Transistor Circuits VI

Two-Transistor Direct-Coupled CE Amplifier / Some basics of troubleshooting CE Amps

The Two-Transistor Direct-Coupled CE circuit configuration

Formulas for same

• 𝐼𝐶1 =𝑉𝐶𝐶−2𝑉𝐵𝐸

𝑅𝐶1+𝑅𝐹𝛽+𝑅𝐸1

• 𝐼𝐶2 =𝑉𝐶𝐶−𝑉𝐵𝐸−𝑅𝐶1𝐼𝐶1

𝑅𝐸2

• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1

• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2

• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1𝐼𝐶1

• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2𝐼𝐶2

First example

• Referring to the figure shown, find the collector current, collector-to-emitter voltage, and collector-to-ground voltage for each BJT. Each BJT has an hFE = 50 and VBE = 0.7V.

12 V

Q1

Q2

68kΩ

330Ω

2.7kΩ

1kΩ

470kΩ

Work for first example

• 𝐼𝐶1 =𝑉𝐶𝐶−2𝑉𝐵𝐸

𝑅𝐶1+𝑅𝐹𝛽+𝑅𝐸1

=12−2 0.7

68k+470k

50+330

=12−1.4

68k+9.4k+330=

10.6V

77.73kΩ= 136.37μA

• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1 =12 − 68k + 330 136.37μA =12 − 68.33kΩ 136.37μA = 12 − 9.318 = 2.682V

• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1𝐼𝐶1 = 12 − 68kΩ 136.37μA =12 − 9.273 = 2.727V

Work for first example (cont.)

• 𝐼𝐶2 =𝑉𝐶𝐶−𝑉𝐵𝐸−𝑅𝐶1𝐼𝐶1

𝑅𝐸2=

12−0.7− 68k 136.37μA

1k=

12−0.7−9.273

1k=

2.027V

1kΩ= 2.027mA

• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2 =12 − 2.7k + 1k 2.027mA =12 − 3.7kΩ 2.027mA = 12 − 7.499 = 4.501V

• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2𝐼𝐶2 = 12 − 2.7kΩ 2.027mA =12 − 5.473 = 6.527V

Second example

• Rework the previous problem using a 680-kΩ resistor in place of the 470-kΩ resistor.

Work for second example

• 𝐼𝐶1 =𝑉𝐶𝐶−2𝑉𝐵𝐸

𝑅𝐶1+𝑅𝐹𝛽+𝑅𝐸1

=12−2 0.7

68k+680k

50+330

=12−1.4

68k+13.6k+330=

10.6V

81.93kΩ= 129.379μA

• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1 =12 − 68k + 330 129.379μA =12 − 68.33kΩ 129.379μA = 12 − 8.84 = 3.16V

• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1𝐼𝐶1 = 12 − 68kΩ 129.379μA =12 − 8.798 = 3.202V

Work for second example (cont.)

• 𝐼𝐶2 =𝑉𝐶𝐶−𝑉𝐵𝐸−𝑅𝐶1𝐼𝐶1

𝑅𝐸2=

12−0.7− 68k 129.379μA

1k=

12−0.7−8.798

1k=

2.502V

1kΩ= 2.502mA

• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2 =12 − 2.7k + 1k 2.502mA =12 − 3.7kΩ 2.502mA = 12 − 9.528 = 2.472V

• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2𝐼𝐶2 = 12 − 2.7kΩ 2.502mA =12 − 6.756 = 5.224V

TROUBLESHOOTING CE CIRCUITS

First example

• If the 180-kΩ (R2) resistor became open in the circuit shown, what would the collector-to-ground voltage be?

