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
+−9 V270 Ω3.3 VIR​ = ?

A 3.3 V Zener diode is fed from 9 V through a 270 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 9 VVZ = 3.3 VR = 270 Ω
IR = mA
2.Easy
+−9 VR = ?3 V5 mA

A white LED with a forward voltage of 3 V runs from a 9 V supply through a series resistor. Find the resistance R that sets the LED current to 5 mA.

Vs = 9 VVF = 3 VI = 5 mA
R = Ω
3.Medium
RC​ = 470 Ωβ = 200IC​ = ?VCC​ = 9 VRB​ = 270 kΩ

An NPN transistor with β = 200 is biased by a base resistor RB = 270 kΩ from the supply VCC = 9 V. The collector resistor is RC = 470 Ω 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 = 270 kΩRC = 470 Ωβ = 200VBE = 0.7 V
IC = mA
4.Medium
RC​ = 5.6 kΩβ = 150RE​ = 1 kΩVCC​ = 20 VR1​ = 47 kΩR2​ = 4.7 kΩ

An NPN transistor with β = 150 has voltage-divider bias from VCC = 20 V: R₁ = 47 kΩ from VCC to the base and R₂ = 4.7 kΩ from the base to ground. The collector resistor is RC = 5.6 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 = 20 VR1 = 47 kΩR2 = 4.7 kΩRC = 5.6 kΩRE = 1 kΩβ = 150
VCE = V
5.Hard
V (V)I (mA)0.7960Q = ?

A diode is modelled as 0.7 V in series with 50 Ω (it conducts above 0.7 V). It is in series with a 150 Ω resistor across a 9 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 9 VR = 150 ΩV0 = 0.7 Vrd = 50 Ω
ID = mA
6.Hard
+−≥ 20 VR = ?3.3 V≤ 25 mA

A 3.3 V Zener regulator must supply up to 25 mA to a load. The input can drop to 20 V, and the Zener needs at least 1 mA to regulate. Find the largest series resistance Rmax that keeps it regulating.

Vin,min = 20 VVZ = 3.3 VIL,max = 25 mAIZ,min = 1 mA
Rmax = Ω

Answer key

  1. 1. IR = 21.11 mA
  2. 2. R = 1200 Ω
  3. 3. IC = 6.148 mA
  4. 4. VCE = 12.86 V
  5. 5. ID = 41.5 mA
  6. 6. Rmax = 642.3 Ω

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