Transistor Circuits IV

Common –emitter

configurations

Three types

• Base-biased

• Two-supply biased

• Voltage divider biased

Base-biased

Minus supply voltage

RCRB

Formulas

• 𝛽dc = ℎ𝐹𝐸 ≅𝐼𝐶

𝐼𝐵

• 𝐼𝐶 ≅ 𝛽dc𝐼𝐵 = ℎ𝐹𝐸𝐼𝐵

• 𝐼𝐵 =𝑉𝐶𝐶−𝑉𝐵𝐸

𝑅𝐵

• 𝐼𝐵 ≅𝑉𝐶𝐶

𝑅𝐵

• 𝐼𝐶 =ℎ𝐹𝐸 𝑉𝐶𝐶−𝑉𝐵𝐸

𝑅𝐵

• If VBE = 0V, then

𝐼𝐶 =ℎ𝐹𝐸𝑉𝐶𝐶

𝑅𝐵

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶

Leakage currents, cutoff and saturation

• These formulas are common to all three types

• 𝐼𝐶 = ℎ𝐹𝐸𝐼𝐵 + 𝐼𝐶𝐸𝑂

• 𝐼𝐶𝐵𝑂 =𝐼𝐶𝐸𝑂

ℎ𝐹𝐸+1

• 𝐼𝐶𝐸𝑂 = ℎ𝐹𝐸 + 1 𝐼𝐶𝐵𝑂

• 𝑃𝐶 = 𝑉𝐶𝐸𝐼𝐶

Saturation = Max Current (no VCE)

Cutoff = Source across transistor (no current)

Two Supply biased

RC

RB RE

VCC

VEE

Formulas (part 1)

• 𝐼𝐶 ≅𝑉𝐸𝐸−𝑉𝐵𝐸

𝑅𝐸+𝑅𝐵 ℎ𝐹𝐸 +

𝑅𝐵

ℎ𝐹𝐸𝑅𝐸+𝑅𝐵𝐼𝐶𝐸𝑂

• Since ICEO is usually negligible, we can write the equation as

• 𝐼𝐶 ≅𝑉𝐸𝐸−𝑉𝐵𝐸

𝑅𝐸+𝑅𝐵 ℎ𝐹𝐸

• If hFE is large (β > 60) and VBE is considered negligible (VBE < 2% of VEE), then the equation simplifies to

• 𝐼𝐶 ≅𝑉𝐸𝐸

𝑅𝐸

Formulas (part 2)

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶

• 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶+𝑉𝐸𝐸

𝑅𝐶+𝑅𝐸

• 𝑉𝐶𝐸(𝑐𝑢𝑡𝑜𝑓𝑓) = 𝑉𝐶𝐶 + 𝑉𝐸𝐸

Voltage divider biased

RC

R2 RE

R1

-VEE

Formulas (part 1)

• 𝑉𝐶 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶

• 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶

𝑅𝐸+𝑅𝐶

• 𝑉𝐶𝐸(𝑐𝑢𝑡𝑜𝑓𝑓) = 𝑉𝐶𝐶

• 𝑃𝐶 = 𝑉𝐶𝐸𝐼𝐶

• 𝑉2 = 𝑉𝐵 ≅𝑉𝐶𝐶𝑅2

𝑅1+𝑅2

• 𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 ≅𝑉𝐶𝐶𝑅2

𝑅1+𝑅2− 𝑉𝐵𝐸

• 𝐼𝐶 ≅ 𝐼𝐸 =𝑉𝐸

𝑅𝐸

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 −𝑅𝐶 + 𝑅𝐸 𝐼𝐶

Formulas (part 2)

• Since the ratio of R1R2/β and the leakage current ICEO are usually negligible, we find the formula becomes

• 𝐼𝐶 ≅𝑉𝐶𝐶𝑅2−𝑉𝐵𝐸 𝑅1+𝑅2

𝑅𝐸 𝑅1+𝑅2

• If R1/R2 and/or ICEO are relatively large or β is relatively small, we should use this equation for IC

