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 = 220 ΩD​I = ?

A 5 V DC source drives current through a 220 Ω 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 = 220 ΩVD = 0.7 V
I = mA
2.Easy
+−24 V100 Ω5.1 VIR​ = ?

A 5.1 V Zener diode is fed from 24 V through a 100 Ω resistor, with no load connected. Find the current IR in the resistor.

Vin = 24 VVZ = 5.1 VR = 100 Ω
IR = mA
3.Medium
RC​ = 2.7 kΩβ = 120IC​ = ?VCC​ = 15 VRB​ = 470 kΩ

An NPN transistor with β = 120 is biased by a base resistor RB = 470 kΩ from the supply VCC = 15 V. The collector resistor is RC = 2.7 kΩ and the emitter is grounded. Take VBE = 0.7 V and assume the transistor is in the active region. Find the collector current IC in mA.

VCC = 15 VRB = 470 kΩRC = 2.7 kΩβ = 120VBE = 0.7 V
IC = mA
4.Medium
RC​ = 5.6 kΩβ = 150RE​ = 2.2 kΩVCC​ = 15 VR1​ = 56 kΩR2​ = 6.8 kΩ

An NPN transistor with β = 150 has voltage-divider bias from VCC = 15 V: R₁ = 56 kΩ from VCC to the base and R₂ = 6.8 kΩ from the base to ground. The collector resistor is RC = 5.6 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 = 15 VR1 = 56 kΩR2 = 6.8 kΩRC = 5.6 kΩRE = 2.2 kΩβ = 150
VCE = V
5.Hard
RC​ = 1 kΩβ = 150VCC​ = 9 V5 VRB​ = ?

An NPN transistor with β = 150 switches a 1 kΩ load resistor connected between VCC = 9 V and its collector. The emitter is grounded. The base is driven from a 5 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 2 times the minimum needed for saturation. Find RB in kΩ.

VCC = 9 VRC = 1 kΩβ = 150Vin = 5 VVBE = 0.7 VVCE(sat) = 0.2 Voverdrive = 2
RB = kΩ
6.Hard
V (V)I (mA)0.65515.15Q = ?

A diode is modelled as 0.65 V in series with 25 Ω (it conducts above 0.65 V). It is in series with a 330 Ω resistor across a 5 V supply. Find the operating current ID where the diode line meets the load line.

Vs = 5 VR = 330 ΩV0 = 0.65 Vrd = 25 Ω
ID = mA

Answer key

  1. 1. I = 19.55 mA
  2. 2. IR = 189 mA
  3. 3. IC = 3.651 mA
  4. 4. VCE = 11.8 V
  5. 5. RB = 36.65 kΩ
  6. 6. ID = 12.25 mA

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