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
+−Vs​5 VR = 4.7 kΩD​I = ?

A 5 V DC source drives current through a 4.7 kΩ resistor in series with a forward-biased silicon diode. Use the constant-drop model with 0.7 V per conducting diode. Find the current I in mA.

Vs = 5 VR = 4.7 kΩVD = 0.7 V
I = mA
2.Easy
+−5 VR = ?3.2 V5 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 5 mA.

Vs = 5 VVF = 3.2 VI = 5 mA
R = Ω
3.Medium
+−12 VR = ?4 × 2 V10 mA

4 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 10 mA.

Vs = 12 VVF = 2 Vn = 4I = 10 mA
R = Ω
4.Medium
RC​ = 2.2 kΩβ = 120IC​ = ?RE​ = 1.2 kΩVCC​ = 18 VR1​ = 47 kΩR2​ = 12 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 18 V: R₁ = 47 kΩ from VCC to the base and R₂ = 12 kΩ from the base to ground. The collector resistor is RC = 2.2 kΩ and the emitter resistor is RE = 1.2 kΩ. Take VBE = 0.7 V and use the Thévenin equivalent of the divider (do not neglect the base current). Find the collector current IC in mA.

VCC = 18 VR1 = 47 kΩR2 = 12 kΩRC = 2.2 kΩRE = 1.2 kΩβ = 120
IC = mA
5.Hard
RC​ = 680 Ωβ = 150IC​ = ?VCC​ = 9 V3.3 VRB​ = 47 kΩ

An NPN transistor with β = 150 switches a 680 Ω 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 = 47 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 = 680 Ωβ = 150Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 VRB = 47 kΩ
IC = mA
6.Hard
V (V)I (mA)0.61236.36Q = ?

A diode is modelled as 0.6 V in series with 5 Ω (it conducts above 0.6 V). It is in series with a 330 Ω resistor across a 12 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 12 VR = 330 ΩV0 = 0.6 Vrd = 5 Ω
ID = mA

Answer key

  1. 1. I = 0.9149 mA
  2. 2. R = 360 Ω
  3. 3. R = 400 Ω
  4. 4. IC = 2.296 mA
  5. 5. IC = 8.298 mA
  6. 6. ID = 34.03 mA

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