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 = 3.3 kΩD1​D2​I = ?

A 5 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 = 5 VR = 3.3 kΩVD = 0.7 V
I = mA
2.Easy
+−Vs​9 VR1​ = 1.5 kΩR2​ = 1 kΩID​ = ?

A 9 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 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 = 9 VR1 = 1.5 kΩR2 = 1 kΩVD = 0.7 V
ID = mA
3.Medium
RC​ = 4.7 kΩβ = 75VCC​ = 10 VRB​ = 1000 kΩ

An NPN transistor with β = 75 is biased by a base resistor RB = 1000 kΩ from the supply VCC = 10 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 = 10 VRB = 1000 kΩRC = 4.7 kΩβ = 75VBE = 0.7 V
VCE = V
4.Medium
RC​ = 2.2 kΩβ = 80IC​ = ?RE​ = 1.5 kΩVCC​ = 20 VR1​ = 33 kΩR2​ = 15 kΩ

An NPN transistor with β = 80 has voltage-divider bias from VCC = 20 V: R₁ = 33 kΩ from VCC to the base and R₂ = 15 kΩ from the base to ground. The collector resistor is RC = 2.2 kΩ and the emitter resistor is RE = 1.5 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 = 20 VR1 = 33 kΩR2 = 15 kΩRC = 2.2 kΩRE = 1.5 kΩβ = 80
IC = mA
5.Hard
+−≤ 24 V150 Ω5.1 V

A 5.1 V Zener regulator has a 150 Ω series resistor. The input can rise to 24 V, and the load may be disconnected. Find the largest power PZ,max the Zener must dissipate.

Vin,max = 24 VVZ = 5.1 VR = 150 Ω
PZ,max = mW
6.Hard
V (V)I (mA)0.6533Q = ?

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 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 = 0.65 Vrd = 5 Ω
ID = mA

Answer key

  1. 1. I = 1.091 mA
  2. 2. ID = 4.833 mA
  3. 3. VCE = 6.722 V
  4. 4. IC = 3.368 mA
  5. 5. PZ,max = 642.6 mW
  6. 6. ID = 2.338 mA

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