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 VR1​ = 330 ΩR2​ = 680 ΩID​ = ?

A 6 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 = 680 Ω, 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 = 6 VR1 = 330 ΩR2 = 680 ΩVD = 0.7 V
ID = mA
2.Easy
+−5 VR = ?3 V5 mA

A white LED with a forward voltage of 3 V runs from a 5 V supply through a series resistor. Find the resistance R that sets the LED current to 5 mA.

Vs = 5 VVF = 3 VI = 5 mA
R = Ω
3.Medium
RC​ = 3.3 kΩβ = 120IC​ = ?RE​ = 1.2 kΩVCC​ = 15 VR1​ = 68 kΩR2​ = 22 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 15 V: R₁ = 68 kΩ from VCC to the base and R₂ = 22 kΩ from the base to ground. The collector resistor is RC = 3.3 kΩ and the emitter resistor is RE = 1.2 kΩ. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector current IC in mA.

VCC = 15 VR1 = 68 kΩR2 = 22 kΩRC = 3.3 kΩRE = 1.2 kΩβ = 120
IC = mA
4.Medium
RC​ = 5.6 kΩβ = 150RE​ = 1.5 kΩVCC​ = 20 VR1​ = 82 kΩR2​ = 10 kΩ

An NPN transistor with β = 150 has voltage-divider bias from VCC = 20 V: R₁ = 82 kΩ from VCC to the base and R₂ = 10 kΩ from the base to ground. The collector resistor is RC = 5.6 kΩ and the emitter resistor is RE = 1.5 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 = 20 VR1 = 82 kΩR2 = 10 kΩRC = 5.6 kΩRE = 1.5 kΩβ = 150
VCE = V
5.Hard
+−≥ 18 VR = ?3.3 V≤ 40 mA

A 3.3 V Zener regulator must supply up to 40 mA to a load. The input can drop to 18 V, and the Zener needs at least 2 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 18 VVZ = 3.3 VIL,max = 40 mAIZ,min = 2 mA
Rmax = Ω
6.Hard
+−≥ 24 VR = ?3.3 V≤ 10 mA

A 3.3 V Zener regulator must supply up to 10 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 = 3.3 VIL,max = 10 mAIZ,min = 2 mA
Rmax = Ω

Answer key

  1. 1. ID = 15.03 mA
  2. 2. R = 400 Ω
  3. 3. IC = 2.2 mA
  4. 4. VCE = 13.32 V
  5. 5. Rmax = 350 Ω
  6. 6. Rmax = 1725 Ω

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