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​6 VR1​ = 2.2 kΩR2​ = 3.3 kΩID​ = ?

A 6 V DC source is connected through resistor R₁ = 2.2 kΩ to a node X. From node X to the source’s negative terminal there are two parallel paths: resistor R₂ = 3.3 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 = 6 VR1 = 2.2 kΩR2 = 3.3 kΩVD = 0.7 V
ID = mA
2.Easy
+−Vs​10 VR = 220 ΩD1​D2​I = ?

A 10 V DC source drives current through a 220 Ω 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 = 10 VR = 220 ΩVD = 0.7 V
I = mA
3.Medium
RC​ = 2.7 kΩβ = 200RE​ = 680 ΩVCC​ = 12 VR1​ = 68 kΩR2​ = 12 kΩ

An NPN transistor with β = 200 has voltage-divider bias from VCC = 12 V: R₁ = 68 kΩ from VCC to the base and R₂ = 12 kΩ from the base to ground. The collector resistor is RC = 2.7 kΩ and the emitter resistor is RE = 680 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 = 12 VR1 = 68 kΩR2 = 12 kΩRC = 2.7 kΩRE = 680 Ωβ = 200
VCE = V
4.Medium
RC​ = 3.3 kΩβ = 100VCC​ = 15 VRB​ = 390 kΩ

An NPN transistor with β = 100 is biased by a base resistor RB = 390 kΩ from the supply VCC = 15 V. The collector resistor is RC = 3.3 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 = 15 VRB = 390 kΩRC = 3.3 kΩβ = 100VBE = 0.7 V
VCE = V
5.Hard
V (V)I (mA)0.65533.33Q = ?

A diode is modelled as 0.65 V in series with 50 Ω (it conducts above 0.65 V). It is in series with a 150 Ω resistor across a 5 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 5 VR = 150 ΩV0 = 0.65 Vrd = 50 Ω
ID = mA
6.Hard
RC​ = 220 Ωβ = 200VCC​ = 9 V3.3 VRB​ = ?

An NPN transistor with β = 200 switches a 220 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 5 times the minimum needed for saturation. Find RB in kΩ.

VCC = 9 VRC = 220 Ωβ = 200Vin = 3.3 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 5
RB = kΩ

Answer key

  1. 1. ID = 2.197 mA
  2. 2. I = 39.09 mA
  3. 3. VCE = 6.932 V
  4. 4. VCE = 2.9 V
  5. 5. ID = 21.75 mA
  6. 6. RB = 2.6 kΩ

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