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 = 4.7 kΩD1​D2​I = ?

A 6 V DC source drives current through a 4.7 kΩ resistor in series with two forward-biased silicon diodes. Use the constant-drop model with 0.7 V per conducting diode. Find the current I in mA.

Vs = 6 VR = 4.7 kΩVD = 0.7 V
I = mA
2.Easy
+−Vs​6 VR1​ = 330 ΩR2​ = 1.5 kΩID​ = ?

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 = 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 = 6 VR1 = 330 ΩR2 = 1.5 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 3.3 kΩβ = 75IC​ = ?VCC​ = 15 VRB​ = 470 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 470 kΩ from the supply VCC = 15 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 current IC in mA.

VCC = 15 VRB = 470 kΩRC = 3.3 kΩβ = 75VBE = 0.7 V
IC = mA
4.Medium
RC​ = 4.7 kΩβ = 150IC​ = ?RE​ = 1 kΩVCC​ = 18 VR1​ = 82 kΩR2​ = 12 kΩ

An NPN transistor with β = 150 has voltage-divider bias from VCC = 18 V: R₁ = 82 kΩ from VCC to the base and R₂ = 12 kΩ from the base to ground. The collector resistor is RC = 4.7 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 current IC in mA.

VCC = 18 VR1 = 82 kΩR2 = 12 kΩRC = 4.7 kΩRE = 1 kΩβ = 150
IC = mA
5.Hard
+−≥ 24 VR = ?3.3 V≤ 50 mA

A 3.3 V Zener regulator must supply up to 50 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 = 50 mAIZ,min = 2 mA
Rmax = Ω
6.Hard
V (V)I (mA)0.65612.77Q = ?

A diode is modelled as 0.65 V in series with 20 Ω (it conducts above 0.65 V). It is in series with a 470 Ω resistor across a 6 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 6 VR = 470 ΩV0 = 0.65 Vrd = 20 Ω
ID = mA

Answer key

  1. 1. I = 0.9787 mA
  2. 2. ID = 15.59 mA
  3. 3. IC = 2.282 mA
  4. 4. IC = 1.484 mA
  5. 5. Rmax = 398.1 Ω
  6. 6. ID = 10.92 mA

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