Diodes & transistors worksheet with answers

Free printable Diodes & transistors worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
+−Vs​6 VR = 2.2 kΩD​I = ?

A 6 V DC source drives current through a 2.2 kΩ resistor in series with a forward-biased silicon diode. Use the constant-drop model with 0.7 V per conducting diode. Find the current I in mA.

Vs = 6 VR = 2.2 kΩVD = 0.7 V
I = mA
2.Easy
+−Vs​15 VR1​ = 330 ΩR2​ = 1.5 kΩID​ = ?

A 15 V DC source is connected through resistor R₁ = 330 Ω to a node X. From node X to the source’s negative terminal there are two parallel paths: resistor R₂ = 1.5 kΩ, and a silicon diode with its anode at X. Use the constant-drop model with 0.7 V. Find the diode current ID in mA.

Vs = 15 VR1 = 330 ΩR2 = 1.5 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 2.7 kΩβ = 200VCC​ = 9 VRB​ = 1000 kΩ

An NPN transistor with β = 200 is biased by a base resistor RB = 1000 kΩ from the supply VCC = 9 V. The collector resistor is RC = 2.7 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector–emitter voltage VCE.

VCC = 9 VRB = 1000 kΩRC = 2.7 kΩβ = 200VBE = 0.7 V
VCE = V
4.Medium
RC​ = 1 kΩβ = 120IC​ = ?VCC​ = 15 VRB​ = 220 kΩ

An NPN transistor with β = 120 is biased by a base resistor RB = 220 kΩ from the supply VCC = 15 V. The collector resistor is RC = 1 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector current IC in mA.

VCC = 15 VRB = 220 kΩRC = 1 kΩβ = 120VBE = 0.7 V
IC = mA
5.Hard
RC​ = 330 Ωβ = 100IC​ = ?VCC​ = 12 V3.3 VRB​ = 47 kΩ

An NPN transistor with β = 100 switches a 330 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 3.3 V logic output through a base resistor RB = 47 kΩ. Take VBE = 0.7 V and VCE(sat) = 0.2 V. Find the collector current IC in mA when the input is high. Decide first whether the transistor is saturated.

VCC = 12 VRC = 330 Ωβ = 100Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 47 kΩ
IC = mA
6.Hard
+−≥ 18 VR = ?12 V≤ 25 mA

A 12 V Zener regulator must supply up to 25 mA to a load. The input can drop to 18 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 18 VVZ = 12 VIL,max = 25 mAIZ,min = 5 mA
Rmax = Ω

Answer key

  1. 1. I = 2.409 mA
  2. 2. ID = 42.87 mA
  3. 3. VCE = 4.518 V
  4. 4. IC = 7.8 mA
  5. 5. IC = 5.532 mA
  6. 6. Rmax = 200 Ω

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