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 = 1.5 kΩD1​D2​I = ?

A 6 V DC source drives current through a 1.5 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 = 1.5 kΩVD = 0.7 V
I = mA
2.Easy
+−Vs​9 VR1​ = 330 ΩR2​ = 2.2 kΩID​ = ?

A 9 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 = 2.2 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 = 330 ΩR2 = 2.2 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 2.7 kΩβ = 100IC​ = ?RE​ = 680 ΩVCC​ = 15 VR1​ = 56 kΩR2​ = 15 kΩ

An NPN transistor with β = 100 has voltage-divider bias from VCC = 15 V: R₁ = 56 kΩ from VCC to the base and R₂ = 15 kΩ from the base to ground. The collector resistor is RC = 2.7 kΩ and the emitter resistor is RE = 680 Ω. 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 = 56 kΩR2 = 15 kΩRC = 2.7 kΩRE = 680 Ωβ = 100
IC = mA
4.Medium
RC​ = 470 Ωβ = 120VCC​ = 12 VRB​ = 150 kΩ

An NPN transistor with β = 120 is biased by a base resistor RB = 150 kΩ from the supply VCC = 12 V. The collector resistor is RC = 470 Ω 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 = 12 VRB = 150 kΩRC = 470 Ωβ = 120VBE = 0.7 V
VCE = V
5.Hard
+−≤ 27 V150 Ω6.2 V

A 6.2 V Zener regulator has a 150 Ω series resistor. The input can rise to 27 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 27 VVZ = 6.2 VR = 150 Ω
PZ,max = mW
6.Hard
+−≥ 15 VR = ?6.2 V≤ 50 mA

A 6.2 V Zener regulator must supply up to 50 mA to a load. The input can drop to 15 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 15 VVZ = 6.2 VIL,max = 50 mAIZ,min = 5 mA
Rmax = Ω

Answer key

  1. 1. I = 3.067 mA
  2. 2. ID = 24.83 mA
  3. 3. IC = 3.067 mA
  4. 4. VCE = 7.751 V
  5. 5. PZ,max = 859.7 mW
  6. 6. Rmax = 160 Ω

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