12V

12V

33kΩ 180kΩ

2.2kΩ

10kΩ

10.7V

Normal operation

• Normal operation (Q point):

• 𝑉𝐵 = 𝑉𝑅2 =𝑅2

𝑅1+𝑅2𝑉𝐶𝐶 =

180k

33k+180k12 =

180k

213k12 = 0.845 12V = 10.14V

• 𝑉𝐵𝐸 = 𝑉𝐵 − 𝑉𝐸 = 10.17 − 10.7 = −0.56V

• 𝐼𝐶 = 𝐼𝐸 =𝑉𝐶𝐶−𝑉𝐸

𝑅𝐸=

12−10.7

2.2k=

1.3V

2.2kΩ= 590.91μA

• 𝑉𝐶 = 𝐼𝐶𝑅𝐶 = 590.91μA 10kΩ = 5.909V

Circuit evaluation

• If R2 opens, VB ≈12V ∴ VBE is reverse biased.

• If VBE is reverse biased, transistor is cutoff (IC = 0mA).This means VC = 0V as VE = VCC = VCE.

Second example

• If the 33-kΩ (R1) resistor became open instead in the figure shown, what would the collector-to-ground voltage be?

12V

12V

33kΩ 180kΩ

2.2kΩ

10kΩ

10.7V

Circuit evaluation

• If R1 opens, VB = 0V ∴ transistor is biased full on (saturation)

• 𝐼𝐶 = 𝐼𝐸 =𝑉𝐶𝐶

𝑅𝐶+𝑅𝐸=

12

2.2k+10k=

12

12.2kΩ=

983.607μA

• 𝑉𝐶 = 𝐼𝐶𝑅𝐶 = 983.607μA 10kΩ = 9.836V

Third example

• Referring to the figure shown, if the BJT’s hFE = 80, find its IC and VCE. Assume that VBE and ICEO are negligible. Hint: The equation for VCE is the same as the voltage-divider-biased circuits.

VCC = 20V

RC = 5.6k

RB = 390k

RE = 1k

𝐼𝐶 =𝑉𝐶𝐶

𝑅𝐸 + 𝑅𝐶 +𝑅𝐵ℎ𝐹𝐸

If VBE ≈ 0V and ICEO ≈ 0A

Work for third example

• 𝐼𝐶 =𝑉𝐶𝐶

𝑅𝐸+𝑅𝐶+𝑅𝐵ℎ𝐹𝐸

=20

1k+5.6k+390k

80

=

20

1k+5.6k+4.875k=

20V

11.475kΩ= 1.743mA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 + 𝑅𝐶 + 𝑅𝐸 𝐼𝐶 =20 − 5.6k + 1k 1.743mA = 20 −6.6kΩ 1.743mA = 20 − 11.503 =8.497V

Circuit revision

• Work the previous problem using hFE = 160 instead of 80.

Work for revision

• 𝐼𝐶 =𝑉𝐶𝐶

𝑅𝐸+𝑅𝐶+𝑅𝐵ℎ𝐹𝐸

=20

1k+5.6k+390k

160

=

20

1k+5.6k+2.438k=

20V

9.038kΩ= 2.213mA

• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 + 𝑅𝐶 + 𝑅𝐸 𝐼𝐶 =20 − 5.6k + 1k 2.213mA = 20 −6.6kΩ 2.213mA = 20 − 14.606 =5.394V

Common issues (CE Amp – 2 Supply biased)

12V

-12V

RB33kΩ

RC4.7kΩ

RE10kΩ

6.5V

-0.3V

DC collector-to-ground

voltage

DC emitter-to-ground

voltage

Chart of changes/outcomes Change

in

Value

IC VC VE

(1) VCC ↑ ↔ ↑ ↔

(2) VCC ↓ ↔ ↓ ↔

(3) RB ↑ ↔ ↔ ↔

(4) RB ↓ ↔ ↔ ↔

(5) RB ∞ ↓ ↑ ↓

(6) RC ↑ ↔ ↓ ↔

(7) RC ↓ ↔ ↑ ↔

(8) RC ∞ ↓ ↓ ↓

(9) RE ↑ ↓ ↑ ↔

(10) RE ↓ ↑ ↓ ↔

(11) RE ∞ ↓ ↑ ↑

(12) VEE ↑ ↑ ↓ ↔

(13) VEE ↓ ↓ ↑ ↔

Any questions?

• Contact us at:

– 1-800-243-6446

– 1-216-781-9400

• Email:

– faculty@cie-wc.edu