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​15 VR1​ = 680 ΩR2​ = 4.7 kΩID​ = ?

A 15 V DC source is connected through resistor R₁ = 680 Ω to a node X. From node X to the source’s negative terminal there are two parallel paths: resistor R₂ = 4.7 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 = 680 ΩR2 = 4.7 kΩVD = 0.7 V
ID = mA
2.Easy
+−12 VR = ?2 V10 mA

A red LED with a forward voltage of 2 V runs from a 12 V supply through a series resistor. Find the resistance R that sets the LED current to 10 mA.

Vs = 12 VVF = 2 VI = 10 mA
R = Ω
3.Medium
RC​ = 2.7 kΩβ = 200RE​ = 680 ΩVCC​ = 12 VR1​ = 56 kΩR2​ = 15 kΩ

An NPN transistor with β = 200 has voltage-divider bias from VCC = 12 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–emitter voltage VCE.

VCC = 12 VR1 = 56 kΩR2 = 15 kΩRC = 2.7 kΩRE = 680 Ωβ = 200
VCE = V
4.Medium
RC​ = 2.7 kΩβ = 150IC​ = ?RE​ = 680 ΩVCC​ = 12 VR1​ = 82 kΩR2​ = 12 kΩ

An NPN transistor with β = 150 has voltage-divider bias from VCC = 12 V: R₁ = 82 kΩ from VCC to the base and R₂ = 12 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 = 12 VR1 = 82 kΩR2 = 12 kΩRC = 2.7 kΩRE = 680 Ωβ = 150
IC = mA
5.Hard
RC​ = 100 Ωβ = 150IC​ = ?VCC​ = 24 V3.3 VRB​ = 1 kΩ

An NPN transistor with β = 150 switches a 100 Ω load resistor connected between VCC = 24 V and its collector. The emitter is grounded. The base is driven from a 3.3 V logic output through a base resistor RB = 1 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 = 24 VRC = 100 Ωβ = 150Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 1 kΩ
IC = mA
6.Hard
+−≥ 20 VR = ?5.6 V≤ 20 mA

A 5.6 V Zener regulator must supply up to 20 mA to a load. The input can drop to 20 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 20 VVZ = 5.6 VIL,max = 20 mAIZ,min = 5 mA
Rmax = Ω

Answer key

  1. 1. ID = 20.88 mA
  2. 2. R = 1000 Ω
  3. 3. VCE = 3.638 V
  4. 4. IC = 1.103 mA
  5. 5. IC = 238 mA
  6. 6. Rmax = 576 Ω

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