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.

New worksheetPractise online with hints

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
+−24 VR = ?2.1 V25 mA

A yellow LED with a forward voltage of 2.1 V runs from a 24 V supply through a series resistor. Find the resistance R that sets the LED current to 25 mA.

Vs = 24 VVF = 2.1 VI = 25 mA
R = Ω
2.Easy
+−Vs​5 VR1​ = 470 ΩR2​ = 1.5 kΩID​ = ?

A 5 V DC source is connected through resistor R₁ = 470 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 = 5 VR1 = 470 ΩR2 = 1.5 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 2.7 kΩβ = 75IC​ = ?VCC​ = 15 VRB​ = 680 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 680 kΩ from the supply VCC = 15 V. The collector resistor is RC = 2.7 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 = 680 kΩRC = 2.7 kΩβ = 75VBE = 0.7 V
IC = mA
4.Medium
+−15 V270 Ω5.1 V680 ΩIZ​ = ?

A 5.1 V Zener regulator is fed from 15 V through 270 Ω and supplies a 680 Ω load. Find the Zener current IZ.

Vin = 15 VVZ = 5.1 VR = 270 ΩRL = 680 Ω
IZ = mA
5.Hard
RC​ = 680 Ωβ = 100VCC​ = 5 V5 VRB​ = ?

An NPN transistor with β = 100 switches a 680 Ω load resistor connected between VCC = 5 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB. Take VBE = 0.7 V and VCE(sat) = 0.2 V. For reliable switching, design for a base current 5 times the minimum needed for saturation. Find RB in kΩ.

VCC = 5 VRC = 680 Ωβ = 100Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 5
RB = kΩ
6.Hard
+−≤ 24 V150 Ω5.1 V

A 5.1 V Zener regulator has a 150 Ω series resistor. The input can rise to 24 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 24 VVZ = 5.1 VR = 150 Ω
PZ,max = mW

Answer key

  1. 1. R = 876 Ω
  2. 2. ID = 8.682 mA
  3. 3. IC = 1.577 mA
  4. 4. IZ = 29.17 mA
  5. 5. RB = 12.18 kΩ
  6. 6. PZ,max = 642.6 mW

Full step-by-step solutions: practise this topic on examproblems.com.