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
+−5 VR = ?3.2 V10 mA

A blue LED with a forward voltage of 3.2 V runs from a 5 V supply through a series resistor. Find the resistance R that sets the LED current to 10 mA.

Vs = 5 VVF = 3.2 VI = 10 mA
R = Ω
2.Easy
+−24 V330 Ω12 VIR​ = ?

A 12 V Zener diode is fed from 24 V through a 330 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 24 VVZ = 12 VR = 330 Ω
IR = mA
3.Medium
RC​ = 4.7 kΩβ = 100RE​ = 2.2 kΩVCC​ = 20 VR1​ = 22 kΩR2​ = 10 kΩ

An NPN transistor with β = 100 has voltage-divider bias from VCC = 20 V: R₁ = 22 kΩ from VCC to the base and R₂ = 10 kΩ from the base to ground. The collector resistor is RC = 4.7 kΩ and the emitter resistor is RE = 2.2 kΩ. 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 = 20 VR1 = 22 kΩR2 = 10 kΩRC = 4.7 kΩRE = 2.2 kΩβ = 100
VCE = V
4.Medium
+−24 VR = ?5 × 2.1 V25 mA

5 yellow LEDs (forward voltage 2.1 V each) are connected in series with a resistor to a 24 V supply. Find the resistance R that sets the current to 25 mA.

Vs = 24 VVF = 2.1 Vn = 5I = 25 mA
R = Ω
5.Hard
RC​ = 1 kΩβ = 50IC​ = ?VCC​ = 5 V5 VRB​ = 22 kΩ

An NPN transistor with β = 50 switches a 1 kΩ 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 = 22 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 = 5 VRC = 1 kΩβ = 50Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 22 kΩ
IC = mA
6.Hard
+−≤ 12 V330 Ω6.2 V

A 6.2 V Zener regulator has a 330 Ω series resistor. The input can rise to 12 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 12 VVZ = 6.2 VR = 330 Ω
PZ,max = mW

Answer key

  1. 1. R = 180 Ω
  2. 2. IR = 36.36 mA
  3. 3. VCE = 3.229 V
  4. 4. R = 540 Ω
  5. 5. IC = 4.8 mA
  6. 6. PZ,max = 109 mW

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