• 𝐼𝐶 ≅𝑉𝐶𝐶𝑅2−𝑉𝐵𝐸 𝑅1+𝑅2

𝑅𝐸 𝑅1+𝑅2 +𝑅1𝑅2 𝛽 +

𝑅1𝑅2

𝑅1𝑅2+𝛽𝑅𝐸 𝑅1+𝑅2𝐼𝐶𝐸𝑂

Formulas (part 3)

• If VBE ≈ 0V, then the formula becomes

• 𝐼𝐶 ≅𝑅2

𝑅1+𝑅2

𝑉𝐶𝐶

𝑅𝐸

Example 1 (Base-biased)

• If RB = 820kΩ, RC = 4.7kΩ and hFE (β) = 80 in the circuit shown, find the quiescent values of IC and VCE and determine the collector power dissipation PC. Assume the leakage is negligible and that VBE = 0.6V. In addition, what are the values for IC(sat) and VCE(cutoff)?

12 V

RCRB

J1

Solution Example 1

• 𝐼𝐵 =𝑉𝐶𝐶−𝑉𝐵𝐸

𝑅𝐵=

12−0.6

820kΩ=

11.4V

820kΩ= 13.902µA

• 𝐼𝐶 ≅ 𝛽dc𝐼𝐵 = ℎ𝐹𝐸𝐼𝐵 = 80 13.902µA = 1.112mA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 12 − 1.112mA 4.7kΩ = 12 −5.227 = 6.773V

• 𝑃𝐶 = 𝑉𝐶𝐸𝐼𝐶 = 6.773V 1.112mA = 6.387mW

• 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶

𝑅𝐶=

12

4.7kΩ= 2.553mA

• 𝑉𝐶𝐸(𝑐𝑢𝑡𝑜𝑓𝑓) = 𝑉𝐶𝐶 = 12V

Example 2 (Base-biased)

• With the switch J1 open and RC = 2.2MΩ in the circuit shown, the collector-to-ground voltage measures 10V. What are the leakage currents ICEO and ICBO if hFE = 100?

12 V

RCRB

J1

Solution Example 2

• 𝐼𝐶 = 𝐼𝐶𝐸𝑂 =𝑉𝐶𝐶−𝑉𝐶𝐸

𝑅𝐵=

12−10

2.2MΩ=

2V

2.2MΩ=

909.091µA

• 𝐼𝐶𝐵𝑂 =𝐼𝐶𝐸𝑂

ℎ𝐹𝐸+1=

909.091µA

100+1=

909.091µA

101=

9.001µA

Example 3 (Base-biased)

• For the circuit shown, if the BJT’s hFE = 80, find the values of RB and RC that will cause IC = 1mA and VCE = 6V. To keep the calculations simple, assume that VBE = 0V (ideal transistor). Hint: Use 𝐼𝐶 ≅

ℎ𝐹𝐸𝑉𝐶𝐶

𝑅𝐵 to solve for RB and

𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 to solve for RC. Once you have those values, and using a list of standard resistors that have 10% tolerance, choose values for RB and RC that are close to the values determined above; recalculate the values of IC and VCE.

12 V

RCRB

J1

Solution Example 3

• 𝑅𝐵 =ℎ𝐹𝐸𝑉𝐶𝐶

𝐼𝐶=

80 12

1mA=

960

1mA= 960kΩ

• 𝑅𝐶 =𝑉𝐶𝐶−𝑉𝐶𝐸

𝐼𝐶=

12−6

1mA=

6V

1mA= 6kΩ

• From 10% table, we need to choose RB = 1MΩ and RC = 5.6kΩ

• 𝐼𝐵 ≅𝑉𝐶𝐶

𝑅𝐵≅

12V

1MΩ≅ 12µA

• 𝐼𝐶 ≅ 𝛽dc𝐼𝐵 = ℎ𝐹𝐸𝐼𝐵 = 80 12µA = 960µA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 = 12V − 5.6kΩ 960µA = 12 −5.376 = 6.624V

Example 4 (Two Supply biased)

• In the circuit shown, RC = 5.6kΩ and RE = 10kΩ. The BJT has an hFE = 100 and negligible leakage. What are the values of (a) quiescent IC and VCE, (b) the collector power dissipation, and (c) the endpoints of the dc load line IC(sat) and VCE(cutoff)? Assume that VBE = 0V.

