of 23/23
Transistor Circuits VI Two-Transistor Direct-Coupled CE Amplifier / Some basics of troubleshooting CE Amps
• date post

08-Feb-2022
• Category

## Documents

• view

2

1

Embed Size (px)

### Transcript of Transistor Circuits V

Transistor Circuits VThe Two-Transistor Direct-Coupled CE circuit configuration
Formulas for same
• 1 = −2
• 1 = − 11
• 2 = − 22
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
77.73kΩ = 136.37μA
• 1 = − 1 + 1 1 = 12 − 68k + 330 136.37μA = 12 − 68.33kΩ 136.37μA = 12 − 9.318 = 2.682V
• 1 = − 11 = 12 − 68kΩ 136.37μA = 12 − 9.273 = 2.727V
Work for first example (cont.)
• 2 = −−11
1kΩ = 2.027mA
• 2 = − 2 + 2 2 = 12 − 2.7k + 1k 2.027mA = 12 − 3.7kΩ 2.027mA = 12 − 7.499 = 4.501V
• 2 = − 22 = 12 − 2.7kΩ 2.027mA = 12 − 5.473 = 6.527V
Second example
• Rework the previous problem using a 680-k resistor in place of the 470-k resistor.
Work for second example
81.93kΩ = 129.379μA
• 1 = − 1 + 1 1 = 12 − 68k + 330 129.379μA = 12 − 68.33kΩ 129.379μA = 12 − 8.84 = 3.16V
• 1 = − 11 = 12 − 68kΩ 129.379μA = 12 − 8.798 = 3.202V
Work for second example (cont.)
• 2 = −−11
1kΩ = 2.502mA
• 2 = − 2 + 2 2 = 12 − 2.7k + 1k 2.502mA = 12 − 3.7kΩ 2.502mA = 12 − 9.528 = 2.472V
• 2 = − 22 = 12 − 2.7kΩ 2.502mA = 12 − 6.756 = 5.224V
TROUBLESHOOTING CE CIRCUITS
First example
• If the 180-k (R2) resistor became open in the circuit shown, what would the collector-to- ground voltage be?
12V
12V
• = − = 10.17 − 10.7 = −0.56V
• = = −
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-k (R1) resistor became open instead in the figure shown, what would the collector-to- ground voltage be?
12V
12V
Circuit evaluation
• If R1 opens, VB = 0V ∴ transistor is biased full on (saturation)
• = =
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
=
Work for third example
11.475kΩ = 1.743mA
• = + + = 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
9.038kΩ = 2.213mA
• = + + = 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