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​10 VR1​ = 1.5 kΩR2​ = 1.5 kΩID​ = ?

A 10 V DC source is connected through resistor R₁ = 1.5 kΩ to a node X. From node X to the source’s negative terminal there are two parallel paths: resistor R₂ = 1.5 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 = 10 VR1 = 1.5 kΩR2 = 1.5 kΩVD = 0.7 V
ID = mA
2.Easy
+−Vs​24 VR = 3.3 kΩD1​D2​I = ?

A 24 V DC source drives current through a 3.3 kΩ resistor in series with two forward-biased silicon diodes. Use the constant-drop model with 0.7 V per conducting diode. Find the current I in mA.

Vs = 24 VR = 3.3 kΩVD = 0.7 V
I = mA
3.Medium
RC​ = 4.7 kΩβ = 200VCC​ = 12 VRB​ = 1000 kΩ

An NPN transistor with β = 200 is biased by a base resistor RB = 1000 kΩ from the supply VCC = 12 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–emitter voltage VCE.

VCC = 12 VRB = 1000 kΩRC = 4.7 kΩβ = 200VBE = 0.7 V
VCE = V
4.Medium
RC​ = 2.2 kΩβ = 80IC​ = ?RE​ = 2.2 kΩVCC​ = 12 VR1​ = 47 kΩR2​ = 6.8 kΩ

An NPN transistor with β = 80 has voltage-divider bias from VCC = 12 V: R₁ = 47 kΩ from VCC to the base and R₂ = 6.8 kΩ from the base to ground. The collector resistor is RC = 2.2 kΩ and the emitter resistor is RE = 2.2 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 = 12 VR1 = 47 kΩR2 = 6.8 kΩRC = 2.2 kΩRE = 2.2 kΩβ = 80
IC = mA
5.Hard
V (V)I (mA)0.65510.64Q = ?

A diode is modelled as 0.65 V in series with 5 Ω (it conducts above 0.65 V). It is in series with a 470 Ω resistor across a 5 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 5 VR = 470 ΩV0 = 0.65 Vrd = 5 Ω
ID = mA
6.Hard
RC​ = 100 Ωβ = 100VCC​ = 9 V3.3 VRB​ = ?

An NPN transistor with β = 100 switches a 100 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. Take VBE = 0.7 V and VCE(sat) = 0.2 V. For reliable switching, design for a base current 3 times the minimum needed for saturation. Find RB in kΩ.

VCC = 9 VRC = 100 Ωβ = 100Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 3
RB = kΩ

Answer key

  1. 1. ID = 5.733 mA
  2. 2. I = 6.848 mA
  3. 3. VCE = 1.378 V
  4. 4. IC = 0.3548 mA
  5. 5. ID = 9.158 mA
  6. 6. RB = 0.9848 kΩ

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