Lecture 9–Bipolar Transistor:

Lecture 9– Bipolar Transistor:

1. NPN Transistor:

An NPN transistor is two PN junctions that are connected but in opposite

Directions like in figure 1.

· In this NPN junction, the outer electrons from the N-type semiconductors

want to go into the p-type semiconductor to combine with holes, but

most of the outer electrons will go with the current from one N-type

through the P-type and through the other N-type.

· So most of the current in the NPN junction is from one N-type to the

other N-type.

· Emitter is one of the N-type semiconductors.

· The Collector is the other N-type semiconductor.

· The Base is the P-type semiconductor, so it needs to have enough

voltage from the base to the emitter to make current flow.

The Base needs to be about Vbe = 0.7V larger than the emitter to

begin current flow.

· The symbol for the NPN transistor is in figure 2.

· The NPN transistor is a current amplifier. Which means that it can be

used to increase the current from input (base) to output (collector or emitter).

· There are two important current equations for bipolar transistors:

1. Current summing: Ie = Ic + Ib
2. Current gain: Ic = β x Ib where the β is set when the transistor is made, and it is usually about 100.

· Since current goes from each of the terminals of the bipolar transistor to another terminal (for example base to emitter current) then there is a

resistance for each of these connections.

· A full model can often be used in the place of a bipolar

transistor. This model is in figure 3.

· Where each terminal has an input resistance, and the

transistors current gain is made using a dependant

current source.

· A simple model can sometimes be used, which removes the

base (rb) and collector (rc) resistances.

Example 1: Given Circuit 1, with Vi = 5V, R1= 200KW,

Vo = 10V, R2= 1KW, and β is 100.

What are the currents Ib, Ic and Ie?

What is Vce (collector to emitter voltage)?

Using Vbe = 0.7V:

Ib = (Vi – Vbe)/ R1 = (5V – 0.7V)/ 200K = 21.5 uA

Ic = β x Ib = 100 x 21.5uA = 2.15mA

Ie = Ic + Ib = 2.15mA + 21.5uA = 2.1715mA

Vce = Vo – R2 x Ic = 10V – 1K x 2.15mA = 7.85V

Example 2: Given Circuit 2, with Vi = 5Vsin(1000t),

Vo = 10V, R1= 500KW, R2= 4KW, and β is 150.

What are the peak currents for Ib, Ic and Ie?

Draw the voltage on R1 and Vce?

Using Vbe = 0.7V:

Ib = (Vi – Vbe)/ R1 = (5V – 0.7V)/ 500K = 8.6 uA

Ic = β x Ib = 150 x 8.6uA = 1.29mA

Ie = Ic + Ib = 1.29mA + 8.6uA = 1.2986mA

Vce(peak) = Vo – R2 x Ic = 10V – 4K x 1.29mA = 4.84V

When Vi is less than 0.7V, then V1=0 and Vce=0

The time period is: T = 2π / w= = 2π/ 1000 = 6.28ms

1. PNP Transistor:

A PNP transistor is also two PN junctions that are connected but in

opposite directions like in figure 4.

· In this PNP junction, the outer electrons from the N-type semiconductor

want to go into the two p-type semiconductors to combine with holes, but

most of the outer electrons will go with the current from one P-type

through the N-type and through the other P-type.

· So most of the current in the PNP junction is from one P-type to the

other P-type.

· The symbol for the NPN transistor is in figure 2

· The Base needs to be about Vbe = -0.7V smaller

than the emitter to begin current flow

· There are two important current equations for bipolar transistors

1. Current summing: Ie = Ic + Ib
2. Current gain: Ic = β x Ib

· Figure 3 can still be used for a PNP transistor, except that the direction of the dependant source for the collector current is reversed.

Example 3: Given Circuit 3, with Vi = -3V, R1= 150KW,

Vo = -9V, R2= 1KW, and β is 200.

What are the currents Ib, Ic and Ie?

What is Vce (collector to emitter voltage)?

Using Vbe = 0.-7V:

Ib = (Vi – Vbe)/ R1 = (-3V – (– 0.7V))/ 150K = -15.33 uA

Ic = β x Ib = 200 x –15.33uA = -3.07mA

Ie = Ic + Ib = -3.07mA + (-15.33uA) = -3.08533mA

Vce = -Vo – R2 x Ic = -9V – 1K x –3.07mA = -5.93V

Example 4: Given Circuit 4, with Vi = 5Vsin(3000t),

Vo = -15V, R1= 600KW, R2= 2KW, and β is 170.

What are the peak currents for Ib, Ic and Ie?

Draw the voltage on R1 and Vce?

Using Vbe = -0.7V: For the negative peak of Vi:

Ib = (Vi – Vbe)/ R1 = (-5V – (-0.7V))/ 600K = -7.17 uA

Ic = β x Ib = 170 x –7.17uA = -1.22mA

Ie = Ic + Ib = -1.22mA – 7.17uA = 1.22717mA

Vce(peak) = Vo – R2 x Ic = -15V – 2K x -1.22mA = -12.56V

When Vi is greater than -0.7V, then V1=0 and Vce=0

The time period is: T = 2π / w= = 2π/ 3000 = 2.09ms

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