RC

15kΩ RE

12V

-10V

Solution Example 4 (part 1)

• 𝐼𝐶 ≅𝑉𝐸𝐸−𝑉𝐵𝐸

𝑅𝐸+𝑅𝐵 ℎ𝐹𝐸 ≅

10−0

10kΩ+15kΩ

100

≅10V

10kΩ+150Ω≅

10V

10.15kΩ≅ 985.222µA

• 𝐼𝐶 ≅𝑉𝐸𝐸

𝑅𝐸≅

10V

10kΩ≅ 1mA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 =12 − 5.6kΩ 985.222µA = 12 − 5.517 =6.483V

Solution Example 4 (part 2)

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 = 12 − 5.6kΩ 1mA =12 − 5.6 = 6.4V

• 𝑃𝐶 = 𝑉𝐶𝐸𝐼𝐶 = 6.483V 985.222µA =6.387mW

• 𝑃𝐶 = 𝑉𝐶𝐸𝐼𝐶 = 6.4V 1mA = 6.4mW

• 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶+𝑉𝐸𝐸

𝑅𝐶+𝑅𝐸=

12+10

5.6kΩ+10kΩ=

22V

15.6kΩ=

1.41mA

• 𝑉𝐶𝐸(𝑐𝑢𝑡𝑜𝑓𝑓) = 𝑉𝐶𝐶 + 𝑉𝐸𝐸 = 12V + 10V = 22V

Example 5 (variation of Example 4)

• Referring to the previous circuit, if the BJT were replaced with one having an hFE = 200 instead of the previously stated value of 100, what are the resulting values for IC and VCE?

Solution Example 5

• 𝐼𝐶 ≅𝑉𝐸𝐸−𝑉𝐵𝐸

𝑅𝐸+𝑅𝐵 ℎ𝐹𝐸 ≅

10−0

10kΩ+15kΩ

200

≅10V

10kΩ+75Ω≅

10V

10.075kΩ≅ 992.556µA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 =12 − 5.6kΩ 992.556µA = 12 − 5.558 =6.442V

Example 6 (Voltage divider biased)

• Consider R1 = 220kΩ. R2 = 39kΩ, RC = 8.2kΩ, and RE = 2.7kΩ in the circuit shown; the BJT has an hFE = 100. If VBE = 0.7V and the leakage is negligible, what are the values of IC, VCE, PC, and the collector-to-ground voltage VC? In addition, calculate IC(sat) and VCE(cutoff).

RC

R2 RE

R1

-24V

Solution Example 6 (part 1)

• 𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 ≅𝑉𝐶𝐶𝑅2

𝑅1+𝑅2− 𝑉𝐵𝐸 =

24 39kΩ

220kΩ+39kΩ−

0.7 =936×103

259×103− 0.7 = 3.614 − 0.7 = 2.914V

• 𝐼𝐶 ≅ 𝐼𝐸 =𝑉𝐸

𝑅𝐸=

2.914V

2.7kΩ= 1.079mA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝑅𝐶 + 𝑅𝐸 𝐼𝐶 =24 − 8.2kΩ + 2.7kΩ 1.079mA = 24 −10.9kΩ 1.079mA = 24 − 11.764 = 12.236V

Solution Example 6 (part 2)

• 𝑉𝐶 = 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 = 24 − 8.2kΩ 1.079mA =24 − 8.484 = 15.152V

• 𝑃𝐶 = 𝑉𝐶𝐸𝐼𝐶 = 12.236V 1.079mA =13.203mW

• 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶

𝑅𝐸+𝑅𝐶=

24V

8.2kΩ+2.7kΩ=

24V

10.9kΩ=

2.202mA

• 𝑉𝐶𝐸(𝑐𝑢𝑡𝑜𝑓𝑓) = 𝑉𝐶𝐶 = 24V

Any questions?

• Contact us at:

– 1-800-243-6446

– 1-216-781-9400

• Email:

– faculty@cie-wc.edu