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
+−Vs​5 VR1​ = 1 kΩR2​ = 3.3 kΩID​ = ?

A 5 V DC source is connected through resistor R₁ = 1 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 = 5 VR1 = 1 kΩR2 = 3.3 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 2.2 kΩβ = 80VCC​ = 12 VRB​ = 390 kΩ

An NPN transistor with β = 80 is biased by a base resistor RB = 390 kΩ from the supply VCC = 12 V. The collector resistor is RC = 2.2 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 = 390 kΩRC = 2.2 kΩβ = 80VBE = 0.7 V
VCE = V
4.Medium
RC​ = 3.3 kΩβ = 120RE​ = 2.2 kΩVCC​ = 20 VR1​ = 33 kΩR2​ = 12 kΩ

An NPN transistor with β = 120 has voltage-divider bias from VCC = 20 V: R₁ = 33 kΩ from VCC to the base and R₂ = 12 kΩ from the base to ground. The collector resistor is RC = 3.3 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–emitter voltage VCE.

VCC = 20 VR1 = 33 kΩR2 = 12 kΩRC = 3.3 kΩRE = 2.2 kΩβ = 120
VCE = V
5.Hard
V (V)I (mA)0.633Q = ?

A diode is modelled as 0.6 V in series with 20 Ω (it conducts above 0.6 V). It is in series with a 1000 Ω resistor across a 3 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 3 VR = 1000 ΩV0 = 0.6 Vrd = 20 Ω
ID = mA
6.Hard
+−≤ 18 V270 Ω3.3 V

A 3.3 V Zener regulator has a 270 Ω series resistor. The input can rise to 18 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 18 VVZ = 3.3 VR = 270 Ω
PZ,max = mW

Answer key

  1. 1. I = 0.9149 mA
  2. 2. ID = 4.088 mA
  3. 3. VCE = 6.901 V
  4. 4. VCE = 8.843 V
  5. 5. ID = 2.353 mA
  6. 6. PZ,max = 179.7 mW

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