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
+−12 VR = ?3.2 V25 mA

A blue LED with a forward voltage of 3.2 V runs from a 12 V supply through a series resistor. Find the resistance R that sets the LED current to 25 mA.

Vs = 12 VVF = 3.2 VI = 25 mA
R = Ω
2.Easy
+−Vs​5 VR1​ = 680 ΩR2​ = 10 kΩID​ = ?

A 5 V DC source is connected through resistor R₁ = 680 Ω to a node X. From node X to the source’s negative terminal there are two parallel paths: resistor R₂ = 10 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 = 680 ΩR2 = 10 kΩVD = 0.7 V
ID = mA
3.Medium
+−12 VR = ?2 × 2 V25 mA

2 red LEDs (forward voltage 2 V each) are connected in series with a resistor to a 12 V supply. Find the resistance R that sets the current to 25 mA.

Vs = 12 VVF = 2 Vn = 2I = 25 mA
R = Ω
4.Medium
RC​ = 1.5 kΩβ = 100IC​ = ?VCC​ = 9 VRB​ = 390 kΩ

An NPN transistor with β = 100 is biased by a base resistor RB = 390 kΩ from the supply VCC = 9 V. The collector resistor is RC = 1.5 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 = 390 kΩRC = 1.5 kΩβ = 100VBE = 0.7 V
IC = mA
5.Hard
RC​ = 150 Ωβ = 50IC​ = ?VCC​ = 9 V5 VRB​ = 10 kΩ

An NPN transistor with β = 50 switches a 150 Ω load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 5 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 = 150 Ωβ = 50Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 10 kΩ
IC = mA
6.Hard
V (V)I (mA)0.65320Q = ?

A diode is modelled as 0.65 V in series with 10 Ω (it conducts above 0.65 V). It is in series with a 150 Ω resistor across a 3 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 3 VR = 150 ΩV0 = 0.65 Vrd = 10 Ω
ID = mA

Answer key

  1. 1. R = 352 Ω
  2. 2. ID = 6.254 mA
  3. 3. R = 320 Ω
  4. 4. IC = 2.128 mA
  5. 5. IC = 21.5 mA
  6. 6. ID = 14.69 mA

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