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Transistor Circuits ICommon-Base, DC operation

The humble transistorCollector (C)Base (B)Q1Emitter (E)

Transistor basics Emitter to base junction is forward biased(normally) Collector to base junction is reverse biased(normally) Transistors are current operated devices, soKCL should be applied first:–IE IC IB

Basics continued Leakage current:ICBO (Emitter open)– Usually is considerednegligible, but can affectthings when IC is small

Basics continued 𝐼𝐶hFB α (α 1)𝐼𝐸𝐼𝐶hFE β (β has no𝐼𝐵definitive limit) The symbols hFB and hFE are BJT parameters;the h means Hybrid parameter. The subscriptFB means Forward current transfer ratio of thecommon, or grounded, Base circuit, whereasFE means Forward current transfer ratio of thecommon, or grounded, Emitter circuit.

Basic configuration of Common-Base

First circuit If VEE 20V and VEB isnegligible, find IE whenRE equals (a) 80kΩ, (b)40kΩ, (c) 20kΩ, (d)10kΩ, (e) 5kΩ, and (f)1kΩ.

Work for first circuitFormula is 𝐼𝐸 𝑉𝐸𝐸𝑅𝐸 20V𝑅𝐸20V(a) 𝐼𝐸 0.25mA 250µA;80kΩ20V(b) 𝐼𝐸 0.5mA 500µA;40kΩ20V(c) 𝐼𝐸 1mA;20kΩ20V(d) 𝐼𝐸 2mA;10kΩ20V(e) 𝐼𝐸 4mA;5kΩ20V(f) 𝐼𝐸 20mA1kΩ

Use approximation for VEBIf VEB 0.7V, rework previous values to determine IE in each instance. Formula is now 𝐼𝐸 19.3V 241.25µA;80kΩ19.3V(b) 𝐼𝐸 482.5µA;40kΩ19.3V(c) 𝐼𝐸 965µA;20kΩ19.3V(d) 𝐼𝐸 1.93mA;10kΩ19.3V(e) 𝐼𝐸 3.86mA;5kΩ19.3V(f) 𝐼𝐸 19.3mA1kΩ (a) 𝐼𝐸 𝑉𝐸𝐸 𝑉𝐸𝐵𝑅𝐸 20V 0.7V𝑅𝐸 19.3V𝑅𝐸

Proving the point(Measuring (a))

Measuring (b)

Measuring (c)

Measuring (d)

Measuring (e)

Measuring (f)

PNP versions

Second circuit18V10kΩ-22V If RE 10kΩ and VEB 0V, what are the valuesof IE, IC and VCB when RCequals (a) 5kΩ, (b) 8kΩ,(c) 10kΩ, (d) 12kΩ.Assume hFB α 1.

Points to ponder Since IE will remain constant, use the valuefound from calculating through the derivation𝑉𝐸𝐸18Vof Ohm’s Law: 𝐼𝐸 1.8mA.𝑅𝐸10kΩAdditionally, we are treating α as 1, which𝐼𝐶means using the formula 1 translates into𝐼𝐸IE and IC being equal. Therefore, the value of ICfor 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 originalformula gives us VCC ICRC VCB. Reworkingthe equation to solve for VCB yields VCB VCC –ICRC (You can use the absolute value for VCCwhen calculating, but remember to includepolarity when writing the final answer).

Work through for circuit (a) 𝑉𝐶𝐵 𝑉𝐶𝐶 𝐼𝐶 𝑅𝐶 22 1.8mA 5kΩ 22 9 13V (Remember the polarity of VCC isnegative ) (b) 𝑉𝐶𝐵 𝑉𝐶𝐶 𝐼𝐶 𝑅𝐶 22 1.8mA 8kΩ 22 14.4 7.6V (c) 𝑉𝐶𝐵 𝑉𝐶𝐶 𝐼𝐶 𝑅𝐶 22 1.8mA 10kΩ 22 18 4V (d) 𝑉𝐶𝐵 𝑉𝐶𝐶 𝐼𝐶 𝑅𝐶 22 1.8mA 12kΩ 22 21.6 0.4V

Third circuit If VCC 24V, find (a) thesaturation current,IC(sat). If VEB 0.7V, whatare 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 ourcalculations (a) 𝐼𝐶(𝑠𝑎𝑡) 𝑉𝐶𝐶𝑅𝐶 24V5kΩ 3mA VRE VEE – VEB 20 – 0.7 19.3V (technicallythis should be written as -19.3V since VEE isnegative in polarity)

Calculating the rest (b) 𝐼𝐸 𝑉𝑅𝐸𝑅𝐸 19.3V20kΩ 965µA 𝐼𝐶 965µA;VCB VCC – ICRC 24 – (965µA)(8kΩ) 24 –7.72 16.28V

And the beat goes on (c) 𝐼𝐸 𝑉𝑅𝐸𝑅𝐸 19.3V10kΩ 1.93mA 𝐼𝐶 1.93mA; VCB VCC – ICRC 24 – (1.93mA)(8kΩ) 24 – 15.44 8.56V

And on (d) 𝐼𝐸 𝑉𝑅𝐸𝑅𝐸 19.3V5kΩ 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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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: –I File Size: 1MBPage Count: 30