (mA) E n 0 0 1 2 3 VCE (V) 4 5 0 0 1 2 3 VCE (V) 4 5 (The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are VBE = 0.656 V and VCB = 4.344 V .) The BJT is in the active mode, and therefore αF IB = 99 × 10 µA = 0.99 mA. IC = β IB = 1 − αF If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and once again, we find that IC = β IB = 0.99 mA. Thus, the plot of IC versus VCE is simply a horizontal line. M. B. Patil, IIT Bombay IC -VCE characteristics 5V IC 0.7 V B IB C n p E n IE αF I′E I′E D1 C IC IB B p 1 1 4.3 V IC (mA) n IE IC (mA) E n 0 0 1 2 3 VCE (V) 4 5 0 0 1 2 3 VCE (V) 4 5 (The actual values for VBE and VCB obtained by solving the Ebers-Moll equations are VBE = 0.656 V and VCB = 4.344 V .) The BJT is in the active mode, and therefore αF IB = 99 × 10 µA = 0.99 mA. IC = β IB = 1 − αF If VCE is reduced to, say, 4 V , and IB kept at 10 µA, our previous argument holds, and once again, we find that IC = β IB = 0.99 mA. Thus, the plot of IC versus VCE is simply a horizontal line. However, as VCE → 0 V , things change (see next slide). M. B. Patil, IIT Bombay IC -VCE characteristics 0.7 V E IC n IE 0.7 V B IB C n 0V p When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V . IC -VCE characteristics 0.7 V E IC n IE 0.7 V B IB C n 0V p 0.3 V E IC n IE IB 0.7 V B C n 0.4 V p When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V . As VCE is reduced further, the B-C junction gets forward biased. For example, with VCE = 0.3 V , we may have a voltage distribution shown in the figure. (The numbers are only representative; the actual VBE and VBC values can be obtained by solving the E-M equations.) IC -VCE characteristics 0.7 V E IC n IE 0.7 V B IB C 0V E p αF I′E I′E n C D1 n IE D2 IC n 0.3 V E IC n IE IB 0.7 V B 0.4 V C n αR I′C I′C IB B p p When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V . As VCE is reduced further, the B-C junction gets forward biased. For example, with VCE = 0.3 V , we may have a voltage distribution shown in the figure. (The numbers are only representative; the actual VBE and VBC values can be obtained by solving the E-M equations.) Now, the component IC0 in the E-M model becomes significant, IC = αF IE0 − IC0 reduces, and IC becomes smaller than βIB . IC -VCE characteristics saturation 0.7 V IC n IE 0.7 V B IB C 0V E p 1 C D1 n IE D2 0.3 V E IC n IE IB 0.7 V B p 0.4 V C n linear αF I′E I′E n αR I′C n I′C IB B IC IC (mA) E p 0 0 1 2 3 VCE (V) 4 5 When VCE ≈ 0.7 V (and IB kept at 10 µA), the B-C drop is about 0 V . As VCE is reduced further, the B-C junction gets forward biased. For example, with VCE = 0.3 V , we may have a voltage distribution shown in the figure. (The numbers are only representative; the actual VBE and VBC values can be obtained by solving the E-M equations.) Now, the component IC0 in the E-M model becomes significant, IC = αF IE0 − IC0 reduces, and IC becomes smaller than βIB . The region where IC < βIB is called the “saturation region.” M. B. Patil, IIT Bombay IC -VCE characteristics saturation linear IC (mA) 2 IB = 20 µA 1 0 IB = 10 µA 0 1 2 3 VCE (V) 4 5 If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear region. Apart from that, there is no qualitative change in the IC − VCE plot. IC -VCE characteristics saturation linear IC (mA) 2 IB = 20 µA 1 0 IB = 10 µA 0 1 2 3 VCE (V) 4 5 If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear region. Apart from that, there is no qualitative change in the IC − VCE plot. Clearly, the IC − VCE behaviour of a BJT is not represented by a single curve but by a family of curves, known as the “IC − VCE characteristics.” IC -VCE characteristics saturation saturation linear 5 IB = 20 µA 1 50 µA 4 IC (mA) IC (mA) 2 linear IB = 10 µA 40 µA 3 30 µA 2 20 µA 1 0 0 1 2 3 VCE (V) 4 5 0 0 IB = 10 µA 1 2 3 VCE (V) 4 5 If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear region. Apart from that, there is no qualitative change in the IC − VCE plot. Clearly, the IC − VCE behaviour of a BJT is not represented by a single curve but by a family of curves, known as the “IC − VCE characteristics.” M. B. Patil, IIT Bombay IC -VCE characteristics saturation saturation linear 5 IB = 20 µA 1 50 µA 4 IC (mA) IC (mA) 2 linear IB = 10 µA 40 µA 3 30 µA 2 20 µA 1 0 0 1 2 3 VCE (V) 4 5 0 0 IB = 10 µA 1 2 3 VCE (V) 4 5 If IB is doubled (from 10 µA to 20 µA), IC = βIB changes by a factor of 2 in the linear region. Apart from that, there is no qualitative change in the IC − VCE plot. Clearly, the IC − VCE behaviour of a BJT is not represented by a single curve but by a family of curves, known as the “IC − VCE characteristics.” The IE − VCB and IC − VBE characteristics of a BJT are also useful in understanding BJT circuits. M. B. Patil, IIT Bombay A simple BJT circuit (revisited) 10 V VCC 1k IC 2V VBB p RC n β = 100 RB I B IE n We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. A simple BJT circuit (revisited) saturation linear 15 10 V VCC IC 2V VBB p RC n β = 100 RB I B IE 130 µA (RB = 10 k) IC (mA) 1k 10 5 13 µA (RB = 100 k) n 0 0 2 4 6 VCE (V) 8 10 We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. VBB − 0.7 V Let us plot IC − VCE curves for IB ≈ for the two values of RB . RB A simple BJT circuit (revisited) saturation linear 15 10 V VCC IC 2V VBB p RC n β = 100 RB I B IE 130 µA (RB = 10 k) IC (mA) 1k 10 load line 5 13 µA (RB = 100 k) n 0 0 2 4 6 VCE (V) 8 10 We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. VBB − 0.7 V Let us plot IC − VCE curves for IB ≈ for the two values of RB . RB In addition to the BJT IC − VCE curve, the circuit variables must also satisfy the constraint, VCC = VCE + IC RC , a straight line in the IC − VCE plane. M. B. Patil, IIT Bombay A simple BJT circuit (revisited) saturation linear 15 10 V VCC IC 2V VBB p RC n β = 100 RB I B IE 130 µA (RB = 10 k) IC (mA) 1k 10 load line 5 13 µA (RB = 100 k) n 0 0 2 4 6 VCE (V) 8 10 We are now in a position to explain what happens when RB is decreased from 100 k to 10 k in the above circuit. VBB − 0.7 V Let us plot IC − VCE curves for IB ≈ for the two values of RB . RB In addition to the BJT IC − VCE curve, the circuit variables must also satisfy the constraint, VCC = VCE + IC RC , a straight line in the IC − VCE plane. The intersection of the load line and the BJT characteristics gives the solution for the circuit. For RB = 10 k, note that the BJT operates in the saturation region, leading to VCE ≈ 0.2 V , and IC = 9.8 mA. M. B. Patil, IIT Bombay