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
+−24 V270 Ω5.6 VIR​ = ?

A 5.6 V Zener diode is fed from 24 V through a 270 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 24 VVZ = 5.6 VR = 270 Ω
IR = mA
2.Easy
+−9 V100 Ω5.1 VIR​ = ?

A 5.1 V Zener diode is fed from 9 V through a 100 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 9 VVZ = 5.1 VR = 100 Ω
IR = mA
3.Medium
RC​ = 2.2 kΩβ = 75VCC​ = 10 VRB​ = 820 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 820 kΩ from the supply VCC = 10 V. The collector resistor is RC = 2.2 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 = 10 VRB = 820 kΩRC = 2.2 kΩβ = 75VBE = 0.7 V
VCE = V
4.Medium
RC​ = 2.2 kΩβ = 120IC​ = ?VCC​ = 15 VRB​ = 820 kΩ

An NPN transistor with β = 120 is biased by a base resistor RB = 820 kΩ from the supply VCC = 15 V. The collector resistor is RC = 2.2 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 = 820 kΩRC = 2.2 kΩβ = 120VBE = 0.7 V
IC = mA
5.Hard
+−≤ 27 V100 Ω6.2 V

A 6.2 V Zener regulator has a 100 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 = 100 Ω
PZ,max = mW
6.Hard
RC​ = 100 Ωβ = 50IC​ = ?VCC​ = 9 V5 VRB​ = 47 kΩ

An NPN transistor with β = 50 switches a 100 Ω load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 47 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 = 9 VRC = 100 Ωβ = 50Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 47 kΩ
IC = mA

Answer key

  1. 1. IR = 68.15 mA
  2. 2. IR = 39 mA
  3. 3. VCE = 8.129 V
  4. 4. IC = 2.093 mA
  5. 5. PZ,max = 1290 mW
  6. 6. IC = 4.574 mA

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