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​12 VR1​ = 330 ΩR2​ = 10 kΩID​ = ?

A 12 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 = 10 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 = 12 VR1 = 330 ΩR2 = 10 kΩVD = 0.7 V
ID = mA
2.Easy
+−Vs​9 VR1​ = 1.5 kΩR2​ = 3.3 kΩID​ = ?

A 9 V DC source is connected through resistor R₁ = 1.5 kΩ to a node X. From node X to the source’s negative terminal there are two parallel paths: resistor R₂ = 3.3 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 = 9 VR1 = 1.5 kΩR2 = 3.3 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 1 kΩβ = 80RE​ = 1 kΩVCC​ = 12 VR1​ = 56 kΩR2​ = 22 kΩ

An NPN transistor with β = 80 has voltage-divider bias from VCC = 12 V: R₁ = 56 kΩ from VCC to the base and R₂ = 22 kΩ from the base to ground. The collector resistor is RC = 1 kΩ and the emitter resistor is RE = 1 kΩ. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector–emitter voltage VCE.

VCC = 12 VR1 = 56 kΩR2 = 22 kΩRC = 1 kΩRE = 1 kΩβ = 80
VCE = V
4.Medium
RC​ = 3.3 kΩβ = 50VCC​ = 9 VRB​ = 680 kΩ

An NPN transistor with β = 50 is biased by a base resistor RB = 680 kΩ from the supply VCC = 9 V. The collector resistor is RC = 3.3 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 = 680 kΩRC = 3.3 kΩβ = 50VBE = 0.7 V
VCE = V
5.Hard
+−≥ 9 VR = ?3.3 V≤ 20 mA

A 3.3 V Zener regulator must supply up to 20 mA to a load. The input can drop to 9 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 9 VVZ = 3.3 VIL,max = 20 mAIZ,min = 5 mA
Rmax = Ω
6.Hard
+−≥ 24 VR = ?12 V≤ 30 mA

A 12 V Zener regulator must supply up to 30 mA to a load. The input can drop to 24 V, and the Zener needs at least 2 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 24 VVZ = 12 VIL,max = 30 mAIZ,min = 2 mA
Rmax = Ω

Answer key

  1. 1. ID = 34.17 mA
  2. 2. ID = 5.321 mA
  3. 3. VCE = 7.535 V
  4. 4. VCE = 6.986 V
  5. 5. Rmax = 228 Ω
  6. 6. Rmax = 375 Ω

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