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 VR = ?3.2 V5 mA

A blue LED with a forward voltage of 3.2 V runs from a 24 V supply through a series resistor. Find the resistance R that sets the LED current to 5 mA.

Vs = 24 VVF = 3.2 VI = 5 mA
R = Ω
2.Easy
+−5 VR = ?2.1 V15 mA

A yellow LED with a forward voltage of 2.1 V runs from a 5 V supply through a series resistor. Find the resistance R that sets the LED current to 15 mA.

Vs = 5 VVF = 2.1 VI = 15 mA
R = Ω
3.Medium
RC​ = 4.7 kΩβ = 100IC​ = ?VCC​ = 9 VRB​ = 1000 kΩ

An NPN transistor with β = 100 is biased by a base resistor RB = 1000 kΩ from the supply VCC = 9 V. The collector resistor is RC = 4.7 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 = 9 VRB = 1000 kΩRC = 4.7 kΩβ = 100VBE = 0.7 V
IC = mA
4.Medium
RC​ = 3.3 kΩβ = 80RE​ = 1 kΩVCC​ = 15 VR1​ = 100 kΩR2​ = 12 kΩ

An NPN transistor with β = 80 has voltage-divider bias from VCC = 15 V: R₁ = 100 kΩ from VCC to the base and R₂ = 12 kΩ from the base to ground. The collector resistor is RC = 3.3 kΩ and the emitter resistor is RE = 1 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 = 15 VR1 = 100 kΩR2 = 12 kΩRC = 3.3 kΩRE = 1 kΩβ = 80
VCE = V
5.Hard
RC​ = 1 kΩβ = 80IC​ = ?VCC​ = 9 V3.3 VRB​ = 10 kΩ

An NPN transistor with β = 80 switches a 1 kΩ load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 3.3 V logic output through a base resistor RB = 10 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 = 1 kΩβ = 80Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 10 kΩ
IC = mA
6.Hard
+−≥ 12 VR = ?9.1 V≤ 25 mA

A 9.1 V Zener regulator must supply up to 25 mA to a load. The input can drop to 12 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 12 VVZ = 9.1 VIL,max = 25 mAIZ,min = 5 mA
Rmax = Ω

Answer key

  1. 1. R = 4160 Ω
  2. 2. R = 193.3 Ω
  3. 3. IC = 0.83 mA
  4. 4. VCE = 11.59 V
  5. 5. IC = 8.8 mA
  6. 6. Rmax = 96.67 Ω

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