Transistor Circuits I
Common-Base, DC operation
The humble transistor
Q1
Emitter (E)
Collector (C)
Base (B)
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
Basics continued
• Leakage current: ICBO (Emitter open)
– Usually is considered negligible, but can affect things when IC is small
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.
Basic configuration of Common-Base
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.
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
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
Proving the point (Measuring (a))
Measuring (b)
Measuring (c)
Measuring (d)
Measuring (e)
Measuring (f)
PNP versions
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
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…)
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).
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
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.
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)
Calculating the rest
• (b) 𝐼𝐸 =𝑉𝑅𝐸
𝑅𝐸=
19.3V
20kΩ= 965µA ∴ 𝐼𝐶 = 965µA;
VCB = VCC – ICRC = 24 – (965µA)(8kΩ) = 24 – 7.72 = 16.28V
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
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
Measure to make sure (a)
(b)
(c)
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