Download - Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Transcript
Page 1: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Transistor Circuits I

Common-Base, DC operation

Page 2: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

The humble transistor

Q1

Emitter (E)

Collector (C)

Base (B)

Page 3: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Transistor basics

• Emitter to base junction is forward biased (normally)

• Collector to base junction is reverse biased (normally)

• Transistors are current operated devices, so KCL should be applied first:

–IE = IC + IB

Page 4: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Basics continued

• Leakage current: ICBO (Emitter open)

– Usually is considered negligible, but can affect things when IC is small

Page 5: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Basics continued

• hFB = α = 𝐼𝐶

𝐼𝐸 (α ≤ 1)

• hFE = β = 𝐼𝐶

𝐼𝐵 (β has no definitive limit)

• The symbols hFB and hFE are BJT parameters; the h means Hybrid parameter. The subscript FB means Forward current transfer ratio of the common, or grounded, Base circuit, whereas FE means Forward current transfer ratio of the common, or grounded, Emitter circuit.

Page 6: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Basic configuration of Common-Base

Page 7: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

First circuit

• If VEE = 20V and VEB is negligible, find IE when RE equals (a) 80kΩ, (b) 40kΩ, (c) 20kΩ, (d) 10kΩ, (e) 5kΩ, and (f) 1kΩ.

Page 8: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Work for first circuit

Formula is 𝐼𝐸 =𝑉𝐸𝐸

𝑅𝐸=

20V

𝑅𝐸

• (a) 𝐼𝐸 =20V

80kΩ= 0.25mA = 250µA;

• (b) 𝐼𝐸 =20V

40kΩ= 0.5mA = 500µA;

• (c) 𝐼𝐸 =20V

20kΩ= 1mA;

• (d) 𝐼𝐸 =20V

10kΩ= 2mA;

• (e) 𝐼𝐸 =20V

5kΩ= 4mA;

• (f) 𝐼𝐸 =20V

1kΩ= 20mA

Page 9: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Use approximation for VEB

If VEB = 0.7V, rework previous values to determine IE in each instance.

• Formula is now 𝐼𝐸 =𝑉𝐸𝐸−𝑉𝐸𝐵

𝑅𝐸=

20V−0.7V

𝑅𝐸=

19.3V

𝑅𝐸

• (a) 𝐼𝐸 =19.3V

80kΩ= 241.25µA;

• (b) 𝐼𝐸 =19.3V

40kΩ= 482.5µA;

• (c) 𝐼𝐸 =19.3V

20kΩ= 965µA;

• (d) 𝐼𝐸 =19.3V

10kΩ= 1.93mA;

• (e) 𝐼𝐸 =19.3V

5kΩ= 3.86mA;

• (f) 𝐼𝐸 =19.3V

1kΩ= 19.3mA

Page 10: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Proving the point (Measuring (a))

Page 11: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Measuring (b)

Page 12: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Measuring (c)

Page 13: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Measuring (d)

Page 14: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Measuring (e)

Page 15: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Measuring (f)

Page 16: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

PNP versions

Page 17: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Second circuit

• If RE = 10kΩ and VEB = 0V, what are the values of IE, IC and VCB when RC equals (a) 5kΩ, (b) 8kΩ, (c) 10kΩ, (d) 12kΩ. Assume hFB = α = 1.

10kΩ

18V -22V

Page 18: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Points to ponder

• Since IE will remain constant, use the value found from calculating through the derivation

of Ohm’s Law: 𝐼𝐸 =𝑉𝐸𝐸

𝑅𝐸=

18V

10kΩ= 1.8mA.

Additionally, we are treating α as 1, which

means using the formula 𝐼𝐶

𝐼𝐸= 1 translates into

IE and IC being equal. Therefore, the value of IC for all changes of RC remains a constant 1.8mA (so long as saturation is avoided…)

Page 19: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Additional points

• Following KVL, VCC = VRC + VCB. Since VRC = ICRC (Ohm’s Law), substituting this into the original formula gives us VCC = ICRC + VCB. Reworking the equation to solve for VCB yields VCB = VCC – ICRC (You can use the absolute value for VCC when calculating, but remember to include polarity when writing the final answer).

Page 20: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Work through for circuit

• (a) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 5kΩ =22 − 9 = −13V (Remember the polarity of VCC is negative…)

• (b) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 8kΩ =22 − 14.4 = −7.6V

• (c) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 10kΩ =22 − 18 = −4V

• (d) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 12kΩ =22 − 21.6 = −0.4V

Page 21: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Third circuit

• If VCC = 24V, find (a) the saturation current, IC(sat). If VEB = 0.7V, what are the values of IE, IC, and VCB when RE equals (b) 20kΩ, (c) 10kΩ, and (d) 5kΩ. Assume αDC = 1.

Page 22: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

IC(sat) and some constants in our calculations

• (a) 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶

𝑅𝐶=

24V

5kΩ= 3mA

• VRE = VEE – VEB = 20 – 0.7 = 19.3V (technically this should be written as -19.3V since VEE is negative in polarity)

Page 23: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Calculating the rest

• (b) 𝐼𝐸 =𝑉𝑅𝐸

𝑅𝐸=

19.3V

20kΩ= 965µA ∴ 𝐼𝐶 = 965µA;

VCB = VCC – ICRC = 24 – (965µA)(8kΩ) = 24 – 7.72 = 16.28V

Page 24: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

And the beat goes on…

• (c) 𝐼𝐸 =𝑉𝑅𝐸

𝑅𝐸=

19.3V

10kΩ= 1.93mA ∴ 𝐼𝐶 =

1.93mA; VCB = VCC – ICRC = 24 – (1.93mA)(8kΩ) = 24 – 15.44 = 8.56V

Page 25: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

And on…

• (d) 𝐼𝐸 =𝑉𝑅𝐸

𝑅𝐸=

19.3V

5kΩ= 3.86mA ∴ 𝐼𝐶 =

3mA (this cannot be exceeded); VCB = VCC – ICRC = 24 – (3mA)(8kΩ) = 24 – 24 = 0V

Page 26: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Measure to make sure (a)

Page 27: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

(b)

Page 28: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

(c)

Page 29: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB

Any questions?

• Contact us at:

– 1-800-243-6446

– 1-216-781-9400

• Email:

[email protected]

Page 30: Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼 𝐼𝐸 (α ≤1) •h FE = β = 𝐼 𝐼 (β has no definitive limit) •The symbols h FB