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 = ?2 V15 mA

A red LED with a forward voltage of 2 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 VI = 15 mA
R = Ω
2.Easy
+−15 V470 Ω5.1 VIR​ = ?

A 5.1 V Zener diode is fed from 15 V through a 470 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 15 VVZ = 5.1 VR = 470 Ω
IR = mA
3.Medium
RC​ = 3.3 kΩβ = 120RE​ = 470 ΩVCC​ = 18 VR1​ = 39 kΩR2​ = 6.8 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 18 V: R₁ = 39 kΩ from VCC to the base and R₂ = 6.8 kΩ from the base to ground. The collector resistor is RC = 3.3 kΩ and the emitter resistor is RE = 470 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 = 18 VR1 = 39 kΩR2 = 6.8 kΩRC = 3.3 kΩRE = 470 Ωβ = 120
VCE = V
4.Medium
RC​ = 1 kΩβ = 100VCC​ = 12 VRB​ = 220 kΩ

An NPN transistor with β = 100 is biased by a base resistor RB = 220 kΩ from the supply VCC = 12 V. The collector resistor is RC = 1 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 = 12 VRB = 220 kΩRC = 1 kΩβ = 100VBE = 0.7 V
VCE = V
5.Hard
RC​ = 100 Ωβ = 100IC​ = ?VCC​ = 12 V5 VRB​ = 220 kΩ

An NPN transistor with β = 100 switches a 100 Ω load resistor connected between VCC = 12 V and its collector. The emitter is grounded. The base is driven from a 5 V logic output through a base resistor RB = 220 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 = 12 VRC = 100 Ωβ = 100Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 220 kΩ
IC = mA
6.Hard
+−≥ 18 VR = ?5.6 V≤ 50 mA

A 5.6 V Zener regulator must supply up to 50 mA to a load. The input can drop to 18 V, and the Zener needs at least 5 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 18 VVZ = 5.6 VIL,max = 50 mAIZ,min = 5 mA
Rmax = Ω

Answer key

  1. 1. R = 200 Ω
  2. 2. IR = 21.06 mA
  3. 3. VCE = 3.744 V
  4. 4. VCE = 6.864 V
  5. 5. IC = 1.955 mA
  6. 6. Rmax = 225.5 Ω

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